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Drilling and Completion Systems
Module 5 Fluids Pressure Control
1
2
Module 5 Fluids Pressure Control
Lesson 1 Functions of Drilling Fluids Lesson 1 Objectives
Functions of Drilling Fluids
Negative Functions of Drilling Mud
Physical Properties of Drilling Mud
Classification of Muds Based on Liquid Phase
Pressurized Mud Balance
Marsh Funnel
Rotational Viscometer
Rotational-Viscometer Geometry
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 2 Drilling Fluid Properties Lesson 2 Objectives
Specific Gravity
Hydrostatic Pressure Calculation
Class Activity Hydrostatic Pressure Examples
Pilot Testing Procedures
Class Activity Pilot Testing Procedure Example
Desired Viscosity
Weight or Density Control
Unit
Lesson 3 Buoyancy and Hook Loads Lesson 3 Objectives
Hook Loads
Buoyancy Example of Archimedes Principle
Hook Load and Buoyancy Calculation Examples
Casing Loads
Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
3
Lesson 1 Functions of Drilling Fluids
4
In this lesson we will
Describe the essential functions a properly designed and maintained drilling fluid performs during well construction
List the properties important to the function of removing cuttings of the drilling mud
Define how the control of oil gas or water formation pressure is accomplished by a hydrostatic pressure
Define how gel helps reduce the power costs of cutting
Define how suspending solids is accomplished
Define how the deposit of cuttings in the mud pit or mud ditch is accomplished
List the negative functions of drilling mud
List the physical properties of drilling mud
List the classification of muds based on liquid phase
Lesson 1 Functions of Drilling Fluids Learning Objectives
5
Functions of Drilling Fluids
A properly designed and maintained drilling fluid performs essential functions during well construction such as
Transporting cuttings to the surface
Preventing well-control issues and wellbore stability
Minimizing formation damage
Cooling and lubricating the drillstring
Providing information about the formation
6
Remove Cuttings
Properties important to the function of removing cuttings of the drilling mud are
Density = rho (ρ)
Viscosity = mu (micro)
Annular Velocity = Va
Type of Flow Size Shape and Density of the Cuttings
Suspension of the Cuttings andor the Gelling Properties
7
Prevent Caving
This important property helps us by
Controlling the hydrostatic head
Consolidating loose or clay type formations by surrounding the particles and holding them in the fluid and gelled mud
Controlling water intrusion prevents clays from swelling and sloughing into the hole mud cake and filtration properties
Source httppetroleumsupportcommechanical-sticking-mechanism-of-stuck-pipe
8
Controlling Oil Gas and Water Formation Pressures
The control of oil gas or water formation pressure is accomplished by
A hydrostatic pressure in this consideration we are worried about
Loss circulation
Gas cut mud
The formations being drilled
Source httpwwwdcmudcleaningequipmentcomHow-Does-Gas-Cut-Affect-Mud-Densityhtml
9
Cooling and Lubricating the Drillstring
Lubrication of the drill pipe the hole wall the casing and the mud pumps is accomplished by
The gel due to clay content
Reducing the power cost to increase the drilling speed
Source httpswwwrigzonecomtraininginsightaspinsight_id=291ampc_id=24
10
Suspending Solids
Suspending of solids is accomplished by
Gel strength thixotropic properties
The holding of cuttings when static
Returning to fluid state when circulation is restored
Sourcehttptrenchlessonlinecomindexwebapp-stories-actionid2489archiveyesIssue2013-04-01titlegel-strengths-for-horizontal-vs-vertical-drilling
11
Deposit of Cuttings
The deposit of cuttings in the mud pit or mud ditch is accomplished by
A careful balance between gel strength and viscosity
Considering velocities as an important factor
The use of the shale shaker and other separation devices at the surface
Source httpindonesiabentoniteblogspotcom201310bentonite-drilling-fluidhtml
Source httpwwwptarmiganservicescomnewsbakken-solids-control-and-recycling
12
Negative Functions of Drilling Mud
Some of the negative functions - which we donrsquot want the drilling mud to do are
Deposit of thick mud cake (reduces the diameter of hole worsens swabbing and further caving)
Fluid loss allowing a harmful amount of water into the formation
Causes swelling
Disintegration of the shales and clays
And may reduce the permeability to hydrocarbons (oil and gas)
Source httpservicepompablogspotcompkendala-kendala-teknishtml
Source httpinibumiblogspotcom201102invasion-drilling-processhtml
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
2
Module 5 Fluids Pressure Control
Lesson 1 Functions of Drilling Fluids Lesson 1 Objectives
Functions of Drilling Fluids
Negative Functions of Drilling Mud
Physical Properties of Drilling Mud
Classification of Muds Based on Liquid Phase
Pressurized Mud Balance
Marsh Funnel
Rotational Viscometer
Rotational-Viscometer Geometry
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 2 Drilling Fluid Properties Lesson 2 Objectives
Specific Gravity
Hydrostatic Pressure Calculation
Class Activity Hydrostatic Pressure Examples
Pilot Testing Procedures
Class Activity Pilot Testing Procedure Example
Desired Viscosity
Weight or Density Control
Unit
Lesson 3 Buoyancy and Hook Loads Lesson 3 Objectives
Hook Loads
Buoyancy Example of Archimedes Principle
Hook Load and Buoyancy Calculation Examples
Casing Loads
Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
3
Lesson 1 Functions of Drilling Fluids
4
In this lesson we will
Describe the essential functions a properly designed and maintained drilling fluid performs during well construction
List the properties important to the function of removing cuttings of the drilling mud
Define how the control of oil gas or water formation pressure is accomplished by a hydrostatic pressure
Define how gel helps reduce the power costs of cutting
Define how suspending solids is accomplished
Define how the deposit of cuttings in the mud pit or mud ditch is accomplished
List the negative functions of drilling mud
List the physical properties of drilling mud
List the classification of muds based on liquid phase
Lesson 1 Functions of Drilling Fluids Learning Objectives
5
Functions of Drilling Fluids
A properly designed and maintained drilling fluid performs essential functions during well construction such as
Transporting cuttings to the surface
Preventing well-control issues and wellbore stability
Minimizing formation damage
Cooling and lubricating the drillstring
Providing information about the formation
6
Remove Cuttings
Properties important to the function of removing cuttings of the drilling mud are
Density = rho (ρ)
Viscosity = mu (micro)
Annular Velocity = Va
Type of Flow Size Shape and Density of the Cuttings
Suspension of the Cuttings andor the Gelling Properties
7
Prevent Caving
This important property helps us by
Controlling the hydrostatic head
Consolidating loose or clay type formations by surrounding the particles and holding them in the fluid and gelled mud
Controlling water intrusion prevents clays from swelling and sloughing into the hole mud cake and filtration properties
Source httppetroleumsupportcommechanical-sticking-mechanism-of-stuck-pipe
8
Controlling Oil Gas and Water Formation Pressures
The control of oil gas or water formation pressure is accomplished by
A hydrostatic pressure in this consideration we are worried about
Loss circulation
Gas cut mud
The formations being drilled
Source httpwwwdcmudcleaningequipmentcomHow-Does-Gas-Cut-Affect-Mud-Densityhtml
9
Cooling and Lubricating the Drillstring
Lubrication of the drill pipe the hole wall the casing and the mud pumps is accomplished by
The gel due to clay content
Reducing the power cost to increase the drilling speed
Source httpswwwrigzonecomtraininginsightaspinsight_id=291ampc_id=24
10
Suspending Solids
Suspending of solids is accomplished by
Gel strength thixotropic properties
The holding of cuttings when static
Returning to fluid state when circulation is restored
Sourcehttptrenchlessonlinecomindexwebapp-stories-actionid2489archiveyesIssue2013-04-01titlegel-strengths-for-horizontal-vs-vertical-drilling
11
Deposit of Cuttings
The deposit of cuttings in the mud pit or mud ditch is accomplished by
A careful balance between gel strength and viscosity
Considering velocities as an important factor
The use of the shale shaker and other separation devices at the surface
Source httpindonesiabentoniteblogspotcom201310bentonite-drilling-fluidhtml
Source httpwwwptarmiganservicescomnewsbakken-solids-control-and-recycling
12
Negative Functions of Drilling Mud
Some of the negative functions - which we donrsquot want the drilling mud to do are
Deposit of thick mud cake (reduces the diameter of hole worsens swabbing and further caving)
Fluid loss allowing a harmful amount of water into the formation
Causes swelling
Disintegration of the shales and clays
And may reduce the permeability to hydrocarbons (oil and gas)
Source httpservicepompablogspotcompkendala-kendala-teknishtml
Source httpinibumiblogspotcom201102invasion-drilling-processhtml
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
3
Lesson 1 Functions of Drilling Fluids
4
In this lesson we will
Describe the essential functions a properly designed and maintained drilling fluid performs during well construction
List the properties important to the function of removing cuttings of the drilling mud
Define how the control of oil gas or water formation pressure is accomplished by a hydrostatic pressure
Define how gel helps reduce the power costs of cutting
Define how suspending solids is accomplished
Define how the deposit of cuttings in the mud pit or mud ditch is accomplished
List the negative functions of drilling mud
List the physical properties of drilling mud
List the classification of muds based on liquid phase
Lesson 1 Functions of Drilling Fluids Learning Objectives
5
Functions of Drilling Fluids
A properly designed and maintained drilling fluid performs essential functions during well construction such as
Transporting cuttings to the surface
Preventing well-control issues and wellbore stability
Minimizing formation damage
Cooling and lubricating the drillstring
Providing information about the formation
6
Remove Cuttings
Properties important to the function of removing cuttings of the drilling mud are
Density = rho (ρ)
Viscosity = mu (micro)
Annular Velocity = Va
Type of Flow Size Shape and Density of the Cuttings
Suspension of the Cuttings andor the Gelling Properties
7
Prevent Caving
This important property helps us by
Controlling the hydrostatic head
Consolidating loose or clay type formations by surrounding the particles and holding them in the fluid and gelled mud
Controlling water intrusion prevents clays from swelling and sloughing into the hole mud cake and filtration properties
Source httppetroleumsupportcommechanical-sticking-mechanism-of-stuck-pipe
8
Controlling Oil Gas and Water Formation Pressures
The control of oil gas or water formation pressure is accomplished by
A hydrostatic pressure in this consideration we are worried about
Loss circulation
Gas cut mud
The formations being drilled
Source httpwwwdcmudcleaningequipmentcomHow-Does-Gas-Cut-Affect-Mud-Densityhtml
9
Cooling and Lubricating the Drillstring
Lubrication of the drill pipe the hole wall the casing and the mud pumps is accomplished by
The gel due to clay content
Reducing the power cost to increase the drilling speed
Source httpswwwrigzonecomtraininginsightaspinsight_id=291ampc_id=24
10
Suspending Solids
Suspending of solids is accomplished by
Gel strength thixotropic properties
The holding of cuttings when static
Returning to fluid state when circulation is restored
Sourcehttptrenchlessonlinecomindexwebapp-stories-actionid2489archiveyesIssue2013-04-01titlegel-strengths-for-horizontal-vs-vertical-drilling
11
Deposit of Cuttings
The deposit of cuttings in the mud pit or mud ditch is accomplished by
A careful balance between gel strength and viscosity
Considering velocities as an important factor
The use of the shale shaker and other separation devices at the surface
Source httpindonesiabentoniteblogspotcom201310bentonite-drilling-fluidhtml
Source httpwwwptarmiganservicescomnewsbakken-solids-control-and-recycling
12
Negative Functions of Drilling Mud
Some of the negative functions - which we donrsquot want the drilling mud to do are
Deposit of thick mud cake (reduces the diameter of hole worsens swabbing and further caving)
Fluid loss allowing a harmful amount of water into the formation
Causes swelling
Disintegration of the shales and clays
And may reduce the permeability to hydrocarbons (oil and gas)
Source httpservicepompablogspotcompkendala-kendala-teknishtml
Source httpinibumiblogspotcom201102invasion-drilling-processhtml
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
4
In this lesson we will
Describe the essential functions a properly designed and maintained drilling fluid performs during well construction
List the properties important to the function of removing cuttings of the drilling mud
Define how the control of oil gas or water formation pressure is accomplished by a hydrostatic pressure
Define how gel helps reduce the power costs of cutting
Define how suspending solids is accomplished
Define how the deposit of cuttings in the mud pit or mud ditch is accomplished
List the negative functions of drilling mud
List the physical properties of drilling mud
List the classification of muds based on liquid phase
Lesson 1 Functions of Drilling Fluids Learning Objectives
5
Functions of Drilling Fluids
A properly designed and maintained drilling fluid performs essential functions during well construction such as
Transporting cuttings to the surface
Preventing well-control issues and wellbore stability
Minimizing formation damage
Cooling and lubricating the drillstring
Providing information about the formation
6
Remove Cuttings
Properties important to the function of removing cuttings of the drilling mud are
Density = rho (ρ)
Viscosity = mu (micro)
Annular Velocity = Va
Type of Flow Size Shape and Density of the Cuttings
Suspension of the Cuttings andor the Gelling Properties
7
Prevent Caving
This important property helps us by
Controlling the hydrostatic head
Consolidating loose or clay type formations by surrounding the particles and holding them in the fluid and gelled mud
Controlling water intrusion prevents clays from swelling and sloughing into the hole mud cake and filtration properties
Source httppetroleumsupportcommechanical-sticking-mechanism-of-stuck-pipe
8
Controlling Oil Gas and Water Formation Pressures
The control of oil gas or water formation pressure is accomplished by
A hydrostatic pressure in this consideration we are worried about
Loss circulation
Gas cut mud
The formations being drilled
Source httpwwwdcmudcleaningequipmentcomHow-Does-Gas-Cut-Affect-Mud-Densityhtml
9
Cooling and Lubricating the Drillstring
Lubrication of the drill pipe the hole wall the casing and the mud pumps is accomplished by
The gel due to clay content
Reducing the power cost to increase the drilling speed
Source httpswwwrigzonecomtraininginsightaspinsight_id=291ampc_id=24
10
Suspending Solids
Suspending of solids is accomplished by
Gel strength thixotropic properties
The holding of cuttings when static
Returning to fluid state when circulation is restored
Sourcehttptrenchlessonlinecomindexwebapp-stories-actionid2489archiveyesIssue2013-04-01titlegel-strengths-for-horizontal-vs-vertical-drilling
11
Deposit of Cuttings
The deposit of cuttings in the mud pit or mud ditch is accomplished by
A careful balance between gel strength and viscosity
Considering velocities as an important factor
The use of the shale shaker and other separation devices at the surface
Source httpindonesiabentoniteblogspotcom201310bentonite-drilling-fluidhtml
Source httpwwwptarmiganservicescomnewsbakken-solids-control-and-recycling
12
Negative Functions of Drilling Mud
Some of the negative functions - which we donrsquot want the drilling mud to do are
Deposit of thick mud cake (reduces the diameter of hole worsens swabbing and further caving)
Fluid loss allowing a harmful amount of water into the formation
Causes swelling
Disintegration of the shales and clays
And may reduce the permeability to hydrocarbons (oil and gas)
Source httpservicepompablogspotcompkendala-kendala-teknishtml
Source httpinibumiblogspotcom201102invasion-drilling-processhtml
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
5
Functions of Drilling Fluids
A properly designed and maintained drilling fluid performs essential functions during well construction such as
Transporting cuttings to the surface
Preventing well-control issues and wellbore stability
Minimizing formation damage
Cooling and lubricating the drillstring
Providing information about the formation
6
Remove Cuttings
Properties important to the function of removing cuttings of the drilling mud are
Density = rho (ρ)
Viscosity = mu (micro)
Annular Velocity = Va
Type of Flow Size Shape and Density of the Cuttings
Suspension of the Cuttings andor the Gelling Properties
7
Prevent Caving
This important property helps us by
Controlling the hydrostatic head
Consolidating loose or clay type formations by surrounding the particles and holding them in the fluid and gelled mud
Controlling water intrusion prevents clays from swelling and sloughing into the hole mud cake and filtration properties
Source httppetroleumsupportcommechanical-sticking-mechanism-of-stuck-pipe
8
Controlling Oil Gas and Water Formation Pressures
The control of oil gas or water formation pressure is accomplished by
A hydrostatic pressure in this consideration we are worried about
Loss circulation
Gas cut mud
The formations being drilled
Source httpwwwdcmudcleaningequipmentcomHow-Does-Gas-Cut-Affect-Mud-Densityhtml
9
Cooling and Lubricating the Drillstring
Lubrication of the drill pipe the hole wall the casing and the mud pumps is accomplished by
The gel due to clay content
Reducing the power cost to increase the drilling speed
Source httpswwwrigzonecomtraininginsightaspinsight_id=291ampc_id=24
10
Suspending Solids
Suspending of solids is accomplished by
Gel strength thixotropic properties
The holding of cuttings when static
Returning to fluid state when circulation is restored
Sourcehttptrenchlessonlinecomindexwebapp-stories-actionid2489archiveyesIssue2013-04-01titlegel-strengths-for-horizontal-vs-vertical-drilling
11
Deposit of Cuttings
The deposit of cuttings in the mud pit or mud ditch is accomplished by
A careful balance between gel strength and viscosity
Considering velocities as an important factor
The use of the shale shaker and other separation devices at the surface
Source httpindonesiabentoniteblogspotcom201310bentonite-drilling-fluidhtml
Source httpwwwptarmiganservicescomnewsbakken-solids-control-and-recycling
12
Negative Functions of Drilling Mud
Some of the negative functions - which we donrsquot want the drilling mud to do are
Deposit of thick mud cake (reduces the diameter of hole worsens swabbing and further caving)
Fluid loss allowing a harmful amount of water into the formation
Causes swelling
Disintegration of the shales and clays
And may reduce the permeability to hydrocarbons (oil and gas)
Source httpservicepompablogspotcompkendala-kendala-teknishtml
Source httpinibumiblogspotcom201102invasion-drilling-processhtml
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
6
Remove Cuttings
Properties important to the function of removing cuttings of the drilling mud are
Density = rho (ρ)
Viscosity = mu (micro)
Annular Velocity = Va
Type of Flow Size Shape and Density of the Cuttings
Suspension of the Cuttings andor the Gelling Properties
7
Prevent Caving
This important property helps us by
Controlling the hydrostatic head
Consolidating loose or clay type formations by surrounding the particles and holding them in the fluid and gelled mud
Controlling water intrusion prevents clays from swelling and sloughing into the hole mud cake and filtration properties
Source httppetroleumsupportcommechanical-sticking-mechanism-of-stuck-pipe
8
Controlling Oil Gas and Water Formation Pressures
The control of oil gas or water formation pressure is accomplished by
A hydrostatic pressure in this consideration we are worried about
Loss circulation
Gas cut mud
The formations being drilled
Source httpwwwdcmudcleaningequipmentcomHow-Does-Gas-Cut-Affect-Mud-Densityhtml
9
Cooling and Lubricating the Drillstring
Lubrication of the drill pipe the hole wall the casing and the mud pumps is accomplished by
The gel due to clay content
Reducing the power cost to increase the drilling speed
Source httpswwwrigzonecomtraininginsightaspinsight_id=291ampc_id=24
10
Suspending Solids
Suspending of solids is accomplished by
Gel strength thixotropic properties
The holding of cuttings when static
Returning to fluid state when circulation is restored
Sourcehttptrenchlessonlinecomindexwebapp-stories-actionid2489archiveyesIssue2013-04-01titlegel-strengths-for-horizontal-vs-vertical-drilling
11
Deposit of Cuttings
The deposit of cuttings in the mud pit or mud ditch is accomplished by
A careful balance between gel strength and viscosity
Considering velocities as an important factor
The use of the shale shaker and other separation devices at the surface
Source httpindonesiabentoniteblogspotcom201310bentonite-drilling-fluidhtml
Source httpwwwptarmiganservicescomnewsbakken-solids-control-and-recycling
12
Negative Functions of Drilling Mud
Some of the negative functions - which we donrsquot want the drilling mud to do are
Deposit of thick mud cake (reduces the diameter of hole worsens swabbing and further caving)
Fluid loss allowing a harmful amount of water into the formation
Causes swelling
Disintegration of the shales and clays
And may reduce the permeability to hydrocarbons (oil and gas)
Source httpservicepompablogspotcompkendala-kendala-teknishtml
Source httpinibumiblogspotcom201102invasion-drilling-processhtml
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
7
Prevent Caving
This important property helps us by
Controlling the hydrostatic head
Consolidating loose or clay type formations by surrounding the particles and holding them in the fluid and gelled mud
Controlling water intrusion prevents clays from swelling and sloughing into the hole mud cake and filtration properties
Source httppetroleumsupportcommechanical-sticking-mechanism-of-stuck-pipe
8
Controlling Oil Gas and Water Formation Pressures
The control of oil gas or water formation pressure is accomplished by
A hydrostatic pressure in this consideration we are worried about
Loss circulation
Gas cut mud
The formations being drilled
Source httpwwwdcmudcleaningequipmentcomHow-Does-Gas-Cut-Affect-Mud-Densityhtml
9
Cooling and Lubricating the Drillstring
Lubrication of the drill pipe the hole wall the casing and the mud pumps is accomplished by
The gel due to clay content
Reducing the power cost to increase the drilling speed
Source httpswwwrigzonecomtraininginsightaspinsight_id=291ampc_id=24
10
Suspending Solids
Suspending of solids is accomplished by
Gel strength thixotropic properties
The holding of cuttings when static
Returning to fluid state when circulation is restored
Sourcehttptrenchlessonlinecomindexwebapp-stories-actionid2489archiveyesIssue2013-04-01titlegel-strengths-for-horizontal-vs-vertical-drilling
11
Deposit of Cuttings
The deposit of cuttings in the mud pit or mud ditch is accomplished by
A careful balance between gel strength and viscosity
Considering velocities as an important factor
The use of the shale shaker and other separation devices at the surface
Source httpindonesiabentoniteblogspotcom201310bentonite-drilling-fluidhtml
Source httpwwwptarmiganservicescomnewsbakken-solids-control-and-recycling
12
Negative Functions of Drilling Mud
Some of the negative functions - which we donrsquot want the drilling mud to do are
Deposit of thick mud cake (reduces the diameter of hole worsens swabbing and further caving)
Fluid loss allowing a harmful amount of water into the formation
Causes swelling
Disintegration of the shales and clays
And may reduce the permeability to hydrocarbons (oil and gas)
Source httpservicepompablogspotcompkendala-kendala-teknishtml
Source httpinibumiblogspotcom201102invasion-drilling-processhtml
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
8
Controlling Oil Gas and Water Formation Pressures
The control of oil gas or water formation pressure is accomplished by
A hydrostatic pressure in this consideration we are worried about
Loss circulation
Gas cut mud
The formations being drilled
Source httpwwwdcmudcleaningequipmentcomHow-Does-Gas-Cut-Affect-Mud-Densityhtml
9
Cooling and Lubricating the Drillstring
Lubrication of the drill pipe the hole wall the casing and the mud pumps is accomplished by
The gel due to clay content
Reducing the power cost to increase the drilling speed
Source httpswwwrigzonecomtraininginsightaspinsight_id=291ampc_id=24
10
Suspending Solids
Suspending of solids is accomplished by
Gel strength thixotropic properties
The holding of cuttings when static
Returning to fluid state when circulation is restored
Sourcehttptrenchlessonlinecomindexwebapp-stories-actionid2489archiveyesIssue2013-04-01titlegel-strengths-for-horizontal-vs-vertical-drilling
11
Deposit of Cuttings
The deposit of cuttings in the mud pit or mud ditch is accomplished by
A careful balance between gel strength and viscosity
Considering velocities as an important factor
The use of the shale shaker and other separation devices at the surface
Source httpindonesiabentoniteblogspotcom201310bentonite-drilling-fluidhtml
Source httpwwwptarmiganservicescomnewsbakken-solids-control-and-recycling
12
Negative Functions of Drilling Mud
Some of the negative functions - which we donrsquot want the drilling mud to do are
Deposit of thick mud cake (reduces the diameter of hole worsens swabbing and further caving)
Fluid loss allowing a harmful amount of water into the formation
Causes swelling
Disintegration of the shales and clays
And may reduce the permeability to hydrocarbons (oil and gas)
Source httpservicepompablogspotcompkendala-kendala-teknishtml
Source httpinibumiblogspotcom201102invasion-drilling-processhtml
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
9
Cooling and Lubricating the Drillstring
Lubrication of the drill pipe the hole wall the casing and the mud pumps is accomplished by
The gel due to clay content
Reducing the power cost to increase the drilling speed
Source httpswwwrigzonecomtraininginsightaspinsight_id=291ampc_id=24
10
Suspending Solids
Suspending of solids is accomplished by
Gel strength thixotropic properties
The holding of cuttings when static
Returning to fluid state when circulation is restored
Sourcehttptrenchlessonlinecomindexwebapp-stories-actionid2489archiveyesIssue2013-04-01titlegel-strengths-for-horizontal-vs-vertical-drilling
11
Deposit of Cuttings
The deposit of cuttings in the mud pit or mud ditch is accomplished by
A careful balance between gel strength and viscosity
Considering velocities as an important factor
The use of the shale shaker and other separation devices at the surface
Source httpindonesiabentoniteblogspotcom201310bentonite-drilling-fluidhtml
Source httpwwwptarmiganservicescomnewsbakken-solids-control-and-recycling
12
Negative Functions of Drilling Mud
Some of the negative functions - which we donrsquot want the drilling mud to do are
Deposit of thick mud cake (reduces the diameter of hole worsens swabbing and further caving)
Fluid loss allowing a harmful amount of water into the formation
Causes swelling
Disintegration of the shales and clays
And may reduce the permeability to hydrocarbons (oil and gas)
Source httpservicepompablogspotcompkendala-kendala-teknishtml
Source httpinibumiblogspotcom201102invasion-drilling-processhtml
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
10
Suspending Solids
Suspending of solids is accomplished by
Gel strength thixotropic properties
The holding of cuttings when static
Returning to fluid state when circulation is restored
Sourcehttptrenchlessonlinecomindexwebapp-stories-actionid2489archiveyesIssue2013-04-01titlegel-strengths-for-horizontal-vs-vertical-drilling
11
Deposit of Cuttings
The deposit of cuttings in the mud pit or mud ditch is accomplished by
A careful balance between gel strength and viscosity
Considering velocities as an important factor
The use of the shale shaker and other separation devices at the surface
Source httpindonesiabentoniteblogspotcom201310bentonite-drilling-fluidhtml
Source httpwwwptarmiganservicescomnewsbakken-solids-control-and-recycling
12
Negative Functions of Drilling Mud
Some of the negative functions - which we donrsquot want the drilling mud to do are
Deposit of thick mud cake (reduces the diameter of hole worsens swabbing and further caving)
Fluid loss allowing a harmful amount of water into the formation
Causes swelling
Disintegration of the shales and clays
And may reduce the permeability to hydrocarbons (oil and gas)
Source httpservicepompablogspotcompkendala-kendala-teknishtml
Source httpinibumiblogspotcom201102invasion-drilling-processhtml
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
11
Deposit of Cuttings
The deposit of cuttings in the mud pit or mud ditch is accomplished by
A careful balance between gel strength and viscosity
Considering velocities as an important factor
The use of the shale shaker and other separation devices at the surface
Source httpindonesiabentoniteblogspotcom201310bentonite-drilling-fluidhtml
Source httpwwwptarmiganservicescomnewsbakken-solids-control-and-recycling
12
Negative Functions of Drilling Mud
Some of the negative functions - which we donrsquot want the drilling mud to do are
Deposit of thick mud cake (reduces the diameter of hole worsens swabbing and further caving)
Fluid loss allowing a harmful amount of water into the formation
Causes swelling
Disintegration of the shales and clays
And may reduce the permeability to hydrocarbons (oil and gas)
Source httpservicepompablogspotcompkendala-kendala-teknishtml
Source httpinibumiblogspotcom201102invasion-drilling-processhtml
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
12
Negative Functions of Drilling Mud
Some of the negative functions - which we donrsquot want the drilling mud to do are
Deposit of thick mud cake (reduces the diameter of hole worsens swabbing and further caving)
Fluid loss allowing a harmful amount of water into the formation
Causes swelling
Disintegration of the shales and clays
And may reduce the permeability to hydrocarbons (oil and gas)
Source httpservicepompablogspotcompkendala-kendala-teknishtml
Source httpinibumiblogspotcom201102invasion-drilling-processhtml
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
13
Physical Properties of Drilling Mud
Physical properties of drilling mud
Density
Viscosity
Filtration properties such as water loss and mud cake
The yield point
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
14
Classification of Muds Based on Liquid Phase
Freshwater
Natural or Native
Nitrate
Phosphate
Organic colloidal
Alkaline (pH gt 10)
Calcium
Lime
Gypsum
Saltwater
Saturated salt
Emulsion
Freshwater oil in water emulsion
Saltwater oil in water emulsion
Oil-based
Note Muds are listed in order of expense from low to high
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
15
Pressurized Mud Balance
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
16
Marsh Funnel
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
17
Rotational Viscometer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
18
Viscometer RevMin
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
19
Problem Solving Class Activity
In pairs solve the following problem
At 200 revmin what is the shear stress
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
20
Rotational-Viscometer Geometry
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
21
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 51 Read Fundamentals of Drilling Engineering pp 87-98
Lesson 1 Wrap Up
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
22
Lesson 2 Drilling Fluid Properties
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
23
In this lesson we will
Calculate specific gravity
Calculate hydrostatic pressure
Demonstrate pilot testing procedures
Calculate weight or density control
Lesson 2 Drilling Fluid Properties Learning Objectives
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
24
Specific Gravity
The ratio of the weight of a given volume of material to the weight of the same volume of water (fresh)
or
SpGr of water = 10 = 10 gmcm3
then
If a fluid weight is 24 gmcm3
SpGr = 24 gmcm3 = 24
Density (ρ) Mass per volume of a material in any units
or
ρ = Mass
Vol
Common units used for drilling fluids
gmcm3 (or SpGr) lbmgal lbmft3 ξ lbmbbl
therefore
Density of fresh water = 1 gmcm3
= 834 lbmgal
= 624 lbmft3
= 350 lbmbbl
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
25
Hydrostatic Pressure Calculation
Force per unit area exerted by a vertical column of fluid
or
Common units gmfcm2 lbfin2 or lbfft2
Using a 1 foot container whose base is 1 ft2 (or 144 in2) and height is 1 ft filled
with water the force exerted on the base will be 624 lbf
therefore
Pressure = P = ForceArea = 624 lbf = 624 lbfft2
10 ft2
or
P = 624 lbf = 0433 lbfin2ft = 0433 psift
144 in2
therefore
Water = SpGr of 10 exerts a pressure of 0433 psifoot of vertical column
Water
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
26
Hydrostatic Head and Hydrostatic Pressure
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
27
Hydrostatic PressuremdashOther Fluids
Other fluids
Wtft3 = (624) (SpGr)
Then
Pressure exerted = (624) (SpGr) lbf = lbfin2ft
144 in2
Or
lbfin2ft = (0433) (SpGr)
Or
lbfin2 = (0433 psift) (SpGr) (Height)
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
28
Class Activity Hydrostatic Pressure Examples
Example 1
What is the SpGr of a fluid whose density is 78 lbmft3
Solution 1
SpGr = 78 lbmft3 = 125
624 lbmft3
Example 2
What is the density in lbmgal of a fluid whose SpGr is 13
Solution 2
ρ= (13) (834 lbmgal) = 1084 lbmgal
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
29
Example 3
What is the density in lbmbbl of a fluid whose density is 115 lbgal 13
Solution 3
ρ = (115 lbmgal) (42 galbbl) = 4830 lbmbbl
Example 4
What is the total weight of 10 bbl of material whose SpGr is 43
Solution 4
Wt = (ρ) (Vol)
lbm = lbm (bbl) = (43) (350 lbmbbl) (10 bbl) = 15050 lbm
bbl
Class Activity Hydrostatic Pressure Examples (Cont)
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
30
Example 5
Calculate the density in all common units of a fluid if 3 ft3 of the fluid weighs 500 lbs
Solution 5
ρ = 500 lbm = 1667 lbmft3
3 ft3
ρ = 1667 lbmft3 = 223 lbmgal
748 galft3
ρ = (1667 lbmft3) (5615 ft3bbl) = 9630 lbmbbl
ρ = (1667 lbm) (454 gmlbm) = 267 gmcm3
(ft3) (28320 cm3ft3)
or
SpGr = 1667 lbmft3 = 267
624 lbmft3
Class Activity Hydrostatic Pressure Examples (Cont)
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
31
Example 6
What pressure will a 94 lbmgal mud exert at a depth of 3500 ft
Solution 6
psi = (SpGr) (0433) (height)
= ( 94 ) ( 0433) (3500) = 1708 psi
834
(Note SpGr = lbmgal
834
and
psi = (SpGr) (0433) (h)
psi = lbmgal (0433) (h) = (lbmgal) ( 0433 ) (h)
834 834
psi = (lbmgal) (0052) (h)
Or
psi = (94) (0052) (3500) = 1711 psi
Note 1 cubic foot contains 748 US gallons a fluid weighing 1 ppg would
weigh 748 pounds per cubic foot The pressure exerted by one foot height of
fluid over the area of the base would be748144 in2=0052 psi
Class Activity Hydrostatic Pressure Examples (Cont)
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
32
Example 7
What density mud is required to exert a pressure of 3000 psi at a depth of 5000 ft
Solution
ρ = psi = 3000 = 1154 lbmgal
(0052) (h) (0052) (5000)
Class Activity Hydrostatic Pressure Examples (Cont)
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
33
Example 8
Point ldquoArdquo is at an elevation of 1200 ft And Point ldquoBrdquo is 2 miles east at an elevation of 900 ft A 2 pipe line is carrying water from ldquoArdquo to ldquoBrdquo What is the difference in hydrostatic pressure
Solution 8
psi = (SpGr) (0433) (h) where h = vertical height
psi = (10) (0433) (1200-900) = 130 psi
Class Activity Hydrostatic Pressure Examples (Cont)
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
34
Pilot Testing Procedures
Water weighs 1 gm per cm3 or 350 gms per 350 cm3 and 1 barrel of water weighs 350 lbs therefore adding 1 gm of material to a 350 gm sample is equivalent to adding 1 lb of material to 350 lbs of the same sample
Or
1gm of material added to 350 cm3 of a sample is equivalent to adding 1 lb of material to 1 bbl of the sample
Therefore a lab barrel will be a 350 cm3 emulating a 350 lb actual barrel
Note Often the density of a fluid is referred to as the weight of the fluid Often the industry and the public do not differentiate between mass and weight (weight is actually mass X acceleration of gravity)
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
35
Pilot Testing Procedures
Another way of looking at it
Mass Volume350 pounds mass 1 blue barrel of water = 1589873 litres
Manipulations explain how many
pounds per barrel of additives needed
to change properties of the fluid
Experiments that are scaled down so
that adding X more pounds to existing
350 pounds is equivalent to adding X
more small units of mass to 350
existing small units of mass
Experiments that are scaled down so
that 1 blue barrel volume is equivalent
another volume for small units of
mass
A small unit is a gram Scale the
volume for the same proportion of
these units1 gram is 1454 of a pound
New volume is 1454 blue barrels
accordingly
1589873 454 = 0350 liters = 350
cubic centimeters
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
36
Pilot Testing Procedure (Cont)
Pilot test procedure must be used to determine the amount of an additive needed to obtain the desired results for viscosity water loss gel strengths etc
hellipbut it is not normally used to determine the amount of additive needed to obtain the desired density
The amount of additive to obtain the desired density can be calculated the amount of additive to control other mud properties cannot be calculated
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
37
Class Activity Pilot Testing Procedure Example
Example 9
How many pounds of bentonite clay must be added to an original system whose viscosity is 5 cP to raise the viscosity to 20 cP
Solution 9 Using a 350 cm3 sample of the original mud the following
laboratory data were obtained
Plot gms of bentonite added to a 350 cm3 sample vs Resulting viscosity in cP
Bentonite added gms Resulting viscosity cP
0 5
4 8
6 12
8 18
16 28
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
38
Desired Viscosity
The desired viscosity of 20 cP can be read from the curve as shown
or
125 gms of bentonite added to the original sample of 350 cmsup3 results in a viscosity of 20 cP
or
125 gms350 cmsup3 _ 125 lbs 1 bbl of the system
0
5
10
15
20
25
30
0 5 10 15 20
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
39
Weight or Density Control
The following relationships are used to calculate mud weighting problems
1Mi + Ma = Mf
2Vi + Va = Vfand
ρ = M M = ρ middotV
Vthen
3 ρi middot Vi + ρa middot Va = ρf middot Vfwhere
Mi = Initial mass Vi = Initial volume ρi = Initial density
Ma = Added mass Va = Added volume ρa = Added density
Mf = Final mass Vf = Final volume ρf = Final density
(Note The above relationships assume no chemical reactions)
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
40
Unit
Any consistent units can be used in Equation 3 if the product of ρ x V is the same in each term
or
M = lbm
ρ = SpGr lbmgal lbmft3 or lbmbbl
V = cm3 gal ft3 or bbl
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
41
Class Activity Unit Example 10
What will be the resulting specific gravity if 01 bbl of clay is added to 10 bbl of water
SpGrrsquos water = 1 and clay = 25
Solution 10
Using Example 3) with ρ = SpGr and V = bbl
Assume water = initial and clay = added
then
SpGri bbli + SpGra bbla = SpGrf bblfand
ρi = 10 Vi = 10 bbl
ρa = 25 Va = 01 bbl
ρf = Vf = (Vi + Va) = (10 + 01) = 101 bbl
then
ρiVi + ρaVa = ρfVf
(10) (10) + (25) (01) = ρf(101)
ρf = (10 + 025) = 101 SpGr
(101)
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
42
Class Activity Unit Example 11
What will be the resulting density in lbmgal if 875 lbs of clay is added to 10 bbl of water SpGrrsquos water = 10 and clay = 25
Solution 11
(Using Example 3) with ρ = lbmgal and V = bbl
Assume water = initial and clay = added
then
lbmgali Vi + lbmgala Va = lbmgalf Vf
and
ρi = (SpGr) (834) = (10) (834) = 834 lbmgal
ρa = (SpGr) (834) = (25) (834) = 2085 lbmgal
ρf =
Vi = 10 bbl
Va = Maρa = Ma = 875 lbm ____ = 01bbl
(SpGr) (350 lbmbbl) 25 times (350 lbmgal)
Vf = (Vi + Va) = (10 +01) = 101 bbl
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
43
Solution 11 continued
then
ρiVi + ρaVa = ρfVf
(834) (10) + (2085) (01) = ρf (101)
ρf = (834 + 2085) = 846 lbmgal
(101)
Note Compare to Example 10 846 = 101 SpGr
834
Class Activity Unit Example 11 (Cont)
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
44
Class Activity Unit Example 12
How many lbs of clay must be added to 6000 gals of water to produce a final density of 65 lbmft3 SpGrrsquos water = 10 and clay = 265
Solution 12
Using Equation 3 where ρ = lbmft3 and V = gals
Assume water = initial and clay = added
and
ρi = 624 lbmft3
ρa = (SpGr) (624) = (265) (624) = 1654 lbmft3
ρf = 65 lbmft3
Vi = 6000 gal
Va = Maρa =
Vf = (Vi + Va) = (6000 + Va)
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
45
Class Activity Unit Example 12 (Cont)
Solution 12 continued
then
ρiVi + ρaVa = ρfVf
(624) (6000) + (1654) (Va) = (65) (6000+ Va)
(1654 - 65) (Va) = (65 - 624) (6000)
Va = 1554 gal
then
Ma = ρaVa = (SpGr) (834) (1554 gal) = (265) (834) (1554)
Ma = 3435 lbs
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
46
Class Activity Unit Example 13
How many bbls of water must be added to an existing system of 400 bbls of 115 lbmgal mud to reduce the density to 105 lbmgal
Solution 13
Using Equation 3) where ρ = lbmgal and V = bbl
Assume 115 lbmgal mud = initial and water = added
then
ρi = 115 lbmgal
ρa = 834 lbmgal
ρf = 105 lbmgal
Vi = 400 bbl
Va =
Vf = (Vi + Va) = (400 + Va)
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
47
Class Activity Unit Example 13 (Cont)
Solution 13 continued
then
ρiVi + ρaVa = ρf (Vi + Va)
(115) (400) + (834) (Va) = (105) (400 + Va)
(834 - 105) (Va) = (105 - 115) (400)
Va = (0463) (400) = 185 bbls
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
48
Class Activity Unit Example 14
How many lbs of barite must be added to a 300 bbl 92 lbmgal system to control a formation pressure of 2550 psi at a depth of 5000 ft
Solution 14
Assume 92 lbmgal = initial and barite = added
and
ρi = 92 lbmgal
ρa = (SpGr) (834) = (42) (834) = 35 lbmgal
ρf = Psi = 2550 = 981 lbmgal
(0052) (h) (0052) (5000)
Vi = 300 bbl
Va = Ma ρa =
Vf = (Vi + Va) = (300 + Va)
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
49
Class Activity Unit Example 14 (Cont)
Solution 14 continued
then
ρiVi + ρaVa = ρfVf
(92) (300) + (35) (Va) = (981) (300 + Va)
(35 - 981) Va = (981 - 92) (300)
Va = 726 bbl
and
Ma = ρaVa = (SpGr) (350) (Va) = (42) (350) (726) = 10672 lbs
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
50
Class Activity Unit Example 15
How many bbls of water and lbs of clay are needed to make 250 bbls of 95 lbmgal mud SpGrrsquos water = 10 and clay = 24
Solution 15
Assume water = initial and clay = added
and
ρi = 834 lbmgal
ρa = (24) (834) = 20 lbmgal
ρf = 95 lbmgal
Vi =
Va = (Vf - Vi) = (250 -Vi)
Vf = 250 bbl
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
51
Class Activity Unit Example 15 (Cont)
Solution 15 continued
then
ρρiVi + ρaVa = ρfVf
(834) (Vi) + (20) (250 - Vi) = (95) (250)
(834 - 20) Vi = (95 - 20) (250)
Vi = 225 bbls (water)
Va = (250 - Vi) = (250 - 225) = 25 bbls
Ma = ρaVa = (24) (350) (25) = 21000 lbs (clay)
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
52
Class Activity Unit Example 16
Drilling at 4400 ft with 915 lbmgal mud A pressure of 3000 psi is expected at a depth of 5600 ft Calculate the required mud treatment
Solution 16
Use barite SpGr = 43 as weighting material
Calculate treatment in bbl barite added per 1 bbl of initial system
Assume 915 lbmgal mud = initial and barite = added
and
ρi = 915 lbmgal
ρa = (43) (834) = 3586 lbmgal
ρf = 3000 = 1032 lbmgal
(00519) (5600)
Vi = 1bbl
Va =
Vf = (Vi + Va) = (1 + Va)
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
53
Class Example Unit Example 16 (Cont)
Solution 16 continued
then
ρiVi + ρaVa = ρfVf
(915) (1) + (3586)Va = (1032) (1+ Va)
(3586 - 1032) Va = (1032 - 915) (1)
Va = 0046 bbl
Ma = ρaVa = (43) (350) (0046) = 6923 lbbbl
Note When additives are added to increase or decrease mud density
other mud properties must be checked to insure they are within operating
limits
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
54
Lesson 2 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 52 Read Fundamentals of Drilling Engineering pp 98 - 119
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
55
Lesson 3 Buoyancy and Hook Loads
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
56
In this lesson we will
List three methods of calculating hook load
Describe buoyancy as an example of the Archimedes Principle
Define basic hook loads
Lesson 3 Buoyancy and Hook Loads Learning Objectives
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
57
Hook Loads
Three methods of calculating hook load
Displacement
Buoyancy Factor
Hydrostatic Pressure
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
58
Basic Hook Loads
The basic hook loads which must be known are
Weight of casing string dead weight or suspended in fluid
Weight of drill string dead weight or suspended in fluid
Weight of drill string less weight on the bit
Weight with pipe or tools stuck in the hole
Hole friction pipe or tools in contact with the hole
Weight with applied pump pressures
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
59
Buoyancy Example of Archimedes Principle
The net force of the fluid
on the cylinder is the
buoyant force FB
Fupgt Fdown because the pressure is
greater at the bottom Hence the
fluid exerts a net upward force
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
60
Archimedesrsquo Principle
Archimedesrsquo Principle
The buoyant force is equal
to the weight of the
displaced water
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
61
Buoyancy Factor
Mud Density ppg Mud Density lbft3
Buoyancy Factor (BF) = (655 ndash mud
density ppg) divide 655
Buoyancy Factor (BF) = (490 ndash mud
density lbft3) divide 490
Example
Determine the buoyancy factor for a
130 ppg fluid
BF = (655 ndash 130) divide 655
BF = 08015
Note 655 ppg is the density of steel
Example
Determine the buoyancy factor for a
9724 lbft3 fluid
BF = (490 ndash 9724) divide 490
BF = 08015
Note 490 is the density of steel
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
62
How to Use the Buoyancy Factor
Buoyed Weight
The air weight of drilling string x the buoyancy factor
= to actual weight in mud
For example determine the string weight in 130 ppg mud Air weight of string is 350000 lbf
The buoyancy factor for a 130 ppg fluidBF = (655 ndash 130) divide 655 BF = 08015
The buoyed weight of drill string in 130 ppg mud = 350 x 08015 = 280000 lbf
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
63
The Buoyant Force
The buoyant force can be expressed as
a a The buoyant force will be equal to the weight of the displaced fluid
b b The buoyant force will be equal to the force of the hydrostatic pressure acting over the exposed area at the bottom of the pipe
c c The buoyant force will be equal to the fraction of the dead weight lost when suspended in a fluid as determined using the Buoyancy Factor
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
64
Hook Load and Buoyancy Calculation Example 1
Example 1
Calculate the effective weight of 1000 ft of 9621 lbft 6 in diameter solid steel rod suspended in water
Solution
Using (a) ndash weight of displaced fluid
Dead weight = (1000) (9621) = 96210 lbf
Volume of displaced fluid = 07854 (6)2 (1000) = 19635 ft3
144
Weight of displaced fluid = (1963) (624) = 12252 lbf
Effective weight = 96210 - 12252 = 83958 lbf
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
65
Using (b) - hydrostatic pressure
Dead weight = (1000) (9621) = 96210 lbf
Hydrostatic pressure = (624144) (1000) = 4333 psi
Area of exposed bottom = (07854) (6)2= 2827 in2
Buoyant force = (4333) (2827) = 12249
Effective weight = 96210 - 12249 = 83961 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
66
Using (c) - Buoyancy factor
Defined as the fraction of the dead weight that an object will weigh when suspended in a fluid or
BF = mft3 in air - mft3 of fluid
mft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
Dead weight = (1000) (9621) = 96210 lbf
then
Effective weight = (Dead weight) (BF) = 96210 08727 = 83960 lbf
Hook Load and Buoyancy Calculation Example 1 (Cont)
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
67
Hook Load and Buoyancy Calculation Example 2
What is the buoyancy factor for oil field steel when suspended in (a) water and (b) 10 lbgal mud
Solution
(a) Density of oil field steel = 490 lbft3
Density of water = 834 lbgal = 624 lbft3
BF = 490 lbft3 - 624 lbft3 = 08727
490 lbft3
(b) BF = 490 lbft3 - (10 lbgal) (748 galft3) = 08473
490 lbft3
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
68
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 1
gt Using weight (wt) = (5000) (2256) = 112800 lbf
Volume (Vol) of displaced (displ) fluid = 07854 (552 - 4672) (5000) =
144
= 2302 ft3
Wt of displ fluid = (2302 ft3) (12 lbgal) (748 galft3) = 20663 lbf
WI = 112800 - 20663 = 92137 lbf
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
69
Hook Load and Buoyancy Calculation Example 3
Calculate the weight indicator reading when 5000 ft of 5 12 in OD 467 in ID 2256 lbft (neglecting tool joints) drill pipe is suspended open ended in a hole filled with 12 lbgal mud
Solution 2
gt Using hyd pressure
Dead wt = 112800 lbf
Ph = (0052) (125000) = 3120 psi
Exposed area = (3120) (663) = 20684 lbf -- =07854(552 - 4672)=
= 663 in2
Buoyant force =pressurearea= (3120) (663) = 20684 lbf
WI = 112800 - 20684 = 92116 lbf
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
70
Hook Load and Buoyancy Calculation Example 3 (Cont)
gt Using BF
Dead wt = 112800 lbf
BF = 490 - (12) (748) = 08168
490
WI = (08168) (112800) = 92135 lbf
Note (748) is gallons per cubic foot
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
71
Hook Load and Buoyancy Calculation Example 3 (Cont)
Displacement Volume
Since there are coupling on tubing tool joints on drill pipe collars on casing etc the volume and weight of these couplings must be considered Oil field tubular goods are described by the outside diameter OD in inches and fractions of an inch and by the weight per foot lbft
(Note This is not always true when describing oil well tubing Tubing is usually described as a nominal diameter which is neither the inside or outside diameter of the tubing When given the nominal diameter of tubing it is necessary to refer to a handbook to determine the OD ID and lbft)
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
72
For the same outside diameter an increase in the weight per foot will decrease the inside diameter The weight per foot also includes the weight of the couplings as a distributed weight
Consider 4 frac12 166 lbft grade D drill pipe
Wall thickness = 0337 and ID = 3826 (pipe body)
(Note ID = OD -2 x wall thickness and that IDs are expressed as inches and decimals of an inch)
The weight of this pipe is 166 lbft which is a distributed weight including tool joints Using 4 frac12 OD 3826 ID and density of steel = 490 lbmft3 the weight per foot of the pipe body is
07854 (452 - 38262)(1)(490) = 15 lbft
144
Hook Load and Buoyancy Calculation Example 3 (Cont)
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
73
Therefore to calculate the volume displaced by this pipe the weight per foot including connections must be used or
Displ volume = weight per foot (length)
density
= lbmft (ft) = ft3
lbmft3
then 166(1) = 00339 ftsup3ft displacement
490
This displacement volume must be used to calculate the buoyant force when using the wt of displaced fluid method
Hook Load and Buoyancy Calculation Example 3 (Cont)
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
74
Hook Load and Buoyancy Calculation Example 4
Calculate the WI reading when 3750 of 2 nominal tubing is suspended in a hole filled with salt water (SpGr = 115)
Solution
2 nom tubing OD = 2375
(H-40) ID = 11995
lbft = 470 lbft
Dead wt = (3750)(470) = 17625 lbf (includes couplings)
Displ volume = 470((3750) = 3597 ft3
490
Wt of displ fluid = (3597) (115)(624) = 2581 lbf
WI = 17625 - 2581 = 15044 lbf
or Dead wt = 17625 lbf
BF = 490 - (115)(624) = 08536
490
WI = (08536)(17625) = 15045 lbf
Note The single quote ( ʹ) means
foot and double quote ( ldquo ) means inches
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
75
Hook Load and Buoyancy Calculation Example 5
A drill string consists of 9000 if 247 lbft drill pipe and 450 of 7 10968 lbft drill collars Hole fluid = 105 lbgal mud Calculate the weight indicator reading when reaching bottom (hole and pipe full of mud)
Solution
Dead wt = (9000)(247) + (450)(10968) = 271656 lbf
BF = 490 - (105)(748) = 08397
490
WI = (08397)(271656) = 228113 lbf
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
76
Bit Weight
Optimum drilling conditions for a particular type formation are a combination of penetration rate (feet drilled per time) weight on the bit (lb) rotation speed (RPM) bit wear (teeth or bearings) and efficient removal of the cuttings (mud properties and circulation rate) The proper combinations are based on manufacturers recommendations experimental data rules of thumb and experience Generally hard formations require high bit weights and low rotation speeds while soft formations require low bit weights and high rotation speeds
The weight on the bit should be applied by the drill collars
(Note Drill pipe should not used to put weight on the bit since torsional properties of the drill pipe are greatly reduced when placed in compression)
A general rule for determining the number of drill collars to be used is that approximately 23 of the total length of collars should be used to put weight on the bit
Hook Load and Buoyancy Calculation Example 5 (Cont)
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
77
Hook Load and Buoyancy Calculation Example 6
How many feet of 6 34 1080 lbft drill collars would be needed to put 20000 lb weight on the bit when drilling in 96 lbgal mud
Solution
Effective wtft of drill collars suspended in mud
BF = 490 - (96)(748) = 08535
490
Eff wtft = (08535)(1080) = 922 lbft
No of feet = 20000 lbf = 217
922 lbft
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
78
Hook Load and Buoyancy Calculation Example 7
If there were 330 of drill collars and 8500 if 4 12 20 lbft drill pipe in Example 6 what would be the weight indicator reading while drilling
Solution
Total effective wt of string = (BF)(dead wt) =
(08535)(8500)(20) + (330)(1080) = 175514 lbf
WI = 175514 - 20000 = 155514 lbf
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
79
Hook Load and Buoyancy Calculation Example 8
Approximately how many drill collars (total) would be needed in Example
7 (1 drill collar = 30)
Solution
217 needed for 20000 lbf bit weight
By general rule this is 23 of total length
Total length = (217) (23) = 3255
No drill collars = 325530 = 1085 or use 11 drill collars (330)
Since the effective wtft of drill pipe and drill collars is constant with a constant mud density regardless of drilling depth the change in total effective weight (Weight indicator) will be due to additional drill pipe added as depth increases
In Example 6 217 of drill collars will be needed to put 20000 lbf on the bit at any depth
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
80
Hook Load and Buoyancy Calculation Example 9
What will be the increase in weight indicator reading when increasing drilling depth to 9730 from Example 8
Solution
Total effective wt at (8500 +330) = 155514 lbf
Eff wtft of drill pipe = (BF)(wtft) = (08535)(20) = 1707 lbft
WI increase = increase in total eff wt =
(1707)(9730 - 8830) = 15514 lbf
or
WI = 155514 + 15363 = 170887 lbf
(Note WI reading increases 1707 lbf for each foot of drill pipe added so long as 20000 lbf is on the bit)
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
81
Casing Loads
Usually the greatest load in the hoisting system will be casing loads (possible exception is stuck pipe)
The weight per foot of casing is higher than most other strings
Casing allows small clearances between the outside diameter of the casing and the hole therefore additional loads due to friction may be added when the casing is hoisted
Frictional loads must be estimated and are usually between 10-25 of the total effective weight (depending on hole condition)
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
82
Casing Load Example 10
Calculate the weight indicator reading when hoisting 6000 of 9 58 435 lbftcasing if the estimated frictional load is 15 Hole and pipe are filled with 10 lbgal mud
Solution
Dead wt = (6000)(435) = 261000 lbf
Eff wt = (BF) (Dead wt+)
= (490 - (10)(748))(261000) = (08473)(261000) = 221145 lbf
490
Total Eff wt = Pipe eff wt + friction load
= Pipe eff wt + (015)(Pipe eff wt)
WI = 115(Pipe eff wt) = 115(221145) = 254317 lbf
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
83
Sometimes it is a practice to run casing in the hole empty (float valve closed) and then fill the pipe with fluid after the casing is landed and before cementing This reduces the load on the hoisting system considerably while running in the hole
The total effective weight of the string decreases due to the buoyant force being increased The string will displace a volume of fluid equal to the total volume of the outside of the casing will be the volume of the casing collars therefore the volume of displaced fluid will be
Volume displaced by the total metal in the pipe (ID is flush or constant) or
Volume displaced by the total metal in the pipe = Wtft(length) = ft3
490
Volume of the ID of the pipe = (7854)(ID)2 = ft3
(144)
Casing Load Example 10 (Cont)
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
84
Hook Load and Buoyancy Calculation Example 11
Calculate the weight indicator reading when reaching landing depth of 8200 using 7 29 lbft (ID = 6184) if the pipe is run empty Hole fluid is 95 lbgal mud
Solution
WI = Dead weight - buoyant force
Dead wt = (8200)(29) = 237800 lbf
Volume of displ fluid = lbft + (7854)(ID)2( length)
490 144
= 29 + (7854)(ID)2 (8200)
490 144
= (0268)(8200) = 2196 ft3
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
85
Hook Load and Buoyancy Calculation Example 11 (Cont)
Wt of displ fluid = (2196) (95)(748) = 156048 lbf
WI = 237800 - 156048 = 81752 lbf or
Since additional volume occupied by the casing collars is small using the OD of the casing shows very small error therefore the casing OD is usually used to calculate the displacement volume of casing
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
86
Hook Load and Buoyancy Calculation Example 12
Calculate the WI reading in Example 11 using the casing OD to determine the displacement volume
Solution
Dead wt = 237800 lbf
Volume of fluid displaced = (7854)(7)2(8200) = 2191 ft3
144
Wt of displ fluid = (2191)(95)(748) = 155692 lbf
WI = 237800 - 155692 = 82108 lbf
(Note Compare to Example 11)
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
87
Hook Load and Buoyancy Calculation Example 13
Calculate the WI reading for Example 13 after the pipe is filled with mud
Solution
Dead wt = 237800 lbf
WI = Eff wt = (BF)(dead wt)
= 490 - (95)(748) (237800) = (08550) (237800) = 203319 lbf
490
(Note Compare to Example 12)
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
88
Hook Loads
Often it is necessary to restring the hoisting system from that used during normal drilling operations to that of a casing block system
The casing block system uses more cables thru the traveling block to increase the mechanical advantage therefore decreasing the load per line
Note Increasing the number of lines and sheave wheels increases friction losses but the decrease in load per line is greatly reduced
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
89
Hook Load and Buoyancy Calculation Example 14
In Section-2 the hoisting system is 6 lines thru the traveling block Using this system calculate the load in the fast line for Example 11 Assume 2 friction per working line (average) and a 1 14 cable
Solution
FLL = HL
(No of supporting lines)(ef)
HL = 254317 lbf
No of supporting lines = 6
ef = 1 - (02)(6) = 088
FLL = 254317 = 48166 lbf
(6)(088)
The recommended maximum load for a 1 14 cable is approximately 40000 lbf therefore the load imposed by the casing is in excess of the recommended load
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
90
Hook Load and Buoyancy Calculation Example 15
The system can be restrung to use 8 lines through the traveling block Assuming the same friction losses calculate the load in the fast line
Solution
FLL = HL
No of supporting lines (ef)
HL = 254317
No of supporting lines = 8
ef = 1 - (02)(8) = 084
FLL = 254317 = 38291 lbf
(8)(084)
With this system the fast line load is less than the maximum recommended load
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
91
Hook LoadsmdashStuck Pipe
When pipe is stuck in the hole the depth at which it is stuck must be determined before any recovery procedure can be used The depth at which the string is stuck or Free Point can be determined by measuring the stretch constants for various sizes and weights of strings are given in handbooks
Note Stretch constants for 4 12 166 lbft Grade D drill pipe is 9722x10-8 inftlb and for 7 35 lbft J-55 casing it is 4545x10-8 inftlb Each foot of free pipe will stretch this amount for each pound of tension
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
92
Hook Load and Buoyancy Calculation Example 16
Calculate the depth ( Free Point) at which 7000 (total) of 7 35 lbft J-55 casing is stuck is at a stretch of 1128 is measured under a tension of 48000 lb above the total effective weight of the string
Solution
Stretch constant = 4545x10-8 inftlb
Feet of free pipe = 1128 in
(4454x10-8 inftlb)(48000 lbf)
= 5171 ft
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
93
For this video there are several important objectives
Describe the various methods that are used to compress gases
List the hazards associated with compressed gases and compressed gas cylinders
Demonstrate proper storage of compressed gas cylinders
Define the safe handling techniques that should be used when working with compressed gas cylinders
Determine what types of fittings and connections are used for most cylinders
Test for leaks within a compressed gas system
93
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory Learning Objectives
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
94
The video covers the following topics
Four major ways to compress gases
Hazards of compressed gases
Proper storage procedures
Markings and labels
Handling cylinders safely
Connections and fittings
Leak detection
94
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
95
Students please watch Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory Complete the assignment as homework Safety video questions will also be on the Module quizzes
To open the video hold the control key down and click the link embedded in the assignment
httpenterprisecoettueduLabSafetyHandling Compressed Gas Cylinderswmv
Safety Video 7 Handling Compressed Gas Cylinders in the
Laboratory
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
96
1 The proper way to move a compressed gas cylinder is by using a hand truck preferably a four-wheeled one
a True
b False
2 Which of the following are ways to store pressurized gases
a ldquoStandard compressionrdquo
b As a liquid
c Dissolved in a solvent
d All of the above
3 ldquoPressure Relief Devicesrdquo (PRDrsquos) control the speed at which gas comes out of the cylinder
a True
b False
96
Safety Video 7 In Class Recap
a True
d All of the above
b False
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
97
4 The purpose of a regulator is to decrease the ldquodelivery pressurerdquo of compressed gases to a usable and safe level
a True
b False
5 A good way to tell what type of gas in a cylinder is by the color the cylinder is painted
a True
6 Cylinders that contain corrosive gases should not be stored for more than how many months
a 3 months
b 6 months
c 9 months
d 12 months
97
Safety Video 7 In Class Recap
a True
b False
b 6 months
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
98
7 Cylinders containing flammable gases and cylinders containing oxidizers can be safely stored together
a True
98
Safety Video 7 In Class Recap
b False
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
99
Lesson 3 Wrap Up
What is still unclear
What questions do you have about the topics we have discussed before we move on
Homework
Assignment 51 Module 5 Self Study Review
Assignment 53 Safety Video 7 Handling Compressed Gas Cylinders in the Laboratory
Assignment 53 Read Fundamentals of Drilling Engineering pp 119 - 133
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
100
Credits
Developer
Lloyd R Heinze PhD Petroleum EngineeringTexas Tech University
Contributors
Rui V Sitoe PhD Department of Mechanical Engineering UEM
Victoria Johnson Instructional Designer
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