weatherford - oilfield maths & hydraulics

W50760 – Rev05 – © 2005 – Weatherford. All rights reserved. i

BASIC OILFIELD MATHS

AND HYDRAULICS

contents

1 BASIC OILFIELD MATHS AND HYDRAULICS 1 1.1 SIGNS AND SYMBOLS 1

1.1.1 Introduction To Signs & Symbols 1 1.2 AREA 2

1.2.1 Cross Sectional Area 4 1.2.2 Annular Area 6

1.3 VOLUME 8 1.3.1 Volume Of Rectangles 8 1.3.2 Volume Of Cylinders 9 1.3.3 Volume Of Cylinder Or Piece Of Pipe 10 1.3.4 Annular Volume 13 1.3.5 Field Volume Calculations (Simplified) 14

1.4 CAPACITIES 15 1.4.1 Annular Capacities 18

1.5 PRESSURE 20 1.5.1 Hydrostatic Pressure 21 1.5.2 Applied Pressure 24 1.5.3 Total Pressure 25 1.5.4 Differential Pressure 26 1.5.5 Tubing/Casing Differential Pressures 28

1.6 MATERIAL STRENGTHS 29 1.7 FORCE 31

1.7.1 Pressure Related Forces 34 1.8 BUOYANCY 36

1.8.1 When Do We Use Buoyancy Calculations? 37 1.9 WEIGHING CASING 39

1.9.1 Guide Shoe 40 1.9.2 Float Shoe 40 1.9.3 Differential Fill Equipment 40

ii W50760 – Rev05 – © 2005 – Weatherford. All rights reserved.


AND HYDRAULICS

W50760 – Rev05 – © 2005 – Weatherford. All rights reserved. 1


AND HYDRAULICS

1 BASIC OILFIELD MATHS AND HYDRAULICS

1.1 SIGNS AND SYMBOLS

Numerous symbols are used to denote functions, of measurement, and/or amounts. The more common ones will be familiar, but some we come across need to be understood.

SYMBOL NAME DEFINITION

π Pi The ratio of the circumference of a circle to its diameter 3.1415+)

% Percent Per 100 (20% of)

> Greater Than 6 > 5

< Less than 5 < 6

µ mu Coefficient of friction (viscosity)

∑ Sigma Sum of

∆ Delta An incremental difference/change

° Degree Temp or longitude/latitude

What is the small figure at the upper right of a number or letter? • It is called an exponent

i.e. a x a = a2

i.e. 106 = one million for example 1,000,000

Sometimes negative exponents are used: a-2

What is the symbol for Square Root? The square root symbol √ i.e. √ 4 = 2

1.1.1 Introduction To Signs & Symbols • Parentheses ( ) • Brackets [ ] • Braces ⎨ ⎬

Rule: work out the formula in the following order • 1st parentheses ( ) • 2nd brackets [ ] • 3rd braces { }

Example: X = 63 ÷ {3 [2 ( 3 x 2 ) – 5]} ??

2 W50760 – Rev05 – © 2005 – Weatherford. All rights reserved.


AND HYDRAULICS

1.2 AREA

Area is the surface within a defined boundary: The boundary can be square, rectangular, circular, or any other closed shape. The units of measure (n) are expressed as square n’s or n2.

In downhole tool operations, the ability to calculate areas properly and to use those area calculations daily during the installations, or when preparing a proposal, leads to a better understanding of the tools and their application.

Area Of A Circle Given the nature of the oil industry and the manner in which operations are performed, we are required to work with items such as tubing, casing and packers which are circular in shape and have various areas.

Of all the calculations performed during a job or for an installation proposal, the two most frequently calculated areas are total surface area and cross sectional area. The key to determining both of these lies in knowing how to calculate the area of a circle.

The area of any circle is determined by multiplying π by the circle’s diameter squared, then dividing by 4, or:

circle area = πD2 ÷ 4

π, an infinite number representing the ratio of the circumference of a circle to its diameter, is rounded to 3.14159 for calculations.

Area is the surface within a defined boundary.



AND HYDRAULICS

Because we know the value of π, we can simplify our circle area calculation by dividing 4 into π to get the constant 0.7854. Our simplified circle area calculation is:

circle area = 0.7854D2

Example: Calculate the area of a circle with a diameter of 2.50 in. Circle area = 0.7854D2

= 0.7854 x (2.50 in.)2

= 0.7854 x 6.25 in2

= 4.91 in2

TRY IT Calculate the area of a circle with a diameter of 2.875 in.

Calculating the surface area of any circle is similar to calculating the area of bull plugged tubing or tubing with a plug in place to determine the amount of area affected by pressure. It would also be similar to the area of a plunger on a triplex pump to determine volume output of the pump.



AND HYDRAULICS

1.2.1 Cross Sectional Area

Because we’re working with tubulars, we need to be able to calculate not only the surface area but also the cross sectional area. The cross sectional area is the area between two circles. The circles can represent the wall thickness of tubing or casing, the area between two seal rings, or the area between the tubing and the casing.

The formula used to calculate any cross sectional area is Pi times (outer diameter squared minus inner diameter squared), all divided by 4, or

cross sectional area = π (OD2- ID2) ÷ 4

We can simplify our calculation by dividing π by 4 to get 0.7854:

cross sectional area = 0.7854 (OD2 - ID2)

REMEMBER: When calculating any cross sectional area, calculate the areas of the two diameters first, then subtract them to find the difference. DO NOT SUBTRACT DIAMETER FROM DIAMETER!!

To find out how buoyancy will affect an open ended pipe run in a well full of fluid, we’d need to calculate the cross sectional area. Tubulars with thicker walls have a greater cross sectional area than tubulars with thinner walls.

When calculating any cross sectional area, the areas of the two

diameters must be calculated separately then the two subtracted

to find the difference.



AND HYDRAULICS

Example: Calculate the cross sectional area of 7 in. casing with an ID of 6.276 in. Cross sectional area = 0.7854 (OD 2 - ID2)

= 0.7854 x (7 in2 - (6.276 in)2)

= 0.7854 x (49 in2 - 39.39 in2)

= 0.7854 x 9.61 in2

= 7.55 in2

TRY IT Calculate the cross sectional area of 5 in casing with an ID of 4.276 in.



AND HYDRAULICS

1.2.2 Annular Area

The annular area is the difference between the area of the casing ID and the area of the tubing OD In open hole situations, the annular area is the difference in areas between the open hole diameter and the pipe (drill pipe or casing).

To calculate the annular area between tubing and casing, the area of the ID of the casing is calculated, then the area of the OD of the tubing is calculated and subtracted from it. The formula for determining annular area is expressed as:

Annular area = A casing ID - A tubing OD

It can be expressed using diameters:

Annular area = π (casing ID2 - tubing OD2) 4

Again, we can simplify our calculation by dividing π by 4 to get 0.7854:

Annular area = 0.7854 (casing ID2 – tubing OD2)

Example: 9 5/8” 47# x 7” 26# casing A = 0.7854 (8.681 in2 – 7 in2)

= 0.7854 (75.35 in2 – 49 in2)

= 0.7854 x 26.35 in2

= 20.6 sq in

TRY IT Calculate the annular area between 7 in. 20 lb/ft casing with a nominal ID 0f 6.456 in and 2 7/8 in tubing.

Annular area is the difference in areas between the area of the

casing ID and the OD area of the tubing



AND HYDRAULICS

In situations where there are two or more tubing strings inside casing the annular area is found by subtracting the total of the tubing OD areas from the ID area of the casing. When written it is expressed as:

Annular area = Area casing ID – (Area tubing 1 OD + Area tubing 2 OD)

Example: Calculate the annular area of a 7 in, 23 lb/ft. casing with one string of 2 3/8 in tubing and one string of 2 7/8 in tubing installed.

1. The ID area for the casing with a nominal ID of 6.366 in is 31.83 in2

2. The OD area for the 2 3/8 in tubing is 4.43 in2

3. The OD area for the 2 7/8 in tubing is 6.49 in2

4. Annular area = 31.83 in2 – (4.43 in2 + 6.49 in2)

= 31.83 in2 – 10.92 in2

= 20.91 in2

The annular area between the OD area of the two strings of tubing and the ID area of the casing is 20.91 in2

TRY IT Calculate the annular area of 9 5/8 in 47# casing with one string 3 ½ in 9.3# tubing and one string 2 7/8 in 6.5# tubing installed.



AND HYDRAULICS

1.3 VOLUME

Volume, one of the most important qualities calculated, is defined as the amount of space something occupies. Different volumes may have to be determined at any point: Tank volumes, tubing or casing volumes, or open hole volumes are often calculated.

Volume is calculated by first determining the area, then multiplying the area by the height over which it extends, or

Volume = area x height

When performing a volume calculation, the units of measurement used to determine the area and the height must always be the same (feet, inches, meters). The answer will be expressed as unit3, (ft3, in3, or m3), or cubic unit.

1.3.1 Volume Of Rectangles

To determine the volume of a rectangular tank, the calculations are based on the area of a rectangle multiplied by the height of the rectangle:

Volume = area of the rectangle x height or

Volume = length x width x height or

V = L x W x H

Example: Calculate the volume of a tank that is 25.0 ft long, 8.0 ft wide and 6.0 ft high

25 ft x 8 ft x 6 ft

= 1200 ft3

TRY IT Calculate the volume of a mud pit 30 ft long, 10 ft wide and 96 ins high.



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1.3.2 Volume Of Cylinders

To calculate the volume of a cylinder, multiply the area of the base by the height of the cylinder:

V = area of the base x height of the cylinder or

V= 0.7854D2 x height

Example: Calculate volume of a cylinder 15 ft in diameter and 32 ft height.

V = 0.7854D2 x height

V= 0.7854 x 152 x 32 ft

V = 176.7 ft2 x 32 ft

V = 5655.6 ft3

TRY IT Calculate the volume of a cylinder with a diameter of 16 ft and a height of 22 ft.



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1.3.3 Volume Of Cylinder Or Piece Of Pipe

• The number 1029.4 is one of the most common and easy to use in the oilfield. • When used it gives the answers in barrels. • Most if not all volumes of tubulars are calculated using the constant in the formula.

The following equations will show where the figures come from: An alternative way to calculate the volume of cylinders/pipe is as follows.

depthxdx4

v 2π=

As depth is normally in feet when using oilfield pipe, the area which is expressed in square inches should be divided by 144 to get square feet.

144depthxdx4

v 2 ÷=π

This gives our answer in cubic feet.



AND HYDRAULICS

In the oilfield most volumes are described in barrels. There are 5.61 cubic feet in a barrel, therefore dividing our answer by 5.61 gives answer in barrels.

5.61x144x4Depthx2dx

v

5.61144Depthx2dx4

v

π

π

=

÷÷=

Example: Pipe ID = 6.094 in. – Well is 10,000 ft deep.

bbls361v

bbl3ft5.613ft2025v

3ft2025144291666v

ft10,000x2in37.136x0.7854v

ft10,000x2in37.136x4

3.14159v

=

÷=

=÷=

=

=

The shaded area is always constant and calculates to: 0.000971423 this can also be expressed as:

1029.430.00097142

1=

Therefore, to find the volume of a piece of pipe in barrels all that is needed is the following formula:

1029.4Depthx2d

v =

Or

bblsDepthx1029.4

2dv ==



AND HYDRAULICS

Example: Piece of pipe with 5.5 in ID – 25 ft long.

bbls0.7345v

ft25xbbls/ft0.02938v

ft25x1029.430.25

v

ft25x1029.4

2ins5.5v

(length)Depthx1029.4

2dv

=

=

=

=

=

TRY IT Piece of pipe 4.65 in ID with length of 3724 ft.



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1.3.4 Annular Volume

Annular volume is the difference in volume between a large cylinder and a small cylinder. The formula for annular volume is:

Annular volume = V large cylinder - V small cylinder or

Annular volume = (0.7854 x height) x (D2 large cylinder – d2 small cylinder)

Example: 17 ft diameter cylinder with, 11.5 ft diameter cylinder inside it 250 ft high. (0.7854 x height) x (D2 - d2)

(0.7854 x 250 ft) x (172 - 11.52)

= 196.35 ft x (289 ft2 - 132.25 ft2)

= 196.35 ft x 156.75 ft2

= 30777.86 ft3

TRY IT Calculate the annular volume between two cylinders 28.0 ft high. The large cylinder has a diameter of 7.5 ft and the small cylinder has a diameter of 3.75 ft.



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1.3.5 Field Volume Calculations (Simplified)

To calculate capacity of an annular space:

D2 – d2 ÷ 1029.4 = bbls/ft

Where: D = inside diameter of wellbore d = outside diameter of pipe in hole



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1.4 CAPACITIES

In most operations performed during installation operations, be it production tools or liners, knowing how to calculate capacities is necessary. When circulating cement or acid for squeezes or displacing the plugs on a liner job, it is important to know what volume of fluid must be pumped at a given point during the operation.

Tubing or casing capacity is the internal volume. The Volume or capacity may be calculated over the entire length of the string or a portion of it. Most engineering handbooks provide a volume factor for the particular size and weight of tubing or casing. This factor is expressed (bbls/ft). Once the correct factor is determined, it is multiplied by the length or portion to be calculated to obtain the capacity.

From the capacity tables for tubing, we know that the capacity of 2 7/8 in EUE 6.5 lb/ft is 0.00579 bbls/ft.

The capacity of 1700 ft of tubing would be:

1700 ft x 0.00579 bbls/ft = 9.84 bbls

When a capacity must be calculated for plastic lined tubing and an accurate capacity factor for the lined tubing is not available, use this formula to calculate a capacity factor:

bbls/ft = 0.0009714 x ID2

or divide by it’s reciprocal 1029.4

The constant 0.0009714 is derived using this formula: bbl/ft = in2 x 0.7854 x 12 in/ft x 1 gal/231 in3 x 1 bbl/42 gal

bbl/ft = (0.7854 x 12) x in3 x gal x bbl 231 x 42 x ft x in3 x gal

bbl/ft = 9.4248 in3 gal bbl 9702 ft in3 gal

= 0.0009714 bbl/ft



AND HYDRAULICS

Example: A string of 7 in 26# casing is run to 8955 ft. There is a float shoe at the bottom of the string, a float collar at 8820 ft, and a 2 stage cementing collar at 4320 ft.

Find the following: 1) Capacity from float collar to stage collar.

2) Capacity from the stage collar to surface.

3) Total capacity of the string of 7 in 26# casing.

From Tech Resource Book 0.03826 bbls/ft.

Cap = 8820 ft – 4320 ft x 0.03826 bbls/ft = 172.17 bbls

Cap = 4320 ft x 0.03826 bbls/ft = 165. bbls

Total Cap = 8955 ft x 0.03826 bbls/ft = 342.6 bbls

Or we could use Volume or Cap = d2 x Depth 1029.4

ID 6.276 in 6.276 in2 x Depth 1029.4

1) 6.276 in2 x 4500 = 39.38 in2 x 4500 ft = 172 bbls 1029.4 1029.4

2) = 39.382 x 4320 ft = 165 bbls 1029.4

3) = 39.382 x 8955 ft = 342.6 bbls 1029.4



AND HYDRAULICS

TRY IT A string of 5 ½ in 14.0 lb/ft J-55 casing is run to a depth of 6815 ft. The float collar is at 6785 ft, and a two stage cementing collar is at 5010 ft.

1) What is the capacity from the float collar to the stage collar? (Answer in bbls)

2) What is the capacity from the stage collar to surface? (Answer in bbls)

3) What is the total capacity of the string? (Answer in bbls)

4) Calculate the capacity factor for 2 7/8 in OD plastic lined tubing with an ID of 2.292 in. (Answer in bbls/ft)



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1.4.1 Annular Capacities

Annular capacity is the capacity between the tubing and casing or casing and open hole. The annular capacity in any well is dependent on the OD of the tubing or pipe string(s) and the ID of the casing(s) or open hole.

It is necessary to be able to calculate the annular capacity when circulating out cement after a squeeze, back washing at a liner top or inhibiting the annulus in an injection well.

Most engineering handbooks have reference tables for annular capacity: Again it is a matter of using the book to obtain the correct calculation. As with tubing or casing capacity factors the factor for annular capacity is expressed the same way: volume/length or length/volume.

There can be more than one annulus in a well bore . On a liner job, for example, we would have to be aware of three annular capacities:

• casing in open hole • casing in casing • drill pipe in casing.

Each of these capacities would have to be calculated separately and then added together to obtain the annular capacity, as in this formula:

bbl/ft = 0.0009714 x (ID2 – OD2) or ID2 – OD2 1029.4

Example: Calculate annular capacity of a well 8.640 ft. deep with 9 5/8 in 47# casing inside 13 3/8 in. 72# casing.

0.000974 x (12.347 in2 - 9.625 in2) = 0.0009714 x (59.80 in2)

= 0.05809 bbls/ft

= 8640 ft x 0.05809 bbls/ft

= 501.9 bbls



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TRY IT Calculate the annular capacity of a well with 6400 ft of 2 3/8 in tubing inside 4 ½ in 9.5 lb/ft casing. (Answer in bbls)

Calculate the total annular volume of a well with a tapered string of tubing run as follows: (Answer in bbls - show workings)

• 4030 ft of 2 7/8 in EUE tubing • 2790 ft of 3 ½ in EUE tubing • 7 in 26 lb/ft casing set at 6975 ft.



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1.5 PRESSURE

Pressure is the force per unit area exerted on a surface. It is expressed as pounds (f) per square inch, or psi.

If, for example a 100 pound force pushes on a one square inch sealed piston, a pressure gauge would show the resulting pressure as 100 psi.

Using these definitions, pressure can be expressed using this formula:

Pressure = Force/Area

Two types of pressure are dealt with on a daily basis in field operations. Hydrostatic pressure is created by the weight of a column, of fluid. Applied pressure originates in the formation, or from an external source such as a pump. The pressure within a well occurs as a result of either hydrostatic or applied pressure, or from a combination of both.

P = F/A

Example: 100 lbs ÷ 1 in2 = 100 psi

TRY IT 1150 lbs force pushes onto a piston area of 1.75 in2 what would the pressure gauge read?

Pressure is the force per unit area that is exerted on a

surface.



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1.5.1 Hydrostatic Pressure

Hydrostatic pressure is the pressure created by the weight of a column of fluid acting on a defined area.

Hydrostatic pressure is changed by either changing the weight of fluid or the height of the fluid column. The formation in any perforated well or a well with open hole is supported by the pressure from the fluid column.

The fluid column height is calculated on the true vertical height of the column in the well and not on the total length of the column. A horizontal well, for example, might be drilled to a measured depth of 10,000 ft. but its true vertical height might be only 7,000 ft. The hydrostatic pressure of this well would be calculated on the true vertical height of 7,000 ft. which is the true vertical height of the fluid column.

Hydrostatic pressure is a result of the earth’s gravitational pull which acts perpendicular to the earth’s surface. The fluid column is essentially being “pulled” downward toward the centre of the earth. As a result, the hydrostatic pressure is greatest at the deepest vertical point.

Hydrostatic pressure exists at every point in the well. If there is no means of isolation, it acts on all the surrounding areas equally.

When two or more columns of different weighted fluid are present in the well, each column’s hydrostatic pressure has to be calculated separately then added together to arrive at the total hydrostatic pressure.

Remember the two different “measurements” which are important.

The True Vertical Depth (TVD) is used for all fluid pressure/hydrostatic calculations.

The Measured Depth (MD) is the actual distance to the bottom of the well which, in a horizontal well, can be quite different to the TVD.

When calculating volumes it is important to use MD and not TVD unless the two are the same.

Measured Depth can relate to length of drill pipe, tubing or casing in the wellbore.



AND HYDRAULICS

The formula for calculating hydrostatic pressure is:

Hp = Ht of fluid column x wt of fluid x 0.052

Where: Hp = pounds per square inch, or psi

Height of fluid column = feet

Fluid weight = pounds per gallon

0.052 – constant used to convert lbs/gal to psi/ft.

The constant 0.052 is derived by determining the psi/ft for 1 lb/gal of fluid:

psi = (lbs/gal) x (ft) x (12 in/ft) x (1 gal/231 in3)

psi/ft = lbs/gal x 12 in/ft x 0.004329 gal/in3 1 lb/gal

= 0.051948 lbs. gal/in2 ft. lb.

= 0.052 psi gal/ft. lb.

Example Well Info

TD 10,000 ft float shoe on bottom, top of cement 8500 ft.

Water to surface Weight of Cement 15.8 ppg.

Weight of water 8.34 ppg.

1) Find hydrostatic of Cement in tubing

2) Find hydrostatic of Water in tubing

3) Find total hydrostatic in Well in tubing

• CMT = 10.000 ft – 8500 ft = 1500 ft x 15.8 ppg x 0.052 = 1232 psi • Water = 8500 ft = 8500 ft x 8.34 ppg x 0.052 = 3686 psi • A + B = 1232 psi + 3686 psi = 4918 psi



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TRY IT What is the total hydrostatic pressure in a vertical well with a measured depth of 6000 ft with a column of mud 800 ft high and weighting 16.0 lbs/gal and 5200 ft of produced water weighing 9.5 lbs/gal? The mud is at the top of the hole.

What is the hydrostatic pressure in the same well if the top of the mud is at 1230 ft from surface?



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1.5.2 Applied Pressure

Applied pressure is any pressure applied to a system typically, the system is the well and the pressure is applied by pump or by the formation. Any applied pressure affects the entire well equally if there is no isolation (a packer)

For example, if 500 psi is applied at surface in a well full of fluid that is 10.000 ft deep, the pressure at the bottom of the hole will increase by 500 psi as will the pressure at 4,000 ft. This is because, for all intents and purposes, fluids are incompressible and an incompressible fluid will not experience a volume change when pressure is applied. When a fluid is pumped under pressure, it pushes on the next volume and the process repeats itself until the applied pressure affects the entire fluid system equally.

When fluid is being pumped and is moving, the applied pressure at surface is greater than the applied pressure at the bottom of the hole. This is because as fluids move there is friction generated between the fluid and the inside diameter of the pipe.

As the rate of the fluid flow increases, the amount of friction between the fluid and the surface of the pipe also increases. The pump at surface not only causes the fluid to move but must also generate enough force to overcome the total friction in the system. The loss in applied pressure due to the friction of moving fluids is referred to as friction pressure loss.

This pressure loss or reduction is dependent on the type of fluid being pumped, its temperature, its flow rate through the pipe, and the inside diameter of the pipe. Remember that friction pressure loss occurs only when fluid is moving, If a blockage occurs (a sand off at the perforations during a frac, for example), friction pressure loss becomes zero and the full surface applied pressure is felt at the bottom of the well. For this reason, friction pressure loss is typically neglected in calculations so that a worst case design is obtained.

Applied pressure can also be present due to the formation pressure or well head pressure. During the production phase of the well. the formation will create a pressure within the wellbore and this pressure must be considered during the installation of downhole tools. In certain situations the formation pressure can and will create a situation where the pressure capabilities of a tool may be exceeded.

Friction loss will also occur in any producing well. As the fluid moves up the tubing, friction is generated between the fluid and the inside surface area of the tubing.

How much friction loss occurs during production is also dependent on the type of fluid being produced, the temperature of the fluid, the production rates and the size of the production tubing.

Applied pressure is any pressure applied to a system



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1.5.3 Total Pressure

Total pressure is the sum of the hydrostatic pressure and the applied pressure at any point in the system.

The combination of the two can result in a failure in the system.

As stated previously. It is generally safer to assume that applied pressure is equal at all points and not reduced by pressure.

Example Well is full of produced water at 9.35 ppg.

It is 15,000 ft deep. We are going to apply 525 psi at surface.

Find pressure at top and bottom of the well

Hydrostatic at bottom of the hole prior to applying pressure.

= 15,000 ft x 9.35 ppg x 0.052 = 7293 psi

Pressure up to 525 psi. with CMT unit – Surface pressure 525 psi

Pressure at bottom of hole = 7293 psi + 525 psi = 7818 psi

TRY IT Well is 12500 ft deep

Fluid is 11.5 ppg mud

Applied pressure is 630 psi.

Find total pressure at bottom of well with pressure applied.

Total pressure is the sum of the hydrostatic

pressure and the applied pressure



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1.5.4 Differential Pressure

Differential pressure is the difference in pressures acting across an area, and is calculated by subtracting the pressure on one side of the surface from the pressure on the other side.

• Differential pressure = total pressure A - total pressure B • A substantial differential pressure will result in material failure in the tubular or the

downhole tools.

When calculating differential pressures it is important to specify the direction in which the pressures are acting. There are four particular areas in a packer type completion that differential pressure calculations apply to:

• Tubing differential pressure at surface or at the packer.

Differential pressures across particular metal components of the packer.

• Differential pressure across the sealing element of the packer. • Differential pressure at the casing.

Example: The well is 10,000 ft. deep

A packer is set at 8500 ft. The fluid in the tubing below the packer is 10.5 ppg water. The annular fluid is changed out and is 8.34 ppg.

1) What is the differential pressure at the packer?

2) What direction is the differential in favour of?

Tubing packer = 10,000 ft x 10.5 ppg x 0.52 = 5,460 psi.

Annulas = 8,500 ft x 8.35 ppg x 0.52 = 3,686 psi.

Differential = 1774 psi in favour tubing (under packer).



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TRY IT WELL INFO

TD 12,570 ft

Perfs at 12,550 ft

Packer Set at 11,000 ft

Fluid in well 10.5 ppg water

Applied pressure 2500 psi from the formation.

1) Find pressure above packer

2) Find pressure below packer

3) Differential pressure & in favour of what? Tubing/Annular.



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1.5.5 Tubing/Casing Differential Pressures

A substantial differential in pressures on tubing or casing can result in material failure, Depending on what side the differential occurs, the end result will either be burst or collapse of the pipe.

Burst and collapse pressures are available from an engineering handbook. The data given in the hand book is based on API (American Petroleum Institute) minimum yields. When API rates tubing or casing, no safety factor is used; Instead a calculation factor of 0.875 is used.

The API calculation for burst and collapse is based on wall thickness derived from subtracting the nominal ID from the nominal OD and using the 0.875 calculation factor to allow for variances in the wall thickness of the pipe. An API casing dimensional data chart shows the variances in dimensions for casing. For example, 5 ½ in. 15.5 lb/ft casing has:

• Maximum OD = 5.555 in • Minimum OD = 5.473 in • Nominal ID = 4.950 in • Minimum ID = 4.923 in • Maximum ID = 5.074 in

Because of this variance in dimensions, you will occasionally encounter situations where an excess of the API minimum pressures has been applied and no damage to the pipe has occurred. Remember: API numbers are minimums only.

When using an engineering handbook for reference, the pressure ratings for tubing and casing are listed as collapse pressure and internal yield pressure. Remember that the pressure ratings are based on API calculations for that particular type of steel.

Look up Burst & Collapse pressure in Tech Resource Book.

Section 1 Page 50 Drill Pipe

Section 1 Page 28 – 39 Casing

Section 1 Page 5 – 6 Tubing



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1.6 MATERIAL STRENGTHS

The tensile rating of any tubing or casing is based on the grade of the material from which the pipe is manufactured and the cross sectional area of the material.

The API grade designation specifies the process of manufacture, the chemical composition and the mechanical properties requirements necessary to meet a certain specification. These specifications are outlined in API Specification 5 CT. Examples of the API

Designations for pipe grades are J-55, N-80 and L-80.

The letter in the API Pipe Grade Specification refers to the chemical composition, or the blend of elements, of which the steel is comprised. Typically, iron, carbon manganese, phosphorus, and sulphur are always present.

In addition, molybdenum, chromium, nickel, and copper can comprise a percentage of a blend. By varying the composition of the alloys and selecting an appropriate manufacturing process (particularly in the heat treating of steel) the manufacturer can produce a variety of API grades of steel. This variety allows for selection and appropriate use in a range of environments and loading conditions.

The number in the API Pipe Grade Specification identifies the minimum yield strength in pounds for one square inch of that particular grade of material to yield. Minimum yield is the point at which permanent material deformation occurs. When any steel yields the “memory” is lost and it will not return to its original shape.

The number 55 in the API designation J-55 is an abbreviation for 55,000 pounds per square inch. A pull of 55,000 pounds exerted on one square inch of J-55 material is the minimum required pull for that one square inch of material to yield.

API has set an acceptable range for the yield rating for a material grade. The tensile rating of any material (also referred to as ultimate yield) has a specified minimum. Tensile or ultimate yield is the point at which material parts. For example, L–80 has a yield strength of between 80,000 – 95,000 psi. The tensile strength is specified as a minimum of 95,000 psi.

Although N–80 and L–80 material have the same minimum yield rating the chemical composition of the L–80 material makes it softer and better suited for use in a sour service environment.

REMEMBER: MINIMUM YIELD IS THE POINT AT WHICH MATERIAL DEFORMATION BEGINS. TENSILE RATING IS THE POINT AT WHICH THE MATERIAL PARTS.

Use this to calculate material strengths: 1) Determine the cross sectional area of the material.

(OD2 - ID2) x 0.7854 = As

2) Multiply the cross sectional area of the material by the minimum yield in psi (lbs/in2) to determine the material strength in pounds.

Minimum yield strength is the point at

which permanent material deformation

occurs

The ultimate tensile strength is the point at which the material will

part.



AND HYDRAULICS

Example: Using an engineering handbook, we find that the yield rating of 2 7/8 in 6.5 lb/ft J-55 tubing is 99,660 lbs. The yield point, though, is based on psi, and the value is expressed in pounds (force). What is the material yield of the tubing?

1) Calculate the cross sectional area of the tubing in this case, it is 1.812 in2

2) Multiply the area by the minimum yield: 1.812 in2 x 55,000 lb/in2 = 99,661 lbs.

The minimum body yield of a joint of 2 7/8 in 6.5 lb/ft J–55 tubing is 99,661 lbs.

TRY IT Calculate the minimum body yield for 5 ½ in 17.0 lbs/ft P110 casing with an ID of 4.892 in.



AND HYDRAULICS

1.7 FORCE

Force is the total amount of push or pull acting on a object. Force on a tubing or casing string is either tension or compression. Tension is created by anchoring the bottom of the string and pulling at surface. It should be remembered however that tubing or casing hanging under its own weight is by definition in tension throughout its length. Each joint must support the weight of the joints of tubing or casing below it. Tubing or casing forces are typically known as string weights.

To calculate the string weight of a tubing string, multiply the tubing weight by its length.

For example, a string of 4.5 in casing has a given weight of 11.6 lb/ft. To determine the string weight in air of 4000 ft of casing, multiply the total length of the casing by its weight.

Weight in air = 11.6 lb/ft x 4000 ft.

= 46,400 lbs.

If it was measured in air, the casing string would weigh 46,400 lbs, or exert a downward force of 46,400 lbs, on the top joint.

The instrument used to measure the force on the top Joint of the string (commonly referred to as string weight) is the rig weight indicator. This device simply displays the force or load carried by the rig (hook load). In addition to the force of the tubing or casing weight indicator also shows the load of the travelling blocks. To determine the string weight of tubing or casing the weight of the blocks must be subtracted from the total value shown on the weight indicator.



AND HYDRAULICS

When setting a packer or a tubing string in tension the setting force is generated by pulling on the tubing string.

The force of the tubing string suspended from the travelling blocks is increased by anchoring the tubing string at the bottom and pulling on it. As the pull increases at surface so does the reading on the weight indicator.

Compression in a tubing string, for example is created by lowering the tension carried at surface so that the bottom of the string supports some of the weight. A tubing string in compression is not fully in compression throughout its length, but rather is in compression at the bottom, neutral at some point part way up the string and in tension at surface, Which is the weight supported by the bottom of the string.

When calculating string weights of two different sizes of tubing or casing, calculate each size separately then add them together to obtain the total string weight.



AND HYDRAULICS

Example: Total string weight of a Dual String of 10 ¾ in 76# and 9 5/8 in 53.5# the 10 ¾ in is 5300 ft and the 9 5/8 in. is 5725 ft.

10 ¾ = 5300 ft x 76# = 402,800 lbs

9 5/8 = 5725 ft x 53.5# = 306,287 lbs

Total weight = 709,087 lbs

TRY IT What is the total string weight of 2500 ft of 2 3/8 in 4.7 lb/ft tubing and 4050 ft of 2 7/8 in 6.5 lb/ft tubing?



AND HYDRAULICS

1.7.1 Pressure Related Forces

Pressure was defined earlier as the force acting on a unit of area typically one square inch or one square meter. There is a direct relationship between force, pressure and area, and it is expressed by the formula:

F = p x a

Where, F = force (pounds)

p = pressure (psi)

a = area (in2)

If a pressure of 50 psi acts on an area of 10 in2 the resulting force would be 500 lbs An example of how the effects of pressure and area result in a force is a hydraulic lift system on a dump truck. In order for the box of the truck to be raised to a sufficient height to dump its load the downward force from weight of material in the box must be overcome. The hydraulic pressure system on the truck generates a pressure that acts on the area of the lift cylinder and the combination of the two creates a force sufficient to raise the box to dump the load.

If the maximum weight of the load to be lifted is 29,000 lbs. and the lift cylinder has an area of 24 in2 what pressure is needed to raise the box?

F= p x a

29,000 lb = p x 24 in2

p = 29,000 lb 24 in2

p = 1208.33 psi

To raise the truck box, the pressure must be greater than 1208 psi.

If it’s not, the box will remain stationary or in a neutral point.



AND HYDRAULICS

When calculating forces it is necessary to keep track of and illustrate the direction in which the forces are acting to determine the net force. In the previous example the force of the load is down and is illustrated with a down arrow. The Force created by the hydraulic system on the truck would be acting upward. The net force acting on any object is determined by keeping track of the direction (identifying which is positive or negative) of all the forces and if the net force is not zero then the object must move.

EXAMPLE: (A) F = P X A A packer has an area of 12.75 in2 The pressure from below the packer acting on the area is 2,500 psi. What force is acting on the packer?

F = P X A 2500 psi x 12.75 in2 = 31875 lbs↑

EXAMPLE: (B) P = F/A What pressure is required to overcome an upward force of 31,875 lbs acting on a packer with an area of 12.75 in2

P = F/A 31875 lbs ÷ 12.75 in2 = 2500 psi

EXAMPLE: (C) A = F / P What is the packer area being acted on if we have an upward force of 31875 lbs and 2500 psi on the pressure gauge.

A = F / P 31875 lbs ÷ 2500 psi = 12.75 in2

TRY IT A packer is set in the hole it is sealed/plugged. The size of the packer is 9 5/8. We have 1500 psi pressuring up below the packer from the well.

1) What is the area in square inches that the pressure is acting on?

2) What is the force in lbs being created by the pressure acting on the packer?

3) If we see the force acting on the packer reduce what could have happened?



AND HYDRAULICS

1.8 BUOYANCY

Buoyancy is defined as the tendency for a submerged object to float in the fluid in which it is immersed. An object will become buoyant when the pressure created by the fluid acts on the object’s area. For example a block of wood will float when the buoyancy force created by the water is greater then the weight of the wood. As the block of wood rises in the water the volume of wood above the water is no longer affected by buoyancy and the wood reaches a balance point and floats.

In situations where the object’s weight is greater than the buoyancy force, the object will remain submerged however it will weigh less. An example is a large rock on the bottom of a lake, if we pick up the rock and walk with it to the shore the shallower the water becomes the heavier the rock becomes. If the weight of the rock is great enough, we may reach a point at which we can no longer carry the rock.

The buoyancy force is equal to the weight of fluid displaced by the submerged object. A rock with a volume of 1 cubic ft weighing 164 lbs displaces 62 lbs of fresh water. The weight of the rock in fresh water would be 164 lbs – 62 lbs = 102 lbs.

In an oilwell, the tubing string is affected by buoyancy due to the wellbore fluid, Rather than calculate tubing displacement volumes to determine the buoyancy force we can approximate the buoyancy force as if all is present at the end of the tubing and pushing up on the cross sectional area of the tubing wall. This buoyant force is always an upward direction and will result in a reduction of the string weight and a reduced weight indicator reading.

To accurately calculate how buoyancy affects the string weight of pipe in fluid, the string weight in air must be calculated first and then the buoyancy force calculated to obtain the buoyant string weight of the pipe.



AND HYDRAULICS

1.8.1 When Do We Use Buoyancy Calculations?

In an oil well, the tubing string is affected by buoyancy due to the wellbore fluid.

To accurately calculate how buoyancy affects the string weight of pipe in fluid, the string weight in air must be calculated first, and then the buoyancy force calculated to obtain the buoyant string weight of the pipe.

Buoyancy Factor = 65.5 – MW (ppg)

65.5 Or = 7.84 – MW (SG)

7.84 Example: Calculate the buoyant string weight of 7 in 26 lb/ft

Casing run in 12.6 lb/gal mud to a depth of 8400 ft

1) Calculate the string weight in air:

26 lb/ft x 8400 ft = 218,400 lbs

2) Calculate the cross sectional area of the casing:

7.54 in2

3) Calculate the hydrostatic pressure in the well:

Hp = 12.6 lb./gal x 0.052 x 8400 ft

= 5504 psi

4) Calculate the upward force on the pipe caused by the hydrostatic pressure using the formula:

F = p x a

F = 5504 psi x 7.54 in2

F = 41,500 lbs

5) Calculate the net force (buoyant string weight):

218,400 lbs

- 41,500 lbs

176, 900 lbs

6) Quick calculation for buoyancy factor:

(65.5 – mud weight ppg) ÷ 65.5

Example: Mud weight of 9.7 ppg (65.5 – 9.7 ppg) ÷ 65.5 = 0.85 Therefore buoyancy factor = 0.85



AND HYDRAULICS

TRY IT Calculate the buoyant string weight of 3500 ft of 4½ in 12.6 lb/ft casing run in well full of 9.6 ppg drilling mud.



AND HYDRAULICS

1.9 WEIGHING CASING

To find the hook load of different strings of casing, you cannot simply multiply the “weight per foot” by the length of the string. Casing weighs less in fluid than it does in air, where its “weight” rating is calculated. The difference is the buoyant force which is equal to the weigh of the volume of fluid displaced by the metal in the casing.

Some engineering tables give the displacement of the metal in the string along with the displaced fluid per 100 couplings. While this rather exacting method is technically correct, a very close approximation can be made for straight holes by using the hydrostatic head of the fluid at the bottom of the string, acting against the cross sectional end area of the casing. In highly deviated and horizontal completions there are many other varieties affecting string weight, such as friction in areas of high build angle.

Calculations are done by first obtaining the casing inside and outside diameters, from the Engineering tables, and calculating the end area of the string. The hydrostatic head of the Column of fluid is then calculated and the resultant force applied as to lift the string from the well.



AND HYDRAULICS

1.9.1 Guide Shoe

With a guide shoe, the only area affected is the cross sectional area of the casing itself; and the casing is being “pushed” upward, or lightened, by the hydrostatic pressure at the bottom of the shoe acting on this end area. Everything inside the shoe will cancel out. To calculate casing weight, multiply the # ft by the depth of that weight casing, then subtract your estimate of the buoyant force. The buoyant force is calculated by multiplying the end area of the casing by the hydrostatic pressure at the shoe. If casings of different weight (and ID’s) are used, you can add the number of feet casing times the weight per foot for each section.

However, changes in casing ID affect the buoyant force and its calculation. The required calculation method will be covered later in this section.

1.9.2 Float Shoe When a float shoe is run on the end of the casing string, none of the wellbore fluid is allowed to enter the casing and the string acts as if it were plugged at the bottom. Casing weight is estimated by calculating the end area of the casing using casing OD then multiplying by the hydrostatic pressure at the shoe. Any fluid inside the casing (sometimes the rig crew will “fill” the casing using a mud hose) should be taken into account by first estimating the amount of fluid inside the casing in (ft), then estimating the hydrostatic pressure at the shoe (remember the fluid inside the casing may be of a different density than that outside), and multiplying by the inside area of the casing. Remember to add the casing weight in the air to obtain correct hook load. When different weights of casing are used, the same rules apply as those used in the Guide Shoe example. Also, remember that the buoyant force estimated cannot exceed the weight of casing in air, that is, there can be no negative hook load.

1.9.3 Differential Fill Equipment The general rule for running differential fill float equipment states that when one piece of differential equipment is used, the casing will be 90 0/0 full, with two pieces (collar and shoe) 810/0 full. Practice is to calculate the casing weight when casing has stopped filling and the valve has closed. At this point, your calculations should be made exactly as you would a float shoe with the casing partially filled. The only difference is that you have a somewhat more accurate estimate of the height of the fluid column inside the casing; you should know that the fluid is the same as the fluid outside the casing.



AND HYDRAULICS

Guide Shoe Calculations: Using the well schematic on the right:

• 8700 feet of 5 ½ in 23 # casing • Mud at 12.6 # gal

Example calculation of hook load: Calculate fluid gradient = 12.6 x 0.052

= 0.655 psi/ft

Calculate the hydrostatic pressure = 8700 x 0.655

= 5698.5 psi

Calculate the end area of casing = 0.7854 x (5.52 – 4.6702)

= 6.630 in2

Estimate the buoyant force = 5698.5 x 6.630

= 37,781 # ↑

Cal. the weight of casing in air = 23 x 8700

= 200, 100 # ↓

Hook load of casing = 200,100 - 37,781

= 162,319 # ↓

TRY IT Using well schematic on the right:

9650 ft of 7 in 26# casing mud weight of 13.1 ppg



AND HYDRAULICS

Float Shoe Calculations Using the well schematic on the right:

• 4800 feet of 7 in 20# casing • Mud at 9.2#/gal

Example calculation of hook load:

Calculate the fluid gradient: = 9.2 x 0.052

= 0.478 psi/ft

Calculate the hydrostatic pressure: = 4800 x 0.478

= 2294 psi

Calculate the end area of the casing: = 0.7854 x 7.0002

= 38.485 in2

Estimate the buoyant force: = 2294 x 38.485

= 88,285#↑

Calculate the weight of casing in air: = 20 x 4800

= 96,000#↓

Hook load of casing: = 96,000 - 88,285

= 7,715#↓

TRY IT Using well schematic on right

5620 ft of 7 in 29# casing weight of mud 9.0 ppg.



AND HYDRAULICS

Differential Fill Calculations Using the well schematic on the right

• 4800 feet of 7 in. 20# casing • Mud at 10.8 #/gal • Differential shoe on bottom

Example calculation of hook load

Calculate the fluid gradient: = 10.8 x 0.052

= 0.562 x psi/ft

Calculate the depth of fluid inside the casing: Using a single piece of differential equipment means that 90% of the casing will be filled.

Fluid Depth in casing: = 0.90 x 4800

= 4320 feet

Calculate the hydrostatic pressures: Inside casing = 4320 x 0.562

= 2428 psi

Outside casing = 4800 x 0.0562

= 2698 psi

Calculate the areas of the casing: Inside casing = 0.7854 x 6.4562

= 32.735 in2

Outside casing = 0.7854 x 7.0002

= 38.485 in2

Calculate the weight of the casing in air: = 4800 x 20

= 96,000 # ↓

Balance forces: Inside casing = 2428 x 32.735

= 79,481#↓

Outside casing = 2698 x 38.485

= 103,833 # ↑

Casing weight in air: = 96,000 #↓

Hook load: = 96,000 + 79,481 – 103,833 = 71,648 #↓

TRY IT Using schematic 5650 ft of 5 ½ in 20# casing mud at 11.2 ppg. Differential fill shoe on bottom



AND HYDRAULICS

Calculation Using More Than One casing Weight in the String

Example: Using the “Guide Shoe” example from previous page, modify

The casing schedule to read:

• 1500 ft of 5½“ 23# • 2500 ft of 5½“ 20# • 4700 ft of 5½“ 15.50# • Mud at 12.6# /gal

Remember: Casing schedules read from the bottom up:

Calculate hydrostatic pressure: At 4700 ft: = 4700 x 0.655

= 3078 psi At 7200 ft: = 7200 x 0.655

= 4716 psi At 8700 ft: = 8700 x 0.655

= 5699 psi

Calculate areas at casing I.D. changes: At 4700 ft: = 0.854 x (4.9502 – 4.7782)

= 1.314 in2 At 7200 ft: = 0.7854 x (4.7782 – 4. 6702)

= 0.801 in2 At 8700 ft. – end area of casing:

= 0.7854 x (5.52 – 4.6702) = 6.630 in2

Calculate forces acting at casing I.D. changes: At 4700 ft: = 1.314 x 3078

= 4044 #↓ At 7200 ft: = 0.801 x 4716

= 3776 #↓ At 8700 ft: = 6.630 x 5699

= 37,784 #↑

Calculate weight of casing in air: = (1500 x 23) + (2500 x 20) + (4700 x 15.50) = 34,500 + 50,000 + 72,850 = 157,350 #↓

Hook load of casing: = 157,350 + 4044 + 3776 – 37,784

= 127,386 #↓

TRY IT Using same schematic as on this page.

• 1650 ft of 7” 35# casing • 2550 ft of 7” 26# casing • 4850 ft of 7” 20# casing • Mud weight 10.2 ppg

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