drillling tech
DESCRIPTION
fundamentals of oil well drilling, Casing design, drill fluid,TRANSCRIPT
CHAPTER-1
INTRODUCTION TO WELL PLANNING
Drilling or digging for oil has occurred in one way or another for hundreds
of years. The Chinese, for instance, invented a bamboo rig to obtain oil and gas
for lighting and cooking. In 1859, oil came spurting out of the ground from a well
69.5 feet deep in Titusville, Pennsylvania. Colonel Drake had just gone down in oil
prospecting history. But although this initiated industrial oil well drilling, a large
number of wells had been drilled long before to produce water, brine and even
naphtha for caulking boats, and for lighting and medicinal purposes.
By the name of a well (borehole) is meant a cylindrical mine opening made
too small forman’s access there to, the diameter of the opening being many
times less than its length. The beginning of the well is called its mouth, collar or
well-head, the cylindrical surface is termed the wall or hole shaft (bore), and its
floor the bottom hole. The distance from the mouth to the bottom hole along
the axis of the borehole shaft is the length of the well, while the projection of
its axis onto the vertical plane represents the depth of the well. The wells are
sunk as straight, slanted or horizontal boreholes. As regards their purpose,
boreholes drilled for geological exploration of the region, search for,
prospecting and exploitation of deposits are classified into key or stratigraphic,
extension or-outpost, structure-exploratory, reconnaissance, prospecting
production and special boreholes.
Structure-exploratory boreholes serve the purpose of a thorough
investigation into the structures encountered in drilling of key and extension
holes and of drawing up a program for exploratory-prospecting drilling into
these structures. The results of the structure-exploratory drilling and of
geophysical investigations are utilized in studying the mode of occurrence,
determining the age and physical properties of the rocks making up the column,
precisely marking the reference or key horizons, and in compiling structural
(subsurface) maps.
Producing wells are drilled into a completely prospected deposit developed
for exploitation. The category of producing wells includes not only the wells
through which oil or gas is recovered (producing wells proper), but also the wells
which help to effectively develop the defining more exactly the reservoir
behavior (drive) and the extent of possible recovery of oil from individual
sections of the pool, ascertaining and accurately delimiting the boundaries of
producing fields.
Injection wells serve the purpose of edge and inter-field boundary
injection of gas or air into the producing reservoir in order in the formation
pressure. Observation wells are put down to affect a regular control over
changes in the pressure, over the position of the water-oil, gas-water and gas-oil
contacts during the operation of the reservoir.
Drilling Methods in Oil Industry
Rocks can be broken up by applying mechanical, thermal, pyhsico-chemical,
electric-spark and other methods. Practical applications in industry have found,
however, only methods involving mechanical disintegration of the rock, the
others continuing so far to be at the stage of development.
In oil industry, mechanical drilling is affected by employing percussive
(cable-tool) drilling and rotary drilling methods. Cable-tool drilling has been
abandoned about 30 years ago, and no longer employed, with some minor
exceptions. However, this method is still used for coal and ore mining industries,
geological engineering, and in drilling of wells for water. Different drilling
techniques are used in oil industry. Early days, cable-tool drilling was the major
method. Recently, rotary drilling systems are widely used. For the last two
decades, downhole motor systems are preferred as directional and horizontal
drilling needs are increasing. Cable-tool drilling will be emphasized in the next
chapter, and the rest of the course will be focused on rotary drilling technique.
Auger Drilling
In order to start an oil well drilling operation, hydraulic hammering or
auger drilling techniques are used for the conductor casing installations. Auger
drilling can be a best solution for dump leaching where dumps are in thickness of
30-60 ft or a bit more, and being in the dump site for long period of time. This
kinds of dumps may be squeezed and show some difficulties to passage of leach
solutions. Portable auger drills can drill fast and adequate sizes of holes which
leach solutions can be poured or dumped in. Drill string components of an auger
drilling system are composed of bits, augers and universal joints and subs.
Augers are usually 5 to 20 ft long. Hexagonal, circular or hollow augers are
available. Rotation of an auger system is achieved by using a top drive system or
a bevel gear. Load on the bit is determined from cable feed, chain feed,
hydraulic feed or combination of these. Cuttings are removed mechanically.
Cable Tool Drilling
The first oil well in the United States was drilled with cable tools in 1859
to a depth of 65feet. This was the historic Drake well located near Titusville,
Pennsylvania; it is credited with having started the American petroleum industry.
The cable tool (also called churn or percussion) drilling method, however, did not
originate in this country, but is believed to have been employed first by the
early Chinese in the drilling of brine wells. In this method, drilling is
accomplished by the pounding action of a steel bit which is alternately raised by
a steel cable and allowed to fall, delivering sharp, successive blows to the
bottom of the hole. This principle is the same as that employed in drilling
through concrete with an air hammer, or in driving a nail through a board. The
original percussion drilling apparatus consisted of a spring pole anchored into the
ground at an angle, with the bit suspended from the free end by a rope. To
impart the necessary reciprocating action to the bit, the Chinese employed a
number of men who alternately jumped on and off the spring pole beam from a
ramp. Many early brine wells in the United States were drilled in the same
manner, except that the spring pole was equipped with stir ups where two or
three men stood and literally kicked the well down. As more and deeper wells
began to be drilled efforts were made to improve the drilling equipment. Steam
engines began to be used; walking beams replaced the spring pole; steecables
replaced manila ropes; and other improvements followed. Although the modern
cable tool rig is a far cry from the ancient Chinese model, the changes have been
in materials and equipment, for the basic operating principle is unchanged
problems previously unsolved except by rig floor trial and error. Further field
verification of the theory is needed; such experiments could possibly aid the
economic application of cable tools.
Rotary Drilling System
Making a hole for the recovery of underground oil and gas is a process
which requires two major constituents; i) man-power, and ii) hardware systems.
The man power includes a drilling engineering group and a rig operator group.
The first provides engineering support for optimum drilling operations, including
rig selection, design of mud program, casing and cement programs, hydraulic
program, drill bit program, drillstring program and well control program. After
drilling begins, the daily operations are handled by a rig operator group which
consists of a tool pusher and several drilling crews. The hardware systems which
make up a rotary drilling rig are :
i) power generation system,
ii) hoisting system,
iii) drilling fluid circulation system,
iv) rotary system,
v) well blowout control system, and
vi) drilling data acquisition system and monitoring system.
Basic Principles of Rotary Drilling
The rotary method uses tricone-type toothed bits or one-piece bits such
as diamond or PDC bits. While the bit is being rotated, a force is applied to it by
a weight. The advantage is that a fluid can be pumped continuously through the
bit, which is crushing the rock formation, and carry cuttings up out of the hole
to the surface with the rising fluid flow. The rotary drilling rig is the apparatus
required to fulfill the following three functions:
-Put weight on the bit
-Rotate the bit
-Circulate a fluid
It is the drill collars, screwed onto the bottom of the drillpipe assembly
just above the bit, that provide the necessary weight, and prevent buckling of
the drillpipes above them. Drill collars, along with drillpipe and bit all make up
the drillstring, which is rotated by the rotary table and the kelly. The drillstring
component parts are hollow down the middle so that the drilling fluid can be
circulated down to the bit. A fluid-tight rotary joint, the swivel, is located at
the top of the kelly and provides a connection between the mud pump discharge
line and the inside of the drillstring. A hoisting system is required to support
the weight of the drillstring, lower it into the hole and pull it out. This is the
function of the derrick, the hook and the drawworks.
The drilling rig is complete with facilities to treat the drilling fluid when
it gets back to the surface, a storage area for tubular goods, shelters and
offices on site. In addition, when a well is being drilled, it is regularly cased. It
is lined with steel pipe, or casing, which is lowered into the hole under its own
weight in smaller and smaller diameters as the hole gets deeper. The first
length of pipe is run in as soon as the bit has drilled the surface formation and is
then cemented in the hole. A casing housing is connected to the top of the
surface casing. All the following lengths of pipe are hung on the casing housing
and cemented at the base to the walls of the hole. After the first drilling phase
is cased, drilling will be resumed with a bit with a diameter smaller than the
inside diameter of the casing string that was run in and cemented. The deeper
the borehole gets and the more casings are set in the well, the smaller the
diameter of the bit must be. The casing housing also serves to hold the safety
equipment, such as blowout preventers.
Drilling Crew
Besides the drilling engineers, geologist and other specialized technical
personnel, there are other men working on a drilling operation those have various
duties. The personnel on an oil or gas well drilling rig consists of a group of
operators who have certain skills and functions which are uniform in the
business of drilling a well. These operators are artisans of a sort who ply their
trade as do carpenters, bricklayers, etc, not caring much for whom they work
for as they do for the opportunity to work on a rig.
A drilling rig operates 7 days a week, 24 hours a day, and a full staff must
be on the job at all times in order that the rhythmical teamwork-type operation
may be accomplished without delays. Drilling rigs are generally staffed for a
tour or shift, by a crew of minimum 5 men. These men consist of a “driller”, a
“derrickman”, a “catheadman”, and two “floormen” (rotary helpers).The driller is
in charge of his crew, and in charge of the rig’s operations during his tour.
Hereports to a foreman who is usually called a “toolpusher”. In his position as
boss of the rig andthe crew, the driller must be responsible for safe, economical
operations at all times. Therelationship between the driller and his crewmen is
such that the driller will often take hiscrew with him when he changes jobs.
Good judgement and alert attention to detail areprobably the two prime
requisites of a driller. Specifically, it is the driller’s function tooperate the
drawworks in the hoisting of the drilling string from the hole and in running it
back into the hole, as well as to maintain the proper characteristics of the
machinery whiledrilling with a bit on bottom. The derrickman works high in the
derrick or mast while making a trip. Stationed on thederrick platform known as
the monkey board, which is 90 ft above the derrick floor, it is hisfunction among
others to seize the upper end of the 90 ft stand of pipe as it comes out of
thehole and pull it over into the rack. He also tests the circulating fluid to
maintain its properties in accordance with the mud program. It is his
responsibility to assure the proper functioningof the mud pumps. The term
catheadman, sometimes referred as “motorman”, is applied to the third ranking
member of the drilling crew. He is usually more experienced in all phases of the
work than the rotary helpers. His job is to operate the manual catheads on the
drawwroks whenever they are used and to supervise service of the motors and
engines for uninterrupted operations. The two rotary helpers are the tong
operators during trip. When the bit is on bottom for drilling, there are
miscellaneous duties for the rotary helper to perform. Maintaining the rig’s
housekeeping is a large order in itself, straightening up the floor, picking up
numerous pieces of equipment, cleaning and painting the machinery, helping the
motorman with the engines, helping the derrickman with the mud treatment, etc.
The toolpusher is actually the manager of his branch of his company’s operation.
During the drilling of the well, he must be available at any time day or night for
consultation with the drillers. Toolpushers must have at least a smattering of
knowledge on labor laws, personnel relationships, of chemistry, hydraulics,
internal combustion engines, etc. He usually has authority to spend money to
keep his rig in operation, cooperating with his supervisor on any items of major
importance. Since he does not drill a foot of hole by himself, it could be said
that his prime function is to assist the drillers and their crews to drill the hole
safely, economically and promptly. The foreman makes his headquarters in an
office at the rig known as “doghouse”, where he must keep accurate records of
daily drilling, performance, purchases of materials, accident reports, and other
paperwork which must go daily to the main office for coordinating with the
other company accounting and records. The foreman’s responsibilities cover
every operation of the rig including clearing and stacking out the drill site,
moving in the rig, and rigging up and tearing it down when the job is finished.
Depending upon the size of the contracting organization, there may be an
assistant drilling superintendent as well as a drilling superintendent. It is the
function of these men primarily to aid the toolpusher in the conduct of their
work. These are the duties of the principal field personnel on a drilling rig.
Behind them in the administrative end of the business, an office force must be
maintained to solicit business, contact clients, estimate for bids, maintain an
equipment yard and repair shop and figure what the company has left after
taxes.
Rig Components
Rotary drilling rigs are used almost all drilling done today. The hole is drilled by
rotating a bit to which a downward force is applied. Generally, the bit is turned
by rotating the entire drillstring, using a rotary table at the surface, and the
downward force is applied to the bit by using sections of heavy thick-walled
pipes, called drill collars, in the drillstring above bit. The cuttings are lifted to
the surface by circulating a fluid down the drillstring, through the bit, and
up the annular space between the hole and the drillstring. The cuttings are
separated from the drilling fluid at the surface.
Rigs
The function of rotary drilling is to allow the drilling and completion of a hole at
the lowest cost possible. The process of making a hole is made by rotating a rock
cutting tool called a bit, upon which a downward force is applied. The generated
cuttings (rock fragments) at the hole bottom are transported by a circulating
drilling fluid down through the drillstring and up the annular space to the
surface.
Rotary drilling rigs are classified as onshore (land) or offshore (marine)
rigs. Their main design features are mobility, flexibility and maximum depth of
operation. Rig components, in general, are common to both onshore and offshore
rigs. The only major difference is the utilization of motion compensation
equipment and an extension pipe (drilling riser between rig floor and seabed
when drilling in a marine environment).
Modern land rigs are built in units and are skid-mounted so that they can
be transported from one drilling site to another. Once on location, the rig
components are easily assembled together to drill the well. The derrick is
designed such that it can be assembled and disassembled with ease. Its
functions are to carry loads that are suspended in the wellbore and to provide
space between rig floor working area and crown block such that certain length
of the drillstring can be made. A rig that can handle singles, doubles or triples is
one that allow length of 30 ft, 60 ft or 90 ft sections of drillstring to be made,
respectively. These sections are called stands. For moderate depth wells, carrier
mounted rigs are used. These portable masts are mounted on flat-bed vehicles
as self-contained single units.
Offshore drilling is performed from space above water provided by
structures which may be classified as i) mobile platforms and, ii) stationary or
fixed platforms. The selection of a particular rig carrier depends on the
operations to be performed and the water depth on location where such
operations are to be conducted. These operations may consist of exploratory
drilling, development drilling, production, storage or a combination of all.
Exploratory drilling, where hydrocarbons might or might not be available
for production, is done most economically by the utilization of mobile rig
carriers. These are defined as units which can be moved from one location to
another as the need arises. Mobile units may be grouped in three categories: 1)
self-elevating (jack-up), 2) column stabilized and, 3) dynamically positioned ship
units.Self-elevating or what are normally referred to as jack-up units are those
which, in their normal operating state, rest on the sea floor by means of legs
with the deck above the highest water level. In the transit stage, these units
float with the legs in the lifted position. Their use is generally recommended in
water depth ranging approximately from 60 feet to 300 feet of water. Column
stabilized units are supported by either caissons with or without footings or by
lower displacement type hulls by means of columns. These units are classified as
submersibles which operate while resting on the seabed or semi-submersibles
which operate when they are a float. Submersibles are generally used in water
depth up to 60 feet. Semi-submersibles and dynamically positioned ships are
used in moderately to extremely deep water depths. In development drilling
where hydrocarbons are known to exist in readily producing reservoirs, drilling
operations are conducted from stationary units which generally referred to as
permanently fixed platforms. These units are of adequate size that they may
contain drilling equipment, production equipment, storage, separation equipment,
living quarters or a combination of all. Based on their construction these units
are classified as tower type, template, tower/template, tender type, tension leg,
etc. They are designed to remain on location through their entire life span. Some
of these units have been erected in water depth of approximately 600 feet. In
the case of tension-leg platforms (TLP), operations in water depth in excess of
3000 ft have been demonstrated. While selecting a derrick or mast, the
drawworks capacity or safe load capacity from the hook load are usually
considered. As a rule of thumb, every 100 ft of borehole requires 10 HP at the
drawworks. Also, the wind load and compressive loads on the joints of the
derrick should also be considered. API has standards and guides for the
selection of proper derrick for a certain drilling operation.
Rig Power System
The hoisting and fluid circulating systems consume most rig power. Total
power requirements for most rigs are from 1000 to 3000 hp. Modern rigs are
powered by internal-combustion diesel engines and generally sub-classified as:
diesel electric type and the direct-drive type depending on the methods used to
transmit power to the various rig systems. Power system performance
characteristics generally are stated in terms of output horsepower, torque, and
fuel consumption for various engine speeds. As illustrated in Fig.1-1, the shaft
power developed by an engine is obtained from the product of the angular
velocity of the shaft, w, and the output torque T:
P = w T
The overall power efficiency determines the rate of fuel consumption, wf at a
given engine speed. The heating values H of various fuels for internal
combustion engines are shown in the following (Table1-1) The heat energy input
to the engine, Qi,, can be expressed by,
Qi = wf H
Since the overall power system efficiency, Et is defined as the energy output
per energy input, then;
Et = P / Qi
Table-1-1 Heating Value of Various Fuels
Fuel type Density, lbm/gal
Heating Value, Btu/lbm
Diesel 7.2 19000 Butane 4.7 21000 Methane - 24000 Gasoline 6.6 20000
Example-1 . A diesel engine gives an output: torque of 1,740 ft-Ibf at an
engine speed of 1.200 rpm. If the fuel consumption rate was 31.5 gal/hr, what is
the output power and overall efficiency of the engine?
Solution.
The angular velocity, w, is given by
W =2 ππππ (l,200) == 7,539.8 rad/min. The power output can be computed,
P = w.T
P =[7,539.8(1740) ft-lbf/min ] / [33,000ft-lbf/min/hp ]
P = 397.5 hp
Since the fuel type is diesel, the density ρ is 7.2 Ibm/gal and the heating value
H is 19,000 Btu/lbm (Table 1-1). Thus, the fuel consumption rate wf is,
wf =31.5 gal/hr (7.21bm/gal) ( 1 hour / 60 min.)
wf =3.781bm/min.
The total heat energy consumed by the engine is equal to;
Qi = wf H
Qi = 3.78 lbm/min (19000 Btu/lbm) (779 ft-lbf/Btu) / 33000 ft-lbf/min/hp
Qi = 1695.4 hp
Thus the overall efficiency of the engine at 1200 rpm is equal to;
Et = P / Qi
Et = 397.5 / 1695.4 = 0.234
Et = 23.4 %
Hoisting System
The function of the hoisting system is to provide a means of lowering or
raising drill strings, casing strings, and other sub-surface equipment into or out
of the hole. The principal components of the hoisting system are
-the derrick and substructure,
-the block and tackle, and
-the draw-works.
Two routine drilling operations performed with the hoisting system are called:
-making a connection and
-making a trip.
Making a connection refers to the periodic process of adding a new joint of drill
pipe as the hole deepens. Making a trip refers to the process of removing the
drill string from the hole to change a portion of the down hole assembly and
then lowering the drill string back to the hole bottom. A trip is made usually to
change a dull bit.
Derrick or Portable Mast:
The function of the derrick is to provide the vertical height required
raising sections of pipe from or lowering them into the hole. The greater the
height, the longer the section of pipe that can be handled and, thus, the faster
a long string of pipe can be inserted in or removed from the hole. The most
commonly used drill pipe is between 27 and 30 ft long. Derricks that can
handle sections called stands, which are composed of two, three, or four joints
of drill pipe, are said to be capable of pulling doubles, thribbles, fourbles,
respectively.
Block and Tackle:
The block and tackle is comprised of (1) the crown block, (2) the
travelling block, and (3) the drilling line. The arrangement and nomenclature of
the block and tackle used on rotary rigs are shown in Fig. 1.16a. The principal
function of the block and tackle is to provide a mechanical advantage, which
permits easier handling of large loads. The mechanical advantage M of a block
and tackle is simply the load supported by the traveling block, W, divided by the
load imposed on the draw-works, Ff.
M = W / Ff
The load imposed on the draw-works is the tension in the fast line. The
ideal mechanical advantage, which assumes no friction in the block and tackle,
can be determined from a force analysis of the traveling block. Consider the
free body diagram of the traveling block as shown in Figure. If there is no
friction in the pulleys, the tension in the drilling line is constant throughout.
Thus, a force balance in the vertical direction yields;
n Ff = W
where, n is the number of lines strung through the traveling block. Solving this
relationship for the tension in the fast line and substituting the resulting
expression in Equation yields;
Mi = W / (W / n) = n
which, indicates that the ideal mechanical advantage is equal to the number of
lines strung between the crown block and traveling block. Eight lines are shown
between the crown block and traveling block. The use of 6, 8, 10, or 12 lines
is common, depending on the loading condition.
The input power Pi of the block and tackle is equal to the draw-works load Ff
times the velocity of the fast line, Vf:
Pi = Ff Vf
The output power, or hook power, Ph is equal to the traveling block load W
times the velocity of the traveling block, Vb.
Ph = W Vb
For a frictionless block and tackle, W = nFf. Since the movement of the fast
line by a unit distance tends to shorten each of the lines strung between the
crown block and traveling block by only 1/n times the unit distance. Then Vb=Vf /
n. Thus, a frictionless system implies that the ratio of output power to input
power is unity:
E = Ph / Pi = [(nFf) (Vf / n)] / (Ff Vf) = 1
Of course, in an actual system, there is always a power loss due to
friction. Approximate values of block and tackle efficiency for roller-bearing
sheaves are shown in the following table. Knowledge of the block and tackle
efficiency permits calculation of the actual tension in the fast line for a given
load. Since the power efficiency is given by;
E = Ph / Pi = (W Vb) / (Ff Vf) = (WVf / n) / (Ff Vf) = W / Ff n
Table 1-2 Efficiency of the Lines
Number of lines (n) Efficiency (E)
6 0.874 8 0.841 10 0.810 12 0.770 14 0.740
Then the tension in the fast line is;
Ff = W / (En)
However, a safety factor should be used to allow for line wear and shock
loading conditions. The line arrangement used on the block and tackle uses the
load imposed on the derrick to be greater than the hook load. As sown in figure,
the load Fd applies to the derrick is the sum of the hook load W, the tension in
the dead line, Fs and the tension in the fast line, Ff.
Fd = W + Ff + Fs
If the load, W, is being hoisted by pulling on the fast line, the friction in
the sheaves is resisting the motion the fast line and the tension in the drilling
line increases from W / n at the first sheave (deadline) to W / En at the last
sheave (fast line). Substituting these values for Ff and Fs in above equation;
Fd = W + (W / En) + (W / n) = [(1+E+En) / En] W
The total derrick load is not distributed equally over four derrick legs.
Since the draw-works is located one side of the derrick floor, the tension in the
fast line is distributed over only two of the four derrick legs. Also, the dead
line affects only the leg to which it is attached. For the arrangement given in
figure, derrick legs C and D should share the load imposed by the tension in the
fast line and leg A would assume the full load imposed by the tension in the
dead line. The load distribution for each leg is given in the following table.
Table 1-3 Load Distribution for Each Leg
Load Total Load Leg A Leg B Leg C Leg D
Hook Load W W / 4 W / 4 W / 4 W / 4
Fast line W / En - - W / 2 En W / 2 En
Dead line W / n W / n - - - Total = W(n+4)/(4n) W / 4 W(En+2)/(4En) W(En+2)/(4En)
-Schematic Diagram of the Load Distribution-
Note that for E>0.5, the load on leg A is greater than the load on the
other three legs. Since if any leg fails, the entire derrick also fails, it is
convenient to define a maximum equivalent derrick load, Fde, which is equal to
four times the maximum leg load.
Fde = [(n + 4) / n] W
A parameter sometimes used to evaluate various drilling line
arrangements is the derrick efficiency factor, defined as the ratio of the
actual derrick load to the maximum equivalent load. For a maximum equivalent
load given by above equation the derrick efficiency factor is
Ed = Fd / Fde = {[(1+E+En) / En] W} / {[(n + 4) / n] W}
=[E (n+1) + 1] / [E (n+4)]
For the block and tackle efficiency values given in previous table, the
derrick efficiency increases with the number of lines strung between the crown
block and traveling block.
The drilling line is subject to rather severe service during normal tripping
operations. Failure of the drilling line can result in
-injury to the drilling personnel,
-damage to the rig, and
-loss of the drillstring in the hole.
Thus, it is important to keep drilling line tension well below the nominal breaking
strength and to keep the drilling line in good condition.
Drilling line does not tend to wear uniformly over its length. The most
severe wear occurs at the pickup points in the sheaves and at the lap points on
the drum of the draw-works. The pickup points are the points in the drilling line
that are on the top of the crown block sheaves or the bottom of the traveling
block sheaves when the weight of the drill string is lifted from its supports in
the rotary table during tripping operations. The rapid acceleration of the heavy
drill string causes the most severe stress at these points.
Drilling line is maintained in good condition by following a scheduled slip-
and-cut program. Slipping the drilling line involves loosening the dead line
anchor and placing a few feet of new line in service from the storage reel.
Cutting the drilling line involves removing the line from the drum of the draw-
works and cutting off a section of line from the end.
API has adopted a slip-and-cut program for drilling lines. The parameter
adopted to evaluate the amount of line service is the ton-mile. A drilling line is
said to have rendered one ton-mile of service when the traveling block has
moved 1 U.S. ton a distance of 1 mile. Note that for simplicity this parameter
is independent of the number of lines strung. Ton-mile records must be
maintained in order to employ a satisfactory slip-and-cut program.
Example-2. A rig must hoist a load of 300000 lbf. The drawworks can provide
an input power the block and tackle system as high as 500 hp. Eight lines are
strung between the crown block and traveling block. Calculate:
1-the static tension in the fast line when upward motion is impending,
2-the maximum hook horsepower available,
3-the maximum hoisting speed,
4-the actual derrick load,
5-the maximum equivalent derrick load, and
6-the derrick efficiency factor.
Solution.
1) The power efficiency for n = 8 is given as 0.8 in Table. The tension in the
fast line is;
Ff = W /En = 300000 / 0.841 (8) = 44590 lbf
2) The maximum hook horsepower available is;
Ph = E Pi = 0.841 (500) = 420.5 hp.
3) The maximum hoisting speed is given by;
Vb = Ph / W = 420.5 hp (33000 ft-lbf/min/hp) / 300000 lbf
Vb = 46.3 ft/min. On the other hand, to pull 90 ft stand would require;
T = 90 ft / 46.3 ft/min. = 1.9 min.
4) The actual derrick load is;
Fd = [(1+E+En) / En] W = [(1+0.841+0.841 (8)] / [0.841 (8)] . 300000 =
382090 lbf
5) The maximum equivalent load is;
Fde = [(n + 4) / n] W = [(8+4) / 8] . 300000 = 450000 lbf
6) The derrick efficiency factor;
Ed = Fd / Fde = 382090 / 450000 = 0.849 or 84.9 %
Drawworks:
The drawworks provide the hoisting and braking power required to raise
or lower the heavy strings of pipe. The principal parts of the drawworks are:
-the drum,
-the brakes,
-the transmission, and
-the catheads.
The drum transmits the torque required for hoisting or braking. It also
stores the drilling line required to move the traveling block the length of the
derrick. The brakes must have the capacity to stop and sustain the great
weights imposed when lowering a string of pipe into the hole. Auxiliary brakes
are used to help dissipate the large amount of heat generated during braking.
Two types of auxiliary brakes commonly used are (1) the hydrodynamic type and
(2) the electromagnetic type. The draw-works transmission provides a means for
easily changing the direction and speed of the traveling block. Power also must
be transmitted to catheads attached to both ends of the draw-works.
Circulating System:
A major function of the fluid-circulating system is to remove the rock
cuttings from the hole as drilling progresses. The drilling fluid is most commonly
a suspension of clay and other materials in water and is called drilling mud. The
drilling mud travels;
-from the steel tanks to the mud pump,
-from the pump through the high-pressure surface connections to the drill
string,
-through the drill string to the bit,
-through the nozzles of the bit and up the annular space between the drill
string and hole to the surface, and
-through the contaminant-removal equipment back to the suction tank.
The principal components of the rig circulating system include;
-mud pumps,
-mud pits,
-mud-mixing equipment, and
-contaminant-removal equipment.
With the exception of several experimental types, mud pumps always have
used reciprocating positive-displacement pistons. Both two-cylinder (duplex)
and three-cylinder (triplex) pumps are common. The duplex pumps generally are
double-acting pumps that pump on both forward and backward piston strokes.
The triplex pumps generally are single-acting pumps that pump only on forward
piston strokes. Triplex pumps are lighter and more compact than duplex pumps,
their output pressure pulsations are not as great, and they are cheaper to
operate. For these reasons, the majority of new pumps being placed into
operation are of the triplex design.
The advantages of the reciprocating positive-displacement pump are the;
-ability to move high-solids-content fluids laden with abrasives,
-ability to pump large particles,
-ease of operation and maintenance,
-reliability, and
-ability to operate over a wide range of pressures and flow rates by changing
the diameters of the pump liners (compression cylinders) and pistons.
The overall efficiency of a mud-circulating pump is the product of the
mechanical efficiency and the volumetric efficiency. Mechanical efficiency
usually is assumed to be 90% and is related to the efficiency of the prime
mover itself and the linkage to the pump drive shaft. Volumetric efficiency of a
pump whose suction is adequately charged can be as high as 100%. Most
manufacturers tables rate pumps using a mechanical efficiency, Em, of 90% and
a volumetric efficiency, Ev, of 100%. Generally, two circulating pumps are
installed on the rig. For the large hole sizes used on the shallow portion of most
wells, both pumps can be operated in parallel to deliver the large flow rates
required. On the deeper portions of the well, only one pump is needed, and the
second pump serves as a stand-by for use when pump maintenance is required.
The theoretical displacement from a double-acting pump is a function of
the piston rod diameter dr, the liner diameter dl and the stroke length Ls. On
the forward stroke of each piston, the volume displaced is given by;
ππππ/4 dl2 Ls
Similarly, on the backward stroke of each piston, the lume displaced is given by;
ππππ/4 (dl2 - dr
2) Ls
Thus, the total volume displaced per complete pump cycle by a pump having two
cylinders is given by;
Fp = ππππ/4 Ls (2dl2 - dr
2) Ev (duplex)
where Ev, is the volumetric efficiency of the pump. The pump displacement per
cycle, Fp, is commonly called the pump factor.
For the single-acting (triplex) pump, the volume displaced by each piston
during one complete pump cycle is given by;
ππππ/4 dl2 Ls
Thus, the pump factor for a single-acting pump having three cylinders becomes;
Fp = 3ππππ/4 (2) Ls Ev dl2 (triplex)
The flow rate q of the pump is obtained by multiplying the pump factor by ,N,
the number of cycles per unit time. In common field usage, the terms cycle and
stroke often are used interchangeably to refer to one complete pump revolution.
Pumps are rated for:
-hydraulic power,
-maximum pressure and maximum flow rate.
If the inlet pressure of the pump is essentially atmospheric pressure, the
increase in fluid pressure moving through the pump is approximately equal to the
discharge pressure. The hydraulic power output of the pump is equal to the
discharge pressure times the flow rate. In field units of hp, psi, and gal/min,
the hydraulic power developed by the pump is given by;
PH = ∆∆∆∆p q / 1714
For a given hydraulic power level, the maximum discharge pressure and
flow rate can be varied by changing the stroke rate and liner size. A smaller
liner will allow the operator to obtain a higher pressure, but at a lower rate. Due
to equipment maintenance problems, pressures above about 3,500 psig seldom
are used.
Example-3 Compute the pump factor in units of barrels per stroke for a duplex
pump having 6.5 inch liners, 2.5 inch rods, 18 inch strokes and a volumetric
efficiency of 90 %.
Solution:
The pump factor for a duplex pump can be determined
Fp = ππππ/4 (2) Ls Ev (2dl2 - dr
2) = ππππ/2 (18) (0.9) [2 (6.5)2 – (2.5)2] = 1991.2
inch3/stroke
There are 231 inch3 in a U.S. gallon and 42 US gallons in a US barrel.
1991.2 inch3 / stroke x gal / 231 inch3 x bbl / 43 gal = 0.2052 bbl/stroke
Rotary System
The rotary system includes all of the equipment used to achieve bit rotation.
The main parts of the rotary system are the: (1) swivel (2) kelly (3) rotary drive
(4) rotary table (5) drill pipe and (6) drill collars.
The swivel supports the weight of the drill string and permits rotation.
The bail of the swivel is attached to the hook of the traveling block, and the
gooseneck of the swivel provides a downward-pointing connection for the rotary
hose. Swivels are rated according to their load capacities. .
The kelly is the first section of pipe below the swivel. The outside cross
section of the kelly is square or hexagonal to permit it to be gripped easily for
turning. Torque is transmitted to the kelly through kelly bushings, which fit
inside the master bushing of the rotary table. The kelly must be kept as
straight as possible. A kelly saver sub is used between the kelly and the first
joint of drill pipe.
The openings in the rotary table that accepts the kelly bushings must be
large enough for passage of the largest bit to be run in the hole. The major
portion of the drill string is composed of drill pipe. API has developed
specifications for drill pipe. Drill pipe is specified by its outer diameter, weight
per foot, steel grade and range length. Drill pipe is furnished in the following
API length ranges.
Table 1-4 API Drill Pipe Ranges
Range Length (ft)
1 18 to 22 2 27 to 30 3 38 to 45
Range-2 drill pipe is used most commonly. Since each joint of pipe has a
unique length, the length of each joint must be measured carefully and recorded
to allow a determination of total well depth during drilling operations. The drill
pipe joints are fastened together in the drill string by means of tool joints. The
female portion of the tool joint is called the box and the male portion is called
the pin. The portion of the drill pipe to which the tool joint is attached has
thicker walls than the rest of the drill pipe to provide for a stronger joint. This
thicker portion of the pipe is called the upset.
The lower section of the rotary drill string is composed of drill collars.
The drill collars are thick-walled heavy steel tubulars used to apply weight to
the bit. The smaller clearance between the borehole and the drill collars helps
to keep the hole straight. Stabiliser subs often are used in the drill collar string
to assist in keeping the drill collars centralized.
In many drilling operations, knowledge of the volume contained in or
displaced by the drill string is required. The term capacity often is used to
refer to the cross-sectional area of the pipe or annulus expressed in units of
contained volume per unit length. In terms of pipe diameter, d, the capacity of
pipe, Ap, is given by;
Ap = ππππ/4 d2
Similarly, the capacity of an annulus, Aa, in terms of the inner and outer
diameter, is
Aa = ππππ/4 (d22 – d1
2 )
The term displacement often is used to refer to the cross-sectional area of
steel in the pipe expressed in units of volume per unit length. The displacement,
As, of a section of pipe is:
As = ππππ/4 (d12 – d2 )
Example-4 A drill string is composed of 7,000 ft of 5-in., 19.5-lbm/ft drill pipe
and 500 ft of 8-in. OD by 2.75-in. ID drill collars when drilling a 9.875-in.
borehole. Assuming that the borehole remains in gauge, compute the number of
pump cycles required to circulate mud from the surface to the bit and from the
bottom of the hole to the surface if the pump factor is 0.1781 bbl/cycle.
Solution
For field units of feet and barrels, the above equation becomes:
Ap = (ππππ/4 d2) in.2 (gal / 231 in.3) (bbl / 42 gal) (12 in. / ft) = (d2 /
1029.4) bbl/ft
The inner diameter of 5-in., 19.5 lbm/ft drill pipe is 4.276 in.; thus, the capacity
of the drill pipe is;
(4.2762 / 1029.4) = 0.01776 bbl/ft
and the capacity of the drill collars is;
(2.752 / 1029.4) = 0.00735 bbl/ft
The number of pump cycles required to circulate new mud to the bit is given by;
[0.01776 (7000) + 0.00735 (500) bbl / 0.1781 bbl/cycle = 719 cycles
Similarly, the annular capacity outside the drillpipe is given by;
(9.8752 - 52) / 1029.4 = 0.0704 bbl/ft
and the annular capacity outside the drill collars is;
(9.8752 - 82) / 1029.4 = 0.0326 bbl/ft
The pump cycles required to circulate mud from the bottom of the hole to the
surface is given by;
[0.0704(7,000) + 0.0326 (500)] bbl / 0.1781 bbl/cycle = 2858 cycles.
The Well Control System:
The well control system prevents the uncontrolled flow of formation
fluids from the well-bore. When the bit penetrates a permeable formation that
has a fluid pressure in excess of the hydrostatic pressure exerted by the
drilling fluid, formation fluids will begin displacing the drilling fluid from the
well. The flow of formation fluids into the well in the presence of drilling fluid is
called a kick. The well control system permits :
-detecting the kick,
-closing the well at the surface,
-circulating the well under pressure to remove the formation fluids and
increase the mud density,
-moving the drillstring under pressure, and
-diverting flow away from rig personnel and equipment.
Failure of the well control system results in an uncontrolled flow of
formation fluids and is called a blow-out. This is perhaps the worst disaster
that can occur during drilling operations. Blow-outs can cause loss of life, drilling
equipment, the well, much of the oil and gas reserves in the underground
reservoir, and damage to the environment near the well. Thus, the well control
system is one of the more important systems on the rig. Annular preventers,
sometimes called bag-type preventers, stop flow from the well using a ring of
synthetic rubber that contracts in the fluid passage. The rubber packing
conforms to the shape of the pipe in the hole. Most annular preventers also will
close an open hole it necessary. Annular preventers are available for working
pressures of 2,000, 5,000, and 10,000 psig. Both the ram and annular BOP’s are
closed hydraulically. In addition, the ram preventers have a screw-type locking
device that can be used to close the preventer if the hydraulic system fails.
Well-Monitoring System:
Safety and efficiency considerations require constant monitoring of the
well to detect drilling problems quickly. Devices record or display parameters
such as (1) depth, (2) penetration rate, (3) hook load, (4) rotary speed, (5)
rotary torque, (6) pump rate, (7) pump pressure, (8) mud density, (9) mud
temperature. (10) mud salinity, (11) gas content of mud, (12) hazardous gas
content of air, (13) pit level, and (14) mud flow rate.
In addition to assisting the driller in detecting drilling problems, good
historical records of various aspects of the drilling operation also can aid
geological, engineering, and supervisory personnel. In some cases, a centralized
well-monitoring system housed in a trailer is used This unit provides detailed
information about the formation being drilled and fluids being circulated to the
surface in the mud as well as centralizing the record keeping of drilling
parameters. The mud logger carefully inspects rock cuttings taken from the
shale shaker at regular intervals and maintains a log describing their
appearance. Additional cuttings are labeled according to their depth and are
saved for further study by the paleontologist. The identification of the
microfossils present in the cuttings assists the geologist in correlating the
formations being drilled. The mud logger using a gas chromatograph analyzes gas
samples removed from the mud. This type of analysis often can detect the
presence of a hydrocarbon reservoir. Recently, there have been significant
advances in sub-surface well-monitoring and data-telemetry systems. These
systems are especially useful in monitoring hole direction in non-vertical wells.
One of the most promising techniques for data telemetry from sub-surface
instrumentation in the drill string to the surface involves the use of a mud
pulser that sends information to the surface by means of coded pressure pulses
in the drilling fluid contained in the drill string.
Ton-miles of a drilling line:
The total service life of a drilling line may be evaluated by taking into
account the work done by the line during drilling, fishing, coring, running casing
and by evaluating the stresses imposed by acceleration and deceleration
loadings, vibration stresses and bending stresses when the line is in contact with
the drum and sheave surfaces. Typical round trip operations include running and
pulling drill pipe during drilling.
Work done in round trip operations = work done by travelling assembly +
work done by drilling line + work done by drill collars. So work done in making a
round trip (Tr);
Tr = [D (Ls + D) We / 10,560,000] + [D (M + C / 2) / 2,640,000] ton-miles
where; M = mass of traveling assembly (lb); Ls = length of each stand (ft) ; D =
hole depth (ft); C = effective weight of drill collar assembly in mud – effective
weight of the same length of drill pipe in mud (LWdc – LWdp) x BF; We =
effective weight per foot of drill pipe in mud.
Drilling Operations:
The ton-mile service performed by a drilling line during drilling operations
is expressed in terms of work performed in making round trips, since there is
always a direct relationship as shown on the following cycle of drilling
operations:
-drill ahead of a length of kelly
-pull up length of kelly
-ream ahead a length of kelly
-pull up a length of kelly
-put kelly in rat hole
-pick up a single (or double)
-lower drill string in hole
-pick up kelly and drill ahead
Operations 1 and 2 give 1 round trip of work done (WD), operations 3 and 4 give
1 round trip of work done, operation 7 gives ½ round trip of work done; and
operations 5,6 and 8 gives approximately ½ round trip of work done. Therefore:
Total work done = 3 round trips = 3 Tr
However, in drilling a length of section from depth d1 to depth d2, the work
done, Td, is:
Td = 3 Tr = 3 (Tr at d2 – Tr at d1) = 3 (T2 –T1)
In coring operations:
Total WD in coring = Tc = 2 round trips to bottom
Tc = 2 (T2 –T1)
In casing operations:
Total work done in setting casing = Ts = ½ [4MD+ Wcs (Ls + D)D]
Ts = ½{[D (Ls +D) x Wcs] / 10,560,000} + (MD / 2,640,000) ton-miles
Example-5 The following data refer to a 1 ½ inch block line with 10 lines of
extra improved plough steel wire rope strung to the travelling block.
Hole depth: 10000 ft
Drill pipe: 5 inch OD / 4.276 inch ID, 19.5 lb/ft
Drill collars: 500 ft, 8 x 2 13/16 inch, 150 lb/ft
Mud weight : 75 lb/ft3
Line and sheave efficiency coefficient: 0.9615
Calculate:
a-weight of drill string in air and in mud
b-hook load, assume weight of traveling block and hook to be 23500 lb
c-dead line and fast line loads, assume an efficiency factor of 0.81
d-dynamic crown load
e-wireline design during drilling if breaking strength of wire is 228000 lb
Solution:
a-weight of string in air = weight of drillpipe + weight of drill collars
weight of string in air = [(10000 – 500) x 19.5] + (500 x 150) = 260250 lb
weight of string in mud = weight of string in air x buoyancy factor =
260250 x (1-75 / 489.5) weight of string in mud = 220432 lb
b-hook load = weight of string in mud + weight of travelling block
hook load = 2204312 + 23500 = 243932 lb
c- deadline load = (HL / N) x K10 / EF) = ( 243932 / 10) x [(0.961510) / 0.81]
deadline load = 20336 lb
fastline load = [HL / (N x EF)] = [342932 / (10 x 0.81)] = 30115 lb
d- dynamic crown load = DL + FL + HL
dynamic crown load = 20336 + 30115 + 243932 = 294383 lb
e-design factor = (breaking strength / fast line load)
desigh factor = 228000 / 30115 = 7.6
Example 1-6 Using the data of the above example, determine:
a-round trip ton-miles at 10000 ft
b-casing ton-miles if one joint of casing = 40 ft
c-ton-miles when coring from 10000 ft to 10180 ft
d-ton,miles when drilling from 10000 ft to 10180 ft
Solution:
a-Tr = [D (Ls + D) We / 10,560,000] + [D (M + C / 2) / 2,640,000]
M= 23500 lb; D = 10000 ft; Ls = 93 ft; We = 19.5xBF = 16.52 lb/ft; C = 55267 lb
Tr = [10000 (93 + 10000) 16.52 / 10,560,000] + [10000 (23500 + 55267/
2) / 2,640,000]
Tr = 157.9 + 193.7 = 351.6 ton-miles
b- Ts = ½{[D (Ls +D) x Wcs] / 10,560,000} + (MD / 2,640,000) ton-miles
Wcs = 29x0.847 = 24.56 lb/ft and Ls = 40 ft
Ts = ½{[10000 (40 +10000) x 24.56] / 10,560,000} + (23500x10000 /
2,640,000)
Ts = ½ (233.5 + 89) = 161.3 ton-miles
c-T2 = round trip time at 10180 ft and T1 = round trip time at 10000 ft
T2 = [10180 (93 + 10180) 16.52 / 10,560,000] + [10180 (23500 + 55267
/ 2) / 2,640,000]
T2 = 163.6 + 197.2 = 360.8 ton-miles
T1 = 351.6 ton-miles (part-a)
Therefore: Tc = 2 (360.8 –351.6) = 18.4 ton,miles
d- Td = 3 (T2 – T1) = 3 (360.8 – 351.6) = 27.6 ton-miles
Example 1-7) A 1.25-in. drilling line has a nominal breaking strength of 138,800
lbf. A hook load of 500,000 lbf is anticipated on a casing job and a safety factor
based on loading conditions on static loading conditions of 2.0 is required.
Determine the minimum number of lines between the crown block and traveling
block that can be used?
Solution:
Safety factor = Ffmax / Ffact
2 = 138,800 / Ffact
Ffact = 69,400 lbf
W / E.n = 69400
E.n = 500,000 / 69,400 = 7.205
So E.vn> 7.205
Use Trial and Error procedure:
For n = 6 ; 6 x 0.874 = 5.244
For n = 8 ; 8 x 0.841 = 6.728
For n = 10 ; 10 x 0.81 = 8.100
Since 8.10 > 7.205
Minimum number of lines is = 10
CHAPTER-2
DRILLING FLUIDS
Drilling fluid -mud - is usually a mixture of water, clay, weighing material and
a few chemicals. Sometimes oil may be used instead of water, or oil added to the
water to give the mud certain desirable properties. Drilling fluid is used to raise
the cuttings made by the bit and lift them to the surface for disposal. But equally
important, it also provides a means of keeping underground pressures in check. The
heavier or denser the mud, is the more pressure it exerts. So weighing materials -
barite - are added to the mud to make it exert as much pressure as needed to
contain formation pressures. The equipment in the circulating system consists of a
large number of items. The mud pump takes in mud from the mud pits and sends it
out a discharge line to a standpipe. The standpipe is a steel pipe mounted vertically
on one leg of the mast or derrick. The mud is pumped up the standpipe and into a
flexible, very strong, reinforced rubber hose called the rotary hose or kelly hose.
The rotary hose is connected to the swivel. The mud enters the swivel the swivel:
goes down the kelly, drill pipe and drill collars and exist at the bit. It then does a
sharp U-turn and heads back up the hole in the annulus. The annulus is the space
between the outside of the drill string and wall of the hole. Finally the mud leaves
the hole through a steel pipe called the mud return line and falls over a vibrating,
screen like device called the shale shaker. Agitators installed on the mud pits help
maintain a uniform mixture of liquids and solids in the mud. If any fine silt or sand
is being drilled, then devices called desilters or desanders may be added. Another
auxiliary in the mud system is a device called degasser.
Functions of Drilling Fluids
In the early days of rotary drilling, the primary function of drilling fluids was
to bring the cuttings from the bottom of the hole to the surface. Today it is
recognized the drilling fluid has at least ten important functions:
A- Assists in making hole by:
1. Removal of cuttings
2. Cooling and lubrication of bit and drill string
3. Power transmission to bit nozzles or turbines
B- Assists in hole preservation by:
4. Support of bore hole wall
5. Containment of formation fluids
C-It also:
6. Supports the weight of pipe and casing
7. Serves as a medium for formation logging
D-It must not:
8. Corrode bit, drill string and casing and surface facilities
9. Impair productivity of producing horizon
10. Pollute the environment.
Removal of Cuttings
The removal of cuttings from the face of the well bore is still one of the
most important functions of drilling fluids. Fluid flowing from the bit nozzles
exerts a jetting action that keeps the face of the hole and edge of the bit clear of
cuttings. This insures longer bit life and greater efficiency in drilling.
The circulating fluid rising from the bottom of the well bore carries the
cuttings toward the surface. Under the influence of gravity the cuttings tend to
sink through the ascending fluid; but by circulating a sufficient volume of mud fast
enough to overcome this effect, the cuttings are brought to the surface. The
effectiveness of mud in removing the cuttings from the hole depends on several
factors.
Velocity is the rate at which mud circulates, and the annular velocity is an
important factor in transporting the cuttings to the surface. Annular velocities
between 30-60 m/min. are frequently used. Velocity is dependent upon pump
capacity, pump speed, bore hole size and drill pipe size. Calculations for annular
velocity are made as follows:
Annular Velocity (m/min) = [(Pump output (m3/min.) / Annular Volume (m3/m)]
Density is weight, per unit volume of mud and has a buoyant effect upon the
particles. Increasing mud density increases its carrying capacity both by buoyancy
and particles due to additional solids in interference.
Viscosity is significant in affecting the lifting power of mud. Viscosity
depends upon the concentration, quality, and dispersal of the suspended solids. In
the field it is measured as a timed rate of flow using a Marsh funnel. Viscosity is
also measured with the Fann viscometer. These instruments are valuable aids in
measuring the effectiveness of drilling mud controls, as shown by viscosity changes
in pilot tests.
Cooling and Lubrication
Considerable heat is generated by friction in the bit and where drill string is
in contact-with the formation or casing. There is little chance for this heat to be
conducted away by the formation; therefore, the circulating fluid must remove it.
The heat, having been transmitted from points of friction to the mud, is
transmitted to the atmosphere at the surface. Solids in the mud help to lubricate.
The application of conventional oil emulsion mud coupled with various emulsifying
agents increases this lubricity. This shows up in decreased torque, increased bit
life, reduced pump pressure, etc.
Wall Building
A good drilling fluid should deposit a good filter cake on the wall of the
hole to consolidate the formation and to retard the passage of fluid into the
formation. This property of the mud is improved by increasing the colloidal
fraction of the mud by adding bentonite and chemically treating the mud to
improve de-flocculation and solids distribution. In many cases it may be
necessary to add starch or other fluid loss control additives to reduce the fluid
loss.
Control of Sub-surface Pressures
The proper restraint of formation pressures depends upon density or weight
of the mud. Normal pressure gradient is equal to 0.105 bar/m of depth. This is the
pressure exerted by a column of formation water. Normally the weight of water
plus the solids picked up from drilling is sufficient to balance formation pressures.
However, at times abnormal pressure requires the addition of a higher specific
gravity material, such as barite, to increase the hydrostatic head of the mud
column. The density of mud is measured with a mud balance in g/cm3, lb/gal, lb/cu
ft or psi/1000 ft of depth. The hydrostatic pressure that a column of mud exerts
upon any point in the hole can be calculated as follows:
Hydrostatic pressure:
bar = (depth, m) x (mud weight, g/ml) x (0.1)
psi = (depth, ft) x (mud weight, lb/gal) x (0.052)
psi = (depth, ft) x (mud weight, lb/ft3) x (0.00695)
Suspending Cuttings and Sand and Releasing Them at the Surface
Good drilling fluids have properties that cause the solids particles being
carried to the surface to be held in suspension, due to a gel or thixotropy that
develops after circulation has stopped. Upon resumption of circulation, the mud
reverts to its fluid condition and these coarse particles, together with the sand are
carried to the surface. The sand tube and screen are used to measure the sand
content of the mud. A comparison of the sand content of samples taken at the flow
line and suction will tell whether or not the sand is being properly removed at the
surface or is being re-circulated through the system. High sand content of the flow
line discharge is to be expected if sandy formations are being drilled. Sand is
extremely abrasive and if it is being re-circulated through the system, pumps, and
fittings will be damaged. Regular tests for sand content should be made on the
mud, and the sand content not allowed to run over two percent at the pump suction.
Support of Weight of Drill Pipe and Casing
With increasing depths, the weight supported by the surface equipment
becomes increasingly important. Since a force equal to the weight of mud displaced
buoys up both the drill pipe and casing, an increase in mud density necessarily
results in a considerable reduction in total weight, which the surface equipment
must support. Equally, if the casing is not completely filled up during running, some
of the hook load is alleviated and the string can be "floated in".
Protection of Well Bore and Assurance of Maximum Hole Information
Optimum values of all the properties of drilling fluid are necessary to offer
maximum protection of the formation, yet sometimes these values must be
sacrificed to gain maximum knowledge of the formations penetrated. For example,
salt may upset mud and increase the fluid loss, yet it may be added to control the
resistivity in order to get the proper interpretation of an electric log. Again, oil
may improve the performance of mud and even the production of a well, but if it
interferes with the work of the geologist or ecologist, it may be forbidden for use
in the drilling fluid.
Transmit Hydraulic Horsepower to the Bit
The drilling fluid is the medium for transmitting available hydraulic
horsepower at the surface to bit-. Hydraulics should be considered when planning a
mud program. In general, this means that circulating rates should be such that
utilization of optimum power is used to clean the face of the hole ahead of the bit.
The flow properties of the mud, plastic viscosity and yield point, exert a
considerable influence upon hydraulics and should be controlled at their proper
values.
Clays and Colloid Chemistry
Colloids are not a specific kind of matter. They are particles whose size falls
roughly between that of the smallest particles that can be seen with an optical
microscope and that of true molecules, but they may be of any substance. Colloidal
systems may be consist of solids dispersed in liquids, i.e., clay suspensions, liquid
droplets dispersed in liquids, i.e., emulsions, or solids dispersed in gases, i.e., smoke.
One characteristic of aqueous colloidal systems is that the particles are so small
that they are kept in suspension indefinitely by bombardment of water molecules, a
phenomenon known as Brownian motion. Another characteristic of colloidal systems
is that the particles are so small that properties like viscosity and sedimentation
velocity are controlled by surface phenomena.
Clay minerals which occur in all types of sediments and sedimentary rocks,
constitute the most abundant class of minerals in these rocks, comprising 40% of
all the minerals present. Clay minerals mostly belong to the group of silicates having
layer structures, but differ from most layer silicates by having hydrous nature.
Over 50% of the clay minerals in the earth’s crust are illites. The order of
relative abundance of clay minerals is as follows:
-Illite
-Montmorillonite and mixed-layer illite-montmorillonite
-Chlorite and mixed-layer chlorite-montmorillonite
-Kaolinite and septechlorite
-Attapulgite, palygorskite and sepiolite
Most layer silicates are usually in microscopic size, also occur as
submicroscopic particles. Although initially water was used as the carrying fluid,
the advantages of using suspensions of clay-in-water became apparent as clay
minerals were inadvertently incorporated in the fluid as a result of drilling through
argillaceous strata. Subsequently, it became the usual practice to add surface clays
to water in order to prepare a mud for circulation. As the primary function of a
drilling fluid is to remove cuttings, a thick slurry facilitated this action as
compared to a liquid of low viscosity and devoid of shear strength, i.e., water. Also,
clay-in-water suspensions served to keep the formation fluids confined to their
respective formations during drilling operations.
Properties of Clay Minerals
Two characteristic physical properties of clay minerals are size and shape of
particles. The physicochemical properties of clays which are of great interest to
the petroleum engineers and petroleum geologists are:
-Base exchange capacity
-Adsorption and retention of water
-Deflocculation and flocculation
Swelling of clays
Some clays do not swell upon hydration, e.g., kaolinite clay exhibits little or no
swelling on hydration. Sodium-montmorillonite, on the other hand, swells in water to
many times its dry volume. Swelling properties of different clays are a function of
-Structure
-Chemical composition
-The amount and types of exchangeable cations
In swelling due to hydration, there are two types of swelling; swelling due to the
expansion of the crystal lattice itself, and swelling due to the adsorption of water
on surface of the clays particles.
Flocculation and Deflocculation
Deflocculation is defined as the state of a dispersion in which solid particles
in a liquid remain geometrically independent and unassociated with adjacent
particles. In good drilling fluids, clay is in a state of deflocculation. Flocculation is
defined as the state of dispersion in which there is a formation of clusters of
particles separable by relatively weak mechanical forces. This may be caused by a
change in chemical composition of the dispersion neutralization of negative charges
on clay particles.
Clays Used in Drilling Fluids
The suitability of clays for use in drilling fluids may be determined by
-Yield, i.e., the number of barrels of mud of a given viscosity obtained from a ton of
clay in fresh water
-Suspension capacity in salt water
-Plastic viscosity
-Apparent viscosity
-Yield strength
Thixotropic properties, i.e., the difference in gel strength determined immediately
after agitation and after quiescence (usually 10 min)
-Wall building properties as measured by water loss through a filter paper
-Thickness of filter cake produced
Types of Drilling Fluids
Many types of drilling fluids are used in industry. Major categories include
air, water- and oil base fluids. Each has many subcategories based on purpose,
additives, or clay states.
Water Based Muds
Water based mud’s consist of four basic phases;
-Water
-Active colloidal solids
-Inert solids
-Chemicals
Water is the continuous phase of any water-based mud. Primary function of
the continuous phase is to provide the initial viscosity which can be modified to
obtain any desirable rheological properties. The second function of the continuous
phase is to suspend the reactive colloidal solids, such as bentonite, inert solids,
such as barite. Water also acts as a medium for transferring the surface available
hydraulic horsepower to the bit on the bottom of the hole. Water is also a solution
medium for all conditioning chemicals which are added to the drilling fluid. In water
based mud’s, clay is added to increase density, viscosity, gel strength and yield
point, and to decrease fluid loss. Clays used in water based drilling fluids are mainly
in three groups:
-Montmorillonites (bentonite)
-Kaolinites
-Illites
Chemicals used in water based mud’s can be grouped according to their functions
as:
-Thinners
-Dispersants
-Deflocculants
Types of Water Based Muds: a) Clear Water: Fresh water and saturated brine can be used to drill hard
formations, compacted and near normally pressured formations. This mud is made
by pumping water down to the hole during drilling and letting it react with
formations containing clays or shales. The water dissolves the clays and returns to
the surface as mud. This mud is characterized by its high solids content and a high
filter loss resulting in a thick filter cake.
b) Calcium Muds: Calcium mud’s are superior to fresh water mud’s when drilling
massive sections of gypsum and anhydrite as they are susceptible to calcium
contamination. When calcium is added to a suspension of water and bentonite, the
calcium cations will replace the sodium cations on the clay plates. When calcium
mud comes into contact with shaly formations, the swelling of shale is greatly
reduced in the presence of calcium cations. The major advantage of calcium mud is
their ability to tolerate a high concentration of drilled solids without these
affecting the viscosity of mud. Calcium mud’s are classified according to the
percentage of soluble calcium in the mud.
1- Lime Mud which contains up to 120 ppm of soluble calcium and it is prepared by
mixing bentonite, lime [Ca(OH)2], thinner, caustic soda and filtration control agent.
2- Gyp Mud which contains up to 1200 ppm of soluble calcium. It is similar to lime
mud except that the lime is replaced by gypsum and they have higher temperature
stability.
c) Lignosulphonate Mud: This mud type is considered to be suitable when :
- high mud densities are required,
- working under moderately high temperatures
- high tolerance for contamination by drilled solids
- low filter loss is required.
This type of mud consists of fresh water or salt water, bentonite, ferrochrome
lignosulphonate, caustic soda, CMC or stabilized starch. It is not suitable for
drilling shale sections due to adsorption of water.
d) KCl / Polymer Mud: The basic components of KCl/polymer mud’s are:
-fresh water or sea water
-KCl
-inhibiting polymer
-viscosity building polymer
-stabilized starch or CMC
-caustic soda
Lubricants
This mud is suitable for drilling shale sections due to its superior sloughing-
inhibition properties. It is also suitable for drilling potentially productive sands.
The advantages of this type of mud are:
- higher shear thinning
- high true yield strength
- improved bore hole stability
- good bit hydraulics and reduced circulating pressure losses.
The disadvantage is their instability at temperatures above 250 oF.
Oil Based Muds
Oil based mud’s has been defined as a system the continuous or external phase of
which is any suitable oil. At the present time, there are two mud systems the
external phase of which is oil, i.e., true oil mud’s and invert emulsion mud’s.
True oil mud systems consist of the following components:
-Suitable oil
-Asphalt
-Water
-Emulsifiers
-Surfactants
-Calcium hydroxide
-Weighting materials
-Other chemical additives
Among all of these, only oil and asphalt are necessary for the proper functioning of
oil mud’s. The others are only used for the purpose of enhancing and stabilizing
rheological properties and plastering characteristics.
Different types of oils have been used as the continuous phase in oil mud’s. The
following commonly available oils have gained widespread acceptance;
-Lease crude oil
-Refined oils
The following specifications are used as guidelines for the selection of oil:
-Specific weight (API gravity) – For viscosity purposes
-Aniline point – A measure for the aromatic content of the oil
-Flash point – It is the temperature at which oil vapor ignites upon passing flame
over the hot oil
-Fire point – It is the temperature at which continuous fire is sustained over the oil
surface when flame is passed over it
Although presence of water is not required in oil mud’s, some water is
generally added to react with chemical additives in order to enhance the
rheological properties and plastering characteristics of oil. A number of bodying
agents have been used in oil mud’s to achieve the desired rheological and filtration
loss characteristics. Bodying agents can be classified into two groups:
-Colloidal size materials
-High molecular weight metal soaps
Asphalts which are colloidal-size organophilic materials are used in oil mud have to
impart required properties and control fluid loss, mainly through their absorptive
characteristics. Asphalt work with the same principle as clays in water-based mud’s.
Heavy metal soaps of fatty acids (emulsifiers) are added to the oil mud’s in order
to emulsify the water in oil. The functions of emulsifiers in oil mud’s are as follows:
-Imparting weak gel strength to oil mud’s because gel strength is necessary for
suspension of weighting materials,
-Emulsification of any water picked up during drilling operation,
-Controlling the tightness of any water emulsion resulting from water
contamination, thus, controlling fluid loss
Aerated Muds
Interest in under balanced drilling is increasing worldwide. In under balanced
drilling operations, pressure of the drilling fluid in the borehole is intentionally
maintained below the formation pore fluid pressure, in the open hole section of the
well. As a result formation fluids flow into the well when a permeable formation is
penetrated during under balanced drilling. Usually, aerated fluids are used in under
balanced drilling operations. Most frequently used aerated fluids are air-liquid
mixtures, foams, mist and gas.
Selection of Drilling Fluids
Selection of the best fluid to meet anticipated conditions will minimize well
costs and reduce the risk of catastrophes such as stuck drill pipe, loss of
circulation, gas kick, etc. Consideration must also be given to obtain adequate
formation evaluation and maximum productivity. Some important considerations
affecting the choice of mud’s to meet specific conditions are presented as follows:
-Location : The availability of supplies must be considered, i.e., in an offshore well,
the possibility of using salt water should be considered.
-Mud-making shales : Thick shale sections containing dispersible clays cause a rapid
rise in viscosity as cuttings become incorporated in the mud. When the mud is un-
weighted, it is easy to reduce the excessive viscosity, however, when the mud is
weighted, costly chemicals such as barite should be used to restore the mud
properties.
-Pressured formations : The density of the mud should be adjusted as pressurized
formations are to be drilled. However, high density mud’s increase the cost of
drilling and have risks of stuck pipe, loss circulation, etc.
-High temperature : Most of the mud additives degrade with time and elevated
temperatures, which are higher than degradation temperature. Special additives
must be used to make mud resistive to high temperatures.
-Hole instability : Two basic forms of hole instability are hole contraction and hole
enlargement. If the lateral earth stresses bearing on the walls of the hole exceed
the yield strength of the formation, hole slowly contracts. He density of the mud
should be high enough to resist contracting. Hole enlargement occurs at water-
sensitive shale zones. Shale stabilizers should be used to prevent hole enlargement.
-Rock salt : To prevent the salt from dissolving and consequently enlarging the hole,
either an oil base mud or a saturated brine must be used.
-Hole inclination : In highly deviated holes, torque and drag are a problem because
the pipe lies against the low side of the hole and the risk of pipe stuck is high.
Proper mud’s should be selected to prevent such problems, and keep cutting to be
removed from the well properly.
Formation evaluation : The selected mud should be suitable for logging tools, MWD,
Productivity impairment : Solids control or density adjustments should be
considered properly to keep the formations non-damaged or blocked.
Field Tests on Drilling Fluids Properties
It is necessary to perform certain tests to determine if the mud is in proper
condition to perform the functions previously discussed. The frequency of these
tests will vary in particular areas depending upon conditions. A standard API form
should be provided for reporting the results of these tests:
Density or mud weight: Report in g/ml or in lb/gal, lb/cu ft, or psi/1000 ft of
depth. Instrument should be calibrated once a day when weight is critical.
Viscosity: The Marsh funnel viscosity is used for routine field measurement.
Report viscosity in seconds per quart. Plastic viscosity (cP) and Yield Point
(lb/100 sq ft) are measured with the Fann viscometer.
Gel strength: Report as lb/100 sq ft.
Fluid loss: Report cc of filtrate at the end of thirty minutes, measure and
record cake thickness in mm.
pH: Measure with p-Hydrion paper or pH meter.
Sand content: Report as percent by volume.
Salt content: Report as ppm chloride, or ppm sodium chloride.
Retort Analysis: Determine the liquid and solid content of a drilling fluid. (percent
by volume)
Other tests which may enter into the evaluation of a mud system are as
follows: Apparent viscosity,Pf, Pm, lime content, ppm calcium (hardness), percent
solids by volume, temperature, resistivity, etc. The API has recommended standard
methods of conducting these tests, and detailed procedures may be found in their
publication, "Recommended Practice on Standard Field Procedure for Testing
Drilling Fluids". (API RP 13B).
Rheological Properties:
Determine viscometer readings to calculate the following for a drilling or
completion/ work over fluid: Plastic Viscosity (PV, cp); Yield Point (YP, lbf/100 ft2);
Gel Strength - Max. dial reading at 3 rpm- (Tau, lbf/100 ft2); Apparent viscosity
(AV, cp); Consistency index (K, lbf/secn/cm2); Yield stress (YS, lbf/100 ft2); Flow
index (n, unit-less)
PV = θ600 - θ300
YP = θ600 - PV
AV = θ600 / 2
n = 3.32 log (θ600 / θ300)
K = θ300 / (511n)
Gel strength = Max. dial reading at 3 rpm
Example:
Given the following well data, determine PV, YP, AV, n and K.
θ600 = 36
θ300 = 24
Solution:
PV = 36 -24 = 12 cP
YP = 24 – 12 = 12 lbf/100 ft2
AV = θ600 / 2 = 36 / 2 = 18 cP
n = 3.32 log (θ600 / θ300) = 3.32 log (36/24) = 0.5846
K = θ300 / (511n) = 24 / 5110.5846 = 0.626
Drilling Fluid Additives
There are fundamental aspects that have to be controlled in order to have an
effectively and successfully purposing drilling fluid. These aspects can be
categorized as:
-Viscosity Control
-Fluid Loss Control
-Weight Control
-Corrosion Control
Viscosity Control
Viscosifiers :Many different products are classified as viscosifiers. Bentonite,
attapulgite clays, subbentonites and polymers are most widely used viscosity
builders. Bentonite, attapulgite clays and sub-bentonites all form colloidal
suspensions in water. They increase viscosity, yield point and gel-strength by inter-
surface friction and by chemically binding-water. Polymers are multi-purpose
additives that may simultaneously modify viscosity, control filtration properties,
stabilize shales and create or prevent clay flocculation.
Thinners : Mud thinners or dispersants reduce viscosity by breaking the
attachment of clay plates through the edges and faces. The thinners absorb to the
clay plates, thus disturbing attractive forces between the sheets. Thinners are
added to a mud to reduce viscosity, gel strength and yield point. Most thinners can
be classified as organic materials or as inorganic complex phosphates. Organic
thinners include lignosulfonates, lignins and tannins. Inorganic thinners include
sodium acid pyrophosphate, tetrasodium pyrophosphate, sodium tetraphosphate and
sodium hexametaphosphate. Organic thinners are good for higher temperatures.
Phosphates :
-Useful as effective thinners in most bentonite water-based mud’s at shallow
depths,
-Small amounts of thinner are very effective at temperatures less than 130°F
-pH is around 5, so caustic soda or some other hydroxyl ion containing additive is
required to maintain pH above 7.0,
-They have very low temperature stability,
-They have no fluid loss ability,
-At relatively low temperatures, they revert to orthophosphates, which severly
flocculates clays and increase viscosities and gel strengths,
Lignites :
-They have a temperature stability of 400°F,
-They are organic thinners serving both as dispersants and as fluid loss control
agents due to their colloidal structure,
-They are not suitable for high-salt content fluids due to the insolubility of lignite
in salt,
-They may cause disintegration of active solids,
Tannins :
-They are dual-purpose materials serving as dispersants and fluid loss control
agents,
-They are effective in thinning lime mud’s and cement contaminated mud’s,
Lignosulfonates :
-They are effective for lime mud’s,
-They are effective as general purpose thinners due to the heavy metal-ions
attached,
-They have temperature stability in the range of 300°F to 350°F,
-They are dual-purpose additives serving as both dispersants and fluid loss
additives,
-They may cause disintegration,
Fluid Loss Additives
The reasons for fluid loss control are:
-To maintain hole integrity,
-To protect water sensitive shales,
-To minimize hole washout to achieve better casing cement jobs,
-To reduce fluid loss to productive formation and to minimize formation damage,
-To reduce log analysis problems,
Bentonite :
-A multipurpose additive that aids in fluid loss control, barite suspension, viscosity
generator for hole cleaning purposes,
-It is not suitable for use in environments high in concentration of sodium, calcium
or potassium without pre-hydration,
-It may contaminate formations such as salt or anhydrite,
-Slurries are susceptible to the effect of high temperature gelation which could
cause an increase in the fluid loss,
Starch :
-They work well as fluid-loss agents in the presence of low soluble calcium or sodium
ions,
-They are suitable for salt-water or gyp mud’s,
-An increase in viscosity is observed when it is used,
-A bactericide must be used to prevent digradation and fermantation,
-It degrades at temperatures over 200°F,
CMC :
-It is active in low to moderate contaminating-concentrations, which makes it
suitable for use in inhibited mud’s,
-It has a temperature stability up to 400°F,
-A thinner may be necessary to counteract the viscosity effects of the additive,
-It may cause a thinning effect in some salt mud’s
Cypan :
-It can be used successfully in high-temperature regions due to its stability up to
400°F,
-It is highly sensitive of calcium ion contamination,
-It may cause dynamic filtration
XC Polymer :
-It builds viscosity,
-It increases gel structure,
-It has low viscosity at high shear rates,
-It has high viscosity at low shear rates,
-Suspends barite,
Lignites and Tannins :
-They have good temperature stability in the range of 300°F to 350°F,
-They have colloidal structure that aids in fluid-loss control,
-The dual action of fluid loss control and dispersing tendencies makes these
products suitable for single-product usage in some cases,
-They are susceptible to calcium-ion contamination and subsequent mud flocculation
due to sequestering nature of the additive,
Weighting Materials
They are substances with high specific gravity which can be added to the mud to
increase its density, usually to control formation pressure. Barite is by far the
most common weighting material used in drilling fluids. It has an API defined
specific gravity of 4.2, which makes it possible to increase mud weight up to 21 ppg.
It is cheap and readily available. However, suspension of barite requires high gel
strength and viscosity. Hematite is sometimes used depending on the availability.
Calcium Carbonate is an additive used in drilling mud’s, workover fluids and packer
fluids to increase the fluid density. It has a specific gravity of 2.7, therefore, the
fluid density can be increased up tp 12 ppg. It is more economical than other
agents, and can be suspended easier than barite. Also it is acid soluble.
Lost Circulation Materials
-Fibous materials – for seepage losses and in combination with other materials
-Flake materials – for seepage losses
-Granular – for losses in fractures
-Slurries that develop strength over time
pH Adjusters
Due to the acid pH of some mud additives and to the operating pH of some mud
systems, it may be necessary to add materials to increase the pH of the mud
system. The three most common pH adjusters are:
-Sodium hydroxide (caustic soda)
-Potassium hydroxide
-Calcium hydroxide
They are corrosive, so additional care is required when used.
Mixture of drilling fluids
If two substances having different densities are mixed then the density of
the mixture is a function of the quantity and density of the components of the
mixture. This relationship can be expressed as follows:
V1D1 + V2D2 = (V1 + V2) (DR)
where,V1=Volume of the first substance; V2=Volume of the second substance;
D1=Density of the first substance; D2=Density of the second substance; DR =
Density of the resulting mixture.
Increasing the weight of drilling mud with a material such as barite can be
related in a similar manner.
V1W1 + V2WB = (V1 + V2) (W2)
where,V1=Volume of mud before weighting; V2=Volume of barite added; W1=Initial
mud density; W2= Final mud density of the second substance; DR = Density of
barite, (35.4 ppg). Assuming an average density for barite of 35. 4 ppg (4.25
Sp.Gr.) then a barrel of barite weighs 35.4 x 42 or 1490 lbs. Barite for oil field
applications is packaged in 100 lb sacks.
Therefore, if V1 is 100 bbls and WB is 35.4 ppg, then
Sacks of barite / 100 bbl of mud = [1490 (W2 – W1)] / (35.4 – W2)
Example-1
How much barite is required to increase the density of 300 bbls of mud from 4 ppg
to 15 ppg.
Solution:
Sacks of barite / 100 bbl of mud = [1490 (W2 – W1)] / (35.4 – W2)
Sacks of barite / 100 bbl of mud = [1490 (15 – 14)] / (35.4 – 15) = 72.4
Sacks of barite / 300 bbl of mud = 72.4 x 3 = 217.2
The below formula is called the starting volume formula. It is used to
determine the initial volume of mud to start with in order to obtain a specific
volume of mud after weighting.
SV = [(35.4 – W2) / (35.4 – W1)] DV
Example-2 How much 12 ppg mud is needed to prepare exactly 250 bbls of 14 ppg
mud?
Solution:
SV = [(35.4 – W2) / (35.4 – W1)] DV
SV = [(35.4 – 14) / (35.4 – 12)] 250 = 228 bbls
The below formula is called the starting volume formula. It is used to
determine the initial volume of mud to start with in order to obtain a specific
volume of mud after weighting.
% by volume water = 100 [(W1 – W2) / (W2 – 8.33)]
Example-3 What volume of water will be necessary to reduce the density of 1200
bbls of 15.2 ppg mud to 13.5 ppg?
Solution:
% by volume water = [(15.2 – 13.5) / (13.5 – 8.33)] = 0.327
bbls of water to add = 1200 x 0.327 = 392
The above problem can be solved similarly to obtain the result directly in barrels
rather than percent by volume.
bbl of water to be added = present vol. [(W1 – W2) / (W2 – 8.33)]
bbl of water to be added = 1200 [(15.2 – 13.5) / (13.5– 8.33)] = 392
Example-4 Calculate how much water and barite must be mixed to make exactly
500 bbl of 14 ppg mud?
Solution:
V1W1 + V2W2 = VFWF
V1 + V2 = VF and V2 = VF – V1
V1 (8.33) + V2 (35.4) = 500 (14)
V1 (8.33) + (500- V1) 35.4 = 7000
V1 = 395 bbl of water
V2 = VF – V1
V2 = 500 – 395
V2 = 105 bbl of barite
105 (14.9) = 1565 sx of barite
Example-5 Calculate how much oil, water and barite are required to make exactly
300 bbl of 15 ppg oil mud with 80/20 oil/water ratio.
Solution:
V1W1 + V2W2 + V3W3 = VFWF
V1 + V2 +V3 = VF
The oil and water have a ratio of 80/20 so that they can be considered as one
volume (V) that is 80 % oil and 20 % water.
V1 + V2 +V3 = VF
V + V3 = 300
V3 = 300 - V
V[ (0.8) (6.8) + (0.2) (8.33)] + V3 (35.4) = 300 (15)
V[ (0.8) (6.8) + (0.2) (8.33)] + (300 – V) (35.4) = 300 (15)
V = 216 bbl of water and oil
216 (0.8) = 173 bbl of oil
216 (0.2) = 433 bbl of water
300 – 216 = 84 bbl of barite (84 x 41.9 = 1252 sx of barite)
Example-6 Calculate how much oil will have to be added to change the oil/water
ratio of 100 bbl of 80/20 oil mud to 90/10.
Retort Analysis: Oil : 64 %, Water : 16 % and Solids: 20 %
Solution:
The oil water ratio is changed by the following formula:
{Volume of water / [(Volume of water + Pressure volume of oil + Volume of oil to be
added) ]} = New percent of water in liquid phase
Volume of water = 100 x (0.16) = 16 bbl
Volume of oil = 100 x (0.64) = 64 bbl
[16 / (16 + 64 +V)] = 0.1
V = 80 bbl
Example-7 If the same 80/20 mud is to be changed to 75/25 oil water ratio,
calculate how much water will have to be added
Retort Analysis: Oil : 64 %, Water : 16 % and Solids: 20 %
Solution:
The oil water ratio is changed by the following formula:
{Volume of oil / [(Volume of oil + Pressure volume of water + Volume of water to be
added) ]} = New percent of oil in liquid phase
Volume of water = 100 x (0.16) = 16 bbl
Volume of oil = 100 x (0.64) = 64 bbl
[64 / (16 + 64 +V)] = 0.75 = 5 bbl
CHAPTER-3
DRILL STRING and CASING
The drill string is an important part of the rotary drilling process. It is the
connection between the rig and the drill bit. Although the drill string is often a
source of problems such as washouts, twist-offs, and collapse failures, it is
seldom designed to prevent these problems from occurring. In many cases, a few
minutes of drill string design work could prevent most of the problems.
Purposes and Components:
The drill string serves several general purposes, including the following:
-provide a fluid conduit from the rig to the bit
-impart rotary motion to the drill bit
-allow weight to be set on the bit
-lower and raise the bit in the well
In addition, the drill string may serve some of the following specialized services:
a-provide some stability to the bottom-hole assembly to minimize vibration and
bit jumping,
b- allow formation fluid and pressure testing through the drill string,
c- permit through-pipe formation evaluation when logging tools cannot be run in
the open hole,
The drill string consists primarily of the drill pipe and bottom-hole
assembly (BHA). The drill pipe section contains conventional drill pipe, heavy-
weight pipe and occasionally a reamer. The BHA may contain the following items:
-drill collars (several types and sizes)- stabilizers-jars-reamers-shock subs-bit,
and bit sub.
Special tools in the BHA or drill pipe may include monitor-while-drilling (MWD)
tools, drill stem testing tools, and junk baskets.
Drill Pipe:
The longest section of the drill string is the drill pipe. The BHA is usually no
longer than 1,000 ft. Each joint of drill pipe includes the tube body and the tool
joint, which connects the sections of drill pipe. Although aluminum drill pipe is
sometimes used in special projects, it will not be presented in this section.
However, it does have important applications in remote areas where airfreight is
required and where otherwise the rig would have insufficient hoisting capacity.
Drill pipe is available in several sizes and weights (Table 3-1). Common sizes
include the following:
3 ½ in.-13.30 lb/ft nominal
4 ½ in.-16.60 lb/ft nominal
5 in. -19.50 lb/ft nominal
Various types of tool joints may increase the average weight per foot, i.e.,
16.60-18.60 lb/ft for 4.5-in. pipe. However, it is still termed as 16.60-lb/ft pipe.
The grade of drill pipe describes the minimum yield strength of the pipe. This
value is important because it is used in burst, collapse, and tension calculations.
Common grades are as follows:
Table 3-1 Common Grades of Drill Pipes
Letter Designation Alternate Designation Yield Strength (psi)
D D-55 55000 E E-75 75000 X X-95 95000 G G-105 105000 S S-135 135000
In most drill string design, the pipe grade will be increased for extra
strength rather than increase the pipe weight. This approach differs somewhat
from casing design. Drill pipe is unlike most other oil-field tubular, such as casing
and tubing, because it is used in a worn condition. Casing and tubing are usually
new when installed in the well. As a result “classes” are given to drill pipe to
account for wear. Therefore, drill pipe must be defined according to its nominal
weight, grade and class.
The API has established guidelines given below:
New = No wear and has never been used.
Premium = Uniform wear and a minimum wall thickness of 80%.
Class 2 = Allows drill pipe with a minimum wall thickness of 65% with all wear on
one side so long as the cross-sectional area is the same as premium class; that is
to say, based on not more than 20% uniform wall reduction.
Class 3 = Allows drill pipe with a minimum wall thickness of 55% with all wear on
one side.
Drill pipe classification is an important factor in drill string design and use since
the amount and type of wear affect the pipe properties and strengths. Drill pipe
is available in several length ranges:
Table-3-2 Ranges of Drill Pipes
Range Length, ft
1 18-22 2 27-30 3 38-40
Heavy Weight Pipe:
The use of heavy weight drill pipe in the drilling industry has become a
widely accepted practice. The pipe is available in conventional drill pipe outer
diameters. However, its increased wall thickness gives a body weight 2-3 times
greater than regular drill pipe. Heavy weight drill pipe provides three major
benefits to the user.
-Reduces drilling cost by virtually eliminating drill pipe failures in the transition
zone.
-Significantly increases performance and depth capabilities of small rigs in
shallow drilling areas through the case of handling and the replacement of some
of the drill collars.
-Provides substantial savings in directional drilling costs by replacing the largest
part of the drill-collar string, reducing down hole drilling torque, and decreasing
tendencies to change direction.
Most heavy-wall drill pipe has an integral center upset acting as a centralizer
and wear pad. It helps prevent excessive tube wear when run in compression.
This pipe has less wall contact than drill collars and therefore reduces the
chances of differential pipe sticking.
Drill Collars:
Drill collars are the predominant components of the bottom-hole assembly.
Some of the functions of the collars are as follows:
-provide weight for the bit
-provide strength needed to run in compression
-minimize bit stability problems from vibrations, wobbling, and jumping
-minimize directional control problems by providing stiffness to the BHA
Proper selection of drill collars (and BHA) can prevent many drilling
problems. Drill collars arc available in many sizes and shapes, such as round,
square, triangular, and spiral grooved. The most common types are round (slick)
and spiral grooved. Large collars offer several advantages.
-fewer drill collars are needed for required weight
-fewer drill collar connections are required.
-less time is lost handling drill collars during trips
-factors governing good bit performance favor close fitting stiff members
-straighter holes can be drilled
Drill-Collar Selection:
The drill collars are the first section of the drill string to be designed.
The collars length and size affect the type of drill pipe that must be used.
Drill-collar selection is usually based on:
-buckling considerations in the lower sections of the drill string when
weight is set on the bit or,
-using a sufficient amount of drill collars to avoid running the drill pipe in
compression.
The design approaches that satisfy these design criteria are the
buoyancy factor method and the pressure-area method, respectively. The
drilling engineer must evaluate these approaches and make some design
decisions since significantly different amounts of drill collars arc required
with each method.
Buoyancy Factor Method
Drill string buckling is a potential problem that must be avoided. If buckling
occurs, stresses in the pipe and tool joints will cause pipe failure. The greatest
potential for drill pipe buckling normally occurs when weight is slacked off on the
bit.
Lubinsky et al. have studied buckling in oil-field tubing, casing, and drill-strings.
They proved that buckling will not occur if bit weights in excess of the buoyed
collar weight are not used. Most current industry practices adhere to this
buoyed-weight concept. The buoyed weight of the drill collars is the amount of
weight that must be supported by the derrick when collars are run in the hole.
This load is always less than the in-air weight if mud is used in the well. For
example, collars that weigh 147 lb/ft while sitting on the pipe racks may have a
buoyed weight of 113 lb/ft in 15.0-lb/gal mud. Several methods arc commonly
used to determine the buoyed weight of the drill collars:
-lower the drill collars (bottom-hole assembly) into the hole and read the
weight indicator (less the hook weight)
-calculate the weight of the displaced mud and subtract from the in-air collar
weight
-multiply the in-air weight with a buoyancy factor that is dependent on mud
weight
The widely used buoyancy factor is calculated from the following equation:
BF = 1 – (MW / 65.5 )
where, BF= buoyancy factor, dimensionless; MW = mud weight (lb/gal), and 65.5
= weight of a gallon of steel, lb/gal.
The available bit weight (ABW) with the buoyancy factor method is the
buoyed weight of the drill collars (bottom-hole assembly) in the mud to be used.
It is calculated as follows:
ABW = (in-air collar weight) x (buoyancy factor)
The required collar length to achieve an arbitrary ABW can be calculated as
follows:
Length = [ABW / (BF) (CW)]
where, ABW = desired available bit weight, lb; BF = buoyant factor,
dimensionless; CW = collar weight (in-air), lb/ft; and length = required collar
length to achieve the desired ABW, ft
Operators usually run 10-15% more collars that ABW would indicate. This
gives a safety margin and keeps the buoyancy-neutral point within the collars
when unforeseen forces (bounce, hole friction, deviation) move the buckling
point up into the weaker drill pipe section.
Pressure-Area Method
Drill pipe tool joints are manufactured to be run in tension. According to
industry guidelines relating to drill pipe, they should not be run in compression.
Therefore, some industry operators design the drill-string so only the drill
collars are subjected to compression loading.
A drill string tension analysis determines the amount of weight that can
be put on the bit without causing the tension-compression neutral point to move
into the drill pipe. The tension neutral point, which is different from Lubinski's
neutral point of buckling, is the depth of zero tension loading. The different
definitions for the term "neutral point" have caused significant controversy in
the industry. A tension analysis includes the pipe and collar weights as well as
the vertical forces acting on the pipe. The vertical forces are calculated as the
hydrostatic pressure at the depth of interest acting on the cross-sectional area
of the pipe. The vertical forces, termed buoyant forces, are usually calculated
at the bottom and top of the collars.
The pressure-area method usually requires a larger section of drill collars
to achieve comparable ABW than the buoyancy factor method. In addition, the
pressure-area method is depth dependent since the hydrostatic pressures are a
function of the well depth as well as the mud weight.
Example 3-1 Use the following data to determine the available bit weight with
the pressure area and buoyancy force methods (See Figure 3-1):
Well depth: 13500 ft; MW = 14.8 lb/gal; Drill Collars: 8 x 3 inch 540 ft; Drill
Pipe:5 x 4.276 inch 19.50 lb/ft
Solution:
Buoyancy Force Method:
-The collar weight on a lb/ft basis is computed as,
Weight (lb/ft) = Area / 0.2945
Weight (lb/ft) =ππππ/4 [(82 – 32) / 0.2945] = 147 lb/ft
-Calculate the collar weight in air,
147 lb/ft x 540 ft = 79380 lb
-Determine the buoyancy factor,
BF = 1 – (MW / 65.5) = 1 – (14.8 / 65.5) = 0.774
-The available bit weight (ABW) with the buoyancy factor is calculated as the
product of the buoyancy factor (BF) and the collar weight.
ABW = 0.774 x 79380 = 61440 lb
Pressure Area Method
-Use the following figure to calculate the buoyant forces for the pressure-area
method.
BF1 = -P x A
BF1 = -(0.052 x 14.8 lb/gal x 13500 ft) [ππππ/4 [(82 – 32)]
BF1 = -(10389 psi) (43.10 inch2)
BF1 = -448726 lb
BF2 = (0.052 x 14.8 lb/gal x 12960 ft) [ππππ/4 (82 – 52) + ππππ/4 (4.2762 – 32)
BF2 = (9974 psi) (37.92 inch2)
BF2 = +378214 lb
-The ABW is the sum of buoyant forces acting on the collars and the collar
weight
ABW = BF1 + BF2 collar weight
ABW = -448726 lb + 378214 lb + 79380 lb
ABW = 8868 lb
Drill Pipe Selection
Drill pipe is used for several purposes, including providing a fluid conduit
for pumping drilling mud, imparting rotary motion to the drill bit, and conducting
special operations such as drill stem testing and squeeze cementing. The
controlling criteria for drill string design are collapse, tension, slip crushing and
dogleg severity. Collapse and tension are used to select weights, grades and
couplings. Typically- higher-strength pipe is required in the lower sections of the
string for collapse resistance, while tension dictates the higher strength pipe at
the top of the well.
Drill String Design:
The following design criteria will be used to select a suitable drill string: (a)
tension, (b) collapse, (c) shock loading and (d) torsion.
Tension:
Prior to deriving any equation, it should be observed that only submerged
weights are considered, since all immersed bodies suffer from lifting or
buoyancy forces. Buoyancy force reduces the total weight of the body and its
magnitude is dependent on fluid density. The total weight, P, carried by the top
joint of drill pipe is given by:
P = (weight of drill pipe in mud) + (weight of drill collars in mud)
P = [(Ldp x Wdp – Ldc x Wdc)] x BF
where; Ldp = length of drill pipe; Wdp = weight of drill pipe per unit length; Ldc =
length of drill collars; Wdc = weight of drill collars per unit length and BF =
buoyancy factor.
Drill pipe strength is expressed in terms of yield strength. This is
defined as the load at which deformation occurs. Under all conditions of loading,
steel elongates initially linearly in relation to the applied load until the elastic
limit is reached. Up to this limit, removal of applied load results in the steel pipe
recovering its original dimensions. Loading a steel pipe beyond the elastic limit
induces deformation which ca not is recovered, even after the load is removed.
This deformation is described as yield and results in a reduction in pipe
strength. Drill string design is never based on the tabulated yield strength value
but, instead on 90% of the yield strength to provide an added safety in the
resulting design. Thus, maximum tensile design load,
Pa = Pt x 0.9
where; Pt = drill pipe yield strength.
The difference between Pa and P gives the margin of over pull (MOP)
MOP = Pa - P
The design values of MOP normally range from 50000-100000 lb.
Actual safety factor gives the ratio of Pa /P.
SF = Pa /P = (Pt x 0.9) / [(Ldp x Wdp – Ldc x Wdc)] x BF
Dynamic loading, which arises from arresting the drill pipe by slips must be
considered.
Ldp = {(Pt x 0.9) / [(SF x Wdp x BF)]} – (Wdc / Wdp) x Ldc
The above equation can also be expressed in terms of MOP instead of SF term ;
Ldp = {(Pt x 0.9 - MOP) / [(Wdp x BF)]} – (Wdc / Wdp) x Ldc
The term Ldp is sometimes expressed as Lmax to refer to the maximum length of
a given grade of drill pipe which can be selected for a given loading situation.
The above equations can also be used to design a tapered string, consisting of
different grades and sizes of drill pipe. In this case the lightest available grade
is first considered and the maximum useable length selected as a bottom
section. Successive heavy grades are then considered in turn to determine the
useable length from each grade along the hole depth.
Collapse:
Collapse pressure may be defined as the external pressure required
causing yielding of drill pipe or casing. In normal drilling operations the mud
columns inside and outside the drill pipe are both equal in height and are of the
same density. This results in zero differential pressure across the pipe body
and, in turn, zero collapse pressure on the drill pipe. In some cases, as in drill
stem testing (DST), the drill pipe is run partially full, to reduce the hydrostatic
pressure exerted against the formation. This is done to encourage formation
fluids to flow into the well bore, which is the object of the test. Once the well
flows, the collapsing effects are small, as the drill pipe is now full of fluids.
Thus, maximum differential pressure, ∆P, across the drill pipe exists prior to
the opening of the DST tool, and can be calculated as follows:
∆∆∆∆P = (L ρρρρ1 / 144) – [(L – Y) x ρρρρ2 / 144]
where; Y = depth to fluid inside drill pipe; L = total depth of well (ft); ρ1 = fluid
density outside the drill pipe (lb/ft3); ρ2 = fluid density inside the drill pipe
(lb/ft3).
When densities are expressed in ppg, the above equation becomes,
∆∆∆∆P = (L ρρρρ1 / 19.251) – [(L – Y) x ρρρρ2 / 19.251]
Other variations of the above equation include the following:
(a) Drill pipe is completely empty, Y=0, ρ2 = 0
∆∆∆∆P = (L ρρρρ1 / 144)
(b) Fluid density inside drill pipe is the same as that outside drill pipes:
i.e. ρ1 = ρ2 = ρ
∆∆∆∆P = (Y ρρρρ / 144)
where, ρ = density of mud (lb/ft3)
Once the collapsing pressure, ∆P, is calculated, it can then be compared with the
theoretical collapse resistance of the pipe. A safety factor in collapse can be
determined as follows:
SF = (collapse resistance) / (collapse pressure)
Normally drillpipe is under tension resulting from its own weightand the
weight of drill collars. The combined loading of tension and collapse is described
as biaxial loading. During biaxial loading, the drillpipe stretches and its collapse
resistance of drillpipe can be determined as follows:
-Determine tensile stress at joint under consideration by dividing the tensile
load by the pipe creoss-sectional area.
-Determine the ratio between tensile stress and the average yield strength.
-Use necessary figure to determine the percent reduction in collapse resistance
corresponding to the ratio calculated.
Torsion:
It can be shown that the drill pipe torsional yield strength when
subjected to pure torsion is given by;
Q = (0.096 167 J Ym / D)
where Q = minimum torsion yield strength (lb-ft); Ym == minimum unit yield
strength (psi); J = polar moment of inertia = π/32 (D4 –d4) for tubes =
0.098175(D4 –d4); D = outside diameter (in), d = inside diameter (in).
When drill pipe is subjected to both torsion and tension, as is the case during
drilling operations the above equation becomes;
Qττττ = (0.096 167 J / D) √√√√ Y2m – (P
2 / A2)
where Qτ= minimum torsion yield strength under tension (lb-ft); J = polar
moment of inertia = π/32 (D4 –d4) for tubes = 0.098175 (D4 –d4); D = outside
diameter (in); d = inside diameter (in); Ym = minimum unit yield strength (psi); P =
total load in tension (lb); A = cross-sectional area (in2).
Example 3-2 A drill string consists of 600 ft of 81/4 in x 2 13/16 in drill collars
and the rest is a 5 in, 19.5 lbm/ft, Grade X95 drillpipe. If the required MOP is
100 000 lb and mud weight is 75 pcf (10 ppg), calculate the maximum depth of
hole that can be drilled when;
(a) using new drill pipe and
(b) using Class 2 drill pipe having a yield strength (Pt) of 394600 lb.
Solution:
(a) Weight of drill collar per foot is;
A x 1 ft x ρρρρs = ππππ / 4 [(8 1/4)2 - (2 13/16)2] x 1 ft x 489.5 x 1/144 = 160.6
lbm/ft
where; ρs = density of steel (489.5 lbm/ft) ; A = cross-sectional area (in).
(weight of drill collar = 161 lbm/ft.)
Ldp = {(Pt x 0.9 - MOP) / [(Wdp x BF)]} – (Wdc / Wdp) x Ldc
Pt = 501090 lb (Grade X95 new pipe)
BF = 1 – (ρρρρm / ρρρρs) = 1 – (75 / 489.5) = 0.847 and MOP = 100000 lb
Therefore;
Ldp = {(501090 x 0.9 –100000) / [19.5 x 0.847)} – (160.6 / 19.5) x 600
Ldp = 16309 ft
Therefore, maximum hole depth that can be drilled with a new drillpipe of
Grade X-95 under the given loading condition is 16309 + 600 = 16909 ft
(b) Now Pt = 394600 lb
Ldp = {(Pt x 0.9 - MOP) / [(Wdp x BF)]} – (Wdc / Wdp) x Ldc
Ldp = {(394600 x 0.9 – 100000) / [19.5 x 0.847)} – (160.6 / 19.5) x 600
Ldp = 10506 ft
Maximum hole depth = 10506 + 600 = 11106 ft
Example 3-3 If 10000 ft of the drillpipe in Example-1 is used, determine the
maximum collapse pressure that can be encountered and the resulting safety
factor. The mud density is 75 pcf (10 ppg). If the fluid level inside the drillpipe
drops to 6000ft below the rotary table, determine the new safety factor in
collapse.
Solution:
a) Maximum collapse pressure, ∆P, occurs when the drillpipe is 100% empty.
∆∆∆∆P = (L ρρρρm / 144)
∆∆∆∆P = (10000 x 75 / 144) = 5208 psi
Collapse resistance of new pipe of Grade X95 is 12010 psi.
SF = 12010 / 5208 = 2.3
(b) When the mud level drops to 6000 ft below the surface;
∆∆∆∆P = (L ρρρρm / 144)
∆∆∆∆P = (6000 x 75 / 144) = 3125 psi
SF = 12010 / 3125 = 3.8
Example-3-4 A drill string consists of 10000 ft of drill pipe and a length of drill
collars weighting 80000 lb. The drill pipe is 5 in. OD, 19.5 lb, Grade S 135,
premium class.
(a) Determine the actual collapse resistance of the bottom joint of drill pipe
(b) Determine the safety factor in collapse. (Assume collapse resistance = 10050
psi and mud weight = 75 lb / ft3.
Solution:
For a 5 in. OD new drill pipe, the nominal ID is 4.276 in. (thickness = 0.376 in.).
For a premium drill pipe, only 80% of the pipe thickness remains. Thus reduced
wall thickness for premium pipe is = 0.8 x 0.362 = 0.2896 inch and reduced OD
for a premium pipe;
Reduced OD = nominal ID + 2 x (premium thickness)
Reduced OD = 4.276 + 2 x (0.2896)
Reduced OD = 4.8552 inch
Cross sectional area (CSA) of premium pipe = π / 4 (OD2 - ID2)
Cross sectional area of premium pipe = ππππ / 4 (4.85522 – 4.2762)
Cross sectional area of premium pipe = 4.1538 inch2
Tensile stress at bottom joint of drill pipe = tensile load
Tensile stress at bottom joint of drill pipe = (weight of drill collars / CSA)
Tensile stress at bottom joint of drill pipe = 80000 lb / 4.1538 inch2
Tensile stress at bottom joint of drill pipe = 19259 psi
From the figure given below; a tensile ratio of 13.3% reduces the nominal
collapse resistance to 53%.
(a) Collapse resist. of bottom joint of drill pipe = 0.93 x collapse resist.
under zero load
Collapse resist. of bottom joint of drill pipe = 0.93 x 10050 = 9347 psi
(b) SF = (collapse resistance / collapse pressure)
For worst conditions assume drill pipe to be 100% empty. Hence;
Collapse pressure = (75 x 10000) / 144 = 5208 psi
SF = 9347 / 5208 = 1.8
CASING
Drilling environments often require several casing strings in order to reach
the total desired depth. Some of the strings are as follows (Figure 3-1).
-drive or structural
-conductor
-surface
-intermediate (also known as protection pipe)
-liners
-production (also known as an oil string)
-tubing
Drive Pipe or Conductor Casing:
The first string run or placed in the well is usually the drive pipe or
conductor casing. The normal depth range is from 100-300 ft. In soft-rock
areas the pipe is hammered into the ground with large diesel hammer. Hard-rock
areas require that a large diameter shallow hole be drilled before running and
cementing the well. A primary purpose of this string of pipe is to provide a fluid
conduit from the bit to the surface. An additional function of this string of pipe
is to minimize hole-caving problems.
Structural Casing:
Drilling conditions will require that an additional string of casing be run
between the drive pipe and surface casing. Typical depth range from 600-1000
ft. Purpose of this pipe includes solving additional lost circulation or hole caving
problems and minimizing kick problems from shallow gas zones.
Surface Casing:
Many purposes exist for running surface casing, including:
-cover fresh water sands
-maintain hole integrity by preventing caving
-minimize lost circulation into shallow- permeable zones
-cover weak zones
-provide a means for attaching the blowout preventers
-support the weight of all casing strings (except liners) run below the surface pipe.
Typical casing string relationship
Intermediate Casing:
The primary applications of intermediate casing involve abnormally high
formation pressures. Since higher mud weights are required to control these
pressures, the shallower weak formations must be protected to prevent lost
circulation or stuck pipe. It is used to isolate salt zones or zones those cause
hole problems, such as heaving and sloughing shales.
Liners:
Drilling liners are used for the same purpose of intermediate casing.
Instead of running the pipe to the surface, an abbreviated string is used from
the bottom of the hole to a shallower depth inside the intermediate pipe. Usually
the overlap between the two strings is 300-500 ft. Drilling liners are used
frequently as a cost-effective method to attain pressure or fracture gradient
control without the expense of running a string to the surface. When a liner is
used, the upper exposed casing, usually intermediate pipe, must be evaluated
with respect to burst and collapse pressures for drilling the open hole below the
liner.
Production Casing:
The production casing is often called the oil string. The pipe may be set at
a depth slightly above, or below the pay zone. The pipe has the following
purposes:
-isolate the producing zone from the other formations.
-provide a work shaft of a known diameter to the pay zone.
-protect the producing tubing equipment.
Casing Physical Properties
The physical properties of oil-field tubular goods include grade, pressure,
resistance, drift diameter and weight.
Grade:
The pipe grade is the designation that defines the pipe’s yield strength
and certain special characteristics. The grade usually consists of a letter and a 2
or 3 digit number such as N-80. As the letter proceeds, the pipe increases in
yield strength. N-80 has greater yield strength than H-40. The numerical code
indicates the minimum yield strength of 80,000 psi. The average yield strength
is usually 10,000 psi greater than the minimum yield, 90,000 psi for N-80 pipe.
The minimum value is used in burst and collapse resistance calculations, whereas
the average is used for biaxial evaluation. C pipe is a controlled yield pipe used
primarily in environments.
Weight:
The pipe weight is usually defined in pounds per foot. The calculated
weights, as defined by the API, are determined by the following formula.
WL = (Wpc L ) + ew
WL = calculated weight of a pipe of length L, lb
Wpc = plain-end weight, lb/ft
L = length of pipe, ft
ew = weight gain or loss due to end finishing, lb
The cross-sectional area of the pipe can be approximated from the pipe weight;
Ap = 0.29 Wpc
Ap = cross sectional area, square-inch
Range:
Pipe range is a value for approximating the length of a section of pipe.
Normal range sizes are 1,2 or 3.
Diameter:
The drilling engineer must consider three types of diameter data when
planning the tubular program. These are outer, inner and drift diameter.
Burst:
The burst rating of the casing is the amount of internal pressure that the
pipe can withstand prior to failure. The internal yield pressure for pipe is
calculated from the following equation.
PB = 0.875 [(2Yp t) / OD]
PB = burst pressure rounded to the nearest 10 psi
Yp = specified minimum yield strength, psi
t = nominal wall thickness, inch
OD = nominal outside diameter, inch
Example-3-5:
Calculate the internal yield (burst) pressure for 26.40 lb/ft, N-80, 7.625
inch pipe. Assume it has a wall thickness (t) of 0.328 inch. Use the API minimum
wall thickness factor of 0.875. Recalculate the results and use 95 % wall
thickness.
Solution:
a) The internal yield stress (burst) is calculated as:
PB = 0.875 [(2Yp t) / OD]
PB = 0.875 [2(80000 psi) 0.328 inch) / 7.625 inch]
P = 6020 psi
b) Recalculate the results with a 95 % wall thickness.
PB = 0.95 [2(80000 psi) 0.328 inch) / 7.625 inch]
P = 6540 psi
Example-3-6:
A drilling engineer must design a production casing string for sour gas
service. The maximum anticipated surface pressure for the 5.5 inch OD pipe is
20800 psi. The engineer’s company dictates that pipe used in sour service will
not have yield strength greater than 90,000 psi. Determine the wall thickness
requirements for the pipe. Use the yield strength of 90,000 psi and assume that
the API tolerance of 87.5 % wall thickness. Round up the wall thickness to the
nearest 1/8 inch.
PB = 0.875 [(2Yp t) / OD]
20800 = 0.875 [2 (90000) t) / 5.5]
t = 0.726 inch and nearest 1/8 is : t = 0.750 inch
Collapse:
Unlike internal yield resistance of the pipe, collapse resistance equations
vary depending on the D/t ratio. The collapse resistance is separated into four
categories.
a) yield strength collapse pressure ,
b) plastic collapse
c) transition collapse ,
d) elastic collapse
The D/t range must be evaluated and the proper equation must be selected.
Formula factors must be used in collapse calculations. The yield strength
collapse pressure is not a true collapse pressure, rather the external pressure
(Pyp) that generates minimum yield stress (Yp) on the inside wall of a tube.
Pyp = 2 Yp [ ((D/t) – 1) / (D/t)2]
The formula for yield strength collapse pressure is applicable for D/t values up
to the value of D/t corresponding to the intersection with plastic collapse
formula. The intersection is calculated as follows:
(D/t)yp = SQRT [ (A-2)2 + 8 (B-(C / Yp))] + (A - 2)) / [ 2 (B + C/Yp)]
The applicable D/t ratios for yield strength collapse are given in Table.
The minimum collapse pressure for the plastic range of collapse (Pp) is :
Pp = Yp [ (A / (D/t)) – B ] – C
The formula for minimum plastic collapse pressure is applicable for D/t
values ranging from (D/t)pt to the intersection for (D/t)t, transition collapse
pressure. Values for (D/t)pt are calculated by means of:
(D/t)pt = [Yp (A-F)] / [C + Yp (B-G)]
Example-3-7:
An engineer must calculate the collapse rating for the following section of
pipe. Using the API tables and equations, calculate the collapse pressure to the
nearest 10 psi.
Pipe diameter: 9.625 inch , Wall thickness: 0.472 inch
Grade: N-80 , Weight: 47 lb/ft
Solution:
1-Determine the D/t ratio:
D/t = 9.625 inch / 0.472 inch
D = 20.392
From Table:
A = 3.071 : B = 0.0667: C = 1955
Pp = Yp [ (A / (D/t)) – B ] – C
Pp = 80000 [ (3.071 / (20.392)) – 0.0667 ] – 1955
Pp = 4756 psi
Pp = 4750 – 4760 psi
The minimum collapse pressure for the plastic to elastic transition zone (Pt)
is calculated:
(Pt) = Yp [F /(D/t) – G]
Values for (D/t)te are calculated from the following equation:
(D/t)te = (2 + (B/A)) / (3 (B/A))
The minimum collapse pressure for the elastic range of collapse is calculated as:
Pe = 46.95 x 106 / (D/t) [(D/t)-1]2
Example-3-8
The collapse rating for 47.0 lb/ft, C-95 grade, 9.625 inch pipe must be
calculated. The wall thickness is unknown. Use the API formulas and tables.
Solution:
1. Compute the cross-sectional area of the pipe.
Ap = 0.29 Wp
Ap = 0.29 (47 lb/ft)
Ap = 13.63 inch2
2. Determine the wall thickness of the pipe from the cross sectional area.
Ap = π/4 (OD2 – ID2)
13.63 = π/4 (9.6252 – ID2)
ID = 8.676 inch
t = (OD –ID) / 2
t = (9.625 – 8.676) / 2
t = 0.4745 inch
3. D/t ratio is:
D/t = 9.625 / 0.4745 = 20.284
4. The formula for C-95 pipe with a D/t ratio of 20.284 is:
A = 3.124 B = 0.0743 C = 2404
Pp = Yp [ (A / (D/t)) – B ] – C
Pp = 95000 [ (3.124 / (20.284)) – 0.0743 ] – 2404
Pp = 5168 psi
Axial Stress:
An axial stress is calculated by modifying the yield stress to an axial stress
equivalent grade:
YPA = [SQRT (1 - 0.75 (SA / Yp)2) – 0.5 (SA / Yp) ] Yp
SA = axial stress, psi
Yp = minimum yield strength, psi
YPA = yield strength of axial stress equivalent grade, psi
Example-3-9:
The engineer must calculate the collapse pressure for the following pipe
characteristics. Size: 7 inch OD; Weight : 26 lb/ft; Grade: P-110; SA = 11000
psi; t = 0.362 inch
Solution:
1. Axial stress equivalent grade is:
YPA = [SQRT (1 - 0.75 (SA / Yp)2) – 0.5 (SA / Yp) ] Yp
YPA = [SQRT (1 - 0.75 (11,000 / 110,000)2) – 0.5 (11,000 / 110,000) ) 110,000
YPA = 104,082 psi
2. D/t = ?
D/t = 7 / 0.362 = 19.34
3. A = 3.181 B = 0.0819 C = 2852
Pp = Yp [ (A / (D/t)) – B ] – C
Pp = 104082 [ (3.181 / (19.34)) – 0.0819 ] – 2852
Pp = 5742 psi
Pipe Body Yield Strength:
The pipe body strength is the axial load required to yield the pipe. It is
the product of the cross-sectional area and the specified minimum yield
strength for the particular grade of pipe.
Py = 0.7854 (OD2 – ID2) Yp
Example-3-10:
A section of 10.75 inch, 55 lb/ft N-80 casing is to be run into a well. It
has a wall thickness of 0.495 inch. Determine the pipe body yield strength.
Solution:
1.The ID is computed from:
ID = OD – 2t
ID = 10.75 – 2 (0.495)
ID = 9.76 inch
2.The yield strength is calculated as:
Py = 0.7854 (OD2 – ID2) Yp
Py = 0.7854 (10.752 – 9.762) 80,000
Py = 1,275,000 psi
CHAPTER-4
DRILLING BITS
Drilling Bits
How well the bit drills depends on several factors, such as the condition
of the bit, the weight applied to it, and the rate at which it is rotated. Also
important for a bit performance is the effectiveness of the drilling fluid in
clearing cuttings, produced by the bit away from the bottom.
The aim of drilling is to i) make hole as fast as possible by selecting bits
which produce good penetration rates, ii) run bits with a long working life to
reduce trip time, iii) use bits which drill a full-size or full-gauge hole during the
entire time they are on bottom.
The choice of bit depends on several factors. One is the type of
formation to be drilled, whether it is it hard, soft, medium hard or medium soft.
A second factor is the cost of the bit. Getting the highest possible footage
from the bit cuts down bit costs and minimizes the number of trips needed for
bit changes. It should be stated, however, that continuing to use a bit that is
still drilling but slowly is false economy.
In the shallower part of the hole only one or two bits are needed before
pipe is pulled for logging or running casing and often one bit is sufficient to make
the hole in which the conductor is to be set. As formations near the surface are
usually very soft, one bit may prove sufficient for several wells. But in the
deeper part or the hole, several bits often have to be drilled before casing
depth is reached.
It is normal that the bit used to drill the cement left in the casing is also
used to drill the formation, although in some instances a separate bit is run to
drill the cement and thereafter changed for a more suitable one for the
formation expected deeper down.
Formations vary a lot in hardness and abrasiveness and have a
considerable effect on bit performance. If there were no difference in rock
formations, one type of bit only would be needed which requires standard bit
weight, rotary speed and pump pressure to drill at the maximum rate.
Unfortunately, such a situation does not exist and several bits are required for
the alternating layers of soft material, hard rocks and abrasive sections.
Changing the bit every time as the formation changes is, however, impracticable.
Therefore a compromise has to be made and a bit that performs reasonably well
in all conditions is selected. The choice of bit for a well in a field where the
formations are familiar is obviously easier than for a wildcat.
Bits can generally be classified into two categories; i) roller bits and ii)
drag bits. The following is a description of both.
Roller Cone Bits
The cutting elements of roller cone bits are arranged on “conical”
structures that are attached to a bit body. Typically three cones are used and
the teeth (cutters) may be tungsten carbide that is inserted into pre-drilled
holes into the steel cone shell or steel teeth that are formed by milling directly
on the cone shell as it is manufactured. The length, spacing, shape, and tooth
material are tailored for drilling a particular rock.Insert types used as teeth on
roller-cone bits.
The IADC has developed a standard classification code that is used to
classify bits made by different manufactures according to the rock hardness
that they are designed to drill including the particular design features of the
bit. Each roller bit cone contains a bearing and lubrication system. In some cases
the drilling mud is used as the lubricant (open bearing) and in other cases a
special lubricant is confined inside the case (sealed bearing). The apes bearing
system is used almost exclusively with roller bearings. The sealed bearing
system may be used with either roller or journal bearings. The rock cutting
process of the roller cone bit is either by gauging (digging and shoveling) in soft
formation or by chiseling in hard formation.
A hydraulic cuttings removal system is incorporated in each bit to remove
the cuttings from around the teeth. Typically, a nozzle is placed between each
cone to direct mud at the bottom of the hole and cutters. These nozzles are
usually located at a height approximately equal to the top of the cone, but in
some cases are extended towards the arms where the cutters contact the rock.
The drilling fluid is pumped through the nozzles at relatively high velocity in
order to remove the drilled cuttings.
The three-cone rolling cutter bit is by far the most common bit type
currently used in rotary drilling operations. This general bit type is available
with a large variety of tooth design and bearing types and, thus, is suited for a
wide variety of formation characteristics. The three cones rotate about their
axis as the bit is rotated on bottom. The shape of the bit teeth also has a large
effect on the drilling action- of a rolling cutter bit. Long, widely spaced, steel
teeth are used for drilling soft formations. As the rock type gets harder, the
tooth length and cone offset must be reduced to prevent tooth breakage; the
drilling action of a bit with zero cone offset is essentially a crushing action. The
smaller teeth also allow more room for the construction of stronger bearings.
The metallurgy requirements of the bit teeth also depend on the formation
characteristics. The two primary types used arc (1) milled tooth cutters and (2)
tungsten carbide insert cutters. The milled tooth cutters arc manufactured by
milling the teeth out of a steel cone while the tungsten carbide insert bits arc
manufactured by pressing a tungsten carbide cylinder into accurately machined
holes in the cone. The milled tooth bits designed for soft formations usually are
faced with a wear-resistant material, such as tungsten carbide, on one side of
the tooth. The milled tooth bits designed to drill harder formations are usually
case hardened by special processing and heat treating the cutter during
manufacturing. The tungsten carbide teeth designed for drilling soft formations
are long and have a chisel-shaped end. Rolling cutter bits with the most advanced
bearing assembly are the journal bearing bits In this type bit, the roller
bearings are eliminated and the cone rotates in contact with the journal bearing
pin. This type bearing has the advantage of greatly increasing the contact area
through which the weight on the bit is transmitted to the cone.
Drag Bits
There are two general types of drag bits that are in common usage. The
oldest is the natural diamond matrix bit in which industrial grade diamonds are
set into a bit head that is manufactured by a powdered metallurgy technique.
The size, shape, quantity, quality, and exposure of the diamonds are tailored to
provide the best performance for a particular formation. Each bit is designed
and manufactured for a particular job rather than being mass produced as roller
cone bits are. The cuttings are removed by mud that flows through a series of
water courses. The design of these water courses is aimed at forcing fluid
around each individual diamond. The matrix diamond bit cuts rock by grinding
and thus a primary function of the fluid is to conduct heat away from the
diamonds.
The other type of drag bit is the polycrystalline diamond compact (PDC)
bit that is constructed with cutters comprised of a man made diamond material.
The cutters are generally much larger than natural diamonds and are designed to
cut the rock by shearing, similar to metal machining. PDC bits have proven very
successful in homogeneous and, soft to moderate strength formations. In
formations where they are successful, they can drill two to three times faster
then a roller cone bit and may have an equally long life.
Classification of Bits
A large variety of bits designs are available from several manufacturers.
The IADC (International Association of Drilling Contractors) approved a
standard classification system for identifying similar bit types available from
various manufacturers. The classification system adopted is the three digit
code.
The first digit in the bit classification scheme is called the bit series
number. The letter “D” precedes the first digit if the bit is diamond or PDC drag
bit. Series D1 through D2 are reserved for diamond bits and PDC bits in the
soft, medium-soft, medium, medium-hard and hard formation categories,
respectively. Series D7 through D9 are reserved for diamond core bits in the
soft, medium and hard formation categories. Series 1,2 and 3 are reserved for
milled tooth bits in the soft, medium and hard formation categories,
respectively. Series 5, 6, 7 and 8 are for insert bits in the soft, medium, hard,
and extremely hard formation categories, respectively. Series 4 is reserved for
future use with special categories such as a “universal” bit.
The second digit is called the type number. Type 0 is reserved for PDC
drag bits. Types 1 through 4 designate a formation hardness sub classification
from the softest to the hardest formation within each category. The feature
numbers are interpreted differently, depending on the general type of bit being
described. Feature numbers are defined for diamond and PDC drag bits, diamond
and PDC drag-type core-cutting bits, and rolling cutter bits.
Eight standard diamond and PDC drag bits features are “1”, step-type
profile, “2”, long-taper profile, “3”, short-taper profile, “4”, nontaper profile,
“5”, downhole-motor type, “6”, sidetrack type, “7”, oil-base type, and “8”, core-
ejector type. The remaining feature, 9, is reserved for special features selected
by the bit manufacturer.
There are two standard feature numbers for diamond and PDC drag-type
core-cutting bits. These bits are used to recover a length of formation sample
cored from the central portion of the borehole. The two features are “1”,
conventional core-barrel type, and “2”, face-discharge type. As in the previous
case, feature “9” is reserved for special features selected by the bit
manufacturer.
There are eight standard feature numbers for rolling-cutter bits. The
standard feature numbers are “1”, standard rolling cutter bit (jet bit or
regular), “2”, T-shaped heel teeth for gauge protection, “3”, extra insert teeth
for gauge protection, “4”, sealed roller bearings, “5”, combination of “3” and “4”,
“6”, sealed friction bearing, “and “7”, combination of “3” and “6”. The remaining
features, “8” and “9” were reserved for special features selected by the bit
manufacturer. Feature “8” is often used to designate bits designed for
directional drilling. Some of the main design features of the various rolling-
cutter bit types include some of the tooth design features of the various bit
types and classes. As the class number increases, the cone offset, tooth height,
and amount of tooth hardfacing decreases while the number of teeth and
amount of tooth case hardening increases. An increase in bearing capacity is
possible for the bits with a higher class number. This is possible shorter length
of bit teeth at higher bit class numbers.
Example-1
IADC classification 124E refers to what kind of a bit?
Solution:
1 = Soft formation milled tooth ;2 = Soft to medium within group
4 = Sealed roller bearings ; E = Extended jets (nozzles)
Bit Selection
Bit selection is based on using the bit that provides the lowest cost per
foot of hole drilled. This cost is expressed by the following cost-pet-foot
equation;
Drilling Cost Formula:
The most common application of a drilling cost formula is in evaluating the
efficiency of a bit run. A large fraction of the time required to complete a well
is spent either drilling or making a trip to replace the bit. The total time
required to drill a given depth, ∆D, can be expressed as the sum of the total
rotating time during the bit run, tb, the non-rotating time during the bit run, tc,
and trip time tt. The drilling cost formula is;
Cf = (Cb + Cr ( tb + tc + tt)] / ∆∆∆∆D
where; Cf is drilled cost per unit depth, Cb is the cost of bit, and Cr is the fixed
operating cost of the rig per unit time independent of the alternatives being
evaluated.
Example-2. A recommended bit program is being prepared for a new well using
bit performance records from nearby wells. Drilling performance records for
three bits are shown for a thick limestone formation at 9,000 ft. Determine
which bit gives the lowest drilling cost if the operating cost of the rig is
$400/hr, the trip time is 7 hours, and connection time is 1 minute per
connection. Assume that each of the bits was operated at near the minimum cost
per foot attainable for that bit.
Bit Bit Cost ($) Rota. Time (hr) Conn. Time (hr) Penetration Rate (ft/hr)
A 800 14.8 0.1 13.8 B 4900 57.7 0.4 12.6 C 4500 95.8 0.5 10.2
Solution:
The cost per foot drilled for each bit type can be computed using above
equation. For Bit A, the cost per foot is;
Cf = (Cb + Cr ( tb + tc + tt)] / ∆∆∆∆D
Cf = (800 + 400 ( 14.8 + 0.1 + 7)] / 13.8 (14.8) = $ 46.81 / ft.
For Bit B, the cost per foot is;
Cf = (4900 + 400 ( 57.7 + 0.4 + 7)] / 12.6 (57.7) = $ 42.56 / ft.
For Bit C, the cost per foot is;
Cf = (4500 + 400 ( 95.8 + 0.5 + 7)] / 10.2 (95.8) = $ 46.89 / ft.
The lowest drilling cost was obtained using Bit B.
Drilling Cost Predictions:
Drilling cost depends primarily on well location and well depth. The
location of the well will govern the cost of preparing the well-site, moving the
rig to the location, and the daily operating cost of the drilling operation. For
example, an operator may find from experience that operating a rig on a given
lease offshore Louisiana requires expenditures that will average about
S30,000/day. Included in this daily operating cost are such things as rig
rentals, crew boat rentals, work boat rentals, helicopter rentals, well
monitoring services, crew housing, routine maintenance of drilling equipment,
drilling fluid treatment, rig supervision, etc. The depth of the well will govern
the lithology that must be penetrated and, thus, the time required completing
the well.
Drilling costs tend to increase exponentially with depth. Thus, when curve-
fitting drilling cost data, it is often convenient to assume a relationship between
cost, C, and depth, D, given by;
C = a ebD
where; the constants a and b depend primarily on the well location. When a more
accurate drilling cost prediction is needed, a cost analysis based on a detailed
well plan must be made. The cost of tangible well equipment (such as casing) and
the cost of preparing the surface location usually can be predicted accurately.
The cost per day of the drilling operations can be estimated from considerations
of rig rental costs, other equipment rentals, transportation costs, rig
supervision costs, and others. The time required to drill and complete the well is
estimated on the basis of rig-up time, drilling time, trip lime, casing placement
time, formation evaluation and borehole survey time, completion time and trouble
time. Trouble time includes time spent on hole problems such as stuck pipe, well
control operations, formation fracture, etc. Major time expenditures always are
required for drilling and tripping operations. An estimate of drilling time can be
based on historical penetration rate data from the area of interest. The
penetration rate in a given formation varies inversely with both compressive
strength and shear strength of the rock. Also, rock strength tends to increase
with depth of burial because of the higher confining pressure caused by the
weight of the overburden.
When major unconformities are not present in the sub-surface lithology,
the penetration rate usually decreases exponentially with depth. Under these
conditions, the penetration rate can be related to depth, D, by;
dD / dt = K e-2.303a2D
where K and a2 are constants. The drilling time, td, required to drill to a given
depth can be obtained by separating variables and integrating. Separating
variables gives;
K ∫∫∫∫ dt = ∫∫∫∫ e2.303a2D dD
Integrating and solving for td yields;
td = ( 1 / 2.303 a2 K ) (e
2.303a2D –1)
Example-3: The bit records for a well drilled in the South China Sea are shown
in the following table. Make plots of depth vs. penetration rate and depth vs.
rotating time for this area using semi-log paper. Also, predict drilling time in
this area.
Solution:
The plots obtained using the bit records are shown in the following
figure. The constants K and a2 can be determined using the plot of depth vs.
penetration rate on semi-log paper. The value of 2.303 a2 is 2.303 divided by
the change in depth per log cycle.
2.303a2 = (2.303 / 6770) = 0.00034
The constant 2.303 is a convenient scaling factor since semi-log paper is
based on common logarithms. The value of K is equal to the value of penetration
rate at the surface. From depth vs. penetration rate plot, K = 280. Substitution
of these values of a2 and K in above equation;
td = 10.504 (e0.00034 D –1)
Bit Evaluation
It is important to maintain careful written records of the performance of
each bit for future references. Bits are worn by abrasion and shocks while
drilling. The wear pattern is important, it should be inspected once the bit has
been pulled and its grading should be recorded. Such records indicate the
working life of the bit and aid the selection of the type of bit which may provide
most efficient in a particular formation. The amount of wear on teeth, bearings
andgauge is recorded according to a special coding system.
Wear on Teeth
Teeth wear is graded in eighths of the original tooth height. Using the
letter T to denote teeth, T8 means that the teeth is completely worn out, and
T3 means that 3/8 of the original height has been worn away. If the majority of
the teeth in any row are broken, “BT” is added.
Bearing Wear
Grading a used bearing is the most difficult part of grading dull bits,
because the condition of the bearings can be determined only by “touch”.
Bearing wear is expressed in eighths of bearing life expended. Using the letter
B to denote bearings, B8 means that the bearing is completely worn out, and B6
means that 6/8 of the estimated life has been used. For sealed bearing bits, the
condition of the seal is a better means of grading the bearing life. For
sealedbearings, only three codes are used; B3 means the seal is effective, B5
means the seal is questionable, and B8 means the seal failed.
Gauge Wear
This can be determined by using a ring gauge and ruler. There are two
methods used to measure the wear. In the first and most popular, the ring gauge
is pulled against the gauge points of two cones, and the space between the ring
and third cone is measured. Usually, this measurement is used for the amount of
wear. However, to be exact, the measurement should be multiplied by 2/3. In
the second method, the bit is centered in the gauge ring and the ruler is used to
measure thedistance from the ring to the outermost cutting surface (gauge
surface). This measurement must be multiplied by two to give the loss in
diameter and, thus, the total amount of wear. Using the letter G for gauge, G0
means in gauge, and G5 means bit diameter is 0.625 in. under gauge.
Degree of Tooth Dullness
Tooth Dullness
T1 Tooth height 1/8 gone T2 Tooth height 1/4 gone T3 Tooth height 3/8 gone T4 Tooth height 1/ 2 gone T5 Tooth height 5/8 gone T6 Tooth height 3/4 gone T7 Tooth height 7/8 gone T8 Tooth height all gone
Bearing Conditions
B1 Bearing life used : 1/8 B2 Bearing life used : 1/4 (tight) B3 Bearing life used : 3/8 B4 Bearing life used : 1/2 (medium) B5 Bearing life used : 5/8 B6 Bearing life used : 3/4 (loose) B7 Bearing life used : 7/8 B8 Bearing life used : all gone
BT : broken teeth; I = in gauge; 0 = out of gauge
Example-3: T2-B4-I; (Teeth 1/4 gone, bearing medium, bit in gauge)
Example-4: T6 BT – B6- 0 1/2 (teeth 3/4 gone, broken teeth, bearing loose,
bit out of gage of 1/2 inch)
Example-5: T5 BT – B2- 0 5/8 (teeth 5/8 gone, broken teeth, bearing tight,
bit out of gage of 5/8 inch)
Drill-off Tests
The drill-off test is performed in order to ascertain combination of
weight on bit (WOB) and rotary speed to maximise penetration rate (PR). Drill-
off tests should be done:
1- at the start of the bit run,
2- on encountering a new formation,
3- if a reduction in rate of penetration (ROP) occurs.
Drill-off Test Procedure
1- Maintain a constant rpm. Select a WOB (near to maximum allowable).
2- Record the time to drill off a weight increment (5000 lb)
3- Record length of pipe drilled during step-2.
4- From step 2-3, the drill rate in ft/hr may be found.
5- Repeat step 2 and 3 at least four times. The last test should be at the same
value as the first test. This will determine if the formation has changed or not.
6- Select the bit weight which produced the faster ROP. Maintain this WOB
constant and repeat the above procedure for varying RPM values.
Example-6: Find the optimum WOB and RPM values with the given Drill-off test data. Solution:
Bit Weight (lbs) Time (sec) Footage (ft) Drill Rate (ft/hr)
50000 59 0.50 30.5 45000 62 0.50 29.0
40000 68 0.60 31.8 35000 74 0.48 23.4 30000 78 0.45 20.8 50000 60 0.50 30.0
Drill Rate = (Footage x 3600) / Time
RPM Time to Drill -1/2 ft- (sec)
Drill Rate (ft/hr)
100 70 25.7 90 65 27.7
80 60 30.0 70 64 23.1 60 69 26.1 100 72 25.0
Drill Rate = (Footage x 3600) / Time
Optimum WOB = 40000 lbs Optimum RPM = 80 rpm
Factors Affecting Tooth Wear
One purpose for evaluating the condition of dull bit is to provide insight
about the selection of a more suitable time interval of bit use. If the dull bit
evaluation indicates that the bit was pulled green, expensive rig time may have
been wasted on unnecessary trip time. However, if the time interval of bit use is
increased too much, the bit may break apart leaving junk in the hole. This will
require an additional trip to fish the junk from the hole or may reduce greatly
the efficiency of the next bit if an attempt is made to drill past the junk. Thus,
knowledge of the instantaneous rate of bit wear is needed to determine how
much the time interval of bit use can be increased safely. Since drilling
practices are not always the same for the new and old bit runs, knowledge of
how the various drilling parameters affect the instantaneous rate of bit wear
also is needed. The rate of tooth wear depends primarily on:
formation abrasiveness, tooth geometry, bit weight, rotary speed, and the
cleaning and cooling action of the drilling fluid.
Tooth Wear Equation
A composite tooth wear equation can be obtained by combining the
relations approximating the effect of tooth geometry, bit weight and rotary
speed on the rate of tooth wear. Thus the instantaneous rate of tooth wear is
given:
dh / dt = 1 / ττττH (N / 60)H1 {[(W/db)m – 4] / [(W/db)m – (W/db)]} [(1 + H2 /
2) / (1 + H2 h)]
where; h = fractional tooth height that has been worn away, t = time, hours ,W =
bit weight, 1000 lbf units, N = rotary speed, τH = formation abrasiveness
constant, hours
H1, H2 and (W/db)m = constants
Table 3-3 Recommended values of H1, H2 and (W/db)m are given in the following
table for the various rolling-cutter bit classes.
Bit Class H1 H2 (W/db)m
1-1 to 1-2 1.90 7 7.0 1-3 to 1-4 1.84 6 8.0 2-1 to 2-2 1.80 5 8.5 2-3 1.76 4 9.0 3-1 1.70 3 10.0 3-2 1.65 2 10.0 3-3 1.60 2 10.0 4-1 1.50 2 10.0
The tooth wear rate formula given above has been normalised so that the
abrasiveness constant, τH, is numerically equal to the time in hours required to
completely dull the bit teeth of the given bit type when operated at a constant
bit weight of 40000 lbf/inch and a constant rotary speed of 60 rpm. The
average formation abrasiveness encountered during a bit runs can be evaluated
using the above equation and the final tooth wear, hf, observed after pulling the
bit. If we define the tooth wear parameter, J2,
J2 = {[(W/db)m – (W/db)] / [(W/db)m – 4 ]} x (60 / N)H1 x [1 / (1 + H2 /2)]
and the abrasiveness constant, τH, gives;
ττττH = tb / [J2 (hf + H2 Hf2 / 2)]
Example-7: An 8.5 inch Class 1-3-1 bit drilled from depth of 8179 to 8404 ft in
10.5 hours. The average bit weight and rotary speed was 45000 lbf and 90 rpm,
respectively. When the bit was pulled, it was graded as T5-B4-I. Compute the
average formation abrasiveness for this depth interval. Also, estimate the time
required dulling the teeth completely using the same bit weight and rotary
speed.
Solution:
From the above table, H1 = 1.84 and H2 = 6 and (W/db)m = 8.0
J2 = {[(W/db)m – (W/db)] / [(W/db)m – 4 ]} x (60 / N)H1 x [1 / (1 + H2 /2)]
J2 = {[8 – (45 / 8.5)] / (8 – 4)} (60 / 90)1.84 [1 / [1 + (6/2)]] = 0.08
Solving for the abrasiveness constant using a final fractional tooth dullness of
5/8 (T5-0.625-) gives;
τH = tb / [J2 (hf + H2 Hf2 / 2)]
ττττH = 10.5 / [0.080 (0.625+ 6 (0.6252)/ 2)] = 73 hours
The time required to dull the teeth completely ( hf = 1.0);
tb = {J2 (τH) [hf+ H2 (hf)2 / 2]}
tb = {0.08 (73.0) [1+ 6 (1)2 / 2]}= 23.4 hours
Bearing Wear Equation
A bearing wear formula frequently used to estimate baring life is given by:
db / dt = 1 / ττττH (N / 60)B1 (W / 4 db)
B2
where: b = fractional bearing life that has been consumed, t = time, hours, N =
rotary speed, rpm, W = bit weight, 1000 lbf, db = bit diameter, inch, B1,B2 =
bearing wear exponent, τH = bearing constant
Table 3-4 Recommended values of bearing wear exponents are given below
Bearing Type Drilling Fluid Type B1 B2
Non sealed Barite mud 1.0 1.0 Non sealed Sulfide mud 1.0 1.0 Non sealed Water 1.0 1.2 Non sealed Clay/water mud 1.0 1.5 Non sealed Oil-base mud 1.0 2.0 Sealed roller bearings - 0.70 0.85 Sealed journal bearings - 1.60 1.00
Bearing constant, , is numerically equal to the life of bearings if the bit is
operated at 40000 lbf and 60 rpm. If we define a bearing wear parameter,J3,
J3 = (60 / N)B1 (4 db / W)B2
The time required to dull the bearing completely ( bf = 1.0);
tb = J3 (ττττB) bf
and, bearing constant is equal to;
ττττB = tb / J3 bf
Example-8: Compute the bearing constant for a 7.875 inch, Class 6-1-6 (sealed
journal bearings) bit that was graded T5-B6-I after drilling 64 hours at 30000
lbf and 70 rpm.
Solution:
From the above table, B1 = 1.6 and B2 = 1.0
J3 = (60 / N)B1 (4 db / W)
B2
J3 = (60 / 70)1.6 (4 (7.875) / 30)1.0 = 0.82
For the bearing constant of B6 ( bf = 6/8),
ττττB = tb / J3 bf = [64 / 0.82 (0.75)] = 104 hours
Penetration Rate Equations
Penetration rate equations for rolling cutter bits have been proposed by
various authors. The approach usually taken is to assume that the effects of bit
weight, rotary speed, tooth wear, etc., on penetration rate arc all independent
of one another and that the composite effect can be computed using an equation
of the form:
R = (f1) (f2) (f3) (f4)…..(fn)
where, f1, f2, f3, f4 , etc, represent the functional relations between penetration
rate and various drilling variables. The functional relations chosen usually are
based on trends observed in either laboratory or field studies. Some authors
have chosen to define the functional relation graphically, while others have used
curve-fitting techniques to obtain empirical mathematical expressions. Some
relatively simple mathematical equations have been used that model only two or
three of the drilling variables. Perhaps the most complete mathematical drilling
model that has been used for rolling cutter bits is the model proposed by
Bourgoyne and Young. They proposed using eight functions to model the effect
of most of the drilling variables. The Bourgoyne-Young drilling model can be
defined by the above with the following functional relations.
f1 = e 2.303a
1 D = true vertical depth- ft
f2 = e 2.303a
2 (10000 – D) gp = pore pressure gradient, lbm/gal
f3 = e 2.303a
3 D0.69 (gp – 9.0) ρc = equivalent circulating density, lbm/gal
f4 = e 2.303a
4 D (gp – ρρρρc) h = fractional tooth dullness
f5 = {[(W/db) – (W/db)t] / (4 - (W/db)t)]}a5
f6 = ( N / 60) a6 Fj = hydraulic impact force beneath the bit, lbf
f7 = e –a7h (W/db)t = threshold bit weight per inch
f8 = (Fj / 1000) a8 a1 to a8 = constants
Example-9: A 9.875-in. milled tooth bit operated at 40,000 lbf/in, and 80 rpm
is drilling in a shale formation at a depth or 12,000 ft at u penetration rate of 15
ft/hr. The formation pore pressure gradient is equivalent to a 12.0 lbm/gal mud
and the equivalent mud density on bottom is 12.5 lbm/gal. The computed jet
impact force beneath the bit is 1,200 lbf and the computed fractional tooth
wear is 0.3. Compute the apparent formation drillability, f1, using a threshold bit
weight of zero and the following values of a1 through a8.
a2 a3 a4 a5 a6 a7 a8
0.00007 0.000005 0.00003 1.0 0.5 0.5 0.5
Solution:
-The multiplier f2 accounts for the normal decrease in penetration rate with
depth from a reference depth of 10000 ft.
f2 = e 2.303a2 (10000 – D)
f2 = e 2.303 (0.00007) (10000 – 12000) = 0.724
-The multiplier f3 accounts for the increase in penetration rate due to under-
compaction.
f3 = e 2.303a
3 D 0.69 (gp – 9.0)
f3 = e 2.303 (0.000005) 120000.69 (12 – 9.0) = 1.023
-The multiplier f4 accounts for the change in penetration rate with overbalance
assuming a reference overbalance of zero.
f4 = e 2.303a
4 D (gp – ρc)
f4 = e 2.303 (0.00003) 12000 (12 – 12.5) = 0.6606
-The multiplier f5 accounts for the change in penetration rate with bit weight
assuming a reference bit weight of 4000 lbf/inch.
f5 = {[(W/db) – (W/db)t] / (4 - (W/db)t)]}a5
f5 = {[(40 / 9.875] / 4] 1.0 = 1.013
-The multiplier f6 accounts for the change in penetration rate with rotary speed
assuming a reference rotary speed of 60 rpm.
f6 = ( N / 60) a6
f6 = ( 80 / 60) 0.5 = 1.155
-The multiplier f7 accounts for the change in penetration rate with tooth
dullness assuming a zero tooth wear as a reference.
f7 = e –a7h
f7 = e -0.5 (0.3) = 0.861
-The multiplier f8 accounts for the change in penetration rate with jet impact
force using an impact force of 1000 lbf as a reference.
f8 = (Fj / 1000) a8
f8 = (1200 / 1000) 0.5 = 1.095
Substituting the values of f2 to f8 into penetration rate equation and solving for
formation drillability yields;
R = (f1) (f2) (f3) (f4)…..(fn)
15 = f1 (0.724) (1.023) (0.6606) (1.013) (1.155) (0.861) (1.095) f1 = 15 /
0.54 =
f1 = 27.8 ft/hr
CHAPTER-5
HYDRAULICS
The hydraulic system serves many purposes in the well. Since it is centred
on the mud system, the purposes of mud and hydraulics are often common to
each other. The hydraulics system has many effects on the well. Therefore, the
reasons for giving attentions to hydraulics are abundant. The more common
reasons are as follows:
-Control sub-surface pressures,
-Provide a buoyancy effect to the drill string and casing,
-Minimize hole erosion due to the mud's washing action during movement,
-Remove cuttings from the well, clean the bit, and remove cuttings from below
the bit,
-Increase penetration rate,
-Size surface equipment such as pumps,
-Control surge pressures created by lowering pipe into the well,
-Minimize well bore pressure reductions from swabbing when pulling pipe from
the well,
-Evaluate pressure increases in the well bore when circulating the mud,
-Maintain control of the well during kicks,
Hydrostatic Pressure
The hydrostatic pressure of the drilling fluid is an essential feature in
maintaining control of a well and preventing blow-outs. It is defined, in a
practical sense, as the static pressure of a column of fluid. Although the fluid is
generally mud, it can include air, natural gas, foam, mist, or aerated mud. Only
liquid-based systems such as mud will be considered in this text. The
hydrostatic pressure of a mud column is a function of the mud weight and the
true vertical depth of the well. It is imperative that attention be given to the
well depth so that the measured depth, or total depth, is not used inadvertently.
Since mud weights and well depths are often measured with different units, the
equation constants will vary. Common forms of the hydrostatic pressure equation
are as follows:
PH = 0.052 (mud weight, lb/gal) (depth, ft), PH = psia
PH = 0.00695 (mud weight, lb/cu ft) (depth. ft), PH = psia
PH = 9.81 (mud weight, g/cm3) (depth, m), PH = kPa
If a column of fluid contains several mud weights, the total hydrostatic pressure
is the sum of he individual sections:
PH = ∑∑∑∑ c ρρρρi Li
c = conversion constant
ρ = mud weight for the section of interest
L = length for the section of interest
Equivalent Mud Weight
Drilling operations often involve several fluid densities, pressures
resulting from fluid circulation, and perhaps applied surface pressure during kick
control operations. It is useful in practical applications to discuss this complex
pressure and fluid density arrangement on a common basis. The approach most
widely used is to convert all pressures to an "equivalent mud weight" that would
provide the same pressures in a static system with no surface pressure.
Suppose a 10,000-ft well has two mud weights. It contains 5,000 ft of 9.0
lb/gal mud and 5,000 ft of 11.0-lb/gal mud. The equivalent mud weight at 10,000
ft is 10.0 lb/gal, even though the well does not contain any real 10.0 lb/gal mud.
Another term commonly used to describe the equivalent mud weight concept is
ECD, or equivalent circulating density. The ECD usually considers the
hydrostatic pressures and the friction pressure resulting from fluid movement.
For example, a 12.0 lb/gal mud may act as if it is 12.3-lb/gal mud (due to the
friction pressure) while it is pumped. Some drilling engineers may refer to the
ECD in this case as 0.3 lb/gal. Typical ranges for the ECD additive factor are
0.1-0.5 lb/gal.
EMW = (total pressure x 19.23) / true vertical depth
EMW = equivalent mud weight, lb/gal
19.23 = reciprocal of the 0.052 constant
Example-1: An intermediate casing string (see the figure) will be cemented as
shown. Calculate the hydrostatic pressure at 12,000 ft. Convert the pressure at
12,000 ft to an equivalent mud weight and determine if it will exceed the
fracture gradient of 14.2 lb/gal.
Solution:
The hydrostatic pressures are computed as follows:
0.052 x fluid weight x depth = pressure
0.052 x 11.4 lb/gal x (7,000ft) = 4,149 psi
0.052 x 15.4 lb/gal x (9,000 - 7,000ft) = 1,602 psi
0.052 x 16.61b/gal x (12,000 - 9,000 ft) = 2,589 psi
Total hydrostatic pressure = 8340 psi
The equivalent mud weight is calculated as:
EMW = (total pressure x 19.23) / true vertical depth
EMW = (8340 x 19.23) / 12000 = 13.36 lb/gal (ppg)
Therefore, the static hydrostatic pressure with a 13.36-lb/gal EMW will not
exceed the fracture gradient of 14.2 lb/gal.
Buoyancy
The drilling fluid provides a beneficial effect relative to drill string
weight or hook load. When pipe is lowered into the well, the mud system will
support, or buoy, some of the pipe weight. This effect is termed buoyancy, or
buoyant forces. The buoyed weight of the drill string will be less than the in-air
weight of the pipe. Buoyant forces are a function of the volume and weight of
the displaced fluid. Heavier mud has greater buoyant forces than low-density
mud.
BW = BF x (in-air weight)
BW = buoyed weight
BF = buoyancy factor
BF = 1 – (ρρρρm / 65.5)
ρm = mud density, lb/gal and 65.5 is the density of a gallon of steel.
Example-2: Casing string will be run into a well that contains 11.7 ppg mud.
Assume the casing will be filled with mud as it is run. If the engineer uses a
derrick safety factor of 2, will the 1,000,000 lb derrick capacity be
satisfactory?
Casing weight, lb/ft Section length, ft
47.0 4500
53.0 5500
47.0 3000
Solution:
1. The casing string weight, in air, is:
4500 ft x 47.0 lb/ft = 211500 lb
5500 ft x 53.0 lb/ft = 291500 lb
3000 ft x 47.0 lb/ft = 141000 lb
Total casing string weight in air is = 644000 lb.
BF = 1 – (ρm / 65.5)
BF = 1 – (11.7 / 65.5) = 0.82
Buoyed weight = 0.82 x 644000 lb = 528964 lb
Applying a derrick safety factor of 2.
2 x 528964 > 1000000 lb
Therefore, the derrick load will exceed the design criteria if a factor of 2 is
used. He actual design factor is,
1000000 / 528964 = 1.89
Flow Regimes
While drilling fluids are flowing in a well, the manner in which the fluid
behaves may vary. This behavior is often termed the flow regime. The most
common regimes are laminar, turbulent, and transitional. Unfortunately, it is
impossible to clearly define each type in the well. As an example, mud flow may
be predominantly laminar, although the flow near the pipe walls during pipe
rotation may be turbulent.
Laminar Flow
The most common annular flow regime is laminar. It exists from very low
pump rates to the rate at which turbulence begins. Characteristics of laminar
flow useful to the drilling engineer are low friction pressures and minimum hole
erosion. Laminar flow can be described as individual layers, or laminae, moving
through the pipe or annulus. The center layers usually move at rates greater
than the layers near the well bore or pipe. The flow profile describes the
variations in layer velocities. These variations are controlled by the shear-
resistant capabilities of the mud. A high yield point for the mud tends to make
the layers move at more uniform rates. Cuttings removal is often discussed as
being more difficult with laminar now. The cuttings appear to move outward
from the higher-velocity layers to the more acquiescent areas. These outer
layers have very low velocities and may not be effective in removing the
cuttings. A common procedure for minimizing the problem is to increase the
yield point, which decreases layer velocity variations. An alternative is to pump a
10-20-bbl high-viscosity plug to "sweep" the annulus of cuttings.
Turbulent Flow
Turbulence occurs when increased velocities between the layers create
shear strengths exceeding the ability of the mud to remain in laminar flow. The
layered structure becomes chaotic and turbulent. Turbulence occurs commonly in
the drill string and occasionally around the drill collars. Much published
literature suggests that annular turbulent flow increases hole erosion problems.
The flow stream is continuously swirling into the walls. In addition, the velocity
at the walls is significantly greater than the wall layer in laminar flow. Many
industry personnel believe that turbulent flow and the formation type are the
controlling parameters for erosion.
Transitional Flow
Unfortunately, it is often difficult to estimate the flow rate at which
turbulence will occur. In addition, turbulence may occur in various stages. It is
convenient to describe this "grey" area as a transitional stage.
Turbulence Criteria
Several common methods can be used to establish turbulence criteria.
The most common approach is the Reynolds number. Others include 1)
intersection of the flow rate vs. pressure loss calculations for laminar and
turbulent flow and 2) calculation of a z-value. The Reynolds number approach is
used almost exclusively in the industry. Turbulence occurs when the ratio of the
momentum of the liquid to the viscosity ability of the liquid to dampen
permeations exceeds some empirically determined value. The momentum force
of the liquid is its velocity times its density. The viscous ability of the liquid to
damp out permeations is the internal resistance against change and the effects
of the walls of the borehole. For the simple case of Newtonian, non elastic liquid
flowing in a pipe dampening effect is the quotient of the viscosity and the
diameter of the well bore.
NR = ρρρρ V D / µµµµ
NR = Reynolds number
ρ = density D = diameter
µ = viscosity
A simpler equation used in the literature to predict the Reynolds number at the
upper limit of laminar flow is as follows:
NR = 3,470 - 1,370 n
The relation for the Reynolds number between the transition and turbulent flow
regimes is
NR = 4,270 - 1,370 n
It is obvious from equations that the Reynolds number is sliding, with its
dependency on the flow behavior index (n). The position of intersection between
the laminar and turbulent flow pressure losses depends on the equations being
used. The Reed slide rule or the Hughes tables can give errors if the mud is
quite non-Newtonian at the applicable shear rate.
Critical Velocity
The term critical velocity is used to define the single velocity at which the
flow regime changes from laminar to turbulent. This variable is the most
important since all other members are considered constant in a typical equation.
Since no single Reynolds number defines the transitional zone, it follows that a
range of critical velocities may be necessary to determine the flow regime.
In practical applications, a critical velocity (Vc) and an actual velocity (Val)
are calculated. If Val < Vc the flow is laminar. If Vic < Va the flow is turbulent. If
Val ≅ Vc calculations are made with both flow regimes and the larger pressure
losses are used.
Flow (Mathematical) Models
A mathematical model is used to describe the fluid behavior under dynamic
conditions. The model can be used to calculate friction pressures, swab and
surge pressures, and slip velocities of cuttings in fluids. The models most used in
the drilling industry are Newtonian, Bingham Plastic, and Power Law.
Terms used in mud models are shear stress and shear rate. They can be
described by considering two plates separated by a specified distance with a
fluid. If a force is applied to the upper plate while the lower plate is stationary,
a velocity will be reached and will be a function of the force, the distance
between the plates, the area of exposure, and the fluid viscosity:
F / A = µµµµ (V / X)
F = force applied to the plate; A = contact area; V = plate velocity X = plate
spacing µ = fluid viscosity
The quotient of F/A is termed the shear stress (ττττ), while V/X is shear rate (γγγγ)
ττττ = µµµµ γγγγ
In drilling operations, the shear stress and shear rate are analogous to pump
pressure and rate, respectively.
Newtonian Fluids
The model used initially to describe drilling mud was the Newtonian model,
ττττ αααα γγγγ
It stated that pump pressure (shear stress) would increase proportionally to
shear rate. If a constant of proportionality is applied to represent fluid
viscosity,
ττττ = µµµµ γγγγ
Unfortunately, drilling mud usually cannot be described by a single
viscosity term. They require two or more points for an accurate representation
of behavior. As a result, the Newtonian model generally is not used in hydraulics
plans.
Bingham Plastic
The Bingham model was developed to describe more effectively drilling mud
presently in use. Bingham theorized that some amount of stress would be
required to overcome the mud's gel structure before it would initiate movement
ττττ = µµµµ p γγγγ + ττττy
τy = yield stress, µ p = fluid viscosity
In practical terms, the equation states that a certain pressure would be
applied to the mud to initiate movement. Flowing mud pressures would be a
function of the initial yield pressure and the fluid viscosity.
Shear rates are normally taken at 300 and 600 rpm rates on the viscometer.
The fluid viscosity (µp) and the yield stress (τy) are calculated as follows:
µµµµp = θθθθ600 - θθθθ300
θ600 , θ300 = readings at 600 and 300 rpm, respectively.
ττττy = θθθθ300 - µµµµp
The fluid viscosity is termed plastic viscosity (PV) due to the plastic nature
of the fluid and is measured in centi-poise (cp). The size, shape, and
concentration of particles affect the plastic viscosity in the mud system. As
mud solids increase, the plastic viscosity increases. The plastic viscosity is a mud
property that is not affected by most chemical thinners and can be controlled
only by altering the state or number of solids.
The yield stresses ττττy, is given the name of yield point and is measured in
lb/100 ft2. It is a function of the inter-particle attraction of the solids in the
mud. Chemical thinners, dispersants, and viscosifiers control the yield point.
Power Law
The Power Law model is a standard mathematical expression used to describe a
non-linear curve. The equation for drilling fluids is ;
ττττ = K (γγγγ)n
K = consistency index; n = flow behavior index
The flow behavior index is descriptive of the degree to which the fluid is non-
Newtonian.
n = 3.32 log (θθθθ600 / θθθθ300)
K = θθθθ300 / 511n
Example-3: Use the following data to compute PV, YP, n and K.
θ600 = 64, θ300 = 35,
Solution:
PV = θ600 - θ300 = 64 - 35 = 29 cp
YP = θ300 - PV = 35 – 29 = 6 lb/100ft2
n = 3.32 log (θ600 / θ300)
n = 3.32 log (64 / 35) = 0.87
K = θ300 / 511n
K = 35 / 511 0.87 = 0.154
Friction Pressure Determinations
Pumping a drilling fluid requires overcoming frictional drag forces from
fluid layers and solids particles. The pump pressure (Pp) can be described as the
summation of the frictional forces in the circulation system:
Pp = PDS + PB + PA
Pp = pump pressure, psi;
PDS = drill string friction pressure, psi;
PB = bit pressure drop, psi;
PA = annulus pressure, psi
The pressure drop in the bit results from fluid acceleration and not solely
friction forces. As a result, it will be discussed in a separate section. Equations
to determine friction pressures vary according to the flow regimes, such as
laminar and turbulent. In addition, Bingham Plastic and Power Law models differ
in form. Since these models are frequently used in drilling applications, they will
be presented in the following sections. Newtonian-based equations will not be
presented.
Bingham Plastic Friction Pressures
The Bingham Plastic model is used primarily to compute friction pressure
associated with laminar flow. This restriction is based on its inability to
accurately describe shear stresses associated with high shear rates. Laminar
and turbulent flow calculations will be presented, however, since they are
frequently used in the drilling industry.
The velocity of the fluid in the drill string is described as;
V = Q / (2.448 d2)
V = fluid velocity, ft/sec;
Q = flow rate, gal/min;
d = pipe diameter, inch
The critical velocity (Vc) for laminar and turbulence determination is computed;
Vc = [1.08 PV + 1.08 √√√√(PV)2 + 12.34d2 YP ρρρρ ] / ρρρρd
ρ = mud weight, ppg
Friction pressures for laminar flow can be calculated as follows:
Pp = [(PV L V) / 1500 d2)] + [(YP L) / 225 d)]
L = section length, ft
Turbulent flow is calculated as;
Pp = (ρρρρ0.75 V1.75 PV0.25 L) / 1800 d1.25
In the annulus, same series of operations is performed but slightly different
equations to account for the hole geometry,
V = Q / [2.448 (dh2 - dp
2)]
dh = casing or hole ID, inch ; dp = pipe or drill collar OD, inch
Vc = [1.08 + 1.08 √√√√(PV)2 + 9.26(dh – dp)2 YP ρρρρ ] / ρρρρ ( dh – dp)
For laminar flow;
PA = [(PV L V) / 1000 (dh – dp)2)] + [(YP L) / 200( dh – dp)]
For turbulent flow;
PA = (ρρρρ0.75 V1.75 PV0.25 L) / 1396 (dh – dp)1.25
Example-4: Calculate friction pressures for flow rates of 100 and 200 gpm.
Use the Bingham model.
Pipe ID = 3.5 inch; MW = 12.9 ppg; PV = 29 cp;YP = 6 lb/100ft2 ; Length = 10000 ft
Solution:
V = Q / (2.448 d2)
V = 100 / [2.448 (3.5)2] = 3.33 ft/sec (at 100 gal/min)
V = 200 / [2.448 (3.5)2] = 6.66 ft/sec (at 200 gal/min)
Vc = [1.08 PV + 1.08 √(PV)2 + 12.34d2 YP ρ ] / ρd
Vc = [1.08 (29) + 1.08 √(29)2 + 12.34(3.5)2 (6) (12.9) ] / (12.9) (3.5) = 3.37 ft/sec
For the flow rate of 100 gal/min, the actual velocity (Va) is slightly less
than the critical velocity (Vc) of 3.37 ft/sec. Use the laminar flow equation.
(Note that the difference between V, and V, is small. Therefore, it might be
advisable in some cases to consider calculating pressure losses for laminar and
turbulent flow and use the larger value.)
PDS = [(PV L V) / 1500 d2)] + [(YP L) / 225 d)]
PDS = [(29)(10000) (3.33) / 1500 (3.5)2)] + [(6) (10000) / 225 (3.5)] = 128.6 psi
At a flow rate of 200 gal/min, the actual velocity of 6.66 ft/sec is
significantly greater than the critical velocity of 3.37 ft/sec. Therefore, use
the turbulent flow equation;
PDS = (ρ0.75 V1.75 PV0.25 L) / 1800 d1.25
PDS = (12.9)0.75 (6.66)1.75 (29)0.25 10000) / 1800 (3.5)1.25 = 505.7 psi
The laminar and turbulence equations can be used to illustrate the basic
difference between these two flow systems. In the laminar equations, a value
for the yield point (YP) is a significant part of the pressure loss, particularly
when it is observed that the PV value is divided by a squared diameter. The
turbulent flow equations do not contain a YP term. The yield point is one of the
forces creating the inter-particle attractions, causing the mud to move in
laminae. When the shear force exceeds the yield stress, turbulence begins and
the yield point is not a factor thereafter.
Power Law Friction Pressures
Power Law calculations follow the same sequence as the Bingham model.
Actual and critical velocities are compared to determine the flow regime before
calculating the pressure loss. If Va and Vc differ significantly, choose the
appropriate flow equation. When Va ≅ Vc makes both pressure loss computations
and chooses the larger.
A word of caution must be given at this point relative to Bingham and
Power Law equations. Many forms of these computations exist in the industry
with units that differ slightly. Velocity can be expressed in ft/sec or ft/min,
which obviously would make a significant error in the calculations, particularly
when V is in exponent form. The Power Law model demands additional attention
because several methods exist for computing the basic parameters of n and K.
This is not the case for the Bingham model because only one accepted method is
used for PV and YP calculations. The equations presented in this text are those
of Moore et al.
Calculating friction pressures in the drill string using the Power law
equations for laminar and turbulent flow are accomplished respectively:
PDS = [(1.6V / D) ((3n + 1) / 4n)]n (KL / 300d) -laminar-
PDS = [2.27 (10-7) ρρρρ0.8 V1.8 PV0.2 L] / d1.2 -turbulent-
For computation simplicity, NR = 3000 is assumed for turbulence criteria.
Basic assumptions for friction factor correlations result in the critical velocity
equation:
Vc = [5.82 (104) K / ρρρρ]1/2-n [(1.6/d) ((3n+1) / 4n)]n/2-n
Annular flow equations follow the same pattern as drill string calculations.
PA = [(2.4V / (dh - dp)) ((2n + 1) / 3n)]n [(KL / 300 (dh - dp)] -laminar-
PDS = [7.7 (10-5) ρρρρ0.8 V1.8 PV0.2 L] / [(dh - dp)2 (dh + dp)
1.8] -turbulent-
Vc = [3.878 (104) K / ρρρρ]1/2-n [(2.4/ (dh - dp)) ((2n+1) / 3n)]n/2-n
Example-5: Refer to previous example and compute the friction pressures for
the system. Use the Power law model and flow rate of 125 gal/min. If Va ≅≅≅≅ Vc,
computes the pressure drop for laminar and turbulent flow and choose the
larger value.
n = 0.87; K = 0.154; Pipe ID = 3.5 inch; MW = 12.9 ppg; Length = 10000 ft;PV = 29 cp;
Solution:
-Determine the actual velocity at 125 gal/min;
Va = [125 / 2.448 (3.5)2] = 4.168 ft/sec (250 ft/min)
-Compute critical velocity;
Vc = [5.82 (104) K / ρ]1/2-n [(1.6/d) ((3n+1) / 4n)]n/2-n
Vc = [5.82 (104) 0.154 / 12.9]1/2-0.87 [(1.6/3.5) ((3(0.87)+1) / 4(0.87))]0.87/2-0.87
Vc = 183 ft/min
-For purpose of illustration in this example, assume Va ≅≅≅≅ Vc ( 250 ft/min ≅≅≅≅ 183
ft/min)
-Calculate laminar flow pressure losses;
PDS = [(1.6V / D) ((3n + 1) / 4n)]n (KL / 300d)
PDS = [(1.6 x 250 / 3.5) ((3(0.87) + 1) / 4(0.87))]0.87 ((0.154) (10000) / 300(3.5))
= 95.4 psi
-Calculate turbulent flow pressure losses;
PDS = [2.27 (10-7) ρ0.8 V1.8 PV0.2 L] / d1.2
PDS = [2.27 (10-7) 12.90.8 2501.8 290.2 10000] / 3.51.2 = 158.6 psi
-Since 158.6 psi > 93.4 psi, assume the pressure loss is the greater value.
Optimization of Drilling Hydraulics
The objective of a drilling hydraulics program is to specify the
operating conditions which will maximize the bottom hole cleaning effect and
hence penetration rate; while effectively removing the drilled cuttings from
the hole. A number of parameters have to be considered when designing the
hydraulics program:
1. Mud pump Output Volume (and Volumetric Efficiency)
2. Mud pump Output in terms of Hydraulic Power
3. Pressure losses in:
-Surface connections
-Drill pipe (inside and outside)
-Drill collars (inside and outside)
-Bit Nozzles
4.Velocity of mud passing through the bit nozzles (Nozzle velocity, Vn)
5.Velocity of mud rising in the annulus (Annular Velocity, Vann )
Mud Pump Output Volume:
The output volume flow rate, Q, depends on:
-pump line size
-pump stroke length
-pump speed (strokes/min.)
-pump volumetric efficiency
Two types of mud pump are in general use:
a) The Double Acting Duplex Pump
Pump Output, Q = 0.0068 x L x (2D2 – d2) x SPM x (Vol. Eff. /100) -gal/min-
b) The Single Acting Triplex Pump
Pump Output, Q = 0.0102 x L x D2 x SPM x (Vol. Eff. /100) -gal/min-
L = stroke length, inch
D = inside diameter of liner, inch
d = rod diameter, inch
SPM = stroke per minute
Pumping a known value of mud or slurry and noting the SPM can determine
volumetric Efficiency of the mud pumps. For the duplex pump the volumetric
efficiency will usually be in the region of 90 % or more. The triplex pump will
usually have a volumetric efficiency greater than 95 %.
Hydraulic Power
The pump output in hydraulic power is generally assumed to equal 85 % of
mechanical or electrical input power.
HHP = (Pt x Q) / 1714
Pt = circulating pressure, psi
Q = pump output volume, gal/min.
Pt = Ps + Pbit Ps = pressure losses in the system, psi
Pbit = pressure losses over the bit nozzles, psi
Total HHP = HHPsystem + HHPbit
Total HHP = [(Ps x Q) / 1714] + [(Pbit x Q) / 1714] or,
HHPbit = HHPtotal – HHPsystem
HHPbit = [(Pt x Q) / 1714] - [(Ps x Q) / 1714]
HHPbit = Q / 1714 (Pt - Ps)
Pressure Losses in the System
The system consists of all items causing pressure losses except the bit nozzles.
a)Surface Connections: from pump to drill pipe. Pressure losses are minimized by
large internal diameters.
b) Drill Pipe: large ID and internal plastic coating reduce pressure drop.
c)Drill Collars: Large ID reduces pressure loss, but also reduces the useful
weight per unit length.
d) Annulus: Pressure losses depends on the ratio of the drill string outside
diameter and the hole size. It is desirable to reduce the back pressure on the
formations, but annular losses are usually small.
As the rate of volume pumped varies, pressure losses in the system change as
follows:
Psystem = c Qn
Q = flow rate (gal/min)
n = a variable power (1.86)
c = a constant
Pressure Drop Across Bit Nozzles
As the rate of volume pumped (Q) varies, the pressure losses at the bit
change as follows:
Bit Pressure Drop, Pbit = (156 x W x Q2) / (Dn2 + Dn
2 + Dn2)2 –psi-
Dn2 = nozzle size number, 1/32 inch
Q = flow rate, gal/min and W = mud weight, ppg
The jet velocity in the nozzles, Vn, can be calculated from
Jet velocity, Vn = (418.3 x Q) / (Dn2 + Dn
2 + Dn2) -ft/sec-
Therefore it can be shown as;
Pbit = ρρρρVn2 / 58.26 -psi-
Annular Velocity
The minimum annular velocity required cleaning the hole. It is important to
avoid solids build-up and increasing the hydrostatic head, which might cause mud
losses to the formation.
Annular Velocity , Vann = (24.5 x Q) / (D22 + D1
2)
D2 = ID of hole, inch and D1 = OD of pipe, inch
Optimization
Two principle approaches are adopted to achieve efficient removal of
cuttings below the bit and so the best penetration rate:
-To maximize hydraulic power expended at the bit; assuming that cuttings
removal depends on the fluid energy dissipated.
-To maximize the hydraulic impact force; assuming that formation is best
removed when the mud hits the bottom of the hole with the greatest force.
It is also important to be able to quickly obtain a value of system pressure loss
in the event of a kick, as this can be used to calculate circulating pressures at
various killing speeds.
System Pressure Loss, Ps = c Qn
-By knowing the value of "n", the proportion of the circulating pressure which
will be lost in the drill string under optimum hydraulics conditions (Ps,opt) can be
determined.
,a) For maximum Hydraulic Power at the Bit
Ps,opt = [1 / (n + 1)] Pt
b) For maximum Impact Force
Ps,opt = [2 / (2 + 1)] Pt
-By knowing the value of “c” the corresponding optimum circulation rate can be
determined.
-Once the optimum circulation rate and pressure are known, bit nozzle sizes can
be selected to obtain the correct Pbit ( Pbit = Pt - Ps,opt)
Theoretically “n” and “c” can be calculated from two pressures observed at two
different pump rates.
n = log (Ps1 / Ps2) / log (Q1 / Q2)
c = Ps1 / Q1n
Small inaccuracies in pressure readings and pump stroke counts can result in
considerable errors in “n” and “c”. It is known that,
Ps = c Qn
log Ps = log c + n log Q
which (on log-log paper) is represented by a straight line with slope “n”.
“c” can be calculated by solving, c = Ps / Qn for any combination of Ps and Q.
Example-6: Before starting a trip out at 9460 ft, to change the bit (12 1/4
with 3 x 14 nozzles), the following readings were taken:
Pump Rate (spm) Circulating pressure (psi)
160 3480
135 2640
110 1810
88 1230
Mud pumps: two single acting triplex pumps, with 6" liner, 10" stroke, 97.5%
volumetric efficiency. The mud gradient in use is: 0.676 psi/ft.
1. Calculate Ps for each circulation rate (Q).
2. Plot Ps - Q on log - log paper and determine n and calculate c.
3. For max. Hydraulic horsepower at the bit, calculate Ps opt and Qopt. assuming the max. pump pressure to be used is 3250 psi.
4. Determine the size of the nozzles for the next bit, to obtain optimum
hydraulics.
5.Calculate bit and total hydraulic horse power for the next bit run.
6. Calculate annular velocity around 5" drill pipe.
7. Calculate nozzle velocity.
Solution:
1. For a triplex pump
Q = 0.0102 x L x D2 x SPM x Vol. Eff.
Q1 = 0.0102 x 10 x 62 x 160 x 0.975 = 573 gal/min
Q2 = 0.0102 x 10 x 62 x 135 x 0.975 = 483 gal/min
Q3 = 0.0102 x 10 x 62 x 110 x 0.975 = 394 gal/min
Q4 = 0.0102 x 10 x 62 x 88 x 0.975 = 315 gal/min
To obtain Ps it is necessary to calculate Pbit for each circulating rate, from
Pbit = (3013 x W x Q2) / (Dn2 + Dn
2 + Dn2)2 –psi- (valid when mud weight is in psi/ft)
Pbit,1 = (3013 x 0.676 x 5732) / (142 + 142 + 142)2 = 1928 psi
Pbit,2 = (3013 x 0.676 x 4832) / (142 + 142 + 142)2 = 1552 psi
Pbit,3 = (3013 x 0.676 x 3942) / (142 + 142 + 142)2 = 899 psi
Pbit,4 = (3013 x 0.676 x 3152) / (142 + 142 + 142)2 = 647 psi
And because, Ps = Pt – Pb, it follows that,
Ps1 = 3480 – 1928 = 1552 psi
Ps2 = 2640 – 1370 = 1270 psi
Ps3 = 1810 – 911 = 899 psi
Ps4 = 1230 – 583 = 647 psi
2. On the graph these values of Ps are plotted against Q.
Ps = c Qn
log Ps = log c + n log Q
which is a straight line on log-log paper with slope equal to “n”.
Measured from the graph; n=1.50
Ps = c Qn
Q3 = 394 gal/min and Ps3 = 899 psi
C = Ps3 / Qn = 899 / 394 1.5 = 0.115
3. For maximum hydraulic pressure at the bit.
Ps,opt = [1 / (n + 1)] Pt
Ps,opt = [1 / (105 + 1)] Pt = 0.4 Pt
As maximum pump pressure is given as 3250 psi,
Ps,opt = 0.4 x 3250 = 1300 psi
Qopt = (Ps / c) 1/n = (1300 / 0.115) 1/1.5 = 504 gal/min
4. Pbit = Pt – Ps = 3250 – 1300 = 1950 psi
Pbit = (3013 x W x Q2) / (3 x Dn2)2
(3 x Dn2)2 = 3013 x 0.676 x 5042 / 1950
(3 x Dn2)2 = 514
Dn2 = 171
Dn = 13
5. General horse power equation;
HHP = (P x Q) / 1714
HHP = (3250 x 504) / 1714 = 955 hp
Bit HHP = (1950 x 504) / 1714 = 573 hp
6. Van = (24.5 x Q) / (D2 – d2)
Van = (24.5 x 504) / (12.252 – 52) = 99 ft/sec
7. Vn = (418.3 x Q) / (3 x Dn2)
Vn = (418.3 x 504) / (3 x 132) = 416 ft/sec