MINISTRY OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF
TECHNICAL AND VOCATIONAL EDUCATION
Sample Questions & Worked Out Examples
For
PE-03032
DRILLING ENGINEERING (I)
B.Tech(First Year)
Petroleum Engineering
29
Ministry of Science and Technology
Department of Technical and Vocational Education
Petroleum Engineering
Sample Questions for
PE 03032 DRILLING ENGINEERING Part ( I )
Chapter 1 Rock Properties and Principles of Rock Breaking in Drilling
1.*
2.**
3.***
4.*
5.**
6.***
7.*
8.**
9.***
10.*
11**
12***
13*
14**
15***
16*
17**
18***
Explain the necessity of study of rock properties and principles of rock breaking in drilling.
(10 marks)
Discuss briefly about the general data on rocks.
(15 marks)
Explain briefly about the main rock characteristics. (15 marks)
Discuss briefly on the determination of rock characteristics. (15 marks)
Explain the classification of rocks according to their hardness.
(10 marks)
Explain the classification of rocks according to their modulus of elasticity. (5 marks)
Explain the classification of rocks according to their coefficient of plasticity.(5 marks)
Discuss briefly rock breakage under the impact of single blows.
(15 marks)
Explain the rock breakage under cycle (repeated) stresses.
(15 marks)
What are the influence of main factors on mechanical properties of rocks. (15 marks)
How many modulus of elasticity of rocks are distinguished.
( 5 marks )
Explain the poisson's ratio of some rocks.
( 5 marks )
What are dependent upon specific work of rock disintegration?
( 10 marks )
What are dependent upon specific work of contact of work failure? ( 5 marks )
Discuss briefly abrasive properties of rocks.
( 15 marks )
Explain the coefficient of friction of some rocks.
( 10 marks )
Explain the classification of the rocks according to their abrasiveness.
( 10 marks )
Discuss about the drillability of rocks. ( 10 marks )
Chapter 2 Subsurface Pressure and Subsurface Temperature
19*
20**
21***
22*
23**
24***
25*
26**
27***
28*
29**
30***
31*
32**
33***
Explain about the subsurface pressures.
( 15 marks )
Explain the relative reservoir pressure of oil and gas deposits of different geological ages. (5
marks )
Explain about the subsurface temperature.
(15 marks)
Explain the reservoir temperature and earth geothermal gradients.
(10 marks )
What are relationship commonly a linear function for temperature and depth? (5 marks )
What are the slim hole drilling?
( 5 marks )
What are making a connection?
( 5 marks )
Explain the basic components of a rotary drilling rig. ( 5 marks )
Write short notes of standard derricks?
( 10 marks )
Write short notes of portable derricks. ( 5 marks )
What are the choice of derrick type and size? ( 10 marks )
Write short notes of substructures.
( 5 marks )
Explain briefly drawworks (hoist).
( 10 marks )
What do you understand drawwork horsepower output?
( 5 marks )
Explain about mud pumps.
( 10 marks )
34*
35**
36***
37*
38**
39***
40*
41**
42***
43*
44**
45***
46*
47**
48***
49**
50*
30
What are the superiority of the piston type pump for drilling service. ( 5 marks )
Discuss pump ratings and capacities. ( 15 marks )
Explain briefly the prime movers in common use.
( 10 marks )
What do you understand steam engines?
( 5 marks )
What do you understand electric motors?
( 5 marks )
What do you understand internal combustion engines?
( 5 marks )
Explain advantages disadvantages of rotary drilling prime mover types and principle areas of
use.
(15 marks )
Define the term of drill string. ( 5 marks )
Define the term of kelly joint.
( 5 marks )
Define the term of drill pipe tool joints.
(10 marks )
What are physical properties of API drill pipe?
( 5 marks )
What do you understand drill collars? ( 5 marks )
Discuss about rotary bit.
( 5 marks )
Define the terms of drag bit diamond bit.
( 10 marks )
Discuss about drilling line.
( 5 marks )
Define terms of blowout preventers. ( 10 marks )
Discuss about offshore drilling equipment. ( 15 marks )
* = Must know, ** = Should know, *** = Could know
31
Ministry of Science and Technology
Department of Technical and Vocational Education
Petroleum Engineering
Worked Out Examples for
PE 03032 DRILLING ENGINEERING Part ( I )
1. Explain the necessity of study of rock properties and principles of rock breaking in drilling.
Necessity of Study of Rock Properties and Principles of Rock Breaking in Drilling
Of the factors affecting the performance of the bit and some other drilling equipment the
basic one is the type of a formation drilled. This factor is, of course beyond the control of the
operator. The only way to meet the requirements imposed is by studying the formation
characteristics, using the bit that is best adapted to conditions the volume rate of the drilling fluid
used, the speed of bit rotation, and load on the bit.
The studies of rock characteristics have resulted in the improvement of penetration rate
and bit life, which are of particular importance as in the case of hard formations being now drilled
with rolling cutter bits. Extremely short life of the bits in drilling certain rocks, such as dolomites
or cherts, is one of the most important elements affecting the economics of present day deep well
drilling.
Application of the results of these and other studies in recent years have resulted in
considerable improvement in the per-bit footage in this type of drilling.
2. Discuss briefly on determination of rock characteristics.
Determination of Rock Characteristics
Prof. L. A. Schreiner and his followers has developed a special method of the
determination of the main rock characteristics by means of indenter pressing against the rock
specimen surface. The magnitude of force, corresponding to the local failure of the rock under the
indenter, divided by the area of the contact is taken as a measure of rock hardness. A special
testing machine has been designed for this purpose. This machine allows to get a forcedisplacement diagram.
The indentors are made of a heat-treated steel or a tungsten carbide and have a fixed area
of the base of 1-5 mm2. The rock samples and their surface are specially prepared for testing.
Schematic diagram of static loading tester is shown in Fig 1.1 and a force-strain diagram is
shown in Fig 1.2. The point "B" in the diagram corresponds to the failure of rock under the
indenter. The point "A" corresponds to the highest force at which there is no plastic deformations
yet. If the force applied on the indenter is higher than this force FA, besides elastic deformations,
plastic deformations are also caused. If the force is less than force FA, only elastic deformations
occur.
Fig 1.1 Schematic diagram of static loading tester, to determine mechanical
characteristics of rocks.
1. Dial indicator;
2. Pressure gauge;
3. Indentor (punch);
4. Rock specimen;
5. Ram of hydraulic press.
32
Elastic deformations take place up to rock failure, and it is assumed, that the elastic
deformation-force relationship is commonly a linear function expressed in Fig. 1.2 by line
OM.
Having such diagrams for rocks, we can determine mechanical characteristic of these
rocks in the following way:
(0) Rock hardness
F
..... (1-8)
Hs = B
Ai
Where FB is force at which the failure occurs
Ai is area of the indentor base (cross-sectional area of the punch base.)
Fig 1.2 Force-displacement diagram for so called elastics-viscous rocks.
(0) Plastic limit
F
..... (1-9)
H p= A
Ai
Where FA is load on the indentor at elastic limit (when plastic deformations commence).
(0) Modulus of elasticity (or Young’s Modulus)
F (1 − µ 2 ) FB (1 − µ 2 )
=
..... (1-10)
Es =
d i ξF
d iξOD
Where F is force corresponding to the given elastic deformation ξF
di is diameter of the indentor (punch) base
ξ OD is elastic deformation at the rock failure
µ is Poison’s ratio.
(0) Coefficient of plasticity
W
Cp = t
We
..... (1-11)
Where Wt is total work spent to get the rock failure
We is work spent on elastic deformations (so called elastic work).
(0) Specific work of rock disintegration
W
..... (1-12)
Wv = t
Vc
Where Vc is volume of the crater caused by rock failure (volume of broken rock).
The volume of broken rock can be determined by direct measurement or by filling the
crater with plasticine, which is easily removed and weighted.
(0) Specific work of contact
W
..... (1-13)
WA= t
Ai
Solving equation (1-11) for total breaking work, we get the following formula
..... (1-14)
Wt = C pWe
(0) Specific disintegration, that is volume of rock destroyed per unit of total work spent on
rock failure
V
vw = c
..... (1-15)
Wt
33
3. Discuss briefly rock breakage under the impact of single blows.
Rock Breakage Under The Impact of Single Blows
Bit teeth act on rocks drilled with percussive forces, ie. with impacts. There are important
differences between static and dynamic rock failures. To investigate the phenomena of the rock
failure under the impacts, instruments called “drop testers”, are used.
A drop tester, a sketch of which is presented in Fig.1.6, consists of a rod carrying a
cylindrical mass at the upper end and a punch at the lower and, the assembly being constrained for
essentially frictionless vertical fall. Strain gages are applied close to the punch so that the force
waveform can be recorded by an oscilloscope camera.
Another camera-a cine one-provides high speed motion pictures at the rate of about 3,000
frames per second. The weight of the mass and the height of fall being known, the energy of
impact, velocity, and momentum are easily calculated. The forces throughout the cycle are given
by the force waveform obtained with the oscilloscope camera, and the motion pictures provide
displacement-time and velocity-time curves as well as specific information as to the time and
place of chip formation.
Rock failure under the impact of single blows is studied by L. Schreiner, L. Simonjantz, F.
Pennington and other investigators.
Fig.1.6 Sketch of a drop tester:
1 – rod
2 – bit (punch, chisel)
3 – rock specimen
4 – mass
5 – guides
Fig.1.7 shows two of the resulting curves of force at the bit vs. time, each curve being
typical of a large number of observations carried out by F. Pennington [15] with a 15-1b mass at
various heights of drop. The first peak occurs at a force about 2,000 lb regardless of the energy
level of the drop. Then the force decreases and the second force peak results whose value
increases with energy level of the drop. For drop heights below 9/16 inches, the second peak is
missing and the first peak varies with energy level. Above 6 in. of drop height, the second peak
may also become constant in magnitude but sufficient data are not available to establish this point
definitely.
Forming the creater in the rock under the impact force resembles closely those under the
static load. The chips, however, fly out of the place instead of remaining there as in the static test.
The force-displacement curve is closely similar to that obtained in the static testing, and the first
major fracture occurred at about the same force level (at some larger force). Again, therefore, it
appears that the amount of rock removed is independent on the rate of application of the load.
In one respect the static loading tests give results that are not in line with some of the drop
test results. In the static test it appears that the energy required to fracture out a given amount of
rock is considerably less than for the drop-tests (under impact forces). For the static tests the
loading is usually increased in each test until major chipping occurs, whereupon the test is
terminated. On the other hand, for the drop tests, a given fixed amount of energy, as determined
by the potential energy of the weight may be introduced into the system for each test.
If this amount is more than enough to produce major chipping, the remainder will be
essentially wasted unless it happens to be enough to produce a second major chipping action. If
the energy is insufficient to produce the first major chipping, it will be largely wasted. In practice
it will not be entirely wasted, because the rock may be lift in a weakened condition so as to
contribute to the effectiveness of subsequent blows. If the energy is insufficient to produce the
crater after a blow, the rock failure may be occurred as the result of the rock fatigue under
34
repeated blows. Thus, if a practical means could be devised for removing the debris each time as
soon as chipping occurred, much more efficient drilling would result.
Fig.1.7 Tracings of typical force waveforms for various heights of drop with 0.03-in.
bit and weight of 15 1b for single blows on Indiana limestone.
The volume of rock removed in a single impact varies linearly with the energy of the
blow. The straight-line graph intersects the energy axis at a point that may be called the threshold
energy, below which substantially no chips are formed. Study of the way in which energy is
dissipated over the entire impact cycle is of interest in determining the most effective means for
its utilization in breaking the rock.
In Fig. 1.8, the energy consumed up to any given time is shown superimposed on the force
waveform. It is seen that approximately half the energy is consumed during the first force pulse
for this height of indentor drop. In general, all the energy is used in the first pulse up to a certain
height of drop. Above this height, the energy so used is a constant, the excess energy going into
the second pulse. The high-speed motion pictures show that the spectacular removal of large chips
takes place during the first pulse.
Fig.1.8 Superimposed force-time and energy-time waveforms for a single blow on
Indiana limestone
4. Explain the rock breakage under cycle (repeated) stresses.
Rock Breakage Under Cycle (Repeated) Stresses
It has been discovered that under cyclic stresses (under repeated blows) many rocks can be
crushed although specific contact stresses of blows are less than rock hardness. Rock failure in
this case is a result of rock fatigue. The lower contact stresses the more is the number of blows to
crush the rock and there is a critical stress under which fragile failure cannot occur after any
number of blows as shown in
Fig. 1.11. This critical stress Hf is called rock fatigue strength. When blow force increases,
the needed number of the blows decreases, and at certain blow force rock failure occurs after each
blow. The specific contact stress corresponding to this critical force can be called dynamic rock
hardness Hd. This hardness is larger than Hs at the static test already discussed, but for actual
drilling conditions the difference between Hd and Hs can be neglected.
Fig. 1.11 Relationship between needed number n of blows to break the rock and
contact specific stress Sc of the blows.
Upon careful consideration of rock failure, we can say that there are three kinds of rock
failure in drilling wells and they are:
(0) surface crushing,
(0) fatigue-volumetric fracturing,
(0) volumetric fracturing
Surface crushing takes place when the specific contact stress is less than the fatigue
strength of the rock, while the volumetric fracturing can occur when the contact stases reaches
hardness of the rock. If the magnitude of contact stress is between the fatigue strength and
hardness of the rock, so called fatigue-volumetric fracturing will occur under repeated blows, i.e,
craters will be caused after several impacts. The number of loading cycles necessary to produce
fatigue rock failure is usually not greater than 100-150.
35
B.A. Zhlobinski has established that mechanism of fatigue-volumetric rock breakage
under repeated blows is similar to that of breaking brittle rocks at the static punching tests. A
cycle of fatigue-volumetric rock breakage consists of the four following stages (see Fig. 1.12).
During the first stage a fracture is caused on the surface of the rock, it propagates into the
rock, and the volume of which limited by the punch base and the semisphere under the base
collapses into the fine powder (into mass).
Fig. 1.12 Mechanism of fatigue-volumetric rock breakage.
At the second stage the mass acts on the rock beyond it and causes a rupture propagating to the
surface round the punch base. At the third stage this rupture results in rock failure around the base
of the punch. During the fourth stage the mass compressed under the punch is crushed, and the
crater is formed. Dimensions of the crater are
1
1
b
R ≈ ,a ≈ ,δ ≈ ,
.....(1-26)
2
3
2
where 1 is length of the punch base,
b is width of the base,
δ is depth of crater,
R and a are shown in Fig. 1.12.
5. Explain about the subsurface pressures.
Subsurface Pressures
It is very necessary to know reservoir pressures in order to avoid or overcome different
drilling hazards some of which can be very dangerous. In general, the pressures encountered with
depth are due to hydrostatic pressure and (or) overburden pressure.
Hydrostatic pressure is imposed by the weight of fluid (predominantly water) which fills
the voids of the rocks of the bed in question and (or) by the weight of fluid which fills the voids of
the overlying rocks contiguous with the bed under consideration. Hydrostatic pressure is
calculated by the following equation:
Phs = 0.1pbfA,
.....(2-1)
where Phs = hydrostatic pressure of the bed fluid, kg/cm2
pbf = density of the bed fluid, gm/cm3
A = altitude of the highest point of the bed fluid above the point for which
hydrostatic pressure is calculated, m.
Equation (2- 1) may be written in the following way
.....(2- 2)
Phs = P*hs A,
2
kg / cm
where P*hs is hydrostatic gradient of the given bed,
. It is obvious that
m
.....(2- 3)
P*hs = 0.1 pbf.
In oil and gas deposits density of bed fluids is dependent on genetical features of the fluids
and hydro-geological peculiarity of the deposits. Density in question ranges within 1.01 to 1.23
gm/cm3.
Overburden pressure is due to the weight of the overlying rocks with their fluid content.
This pressure is called geostatic pressure and is calculated by the formula
Pgs = 0.1 pr D,
.....(2- 4)
Where Pgs = overburden (geostatic) pressure at the given point of the formation in question,
kg/cm2.
pr = average density of the rocks and their fluid content existing above the
formation. gm/cm3;
36
D = thickness of overlying rocks (depth of bedding of the point of formation), m.
Equation (2- 4) may be written in the following way
Pgs = Pgs* D,
.....(2- 5)
where Pgs* is gradient of overburden pressure (i.e., overburden gradient)
It is obvious that
Pgs* = 0.1 pr.
kg / cm 2
m
.....(2- 6)
In calculation pr = 2.3 – 2.5 g/cm3 and, consequently, Pgs* = 0.23 – 0.25
kg / cm 2
, are
m
usually taken.
Overburden pressure acts through the skeleton (the solid phase) of rocks, and the skeleton
represents posts or poles which support the weight of the overlying rocks and insulate the
overburden pressure from the hydrostatic pressure.
A great many oil and gas reservoirs are in contact with blanket-water sands which
eventually outcrop to the surface of the earth. Such reservoirs have an initial pressure equal to the
hydrostatic gradient from the outcrop elevation to the reservoir, minus the flowing pressure drop
between the outcrop and the given point of the reservoir. Even many reservoirs that do not have
known direct communication with the surface have pressures that conform to this gradient.
Variations in the depth-pressure gradient are caused by salinity of the water and temperature.
Abnormal pressures are found on both the high and the low sides of the pressure-gradient
curves of the above mentioned oil and gas reservoirs. Abnormal pressures indicate that the
reservoir fluids are not in communication with outcrop formations.
Consequently, it is more common to find initial reservoir pressures varying as a linear
function of depth with a gradient close to the hydrostatic gradient of fresh to moderately saline
water. Departures from this behavior, both higher and lower, are considered abnormal. It is,
however, the abnormally high pressures which are more important as a source of serious drilling
and production hazards.
Abnormally high pressures are most common in soft rock areas such as the ones of the
North Caucasus (in the USSR) and the Gulf Coast (in the USA). Here the sediments are
geologically young and are not as consolidated as the hard rocks of the Tatar Autonomous Soviet
Socialist Republic (in the USSR) and West Texas (in the USA). The young slightly consolidated
rocks are subjected to compaction, repacking, and other sedimentary changes. As the result these
rocks have failed to support the total weight of the overlying rocks, and the rest of this weight is
supported by the fluid enclosed in the reservoir. Therefore the reservoir pressure becomes higher
then the hydrostatic pressure and can exceed the latter by two or even more times.
Sometimes the concept of conditional hydrostatic pressure is used. The conditional
hydrostatic pressure is equal to the pressure of fresh water (with density of 1.0 g/cm3) column, the
height of which is equal to the depth of the reservoir bedding:
Pchs = 0.1 Dr,
.....(2- 7)
kg
where Pchs is so called conditional hydrostatic pressure,
cm 2
Dr is depth of the reservoir bedding, meters.
The ratio of a reservoir pressure to the conditional hydrostatic pressure is called the
relative reservoir pressure:
P
Prr = r ,
.....(2- 8)
Pchs
where Prr is relative reservoir pressure and Pr is the reservoir pressure.
6. Explain about the subsurface temperature.
37
Subsurface Temperatures
The earth is assumed to contain a molten core. Therefore it is logical to assume that
temperature should increase with depth. This temperature-depth relationship is commonly a linear
function of the following form
......(2- 18)
TD = TO + aD
where TD = temperature of the rock at any depth D
TO = average surface temperature
a = temperature gradient for the area in question, degrees per 100 ft.
D = depth of the rock bedding, hundreds of feet.
A normal temperature gradient is some 1.6°F per 100 ft, although it should be noted that
considerable variations (from 0.6 to 3° F per 100 ft. of depth) occur in various areas. The
temperature near the surface, usually taken at 100 ft, approximates the mean annual atmospheric
temperature. Since the earth is cooling, the temperature at 100 ft. should exceed the mean average
atmospheric temperature, and is observed to exceed it by from 2 to 10° F.
The equilibrium earth temperature is of interest since it represents reservoir temperatures.
By injecting gas or water into a reservoir at a temperature below the reservoir temperature, it is
possible to change the formation temperature from the equilibrium earth gradient. The circulation
of drilling fluid during drilling or producing of oil and gas up the well bore will upset the
equilibrium earth gradient.
When a drilling fluid is circulated down the inside of the drill string it receives heat from
its ascending flow through the string walls and so is heated gradually: temperature of the
descending flow of the drilling fluid is minimum at the drill string top and temperature of this
flow is maximum at the drill string bottom; both these temperatures (at the top and at the bottom
of the drilling string) increase when the length of the drilling string (or when the well depth)
increases.
Temperature of the drilling fluid at the drill string bottom is lower than temperature of the
rocks at the same well depth. Ascending flow temperature at the well head is higher than rock
temperature at the same place. Consequently, there is a point at which temperature of the
ascending flow is equal to temperature of the rocks.
The ascending flow of a drilling fluid receives the heat from the rocks when the drilling
fluid is circulated from the drill string bottom to the mentioned point; and then this flow returns a
part of the heat to the rocks when the drilling fluid is circulated to the well head. The flow in
question gives heat to the descending flow of the drilling fluid during the whole way from the drill
string bottom to the well head.
Ascending flow temperature at the well head is higher than descending flow temperature
at the same point and the longer the drill string (the longer the well depth) the higher is this
temperature.
The temperature-depth relationship for circulated fluid differs from equation (2- 18) for
the earth crust and can be written in the following way.
.....(2- 19)
ThL = To + ah + bL,
where ThL = temperature of the circulated drilling fluid at any well depth h under the
drilling string length L
h = depth of the well point for which the temperature of the drilling fluid is
calculated, m
L = length of the drill string (length of the circulating drilling fluid flow), m
a and b = coefficients dependent on some factors, degrees/m.
7. Write short notes of standard derricks.
38
Standard Derricks
A standard derrick cannot be raised to a working position as a unit, i.e., it is of bolted
construction and must be assembled part by part. Likewise, it must be disassembled if it is to be
transported. Exceptions to this are those relatively short moves in which the entire rig is skidded
to a nearby location. Detailed specifications for the nine API standard derricks may be found in
the API Std. 4A.
Derricks are rated according to their ability to withstand two types of loading:
(0) Compressive loads
(0) Wind loads
The allowable compressive load of a derrick is computed as the sum of the strengths of the
four legs. In these calculations, each leg is treated as a separate column and its strength computed
at the weakest section. This load, (excluding the weight of the derrick) with its safety factor of 2,
is the API safe load capacity. Capacities for reinforced derricks are computed in a similar manner
by the formulae of API Std. 4B. Derricks with load capacities from approximately 86,000 to
1,400,000 lb, depending on steel grade and leg size, are available.
Allowable wind loads for API derricks are specified in two ways, with or without pipe
setback. For the drill pipe to stand vertically stable during a trip, the tops of the stands must lean
outward against the fingers at the pipe racking platform. This results in an overturning moment
applied to the derrick at that point. If the wind is blowing perpendicular to the setback, which is
essentially a pipe wall, a further overturning moment is applied. This is the worst possible
condition, i.e., wind and setback loads acting in the same direction. The minimum allowable wind
loads for 136 ft and shorter derricks is 11.76 lbs/ft2 (54 mph) with pipe setback. Minimum wind
loads for taller derricks are 22.50 lbs/ft2 (75 mph) with setback and 52.90 lbs/ft2 (115 mph)
without setback. Wind loads are calculated by the formula:
ρ = 0.004 V2
.....(3.1)
where ρ = wind load, lb/ft2
V = wind velocity, mph
Calculation of Derrick Loads
The block and tackle arrangement for a rotary rig .If we assume the system as frictionless,
the following relationships are apparent:
n+2
Fd =
W
.....(3.2)
n
where Fd = total compressive load on the derrick
n = number of lines through the traveling block (those supporting W )
W = hook load
Hence it is seen that the derrick load is always greater than the hook load by the factor (n +2) /n
due to the two additional lines (drawworks and anchor) exerting downward pull. Further it may be
noted that during hoisting:
8. Explain about mud pumps.
Mud Pumps
The function of the mud or slush pumps is to circulate the drilling fluid at the desired
pressure and volume. The pump normally used for this service is the reciprocating piston, doubleaction, duplex type shown in Figure 5.7. Piston and valve action are illustrated in Figure 5.8. The
term “double-action” denotes that each side of the piston does work, while “duplex” refers to the
number of pistons two. Triplex (three pistons) types have been used but to no great extent.
The superiority of the piston type pump for drilling service is largely due to the following
features:
39
Ability to handle fluids containing high percentages of solids, many of which are
(0)
abrasive.
Valve clearance will allow passage of large solid particles (typically lost circulation
(0)
materials) without damage.
Ease and simplicity of operation and maintenance. Liners, pistons, and valves may be
(0)
replaced in the field by the rig crew.
Wide range of volume and pressure available by using differend liner and piston sizes.
(0)
The main disadvantage of these pumps is that the discharge flow is pulsation, which
causes periodic impact loads on discharge lines. This effect is minimized by air filled surge
chambers located on the discharge line.
Mud pumps are commonly denoted by bore and stroke. An 8 × 18-in. Power pump has a
piston diameter (line size) of 8in, and a stroke length of 18 in. Direct acting steam pumps require
an extra dimension for their description; hence, a 16 × 8 × 18-in. pump is a steam pump having a
16 in. steam piston opposing the 8 in. mud piston with an 18 in. stroke. These pumps differ from
those in Figure 3.7 in. that the power end of the steam type consists of another set of pistons
against which high pressure steam exerts sufficient force to power the pump. By making the
steam pistons larger than those at the mud end, one can hold the operating steam pressure much
lower than the mud discharge pressure (commonly down at about one-half). The relative merits of
steam pumps versus power pumps are about the same as those of steam engines versus internal
combustion engines, as is discussed in the next section.
9. Explain briefly the prime movers in common use.
Prime Movers
Before discussing the prime movers in common use, let us look more closely at the drilling
services which must be performed. The bulk of rig power is consumed by two operations:
(1) Circulation of the drilling fluid
(2) Hoisting
Fortunately these requirements do not occur at the same time, and the same engines perform both
jobs. The power consumption of the circulation system and rotary table is essentially constant
over reasonable time periods. Such is not the case with hoisting, as is illustrated in Figure 3.9.
Here it may be seen that rotary rig prime movers must be capable of handling highly variable
loads at rapid acceleration over a wide torque and speed range.
Steam Engines
The first rotary drilling prime movers were steam engines. Although there are no longer
dominant in the drilling industry, they are still commonly used in many
Fig 3.8 Schematic valve operation for double acting slush pump.
areas. The common engine for this service is a twin- cylinder, double-acting type, which in basic
principle operate much like a power mud pump in reverse. High pressure steam is furnished by a
boiler plant located near the rig. He desirable steam pressure is governed by the power
requirement (depth, circulation rate, etc.) and the piston size against which the steam works.
Electric Motors
Both direct and alternating current motors have been used for drilling purposes. The
speed-torque range and ease of control of the d-c type make it an ideal prime mover. Its principal
disadvantage is, of course, the in availability of such power. Recent improvements in the design
of diesel-electric power units have overcome many disadvantages of previous motor generator
systems; and these units are finding considerable application, particularly in offshore drilling. The
40
operating characteristics of alternating current motors do not fit drilling requirements sufficiently
well to warrant any particular use. They have been used for special applications such as in city
drilling where power availability and noise restrictions have made it practical.
Internal Combustion Engines
Engines of this type are the most commonly used in the drilling industry. This has not
been the result of any superiority of performance, as they are inferior in this respect to both steam
and D.C. electric. Torque-speed characteristics may be improved by the use of torque converters.
The internal combustion engines used are automotive type (multicylinder, light flywheel) diesel
and gas engines capable of rapid accelerations.
Relative Merits of Prime Mover Types
Table 3.1 is an attempt to consolidate the general advantages and disadvantages of various
prime mover types. The importance of these will vary with the intended area of use, and final
selection is normally based on an economic appraisal.
Fig 3.9 Load-time diagram illustrating variation during hoisting cycle.
10. Discuss about offshore drilling equipment.
Offshore Equipment
Large Self-Contained Platforms
The first offshore (unprotected water) drilling structure was built in 1937 and was located
1
1 miles from shore in 15 feet of water. This was a large self-contained platform supported by
2
numerous creosote treated timber piling. The Creole Field was discovered in 1938 by a well
drilled from this platform; this field eventually contained 19 wells drilled from ten surface
locations. 11 The self-contained platform, with design improvements, is still in use. As the name
implies, these platforms are sufficiently large to house the entire rig, crew quarters and supplies.
The cost of such an installation is enormous, but may be quite economical on a per well basis,
since numerous wells may be directionally drilled from its deck. This practice, however, has the
very dangerous disadvantage that a blowout and / or fire of any well jeopardizes the entire
installation. A more recent self-contained platform is shown is Figure 3.16.
While these structures lend themselves to development drilling (drilling in known fields),
they are poorly adapted to exploratory work. It would be extremely costly to build such a structure
only to find no oil accessible from its surface. Consequently, smaller, less expensive platforms
with floating tenders evolved to offset this and other disadvantages of the self-contained
platforms.
Platform Drilling Tender Installations
In this operation the relatively small platform carries the substructure, drawworks, rotary
table, engines, and other small items. The barge or tender houses the rest which includes the mud
pits, supplies(dry mud, chemicals, etc.), pipe racks, cementing equipment, fuel and water, and
crew quarters. The principal advantages of these installations are greater mobility and lower
installation costs. A disadvantage is the higher percentage of lost time due to high wind and
waves: during periods of rough weather, the tenders must pill away to prevent collision with the
platform. Large self-contained platform operations may proceed in any weather short of a
hurricane.
41
Completely Mobile Offshore Units
The next development in offshore equipment was the completely mobile offshore unit.
This is based on an extension of the inland barge principle, in that the entire rig is floated to the
location as a unit. Figure 3.17 shows the classifications of these units as given by Howe.12 Just as
in land drilling, the economic advantages of complete mobility are considerable; units of this type
are rapidly replacing former installation types.
The offshore drilling topic could be expanded over the entire length of this book if we
were to attempt anything like a complete coverage of the subject. No mention has been made of
such problems as corrosion, wave forces, and meteorological statistics, all of which are extremely
important to offshore equipment design. The offshore oil potential is great; undoubtedly, an
increasingly greater proportion of U.S. production will come from these fields in future years. The
drilling and production problems of these operations offer a challenge to the entire industry.
Fig 3.17 Classification of offshore mobile drilling units.
**********
42
Ministry of Science and Technology
Department of Technical and Vocational Education
Petroleum Engineering
Sample Questions for
PE 03032 DRILLING ENGINEERING Part( II )
Chapter 1 Rotary Drilling Hydraulics
1.*
2.**
3.***
4.*
5.*
6.**
7.***
8.*
9.***
10.*
11.**
Mud is flowing through 4½-in. OD, internal flush drill pipe. Calculate the frictional pressure
drop per 1000 ft of pipe.
(10 marks)
Mud properties:
ρm = 10 lb/gal
Pipe ID = 3.64 in.
Yb = 10 lb/100 ft2
Circulating rate, q
µp = 30 cp
A 10 lb/gal mud is being circulated at the rate of 500 gal/min through a tri-cone bit having
three ½-in. diameter jets. What is the pressure drop across the bit?
(5 marks)
Operating data:
Depth = 6000 ft (5500 ft drill pipe, 500 ft drill collars)
Drill pipe = 4½-in. internal flush, 16.6 lb/ft (ID = 3.826 in.)
Drill collar = 6½-in. (ID = 2.813 in.)
Mud density = 10 lb/gal, µp = 30 cp, Yb = 10 lb/100 ft2
Bit = 7⅞-in., three-cone, jet rock bit
Nozzle velocity = at least 250 ft/sec per inch of bit diameter
What hydraulic (pump output) horsepower will be required for these conditions? (15 marks)
Compute the bottom-hole circulating pressure for the given following data.
Hole size = 9⅞-in
Depth = 10,000 ft
Drill pipe = 4⅛-in.
Drill collars = 500 ft of 8-in.
Circulating rate = 400 gal/min
Mud = 10 lb/gal
Yb = 60 lb/100 ft2
Plastic viscosity = 40 cp
(10 marks)
Explain about hydraulics and rate penetration.
(5 marks)
Discuss pressure surges caused by pipe movement. (10 marks)
Calculate the actual pressure imposed on a formation at 10,000 ft. Assume pipe is closed at
lower end. Given data:
Length of pipe in the hole = 9500 ft µp = 35 cp
ρm = 10 lb/gal
Yb = 15 lb/100 ft2
Hole size = 9 in.
Pipe size =4½-in. OD (ignore drill collars)
The pipe is being lowered at 400 ft/min.
(10 marks)
(a) What air velocity will just float a ⅜-in. diameter, 2.6 specific gravity, spherical particle at
30 psig and 120°F?
(5 marks)
(b) What would be the velocity for a shale cutting of the same density?
(5 marks)
Calculate the circulation rate required to air drill a 7-in. hole with 4½-in. drill pipe at a rate of
90 ft/hr at 12,000 ft depth.
(5 marks)
(a) Estimate the compressor output horsepower, 7-in. hole, 4½-in. drill pipe, rate 90 ft/hr,
depth 12,000 ft. Assume suction conditions of 80°F and an operating altitude of 30 ft above
sea level.
(5 marks)
(b) How many stages will be required for these conditions and what will be the compression
ratio for stage? (5 marks)
Write short notes on aerated mud drilling.
(5 marks)
43
12.*** Compute the pressure drop per 1000 ft, 4½-in. internally flush drill pipe, if water (1 cp) is
being used as drilling fluid. The circulation rate is 500 gal/min. Inside diameter is 3.83 in.
(10 marks)
Chapter 2 Pressure Surges and Anomalies
13.*
14.**
15.***
16.*
17.**
What are causes and effects of pressure anomalies? (5 marks)
Discuss briefly the blowouts caused by pressure reductions during hoisting. (10 marks)
Explain mud losses caused by mud movement.
(15 marks)
Discuss briefly the relationships between pressure and gel strength. (15 marks)
Mud circulation has been shut down on a drilling well for sufficient time for the mud to
develop a gel strength of 40 grams, stormer. If the drill pipe is not moved longitudinally or
rotated, what pump pressure will be required to break (start) circulation?
(10 marks)
Additional data:
Depth = 8000 ft
Drill pipe = 3.8 in. ID, 4.5 in. OD
Drill collar = 2.25 in. ID, 6.25 in. OD, length = 450 ft
Hole diameter = 8.75 in.
18.*** Discuss (briefly) the effects of plastic and turbulent flow on the calculation of pressure
changes.
(15 marks)
Chapter 3 Oil Well Cementing and Casing Practices
19.* Explain the general functions of all casing string.
(5 marks)
20.** Discuss the primary oil well cementing techniques. (15 marks)
21.*** Casing is being cemented in a 10,000 ft well containing 10 lb/gal mud. The slurry density is
14 lb/gal and the plug is to be chased with 0.85 sp.gr. oil. If the anticipated fill up is 1000 ft,
what surface pressure must be held on the casing to prevent back flow at the end of the job in
case the floating equipment fails to hold?
(10 marks)
22.* Explain the six API cement classifications. (5 marks)
23.** Discuss the numerous types of special cement commonly used.
(15 marks)
24.*** Write short note on the pozzolan cements.
(5 marks)
25.* Write short note on diesel-oil-cement. (5 marks)
26.** Write short note on oil-in-water emulsion cement.
(5 marks)
27.*** Explain briefly the slurry density for cement system. (10 marks)
28.* Explain about thickening time required for a cement slurry. (10 marks)
29.** Write short note on the strength-time required for a cement slurry. (5 marks)
30.*** Describe the filtration properties of a cement slurry. (5 marks)
31.* Explain the permeability of set cement.
(5 marks)
32.** Discuss about the auxiliary cementing equipment.
(15 marks)
33.*** Write short notes on two types of cementing plugs. (10 marks)
34.* Write short note on wall scratchers used in cementing operation.
(10 marks)
35.** Explain about the casing centralizers.
(10 marks)
36.*** Explain about the floating equipment used in cementing operation. (10 marks)
37.* Write short note on packer type cementing shoes.
(10 marks)
38.** Explain the cementing volumes in cementing operations.
(10 marks)
39.*** Explain about the squeeze cementing. (15 marks)
40.* Discuss briefly the types and specifications of casing. (15 marks)
41.** Write short notes on surface strings, intermediate string, and oil string.
(15 marks)
42.***
43.*
44.***
45.**
46.*
47.**
48.***
49.*
50.**
44
Casing strings are usually designed to withstand three principal types of loading. Describe
them. (10 marks)
A 10,000 ft of 5½-in., 20 lb casing is freely suspended in 12 lb/gal mud.
Compute (a) the maximum compressive stress in the string, (b) the location of the neutral
zone, and (c) the location of the zero axial stress plane.
(15 marks)
Explain the special considerations during squeeze cementing. (15 marks)
Explain the tubing selection. (15 marks)
What surface pressure must be held to prevent cement backflow under the following
conditions?
(10 marks)
Mud = 9 lb/gal
Cement = 15 lb/gal
Well depth = 10,000 ft Fill up = 4000 ft
Oil = 30°API (used for displacement)
Write short note on design or safety factor for casing operation.
(5 marks)
Explain the most common range of design factors and assumed conditions for casing
operation.
(5 marks)
Explain the procedures for casing string design.
(15 marks)
Design a 14,000 ft oil string under the following assumed conditions. (20 marks)
Design factors:
Tension = 1.80 Collapse = 1.00 (mud density = 10 lb/gal)
Burst = 1.00 Based on known reservoir pressure, Pe = 6300 psi)
Buoyancy is to be used, but design is arbitrarily restricted to 3 API weights and/or grades
with either short (ST and C) or long (LT and C) API couplings.
* = Must know, ** = Should know, *** = Could know
45
Ministry of Science and Technology
Department of Technical and Vocational Education
Petroleum Engineering
Worked Out Examples for
PE 03032 DRILLING ENGINEERING Part ( II )
1.
Mud is flowing through 4 12 inch OD, internal flush drill pipe. Calculate the frictional
pressure drop per 1000 ft of pipe.
Mud properties
p m = 10 lb/gal
Pipe ID = 3.640 in.
2
Yb = 10 lb/100 ft
Circulating rate q = 400 gal min
µ p = 30 cp
Solutions:
(1.08)(30) + (1.08) (30) 2 + (9.3)(10)(3.64) 2 (10)
(10)(3.64)
= 4.3 ft/sec
1 ft 2
1 min
qgal
/
min
×
×
2
Qft / sec
7.48 gal 60 sec
=
(2) v =
2
Aft
(π / 4)(d / 12) 2
q
=
2.45d 2
400
=
= 12.3 ft/sec
2.45(3.64)2
(1) vc =
(3) v > vc , and flow is turbulent.
(2970)(10)(1000)(12.3) 2
= 44,300
30
(b) f = 0.0062 from Curve II, Figure 1.1
(0.0062)(10)(1000)(12.3) 2
(c) ∆p p =
(25.8)(3.64)
= 100 psi/1000 ft
(a) Rc =
2. A 10 lb ft/gal mud is being circulated at the rate of 500 gal/ min through a tri cone bit having
1
three − in diameter jets. What is the pressure drop across the bit?
2
Solution:
2
3
d = 3  = 0.25in (equivalent single nozzle diameter)
6
By Eq. (1.13),
46
∆p =
(500)2 (10)
(7430)(0.95)2 (0.65)4
= 2100 psi
3. Explain about hydraulics and rate of penetration.
Hydraulics and Rate of Penetration
It has been established that the penetration rate in many formations is proportional to the
hydraulic horsepower expended at the bit. Realization of this factor and the subsequent
widespread use of jet bits has reversed the trend to small drill pipe sizes which existed a few years
ago. Pressure drops inside the drill string are a large part of total system losses, and it is obvious
that an increase of inside diameter will greatly improve hydraulic efficiency.
Consequently, tool joints which have little or no internal restriction are most commonly used. The
use of larger drill pipe and drill collars also entails closer hole-pipe clearances; thus the desired
annular velocity can be obtained at lower circulation rates. In addition, annular velocity
requirements have been reappraised and lower figures are now being applied.
4. (a) Given the following data calculate the actual pressure imposed on a formation at 10,000 ft.
Assume pipe is closed at lower end.
Length pipe in hole = 9500 ft
µp = 35 cp
Yb = 15 lb/100 ft2
pm = 10 lb/gal
Hole size = 9 in.
Pipe size = 4 12 in.O.D. (ignore drill collars)
The pipe is being lowered at 400 ft/min.
Solution:
(1.08)(35) + 1.08 (35) 2 + (9.3)(4.5) 2 (15)
(10)(4.5)
= 5.0 ft/sec
From Eq. (1.19):
vc =
400  1
(4.5) 2
 +
60  2 (9) 2 − (4.5) 2
∴flow is turbulent.
for turbulent condition:
pa =
∆p ap

 = 5.6 ft/sec

(297)(10)(5.6)(4.5)
= 21,400
35
f = 0.0084 from Figure 1.1, curve IV
(0.0084)(10)(9500)(5.6) 2
∆p ap =
= 215 psi due to friction
(25.8)(4.5)
The total instantaneous bottom hole pressure is:
28
p = 215 + (0.052)(10,000)(10) = 5415 psi
(b) Rework (a) including the effect of 500 ft of 7 12 inch o.d. drill collars on the bottom of the drill
string.
Rc =
47
Solution:
v ac =
400  1
(7.5)2 
+
= 18.5 ft/sec

2
60  2 (9) − (7.5) 2 
(2970)(10)(18.5)(1.5)
= 23,400
35
f = 0.0083
(0.0082)(10)(500)(18.5) 2
∆p ac =
= 370 psi
(25.8)(1.5)
From part (a),
9000
∆p ap =
× 215 = 203
9500
Pressure surge = 203 + 370 ≅ 570 psi
Total p = 570 + 5200 = 5770 psi
(c) What would the instantaneous bottom hole pressure be if the pipe were being
removed at the same rate as in parts (a) and (b)?
(1) p = 5200 - 215 = 4985 psi
(2) p = 5200 - 570 = 4630 psi
Rc =
5. What are causes and effects of pressure anomalies?
Causes and Effects of Pressure Anomalies
Pressure anomalies are considered to be all of the pressure variations by which the total
pressure in the drilling fluid deviates either from the hydrostatic mud pressure when fluid is not
being circulated or from the average flowing pressure at any point during fluid circulation.
Pressure anomalies occur during a pulsating type of flow and are associated with rapid starting of
drilling-fluid circulation and particularly with pipe movements in the hole.
The rapid raising or lowering of a string of pipe in a hole filled with fluid is perhaps the
most common cause of large pressure surges. The decreases caused in total mud pressure have
been responsible for blow-outs and the loss of wells. The increases in total mud pressure have
been responsible for expensive losses of mud into exposed geological formations. Where such
mud losses have occurred into producing formations, they have undoubtedly in many cases
injured the producing ability of the well.
6. Discuss briefly blowouts caused by pressure reductions during hoisting.
Blow-outs Caused by Pressure Reductions During Hoisting
The first important paper on the subject of pressure changes associated with pipe
movement, "Changes in Hydrostatic Pressure Due to Withdrawing Drill Pipe from the Hole," was
presented before the American Petroleum Institute in 1934 by George E. Cannon. This paper cited
a study of twenty-seven blow-outs which occurred in the drilling of 891 wells. Of these, twentytwo were definitely associated with the withdrawal or subsequent lowering of the drill string into
the hole. A total of eleven wells had to be abandoned.
Since it was known that the hydrostatic head of the mud was sufficient to overbalance the
formation pressures, a review of the statistics had prompted an experimental investigation to
measure pressure reductions which might occur during the raising of the drill pipe. The
measurements were made by means of a subsurface pressure gauge placed in the bottom of the
48
hole or in the lower part of the drill string. Some of the results are illustrated in Figs.2-1, 2-2, and
2-3. This investigation showed that serious pressure reductions can occur when the drill pipe is
being raised and that such reductions are related to the annular area between the drill pipe and the
hole. The only other correlation obtained was that the pressure reductions were directly related to
the gel strengths of the muds. Muds of zero gel strength have subsequently been preferred for
most drilling operations, particularly for drilling in such areas as the Gulf Coast.
The fact that no correlations between pressure reductions and such quantities as the mud
viscosity and rate of pipe movement were obtained has been discussed by Cardwell. He pointed
out that, for the high gel strength muds in use at the time, once fluid (or pipe) movement was
started, no further pressure changes would be expected. The systems were controlled by gel
strength so far as pressure reductions were concerned. Where muds have lower gel strengths, as
recommended from the results of the investigation, other factors become of governing importance
in determining the magnitude of the pressure surges.
Fig. 2-1. Relation of gel strength of mud to pressure drop due to swabbing in 7-in.
casing.
Fig.2-2. Relation of gel strength of mud to pressure drop due to swabbing in 10 34 - in.
casing.
Fig. 2-3. Relation of pressure drop due to swabbing in 7-in. casing to depth.
7. Explain the general functions of all casing strings.
During the course of drilling, it is necessary to run casing at various depth intervals, i.e., to
lower the desired length of casing or pipe into the hole and cement it in place. The number and
size of the casing strings used vary with the area, depth, anticipated producing characteristics of
the well, and the choice of the operator. In addition to the casing, a smaller diameter string called
tubing is used as the actual flow conduit for the produced fluids. The final appearance of a typical
completed well is shown in Figure 3-1. Note also that the tubing-oil string annulus is segregated
from the pay sand by a packer. This is a common arrangement in wells completions in the next
chapter, and further consideration of the subject is postponed until then.
Each casing string is cemented in place by a slurry pumped down the pipe and up the
annulus between the casing and the open hole. The amount of slurry used is predetermined for the
particular annular volume and fill-up height desired. The cement is then allowed to set for several
hours before drilling or other operations are recommended. The general functions of all casing
strings are:
To furnish a permanent borehole of precisely known diameter through which
(1)
subsequent drilling, completion, and producing operations may be conducted.
To allow segregation of formations behind the pipe, which prevents interformational
(2)
flow and permits production from a specific zone.
To afford a means of attaching the necessary surface values and connections to control
(3)
and handle the produced fluids.
Our discussion of casing programs will be enhanced by a consideration of the principles
of cementing.
Fig. 3-1. Sketch of Typical Completed Well.
8. Discuss the numerous special cement types are also in common use.
Numerous special cement types are also in common use. Some of the more important of these are:
49
Pozzolan Cements: Pozzolans are siliceous materials which will react with lime to
1.
form a cementitious material. As such, they may be used as an additive to ordinary cement or
prepared as a limepozzolan blend without portland-type cement. Both natural (volcanic
origin) and synthetic pozzolans are in use. A common form of the latter is called fly ash and is
the combustion product of called fly ash and is the combustion product of called fly ash and is
the combustion product of pulverized coal as obtained from steam generating plants. Limepozzolan cement has proved to be satisfactory deep well cement. Considerable variance in
properties of pozzolans has been observed and individual testing is required.
Diacel Cement Systems: This designation refers to cement systems modified by
0.
one or more of the additives listed as Diacel D,LWL, and A in Table 3-1. Such cements have
a large range of densities and thickening times, which gives them a wide scope of
applicability. Fine sand (95% through 200 mesh) is sometimes added to increase early
strength of the mixture.
Latex-Cement: This is a special cement composed of latex, cement, a surface
0.
active agent, and water. It has proved useful in such special applications as plug-back jobs for
water exclusion. It is especially resistant to contamination with oil and/or mud and exhibits a
high strength bond with other materials (casing, rocks, etc.).
Diesel-Oil-Cements (DOC): Mixtures of portland cement, diesel oil (or kerosene),
0.
and a chemical dispersant have been found useful in well repair (remedial) work to seal off
water-bearing strata. This material does not set until brought into contact with water and has,
therefore, an unlimited pumping time. It has also been used to prevent lost circulation.
* The depth limits are based on the conditions imposed by the casing well-simulation tests, API
RP 10B, schedules 1-10, inclusive, and should be considered as approximate values.
0.
Oil-in-Water Emulsion Cements: Low water loss, low density cements of adequate
strength and thickening time have been prepared from kerosene, water, cement, and 2 to 4%
bentonite. Calcium lignosulfonate is used as emulsifying agent and retarder. Such cements
have applicability in both primary and remedial cementing.
0.
Resin Cements: Proper combination of synthetic resins, water, and portland cement
are often used to provide an improved formation-cement are often used to certain remedial
operations. Cost prohibits use of this material for routine cementing of casing.
Gypsum Cements: These are special mixtures which have high early strength and
0.
easily controlled setting times. Gypsum is the basic ingredient. Their principal use is to
provide temporary plugs during testing and remedial work.
9. Discuss the auxiliary cementing equipment.
Auxiliary Cementing Equipment
A discussion of certain auxiliary equipment items (Figures 3.6, 3.7) will serve to illustrate other
factors influencing attainment for a successful primary cement job.
1.Cementing Plugs. The cementing process in Figure 3.2 showed the use of two cementing plugs.
The main function of the bottom plug is to wipe the casing wall free of mud film and prevent
build up of any thick, tough, cement-mud deposit on the casing wall which may be bypassed by
the top plug. Such deposits can reduce pipe i.d. enough to interfere with subsequent passage of
tools. Contamination of cement by mud at the interface is a lesser consideration. The top plug
serves to signal the proper
Fig.3.6. Typical cementing plugs. Courtesy Halliburton Oil Well Cementing
Company
placement of the slurry, as was mentioned earlier. It also prevents commingling of cement and
displacing fluid; in this case, this is quite important since strong, undiluted cement is especially
desirable in the vicinity of the casing, this is quite cement and displacing fluid; in this case, this is
50
quite important since strong, undiluted cement is especially desirable in the vicinity of the casing
shoe. The volume of cement left inside the pipe is usually small and leaves little safety factor for
contamination or dilution. The proper use of cementing plugs will greatly reduce the frequency of
occurrence of the problems mentioned, and thereby reduce primary cementing failures from such
causes.
2.Wall Scratchers: It is evident that a strong, positive bond between cement and formation is
desirable. Obviously, the cement must contact the formation and hence, the mud cake must have
been removed. Howard and Clark have shown that the erosive or scouring action of the cement
does not adequately perform this function, and that other means are required. The most common
practices is use of mechanical wall scratchers of either the rotating (see Figure 3.7) or
reciprocating type. These are tack-welded onto the string at the desired level(s) as determined
from well logs. The spring loaded steel fingers literally dig into and loosen the wall cake as the
pipe is rotated, allowing intimate contact between cement and the scratched area. Early use of
these devices often resulted in sticking of the pipe indicated by excessive rotating torque), which
was generally due to rapid dehydration (flash set) of the cement as it contacted porous formations.
This can be largely eliminated by reduction of the water loss from the slurry. Sine the pipe is
rotated at the desired setting depth, there is actually no cause for alarm if it does stick, provided,
of course, that reasonable care is exercised in applying torque. Scratchers are generally used on oil
strings only, as it is on these strings that precise segregation of zones is necessary. Canon 24 has
cited a study in which remedial cementing frequency due to primary cementing failure was
reduced from 0.58 to 0.16 remedial jobs per oil string by use of rotation type wall scratchers.
Scratcher use may be advantageous on the strings, depending on the circumstances. Selection of
intervals to be scratched depends on the pay section characteristics. Segregation of water zones,
high gas-oil ratio or gas zones, and strata within the oil zone having greatly different
permeabilities are the usual goals. Common sense along with well log inspection will show the
best scratcher spacing pattern. There is, of course, no point in scratching dense impermeable
zones where no mud cake exists. Caliper logs are also helpful in some instances for showing
enlarged zones in which the scratcher fingers cannot reach the wall and are therefore useless.
Temperature surveys consistently show that lower fill-up height is obtained when scratchers are
used. This clearly implies that mud cake removal occurs.
3.Casing Centralizers: These devices are designed to provide a reasonably uniform cement layer
around the pipe. This is the source of their name: they centralize the pipe in the hole and prevent it
from lying against the wall. Several designs are in use although the hoop skirt type (Figure 3.7) is
quite common. These have hinged end bands which allow them to be readily slipped around the
pipe. Sufficient clearance is allowed so that the pipe may rotate within the centralizer. Either
casing coupling or various types of attachable stops are used to prevent excessive vertical travel of
the centralizer. Centralizer spacing should be sufficiently close to keep pipe-wall clearance at
some acceptable minimum vale. A few, properly spaced centralizers are a valuable aid in
obtaining a uniform cement sheath over critical hole intervals. Again the caliper log may be used
as an additional aid to centralizer location.
4.Floating Equipment. This commonly consists of a guide shoe and float collar (Figure.3.8). The
round nosed guide shoe helps prevent the casing from catching or banging up as it is lowered into
the well. The float collar is generally inserted one to three joint above bottom, where it serves as
back pressure valve preventing backflow of cement after placement Float valves are also
incorporated into the guide shoe, in which case it is called a float shoe. Conventional floating
equipment prohibits mud
Fig 3.7 Auxiliary cementing equipment. (A) Rotation type wall scratchers. (B)
Function of casing centralizer.
Fig. 3.8. Conventional floating equipment. Courtesy Baker Oil Tool. Inc(A) Guide
51
shoe. (B) Float collar.
Fig.3.9. Operating diagram for one type of automatic full-up float shoe. Courtesy
Halliburton Oil Well Cementing Company.
entry into the casing as it is run, hence the term, floating. While this lessens the hook load, it also
increases down hole pressure surges and causes frequent delays in the casing job for filling the
pipe. These disadvantages are overcome by use of automatic fill-up equipment which allows
either partial or complete mud entry into the casing during while still retaining the back pressure
valve feature after cementing. One such device is shown in Figure.3.9. The internal parts of all
floating equipment are drillable.
5.Packer Type Cementing Shoes : In instances where casing is to be set some distance off bottom,
it is often desirable to segregate the open hole zone(s) from the hydrostatic pressure of the cement
column. Packer shoes allow this, as is shown by the operating diagram of Figure.3.10. A casing
shoe swivel run directly above the packer shoe permits pipe rotation if desired.
Fig 3.10 Operating diagram of typical packer type cementing shoe. Courtesy
Halliburton Oil Well Cementing Company. (A) Casing on bottom. Setting
ball falling to the setting cage. (B) Packer Set. Cementing job underway.
(C) Job completed. Top Plug at shut off position in top of packer.
10. Discuss briefly casing types and specifications.
Casing Types and Specifications
Some expansion of the general casing functions as applied to specific strings is in order.
Surface Strings: In hard rock areas, a single string of surface pipe set a few hundred feet
deep is usually adequate to:
(0)
Control caving and washing out of poorly consolidated surface beds
(0)
Furnish a means of handling the return flow of drilling mud
(0)
Protect fresh water sands from possible contamination by drilling mud, or
oil, gas, and/or salt water from lower zones
(4)
Allow attachment of blowout preventors
In soft rock areas, it is often necessary to set both a conductor string and a surface string to
perform these functions. In such instances, the conductor string may be two or three hundred feet
long with the surface pipe extending to a depth of two or three thousand feet. Some surface pipe is
used on all wells of appreciable depth and is in fact required by law in many states. The main
purpose of such legislation is to insure protection of fresh water supplies, as stated in function (3).
Intermediate Strings: The decision to use intermediate strings is largely dependent on well
depth and geologic conditions in a specific area. Wells of moderate depth may use no
intermediate pipe, while deep wells may require one or more protective strings. The principal
function of intermediate casing is to seal off troublesome zones which
(1) contaminate the drilling fluid, making mud control difficult and expensive (salt,
gypsum, heaving shales, etc.)
(2) jeopardize drilling progress with possible pipe sticking, excessive hole
enlargement, or other fishing hazards.
In deep drilling, there is often a reluctance to drill with excessive intervals of open hole even
though no particularly troublesome zones are exposed. In such instances, an intermediate string
may be run as a precautionary measure so that any ensuing problems may be solved more easily.
The intermediate pipe also affords greater safety in case of blowouts.
Oil Strings: This is the last and deepest casing string run. It is set above, (Figure 3.1), part
way through, or completely through the lowermost pay zone, depending on the type of completion
to be performed. The oil string furnishes the means of segregating the pay section from all other
zones and is the work shaft through which access to the producing zone is gained. Normally, the
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produced fluids flow through tubing; sometimes, however, the oil string may actually serve as the
flow conduit.
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