ch2 rig components (student copy)
TRANSCRIPT
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PETROLEUM ENGINEERING DEPARTMENT
UNIVERSITI TEKNOLOGI PETRONAS
DRILLING ENGINEERING by Muhammad Aslam Md Yusof 1
Chapter 2: Rig Components
POWER SYSTEM
ENGINE TERMINOLOGY
Horsepoweris a unit of work established by James Watt in 1870. Watt defined one horsepower as
the ability to do 33,000 ft-lbf of work per minuteor 550 ft-lbf of work per second. This is also the
power necessary to raise one ton a distance of feet in one minute, assuming no friction.Engine horsepoweris a measure of the theoretical characteristics of an engine. Brake horsepower
(BHP) is the actual horsepower of the engine and also the only measurable horsepower, after
deducting the internal friction losses of an engine. It is usually determined from the force exerted on
a friction brake or dynamometer connected to the drive shaft.
Torqueis defined as a force around a given point, F applied at a radius, R from that pointor in much simpler word as the twisting force of the engine. Note that the unit of torque, T is
.
Efficiency factor(describes the power losses from the prime movers to the end-use equipment,which can be expressed as follows:
Where energy output is from the prime mover and energy input is the amount remaining for actual
usage after some losses. The system losses result from friction, gears, and belt and line losses.
Fuel TypeDensity
Heating Value Diesel 7.2 19000
Gasoline 6.6 20000
Butane 4.7 21000Methane - 24000
Table 2.1: Heating value of various fuel types
POWER SYSTEM CALCULATIONS
Generally, the main characteristics of power system performance are engine output or brake
horsepower (BHP), heat energy input, overall power system efficiency and fuel consumption for
different engine speeds.
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From basic calculations, work, W is the product of force, F and distance, D ;
The power, P , can be expressed as;
Where t is time .For a rotating shaft;
Where N is number of revolutions per minute (rpm)
Thus, the horsepower, HP (hp) is given by
Hence, to rearrange the equation it give the engine output or BHP, BHP (hp);
Where
angular velocity
, T is the output torque,
Or
Heat energy input, (hp);
Where is fuel consumption rate in and H is the heat value of fuel in .The overall efficiency factor is defined as the energy output over energy input and yields;
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Example 2.1
Based on the data given, determine the output power, BHP and overall efficiency, Output torque from a diesel engine, Engine speed, Fuel type = DieselFuel consumption rate, Unit conversion factors:
Answer:
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PETROLEUM ENGINEERING DEPARTMENT
UNIVERSITI TEKNOLOGI PETRONAS
DRILLING ENGINEERING by Muhammad Aslam Md Yusof 4
Example 2.2
In a drilling operation major power supply is required by the hoisting system, rotary system,
circulation system and mud cleaning equipment. If the output horsepower required by the top drive
is 2200 hp, the hoisting system is 2000hp, the circulation system 2400hp, cleaning equipment 300hp
and 20% of all is required for additional facilities, calculate the diesel consumption for a 25 day
drilling operation and casing operation (in gallons). The overall engine efficiency,, is 15% andtransmission efficiency, , is 85%. Other data;
Heat value of fuel, Density of fuel,
Answer:
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DRILLING ENGINEERING by Muhammad Aslam Md Yusof 5
EXERCISES
1. Calculate the overall efficiency, , for the following conditions:A diesel engine has an output torque, Engine speed, Density of fuel, Fuel consumption, Heat value of fuel,
Answer: 13.3%
2. A diesel engine gives an output torque of at an engine speed of . Ifthe fuel consumption rate was , what is the output power and overall efficiency ofthe engine?
Answer: 331.29 hp and 19.24%
3. The total power consumed by different equipment on a land rig is 7000hp. If the rig is
operating for a period of 25 days, how much diesel fuel will be required? The overall engine
efficiency and transmission efficiency are 20% and 95% respectively.
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CIRCULATING SYSTEM
MAJOR COMPONENTS
Circulating system can be described as the blood of the drilling operation. This is because, all its
major components especially drilling mud is flowing throughout the entire drilling operation. It
travels from (1) the mud tanks (2) to the mud pumps, (3) through the surface connections, (4) to the
drill string, (5) then to the bit, (6) up the annulus, (7) to the mud cleaners and (8) back to the suction
tank. This is to remove drill cuttings from the hole as the drilling progresses and also to clean the
wellbore, to cool and to lubricate the bit and also to transmit hydraulic horsepower to the bit. The
major components of the circulating system are;
Mixing equipment
Pits
Mud cleaners
Mud pumps
MIXING EQUIPMENT
The drilling mud mixing equipment;
Hopper; mud chemicals and solids are poured through the hopper, which is connected to a
high shear jet. Due to the jet action, the mud chemicals are homogeneously mixed.
Mud gun & Agitators; the resulting mud is again vigorously agitated with a mud gun and
then directed to the suction tank. In the suction tank, mud is further kept in motion with
mechanical agitators. This is done to prevent solid settlement. Before the mud is delivered
to the main pump it is charged by a centrifugal pump to give the mud a pressure of 80-90
psi.
PITS
It is usually a series of large steel tanks, all interconnected and fitted with agitators to maintain the
solids, used to maintain the density of the drilling fluid, in suspension. Some pits are used for
circulating (e.g. suction pit) and others for mixing and storing fresh mud. Most modern rigs have
equipment for storing and mixing bulk additives (e.g. barite) as well as chemicals (both granular and
liquid). The mixing pumps are generally high volume, low pressure centrifugal pumps.
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MUD CLEANERS
Solid control or cutting separation can be done by using one or more methods of solid separation;
Settling
Screening
Hydrocyclone
Rotating centrifuge
While settling and screening methods are the mainly the mechanism of shale shaker by using
gravitational force, hydrcocyclone and centrifuge, however, achieve higher rates of separation by
using centrifugal forces than can be obtained by gravitational settling. Cutting can divide into four
sizes which are;
Category Size Example
Colloidal 2 or less Bentonite, clays and ultra-fine drill solids
Silt 2 to 74 Barite, silt and fine drill solids
Sand 74 to 2000 Sand and drill solids
Gravel More than 2000 Drill solids, gravel and cobble
Shale Shaker
The most important cutting separation equipment is shale shaker, a vibrating screen separators used
to remove drill cuttings from the mud. It is the first line of defense against solids accumulation in
mud cleaning equipment or solid removal chain. A modern shale shaker works by removing about
100% of solids greater than 74microns. The shale shaker cannot remove silt and colloidal size solids.
Thus, it requires dilution and other equipment to control the ultra-fine drill solids.
There are three basic types of shale shakers that are widely use in the industry which are;
The circular-motion shaker; an older design and produces lowest centrifugal force, or G-
force.
The elliptical-motion shaker;a modified circularmotion type in which the center of gravity
is raised above the deck and counter weights are used to produce an egg-shaped motion
that varies in both intensity and throw as solids move down the deck.
The liner-motion shaker;use two circular-motion motors mounted on the same deck. The
motors are set for opposite rotation to produce a downward G-force and an upward G-force
when the rotations are complementary, but zero G-force when the rotations are opposed.
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Sand trap
It is a settling pit located directly after the shale shaker. It can gather the larger particles that can
plug or damage the next solid removal equipment if a screen develops a hole of if the shaker is
bypassed. It use gravitational force acting on the particle for settlement, thus this compartment
should never be stirred or used as suction for discharge for hydrcyclones.
Hydrocyclone (Desander and Desilter)
Hydrocyclone is a static device, closed vessel that applies centrifugal forces to incoming liquid so
that it can classify or separate heavy and light components in the liquid. It is designed to convert
incoming liquid velocity into rotary motion. A centrifugal pump feeds a high volume of mud
through a tangential opening into the large end tunnel-shaped hydrocyclone. When the proper
amount of head pressure is used, it spins the mud in the cyclone and creates centrifugal force in
the mud. A a result, higher mud solids move outward toward the wall of the cyclone where they
agglomerate and spiral down the wall to the outlet at the bottom of the vessel. Light components
like liquid, however, move toward the axis of the hydrocyclone where they move up toward the
outlet at the top. Thus, all hydrocyclones operate in a similar manner, whether they are used as
desanders, desilters or clay ejectors.
Desander is required to remove those abrasive solids in the mud that can be done by the shale
shaker and also to avoid over load on the desilters. It normally use 6-in internal diameter or larger.
Larger desander has the advantage of larger volumetric capacity (flow rate) per hydrocyclone but
has some disadvantage of making wide particle size cuts in range of 45 to 74 micron. In order to get
the best result, it is required to have proper head pressure at the inlet.
Normally, a 4-in internal diameter hydrcyclone is usually used for desilting. Desilter is needed to
remove cuttings with size of 15 to 35 microns. Since barite falls into the same size range as silt
(refer to the cutting size table above), it alos will be separated from the mud by desilter. Because of
this, desilter is rarely used on the mud weight above than 12.5ppg. Moreover, desander and
desilter are widely used while drilling surface hole and when the low density mud is used.
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Mud Cleaner
A mud cleaner, basically will remove sand size drill solids from the mud and it somehow can retain
the barite. Firstly, the mud is passing through the desilter, and then screens the discharge through
a fine-mesh shaker. The larger solids trapped on the screen are discarded leaving the mud and
solids that pass through return into the circulation. Generally, 97% of barite particles are less than
74-micron in size. Thus, most of the barite will be discharged by the hydrocyclone and moving
through the screen and returned into the mud system. In other words, we can say that mud
cleaner is a backup to the shale shakers and in order for a mud cleaner to be an effective cutting
separator equipment, the screen size must be finer than the screen size on the shale shaker.
Centrifuge
Basically, a centrifuge is solid-control equipment that able to remove fine and ultrafine solids. In
normal industrial practice, a decanting-type centrifuge is used to increase the forces causing
separation of the solids by increasing centrifugal force. It consists of a conical, horizontal steel bowl
rotating at a high speed, with a screw shaped conveyor inside. Mud feeds into one end and the
cuttings are separated and moved up the bowl by a rotating scroll to exit at the other end. Even
though it principle is quite similar to hydrocyclone, there is a key difference between these two
equipment. The centrifuge is dynamic separator that is able to apply much more centrifugal force
than hydrocyclone thus remove finer materials while the hydrocyclone is passive separator capable
of applying modest amounts of centrifugal force.
On the rig site, this mud cleaning equipment arrangement is really important to get maximum
separation efficiency. The mechanical equipment is set up based on the particle size that it will
remove. In addition, although a degasser or mud-gas separator is technically a not solid removal
device, it needs to be located after the shale shaker since the centrifugal pumps and other solid
control equipment do not operate efficiently in existence of gas in the mud. Thus, the proper
arrangement for mud cleaning equipment is per below;
1. Shale shaker
2. Sand trap
3. Degasser
4. Desander
5. Desilter
6. Mud Cleaner
7. Centrifuge
8. Cutting dryer (if using oil-based mud, to recover the mud before the dumped the cuttings)
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Figure 1: Example of mud cleaning system (courtesy of
https://reader009.{domain}/reader009/html5/0328/5abba8e0621ff/5abba8e70ac1c.jpg)
MUD PUMPS
At least two slush pumps are installed on the rig. At shallow depths they are usually connected in
parallel to deliver high flow rates. As the well goes deeper the pumps may act in series to provide
high pressure and lower flow rates. Positive displacement pumps are used (reciprocating pistons) to
deliver the high volumes and high pressures required to circulate mud through the drill string and up
the annulus. There are two types of positive displacement pumps which are;
Duplex pump (two cylinders)double acting (pump on the up-stroke and down-stroke)
Triplex pump (3 cylinders)single acting (pump on the up-stroke only)
Triplex pumpor three-cylinder pump is a single acting pump only on forward piston stroke. It means
that when the piston moves forward, it will pump and suction happen when the piston moves in
backward direction. Triplex pump system needs three different cylinders, one each while the duplex
pump requires only single cylinder unit. This pump are the most common type of pump used in well
service activities since it require less maintenance, light, compact, cheaper than duplex pump and
generally is capable of handling many fluid types including corrosive fluids, abrasive fluids and
slurrys containing relatively large particulates.
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Duplex pumpor two cylinders pump is a double acting pump. For every movement of the piston it
will result in pumping and suction. For example, while the piston moves forward, both suction and
pumping occur simultaneously and separately at both ends. Same thing goes when the piston moves
backward. Even though the duplex pump is cheaper and has higher efficiency than triplex pump, it
requires high cost of maintenance and also difficult to maintain.
Figure 2: Triplex (three-cylinder pump) design
Figure 3: Duplex (two-cylinder pump) design
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CIRCULATING SYSTEM CALCULATIONS
Theoretical displacement on the forward stroke, the volume displaced is;
Where is liner diameter and is stroke length . Similarly, on the return stroke;
( ) Where is piston diameter Taking into account of both strokes, the total volume displaced for a complete cycle by a pump,
can be expressed as; ( ) For a triplex pump, the piston diameter does not affect the pump output because the piston
discharges in one direction. The total volume displacement for single-acting triplex pump can be
expressed as;
() Considering a volumetric efficiency, the total volume displacements for both pumps are;
() ( ) () ()
The pump displacement per cycle is also known as pump factor. The pump flow rate, is the product of total volume displacement per cycle, and pump speed, in cyclesper minute (in oilfield, 1 complete pump revolution = 1 stroke, hence pump speed is usually given in
strokes per minute).
Hydraulic horsepower of a pump, is given by;
Where is pump pressure
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Example 2.3
Calculate the pump factor in units of barrel per stroke for a duplex pump below
Stroke length 17
Rod diameter 2.5Liner diameter 6.5
Volumetric efficiency 90%
Unit conversion factors:
Answer:
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Example 2.4
A triplex pump with 6.5 liners, 2.5 rods and 18 strokes is operating at 150 cycles per minute and a
discharge pressure of 4000 psig.
a) Determine the pump factor in gal per cycle at 100% volumetric efficiency
b) Calculate the flow rate in gal per min
c) Compute the pump power developed during the operation
Answer:
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Example 2.5
A double-acting duplex piston pump with 6.5 liner, 2.5 rod and 17 stroke was operated at 4,000
psig and 20 cycles/min for 10 minutes with the suction pit isolated from the return mud flow. The
mud level of the suction pit, which is 6.5 ft wide and 20 ft long, was observed to fall 20 during this
period. Compute (a) volumetric pump efficiency, (b) pump factor and, (c) the hydraulic horsepower
developed by the pump.
Answer:
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EXERCISES
1. A triplex pump with 16 in stroke is operating at 4000psig and 20 cycles per minute. If the
volumetric efficiency of the pump is 80% and the horsepower of the pump is 350hp.
Calculate the piston/liner diameter for this pump.
Answer: 7.579 in
2. A double-acting duplex pump with 6.5 liners, 2.5 rods, and 18 strokes was operated at
3000 psig and 20 cycles/min for 10 minutes with suction pit isolated from the return mud
flow. The mud level in the suction pit, which is 7 ft wide and 20 ft long, was observed to fall
18 during this period. Compute the pump factor, volumetric pump efficiency, and hydraulic
horsepower developed by the pump.
Answer: 7.854 gal/cycle; 0.82; 274.9 hp
3. Consider triplex pump having 6-in liner and 11- in, strokes operating at 120 cycles/min and a
discharge pressure of 3000 psig. Calculate (1) pump factor in units of gal/cycle at 90%
volumetric efficiency (2) flow rate in gal/min (3) hydraulic horsepower developed.
Answer: (1) 3.635 gal/cycle (2) 436.2 gal/min (3) 763.5 hp
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HOISTING SYSTEM
MAJOR COMPONENTS
The main function of hoisting system is provide means of lifting and lowering drill string, casing
string, and other subsurface equipment into the wellbore during drilling activities and also helps the
replacement of the drilling line when it was subjected to wearing. Two routine works carried out by
using the hoisting system are making a connection and making a trip.
The major components of hoisting system are;
Drawworks
Derrick and substructure
Block and tackle
DRAWWORKS
Drawworks, primary hoisting machinery can be described as the heart of the rig. It provides a
means of raising and lowering the travelling block. The main components of modern drawworks are;
Hoisting drum & motor
Brakes (main brake and auxiliary brake)
Transmission
Catheads
The drawworks consist of wire-rope drilling line winds on the large revolving drum and extends over
a set of sheaves in the top of derrick, known as the crown block and down to another sheaves
known as travelling block. A large drilling hookis suspended at the end of the travelling block use to
suspend the drill string. This is allowing the drill string to be moved up and down as the drum turns.
There are two segments of the drilling line known as fast line and dead line. The fast line is the
wire rope drilling line that winds on the drawworks drum to the crown block. The drilling line then
enters the sheaves of the crown block and is makes several passes between the crown block and
traveling block pulleys for mechanical advantage. One end of the line then exits the last sheave on
the crown block and is anchored to a derrick leg on the other side of the rig floor. This section of
drilling line is called the "dead line" because it does not move.
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Figure 4: Schematic design of hoisting system (After Adam T Bourgouyne et al, 1991)
BLOCK AND TACKLE
Block and tackle is the main link between the drawworks and pipe or casing and consists of;
Crown block
Travelling block
Drilling lines (fast line and dead line)
The main principal function of block and tackle is to provide a mechanical advantage, M to allow
easier handling of large loads. It can be defined as;
Where is the load supported by travelling block or hook load and is the load imposed on thedrawworks (tension in the fast line).
Under a static condition, consider a free body diagram in Figure 4, in ideal mechanical advantage
condition where there is no friction in pulleys, the tension in the drilling line is constant. So, hook
load, is supported by the fast lines (i.e. by each of the strung lines through the travelling block). Itgives;
Or
Where is the number of lines strung throughout the crown block and the travelling block. Thenumber of lines is chosen based on the loading condition.
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Table 1: Average efficiency factors for block and tackle system (After Adam T Bourgoyne et al)
Number of lines, Efficiency, 6 0.874
8 0.841
10 0.810
12 0.770
14 0.740
Note that dead-line load is carried by dead-line anchor and the fast-line load is carried by the
hoisting drum.
The input power, of the block and tackle is given as;
Where is the fast-line velocityThe output power or power supplied to the hook, can be expressed as;
Where is the travelling block velocity. The movement of the fast line by a unit distance tends toshorten each of the strung lines between the block and tackle by times the unit distance, thus;
Block and tackle efficiency is the ratio between output power to input power. In an ideal condition
where no friction in the block and tackle system, the efficiency if unity.
However, there is power loss due to friction and other factors in real system. Thus, the approximate
value of average efficiency for the block and tackle system can be referred in Table 1. Using the
value, the actual tension in the fast line for a given load can be determined by using;
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Thus, the fast-line tension gives;
Note that after some or total part of the drill string immersed in the drilling fluid, the weight of the
drill string is reduced because there is upward force that keeps the drill string afloat. Buoyancy
factoris the factor used to consider loss of weight because of the immersion in drilling mud.
Whereis the mud weight in ppg andis the steel density either or .Moreever, under a dynamic condition, the fast-line tension and dead-line tension are illustrated in
Figure 5.
The tension in the successive fast-lines is equal to load imposed on the drawworks, times thesheave efficiency factor,and the tension in the last line is the tension in the dead-line, .The normal value for sheave efficiency, is 0.9615.
So, the load supported by the travelling block, can be expressed as;
Solving this, the tension in the fast-line yields, ;
Considering the block and tackle efficiency, the tension in the dead-line gives;
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Figure 5: Effect of friction on the efficiency of the hoisting system (After H Rabia)
Basically, the number of lines is depending on the load to be supported by the hoisting system and
also the tensile rating of the drilling line used. After certain drilling period, the drilling line becomes
worn and must be replaced. To maintain a good drilling line while drilling, a regular practice known
as slip and cut operation has been practiced in the industry. It is a periodically replacement of a
segment of the drilling line wrapped around the crown block and the travelling block to avoid any
drilling line fatigue. The slipping the drilling lineinvolves losing the dead line anchor and placing a
few feet of new line in service from the storage reel. Cutting the drilling linemeans removing the
line from the drum and cutting off a section of the line from the end.
The slip and cut operation at specific period is determined by the ton-mile calculation. It is a method
introduced to measure the work done by the drilling line by quantifying the cumulative product of
the loaf lifted (in tons) and the distance lifted of lowered (in miles).
Ton-mile is calculated for
Round trip
Drilling or connection
Setting casing
Coring
Short trip
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ROUND TRIP TON-MILE
It is total of the work done by travelling block, drill pipe and drill collar when running and pulling the
drill string from the well.
i. work done by travelling assembly, Where is the weight of travelling assembly, is the length of each stand and is the number ofstand.
Note that to run a stand in the hole, the travelling block has to move a distance of
approximately. Similar amount of work is required to pull the stand out of the hole. Thus totalwork done by travelling assembly is given by;
Assuming , so
ii. work done by drill pipe,
Where
Taking into account the buoyancy factor affecting the weight of drill pipe, the work done by drill pipe
is;
[( ) ( )] Where is buoyant weight of the drill pipe.Assuming the friction loss is the same in going into the hole as in coming out, the work done in lifting
the drill pipe is the same as in lowering, so for a round trip, the work done is equal to
[( ) ( )]
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iii. work done by drill collar,
Where
is the correction for the added work done in lowering and lifting assembly because drill
collars and bit weigh more per foot than drill pipe (previously it is assumed the weight of drill pipe
was run to bottom of the hole).
( )
It yields,
( )
Thus, the total work done for a round trip is given by
( )
Rearrange the equation it gives;
Or
DRILLING OPERATIONS TON-MILE
To calculate the ton-mile in drilling operations,
it has to consider the work performed
during the operation;
a) drill ahead length of the Kelly
b) pull up length of the Kelly
c) ream ahead length of the Kelly
d) pull up length of the Kelly to add single or double
e) put Kelly in rat hole
f) pick up single or double
g) lower drill stem in hole
h) pick up Kelly
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Note that for operations (a) and (b) equal for 1 round trip, operations (c) and (d) equal for 1 round
trip, operations (e), (f) and (h) is equal to round trips, and operation (g) equals to round trips.
If thus the ton-miles for drilling operations;
Where is the ton-miles for one round trip at depth (shallower depth) and is the ton-milesfor one round trip at depth (deeper depth).If operations (c) and (d) are omitted, it gives
If top drive is used,
If reaming is done between connections with a top drive then,
SETTING CASING OPERATIONS TON-MILE
The ton-miles for setting casing operations,is given by;
Since no extra weight from the drill collar, the ton-miles for setting casing operations;
Where is the buoyant weight of casing and is the casing joint length.SHORT TRIP OPERATIONS TON-MILE
The ton-miles for short trip operations, is given by;
Where is the ton-miles for one round trip at depth (shallower depth) and is the ton-milesfor one round trip at depth
(deeper depth).
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PETROLEUM ENGINEERING DEPARTMENT
UNIVERSITI TEKNOLOGI PETRONAS
DRILLING ENGINEERING by Muhammad Aslam Md Yusof 26
DERRICK AND SUBSTRUCTURE
Derrick or Mast is an elevated structure (usually of metal construction) on a drilling rig to provide the
vertical height necessary for the hoisting system to raise and lower the pipe. All the drilling rigs
construction must follow the requirement API standard 4A. This is to ensure the rig is capable to
handle all loads, including maximum drilling loads and weight of pipe set in the derrick. In the
meantime, the derrick also must be able to withstand winds loads acting horizontally and also the
compressive loading. In industrial practice, the selection of drilling rig is based on the rig usage such
as drilling activities, workover or servicing.
Figure 6: API standard 4A derrick size classification
For the normal drilling operation, the pipe is requires to be handled in stands (3 joint of connected
pipe is about 90 ft height). Hence, for older rig usually requires 123 ft of derrick height.
However, in the modern drilling operation nowadays, the height of derrick normally requires
additional height to accommodate the top drive system. Most widely used size is API 19 which
provides 146ft height and 30ft square base with pipe racking capacity of 160 stands of 5ft drill pipe.
The derrick can be classed into two types which are;
Standard derrick
Portable derrick and mast
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PETROLEUM ENGINEERING DEPARTMENT
UNIVERSITI TEKNOLOGI PETRONAS
DRILLING ENGINEERING by Muhammad Aslam Md Yusof 27
The load applied on the derrick,is the total of hook load,, the fast-line tension,and the dead-line tension, and the;
In a static condition, given that the fast-line tension is and dead-line tension is ityields;
Noted that the load on the rig is not distributed equally over all four derrick legs because the dead
line is anchored to one of the leg and the drawworks is situated on one side of the derrick floor. The
load distribution for each derrick leg is shown in the Table 2.
Figure 7: Illustration of drilling lines on the rig floor
The dead-line affects only the leg A to which it is anchored. Besides that, the leg C and leg D would
share the full load of fast-line tension. The leg B only loaded with the hook load during the drilling
operation. It is important to know that at block and tackle efficiency exceed 0.5, the load exerted on
leg A is greater than other legs. Since the derrick will fail if any leg failed, it is convenient to define
the maximum equivalent derrick load,which is equal to four times the maximum leg load (legA),.
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PETROLEUM ENGINEERING DEPARTMENT
UNIVERSITI TEKNOLOGI PETRONAS
DRILLING ENGINEERING by Muhammad Aslam Md Yusof 28
Table 2: Distribution of load on the derrick leg
Load on each derrick leg
Load
sourceLoad Leg A Leg B Leg C Leg D
Hook
Load
Fast-line
- -
Dead-
line - - -
Total load
( )
( )
Noted that derrick efficiency, is used to evaluate various drilling line arrangement.
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PETROLEUM ENGINEERING DEPARTMENT
UNIVERSITI TEKNOLOGI PETRONAS
DRILLING ENGINEERING by Muhammad Aslam Md Yusof 29
Example 2.6
A rig must hoist a load of 450,000. The drawworks can provide an input power to the block andtackle system as high as 450 . Eight lines are strung between the crown block and travelling block.Calculate:
a) The static tension in the fast line when upward motion is impending,b) The maximum hook horsepower available,c) The maximum hoisting speed,d) The actual derrick load,e) The maximum equivalent derrick load,f) The derrick efficiency factor,
Answer:
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PETROLEUM ENGINEERING DEPARTMENT
UNIVERSITI TEKNOLOGI PETRONAS
DRILLING ENGINEERING by Muhammad Aslam Md Yusof 30
Example 2.7
Ten lines are strung between the crown block and travelling block, and the dead line is anchored to a
derrick leg on one side of the v-door, as shown in the figure below. The rig must hoist a load of
500,000, calculate the load distribution on the derrick legs A, B, C and D.Answer:
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PETROLEUM ENGINEERING DEPARTMENT
UNIVERSITI TEKNOLOGI PETRONAS
EXERCISES
1. Calculate the fast-line tension when lifting a 500,000 load for 6, 8, 10, 12 and 14 linesstrung between the crown block and travelling block.
Answer: , , , , ,2. A rig must hoist a load of 200,000. The drawworks can provide an input power to the
block and tackle system as high as 800 . Eight lines are strung between the crown blockand travelling block. Calculate:
a) The static tension in the fast line when upward motion is impending,Answer: 24961
b) The maximum hook horsepower available,(Answer: 648 hp)c) The maximum hoisting speed, (Answer: 106.9 )d) The actual derrick load, (Answer: 244691)e) The maximum equivalent derrick load,(Answer: 280000)f) The derrick efficiency factor, (Answer: 0.874)
3. Eight lines are strung between the crown block and travelling block, and the dead line is
anchored to a derrick leg on one side of the v-door, as shown in the figure below. The rig
must hoist a load of 400,000, calculate the load distribution on the derrick legs A, B, C andD.
Answer: , ,
REFERENCES
1. Bourgoyne, A.T., M. E. Chenevert, K. K. Millheim, and F. S. Young. 1986. Applied Drilling
Engineering. SPE Tectbook Series, Vol. 2. Crittendon, B. C. 1959.2. https://reader009.{domain}/reader009/html5/0328/5abba8e0621ff/5abba8f73e577.jpg