Rig Sizing - Text

Rig Sizing 1.

Fundamentals .................................................................................................................................... 2 1.1. Types of Rigs ............................................................................................................................ 2 2. Hoisting System................................................................................................................................ 3 2.1. Hoisting Design Considerations ............................................................................................... 5 2.2. Power Requirements of the Drawworks................................................................................... 7 3. Drilling Line Design Considerations .............................................................................................. 12 3.1. Ton-Miles (Mega joules) Of A Drilling Line ......................................................................... 12 3.2. Evaluation Of Total Service And Cut-Off Practice................................................................ 15 4. ROTATING EQUIPMENT............................................................................................................ 16 5. Circulating System.......................................................................................................................... 17 5.1. Volumetric Efficiency ............................................................................................................ 18 5.2. HORSEPOWER ..................................................................................................................... 18 5.3. Pump Output ........................................................................................................................... 19 5.4. Pump Factors .......................................................................................................................... 19 5.5. Centrifugal Pumps .................................................................................................................. 20 5.6. Mud Handling Equipment ...................................................................................................... 20 6. Pressure Cont rol Equipment ........................................................................................................... 21 7. Derrick Capacity and Substructure ................................................................................................. 23 8. Total Power Requirements ............................................................................................................. 27

__________________________________________________________________________________ 1 Network of Excellence in Training

Rig Sizing 1.

FUNDAMENTALS

A drilling rig is a device used to drill, case and cement water, oil and gas wells. The correct procedure for selecting and sizing a drilling rig is as follows: 1. Design the well 2. Establish the various loads to be expected during drilling and testing and use the highest. This point establishes the DEPTH RATING OF THE RIG. 3. Compare the rating of existing rigs with the design load 4. Select the appropriate rig and its components.

1.1.

Types of Rigs

Drilling rigs are classified as ♦ Land rigs ♦ Offshore rigs There are two types of offshore rigs: 1. Floating rigs: ♦ Semisubmersible ♦ Drillships 2. Bottom-supported rigs: There are three types: ♦ Jack-ups ♦ Platform ♦ Barge

The major components that need to be selected and sized for the purpose of rig sizing are: 1. Hoisting 2. Rotating Equipment 3. Circulating System 4. Tubular Goods __________________________________________________________________________________ 2 Network of Excellence in Training

5. Pressure Control 6. Derrick Capacity And Substructure 7. Power Requirements

2.

HOIST ING SYST EM

The hoisting system consists of: (Figure 1) ♦ Drawwork: this is an assembly of a rotating drum, a series of shafts, clutches, chains and gears for changing speed and for reversing, figure 2. It also contains the main brake for stopping the drilling line. The drilling line is wound a number of times around the drum, the end of the line then passes on the crown and travelling blocks. ♦ Crown Block: A block located at the top of the derrick. It contains a number of sheaves on which is wound the drilling line. The crown block provides a means of taking the drilling line from the hoisting drum to the travelling block. The crown block is stationary and is firmly fastened to the top of the derrick. Each sheave inside the crown block acts as an individual pulley. The drilling line is reeved round the crown block and travelling block sheaves with the end line going to an anchoring clamp called “ DEAD LINE ANCHOR”. The static line is called the deadline. The line section connecting the drum with the crown block is called the fastline. Figure 1 Schematic Of The Hoisting System Crown Block Fixed sheaves

W/4

W/4

W/4

W/4

W/4 W/4

Fastline Deadline

Travelling Block Hook

Drawworks

Drilling Lines

Deadline Anchor

W


Hence during hoisting operations, if there are 10 lines between the crown and travelling block, the fastline line travels 10 times faster than the travelling block in order to spool or unspool drilling line from the hoisting drum. The DEAD LINE ANCHOR anchors the last line coming from the crown block and also stores drilling line on a reel. This allows new lengths of line to be fed into the system to replace the worn parts of the line that have been moving on the pulleys of the crown block or the travelling block. The worn parts are regularly cut and removed, Slip and Cut Practice. Slipping the line, then cutting it off helps to increase the lifetime of the drilling line.

♦ Travelling Block: a diamond-shaped block containing a number of sheaves which is always less than those in the crown block. The drilling line is wound continuously on the crown and Travelling blocks, with the two outside ends being wound on the hoisting drum and attached to the deadline anchor respectively, figure 3. ♦ The Hook: connects the Kelly or topdrive with the travelling block. The hook carries the entire drilling load, figure 3. ♦ Drilling Line The drilling is basically a wire rope made up of strands wound around a steel core. Each strand contains a number of small wires wound around a central core. The drilling line is of the round strand type with Lang’s lay. The drilling line has a 6x19 construction with Independent Wire Rope Core (IWRC). This construction implies that there are 6 strands and each strand containing 19 filler wires. The size of the drilling line varies from ½ “ to 2 “.

Figure 2 Drawworks, Courtesy of National Oilwell


Figure 3 Travelling Block and Hook, Courtesy of National Oilwell

2.1.

Hoisting Design Considerations

The procedure for carrying out hoisting design calculations are as follows: 1. Determine the deepest hole to be drilled 2. Determine the worst drilling loads or casing loads 3. Use these values the select the drilling line, the derrick capacity and in turn the derrick 1.

Static Derrick Loading

Static derrick loading (SDL)= fast-line load + hook load + dead-line load Referring to figure 1 and for a system consisting of four lines supporting the hook load, then under static conditions: Fast- line load (FL) = Hook load /4 Dead-line load (DL) = Hook load /4

Hl HL 3 + HL + = HL 4 4 2

Equation 1 __________________________________________________________________________________ 5 Network of Excellence in Training SDL =

For N lines, the static derrick load is given by: =

( N + 2) HL N

Equation 2

where N = number of lines strung to travelling block HL = hook load 2.

Efficiency Of The Hoisting Systems

a.

Efficiency Factor (EF) Of The Hoisting System – Hoisting Operations

EF =

K (1 − K n ) N (1 − K )

FL =

HL NxEF

Equation 3 Equation 4

where K = sheave and line efficiency per sheave Deadline- load is given by: HL x KN DL = –––––––––– N x EF

Equation 5

If the breaking strength of the drilling line is known, then a design factor, DF, may be calculated as follows: nominal strength of wire rope (lb) DF = ––––––––––––––––––––––––––––––– fast-line load (lb) B.

Equation 6

Lowering Operations

During lowering of pipe, the efficiency factor and fastline load are given by


N x KN x (1 - K) (EF) LOWERING = ––––––––––––––– (1-KN)

Equation 7

W x KN x (1 - K) (FL) LOWERING = –––––––––––––– (1-KN)

Equation 8

Example 1 Hoisting System Efficiency Factor (Note all design calculations require this number) Calculate the efficiency factor for a hoisting system employing 8 string lines. Assume the value of K to be 0.9615. Solution K x (1 - KN) EF = –––––––––––– N x (1 - K) 0.9615 (1 - 0.9615 8 ) = –––––––––––––––––– 8 (1 - 0.9615) = 0.842 Table 1 can be constructed for different numbers of lines strung between the crown and travelling blocks. TABLE 1 Block And Tackle Efficiency Factors For K = 0.9615 Number of lines strung 6 8 10 12

2.2.

Efficiency factor 0.874 0.842 0.811 0.782

Power Requirements of the Drawworks

As a rule of thumb, the drawwork should have 1 HP for every 10 ft to be drilled. __________________________________________________________________________________ 7 Network of Excellence in Training

Hence for a 20,000 ft well, the drawwork should have 2000 HP. A more rigorous way of calculating the horse power requirements is as follows: a)

Velocity of fastline load (Vf) Vf = N x VL

Equation 9

where VL = velocity of travelling block N = number of lines strung b)

Power output at drum = FL x Vf (HL) x VL P = --------EF

Equation 10

In the Imperial system, power is quoted in horse-power and the above equation becomes: HL x VL Drum output = ------------ horsepower EF x 33,000

Equation 11

Example 2: Hook Loads The following data refer to a 1.5 in block line with 10 lines of extra improved plough steel wire rope strung to the travelling block. hole depth = 10,000 ft drillpipe = 5 in OD/4.276 in ID, 19.5 lb/ft drill collars = 500 ft, 8 in/2,825 in, 150 lb/ft mud weight = 10 ppg line and sheave efficiency coefficient = 0.9615 Calculate: (1) (2) (3) (4) (5) (6)

weight of drill string in air and in mud; hook load, assuming weight of travelling block and hook to be 23,500 lb; deadline and fast-line loads, assuming an efficiency factor of 0.81; dynamic crown load; wireline design factor during drilling if breaking strength of wire is 228,000 lb design factor when running 7 in casing of 29 lb/ft.

Solution (1) Weight of string in air __________________________________________________________________________________ 8 Network of Excellence in Training

=

weight of drillpipe + weight of drill collars

=

(10,000 - 500) x 19.5 + 150 x 500

=

260,250 lb

(Note: Weight of string in air is also described as pipe setback load).

Weight of string in mud =

buoyancy factor x weight in air

= 0.847 x 260,250 = (2)

220,432 lb

Hook load = weight of string in mud + weight of travelling block, etc = 220,432 + 23,500 = 243,932 lb

(3)

Deadline load HL K10 243,932 x 0.961510 = –––––-––– = ––––––––––––––––– N EF 10 x 0.81 = 20,336 lb HL 243,932 Fast-line load = ––––––– = –––––––––– N x EF 10 x 0.81 = 30,115 lb

(4) Dynamic crown load __________________________________________________________________________________ 9 Network of Excellence in Training

= DL + FL + HL = 20,336 + 30,115 + 243,932 = 294,383 lb (5)

breaking strength = ––––––––––––––– fast-line load

Design factor

228,000 = –––––––– 30,115 (6)

= 7.6

Weight of casing in mud = 10,000 x 29 x BF = 245,630 lb HL

= weight of casing in mud + weight of travelling block, etc = 245,630 + 23,500 = 269,130 lb

FL

DF

HL 269,130 = –––––– = ––––––––––– = 33,226 lb N x EF 10 x 0.81 228,000 = –––––––– = 6.9 33,226

Example 3: Power Requirements of The Drawwork The following data refer to an oilwell block-and-tackle system: Number of lines = 10 with EF = 0.81 Maximum expected hook load = 500,000 lbf hook load speed = 120 ft/min Hoisting drum diameter = 32" __________________________________________________________________________________ 10 Network of Excellence in Training

Mechanical efficiency of draw works = 0.88 Calculate (i) (ii) (iii) (iv)

The power at the drawwork The motor power required The fastline Motor to drum gear ratio when pulling out of hole the maximum allowable load.

Note: Use an efficiency factor of 0.81. Solution Vw

= velocity of hook load = 120 ft/min

(i) Power at drum =

HLxVw 1 x( ) EF 33,000

= 500,000 x 120 x 1 0.81 x 33000 = 2245 HP Power at drum 2245

= motor power x mechanical efficiency = Motor Power x 0.88

Motor Power = 2551 HP

Select a motor with 3000 HP rating. Fastline

= 10 x Hook load speed (Vf = N x Vw_) = 10 x 120 = 1200 ft/min


Gear ratio

=

Motor speed –––––––––– Drum speed

fastline speed = drum speed x drum perimeter 1200 (ft/min)

= drum speed (rpm) x 2 x π x (32 in /2) x (1 ft/12 in)

Drum speed

= 143 rpm

Gear ratio

1200 = ––––– 477.5

= 2.5

Assuming the motor speed is 1200 rpm, which is a reasonable speed for a motor rated to 3000 HP.

3.

DRILLING LINE DESIGN CONSIDERAT IONS

3.1.

Ton-Miles (Mega joules) Of A Drilling Line

The drilling line, like any other drilling equipment, does work at any time it is involved in moving equipment in or out of the hole. The amount of work done varies depending the operation involved. This work causes the wireline to wear and if the line is not replaced it will eventually break. The amount of work done need to be calculated to determine when to change the drilling line. The following gives equations for calculating the work done on the drilling line: a)

Work done in round trip operations (Tr) D (LS+D) We D (M+C/2) Tr = ––––––––––– + –––––––––––– ton-miles 10,560,000 2,640,000

Equation 12

where M Ls D We C

= mass of travelling assembly (lb) = length of each stand (ft) = hole depth (ft) = effective weight per foot (or master) of drill pipe in mud = (L x Wdc - L x Wdp ) x BF


Wdc Wdp L b)

= weight of drill collars in air = weight of drill pipe in air = length of drill collars

In drilling a length of section from d1 to depth d2 the work done is given by Td = 3(T2 - T1 )

c)

Equation 13

Total work done (WD) in coring = 2 round trips to bottom Equation 14

Tc = (T2 - T1 ) where

T2 = WD for 1 round trip at d2 where coring stopped before coming out of the hole. T1 = WD for 1 round trip at depth d1 , where coring started

d)

Work done in setting casing (Ts) 1 Ts = ––– 2

D (Ls+D) x Wcs MD [ –––––––––––– + ––––––––– ] 10,560,000 2,640,000

Equation 15

where Wcs Ls M D

= effective weight per unit length of casing in mud = length of casing joint = mass of travelling assembly (lb) = hole depth (ft)

Example 4: Ton- Miles Evaluation Using the data given in Example 3, determine; (a) round trip ton-miles at 10,000 ft; (b) casing ton- miles if one joint of casing = 40 ft; (c) design factor of the drilling line when the 7 inch casing is run to 10,000 ft; (d) the ton- miles when coring from 10,000 ft to 10.180 ft and (e) the ton- miles when drilling from 10,000 to 10,180 ft. Solution (a)

From Equation (12):


D (LS+D) We D (M+C/2) Tr = ––––––––––– + –––––––––––– ton-miles 10,560,000 2,640,000 M C

= 23.500 lb = (L x Wdc - L x Wdp) BF = (500 x 150 - 500 x 19.5) x 0.847 = 55,267

D Ls We

= 10,000 ft = 93 ft =19.5 x BF = 167.52 lb/ft

Therefore,

Tr

10,000 x (93 + 10,000) x 16.52 = ––––––––––––––––––––––––– 10,560,000 10,000 x (23,500 + 55,267/2) + ––––––––––––––––––––––––––––– 2,640,000 = 157.9 + 193.7 = 351.6 ton-miles

(b) Ts

1 = –– 2

D x (Ls + D) x Wcs DxM [–––––––––––––––– + –––––] 10,560,000 2,640,000

Wcs

= Weight of casing in air x BF = 29 x 0.847 = 24.56 lb/ft

Ls

= 40 ft

(b) Casing operations

Ts

1 10,000 x (40 +10,000) = –– [––––––––––––––– + 2 10,560,000

10,000 x 23,500 ––––––––––––––––] 2,640,000


=½

(233.5 + 89.0)

= 161.3 ton-miles (c)

DF

= 5.6 (see Example 3)

(d)

Tc

= 2 (T2 - T1 )

where T2 = round trip time at 10,180 ft, where coring stopped, and T1 = round trip time at 10,000 ft, where coring started. Therefore, T2

10,180 x (93 + 10,180) x 16.52 = –––––––––––––––––––––––––––– 10,560,000 10,180 x (23,500 + 55,267/2) + ––––––––––––––––––––––––––––––– 2,640,000 = 163.6 + 197.2 = 360.8 ton-miles

T1

= 351.6 (from Part a)

Therefore, Tc

= 2 x (360.8 - 351.6) = 18.4 ton-miles

(e)

Td

= 3 x (T2 - T1 ) = 3 x (360.8 - 351.6) = 27.6 ton-miles

3.2.

Evaluation Of Total Service And Cut -Off Practice

Portions of the drilling line on the crown and travelling blocks sheaves and on the hoisting drum carry the greatest amount of work and is subjected to a great deal of wear and tear. These parts must be cut and removed at regular times other wise the drilling line will fail by fatigue. The process is called “slip and cut practice”.


The length of line to be cut is calculated as follows: Length of drum laps = number of laps x drum circumference = number of laps x π x D

4.

Equation 16

ROTATIN G E QUI PM ENT

The main components are: ♦ ♦ ♦ ♦ ♦

Rotary table (figure 4) Kelly (Figure 5) Top Drive (this is equivalent to the Kelly and rotary table, i.e. either top drive or Kelly/rotary table Swivel Rotary hose

Figure 4 Rotary Table, Courtesy Of National Oilwell


Figure 5 Kelly (black), Kelly Bushing (red) and Drawworks (blue), Courtesy of National Oilwell

5.

CIRCULAT ING SYST EM

The heart of the circulating system is the mud pumps. There are two types of pumps used in the oil industry: Duplex and Triplex. A basic pump consists of a piston (the liner) reciprocating inside a cylinder. A pump is described as single acting if it pumps fluid on the forward stroke (Triplex pumps) and double acting if it pumps fluid on both the forward and backward stokes (Duplex). Figure 7 shows a triplex mud pump. Figure 7 Triplex Mud Pump, Courtesy of National Oilwell


Pump Liners: Pump liners fit inside the pump cavity. These affect the pressure rating and flow rate from

the pump. For a given pump, a liner has the same OD but with different Ids. The smaller liner (small ID) is used in the deeper part of the well where low flow rate is required but at much higher operating pressure. The horse power requirements of the pump depends on the flow rate and the pressure. The operating pressure depends on flow rate, depth and size of hole, size of drillpipe and drillcollars, mud properties and size of nozzles used. A full hydraulics program needs to be calculated to determine the pressure requirement of the pump. The size of the pump is determined by the length of its stroke and the size of the liner. 5.1.

Volumetric Efficiency

Drilling mud usually contain little air and is slightly compressible. Hence the piston moves through a shorter stroke than theoretically possible before reaching discharge pressure. As a result the volumetric efficiency is always less than one; typically 95% for triplex and 90% for duplex. In addition due to power losses in drives, the mechanical efficiency of most pumps is about 85%. 5.2.

HORSEPOWER

The following equations can be used to calculate the power output of a mud pump:

Hydraulic horsepower = flow rate (gal/min) x pressure 1713.6

(psi)

Equation 17

Hydraulic horsepower = 0.000584 (gal/min) x pressure (psi)

Hydraulic horsepower = (bbl/min) x pressure (psi)/40.8

Equation 18

Hydraulic horsepower = 0.02448 (bbl/min) x pressure (psi) Hydraulic horsepower = brake horsepower x efficiency of power train to pump x pump efficiency [Power in kW

Equation 19

= 0.01667 Pressure (Kpa) x Flow rate (m3 /min)]


5.3.

Pump Output

Double acting duplex pump gal/min = 0.00679 x L x (2D2 - d2 ) x spm x volumetric efficiency

Equation 20

bbl/min = 0.000162 x L x (2D2 - d2 ) x spm x volumetric efficiency

Equation 21

Single acting triplex pump

5.4.

gal/min = 0.010199 x L x D2 x spm x volumetric efficiency

Equation 22

bbl/min = 0.000243 x L x D2 x spm x volumetric efficiency

Equation 23

Pump Factors

In practice, it convenient to express the pump output in terms of how many gallons or barrels for every stroke of the pump. The equations for the two types of pumps are: For duplex Nc x 1s x (2 x dl2 - dr2 ) x Ev Fp = ––––––––––––––––––––––– 42 x 294

Equation 24

For triplex Fp =

1s x dl2 x Ev ––––––––––– 42 x 98.03

Equation 25

where Nc = number of cylinders 1s = length of stroke, inch dl = liner diameter, inch dr = rod diameter, inch= Ev = volumetric efficiency, fraction __________________________________________________________________________________ 19 Network of Excellence in Training

Fp = pump factor, bbl/stroke

Example 5: Horse Power Requirement of a Mud Pump Calculate the power requirement for the following pump: Flow rate Pressure

= 1200 gpm = 2000 psi

Mechanical Efficiency = 0.85 Solution Hydraulic horsepower = flow rate (gal/min) x pressure (psi) 1713.6 Hydraulic horsepower = 1200 x 2000 1713.6 = 1400.6 HP Power required from motor = 1400.6 / 0.85 = 1648 HP 5.5.

Centrifugal Pumps

This type uses an impeller for the movement of fluid rather than a piston reciprocating inside a cylinder. Centrifugal pumps are used to supercharge mud pumps and providing fluid to solids control equipment and mud mixing equipment. 5.6.

Mud Handling Equipment

Rig sizing must incorporate mud handling equipment as these equipment form the heart of the circulation system and determine the speed of drilling and the quality of hole drilled. The equipment includes: 1. Shale Shakers: size, number of type The type of mud (i.e. oil-based or water-based) determines the type of the shaker required and the motion of the shaker. Deep holes require more than the customary three shakers. __________________________________________________________________________________ 20 Network of Excellence in Training

2. Mud Pits The number and size of pits is determined by the size and depth of hole. Other factors include: size of rig and space available, especially on offshore rigs. The size of a mud pit is usually 8-12 ft wide, 20-40 ft long and 6-12 ft high. Volumes Of Tanks bbl/in in round tank = (diameter in feet) 2 /85.7 bbl/in in square tank = 0.143 (length, ft) x (width, ft) cu.ft/in in square tank = 0.0833 (length, ft ) x (width, ft) m3/cm in round tank

= 0.007854 (diameter , m)3

m3/cm in square tank

= length(m) x width(m) x 0.01

3. Mud degasser 4. Centrifuges and mud cleaners 5. Desanders and desilters The selection of the above equipment determines the loading on the derrick.

6.

PRESSURE CONTROL EQUIPMENT

BOPs equipment are selected base on the maximum expected wellbore pressures. The pressure rating, size and number of BOP components must be determined by the Drilling engineer prior to drilling the well. This is the sizing exercise. Select: 1. Diverter if required, usually for offshore operations during the drilling of top or surface hole. Make sure the diverter discharge line is 12” or above. 2. Annular preventer 3. Ram preventers (determine minimum size of rams required to suit the drillstring) 4. Blind or Shear rams 5. Choke manifold 6. HCR valves 7. Choke and Kill lines 8. Accumulator and BOP Control System (Koomey Unit) 9. Drilling spools: used as an element between rams to provide mud exit lines such as choke and kill lines. Drilling spools can be flanged, studded or clamp-on type. 10. For air drilling, rotating heads are used to allow well control while the pipe is rotating. __________________________________________________________________________________ 21 Network of Excellence in Training

11. Drillpipe Blowout Preventers: include: ♦ ♦ ♦ ♦

Kelly cock Drop in valve (check valve) Float valve (either flapper or spring- loaded ball valve) Full opening safety valves

Figure 8 A Basic BOP Configuration

BOPs are rated by API as 3M (3000 psi), 5M, 10 M and 15 M. For HPHT, BOPS are either 15 M or 20 M. All the above equipment must be rated to the highest pressure to be expected at the well during a kick or during controlled testing and production. In subsea operations, the BOP stack is installed at seabed. The stack has several back up units in case of failure, for example two annulars are used so that if one failed the other can be used. This back-up system principle is applied to all the BOP components. The subsea stack for HPHT operation may not be part of the rig contract and may have to be rented out separately, egg a 20K stack.


7.

DERRICK CAP ACIT Y AND SUBST RUCTURE

The derrick provides the necessary height and support to lift loads in and out of the well. The derrick must be strong enough to support the hook load, deadline and fastline loads, pipe setback load and wind loads. There are two types of derricks: 1. Standard Derrick: is a bolted structure that must be assembled part by part, usually used on offshore platforms. Derricks installed on floating structures such as ships and semisubmersibles are designed to withstand extra dynamic stresses due to rolling, pitching and heaving of the support and due to stresses from winds. The space available between the rig floor and the crown block must be higher to handle the wave- induced vertical movements of the floating support. 2. Mast or Portable derrick: This type is pivoted at its base and is lowered to the horizontal by the use of drawers after completing the well and the rig is ready to move to another location. the mast dismantles into a number of pin-jointed sections, each of which is usually a truck load. The mast is usually used on land operations where the complete rig must be moved between well locations. at the new location, the sections are quickly pined together and the mast is raised to the vertical by the drawworks. The derrick consists of four legs connected by horizontal structural members described as girts. the derrick is further strengthened by bracing members connecting the girts. The derrick sits on a substructure on which drilling equipment is mounted. The substructure is composed of derrick supports and rotary supports. The derrick supports consist of four posts and exterior bracing between the supports. the rotary supports consist of beams and braces to support the rotary table and pipeset back load. The height of the substructure above the ground varies according to the size of the substructure and the size and rating of the wellhead and BOPs. For a base size of 30 ft , the height is 10- 14 ft.

Static Derrick Loading = fast-line load + hook load + dead-line load =

Hl HL 3 + HL + = HL 4 4 2

For N lines, the static derrick load is given by: __________________________________________________________________________________ 23 Network of Excellence in Training

=

( N + 2) HL N

where N = number of lines strung to travelling block HL = hook load The wind load is given by : 0.004 V2 (units lb/ft2 ) where V is wind speed in miles/hour The above result must be multiplied by the WIND LOAD AREA which is given in API 4A for different derrick sizes in order to obtain the load in lb..

Example 6 : Derrick Loading The following data refer to a 1.5 in block line with 10 lines of extra improved plough steel wire rope strung to the travelling block. hole depth = 10,000 ft drillpipe = 5 in OD/4.276 in ID, 19.5 lb/ft drill collars = 500 ft, 8 in/2,825 in, 150 lb/ft mud weight = 10 ppg line and sheave efficiency coefficient = 0.9615 Calculate (1) (2) (3) (4) (5) (6) (7)

weight of drill string in air and in mud; hook load, assuming weight of travelling block and hook to be 23,500 lb; deadline and fast-line loads, assuming an efficiency factor of 0.81; dynamic crown load; wireline design factor during drilling if breaking strength of wire is 228,000 lb (1,010 kN); design factor when running 9 5/8 in casing of 53.5 lb/ft . dynamic derrick load when running the 9 5/8” casing

Solution (1)

Weight of string in air =

weight of drillpipe + weight of drill collars


=

(10,000 - 500) x 19.5 + 150 x 500

=

260,250 lb

(Note: Weight of string in air is also described as pipe setback load). Weight of string in mud

(2)

=

buoyancy factor x weight in air

=

0.847 x 260,250

=

220,432 lb

Hook load = weight of string in mud + weight of travelling block, etc = 220,432 + 23,500

HL

= 243,932 lb

(3)

Deadline load HL K10 = ––– x –––– N EF

=

243,932 x 0.961510 ––––––––––––––10 x 0.81

= 20,336 lb

Fast-line load

HL = ––––– N x EF

=

243,932 ––––––– 10 x 0.81

= 30,115 lb (4)

Dynamic derrick Loading during drilling = DL + FL + HL


= 20,336 + 30,115 + 243,932 = 294,383 lb (5)

Design factor =

breaking strength ––––––––––––––– fast-line load 228,000 = –––––––––- = 7.6 30,115

(6)

Weight of casing in mud = 10,000 x 53.5 x BF = 453,145 lb

HL

= weight of casing in mud + weight of travelling block, etc = 453,145 + 23,500 = 476,645 lb FL

HL = ––––– N x EF

=

476.645 –––––– 10 x 0.81

= 58,845 lb

228,000 DF = ------= 3.9 58,845 (7) Dynamic derrick loading during running casing = FLL + HL + DLL Deadline load

HL K10 = –––––––––N EF

476,645 x 0.9615 10 = ––––––––––––––– 10 x 0.81 = 39,738

Dynamic derrick loading

= 58,845+ 476,645 + 39,737 = 575,228 lb


Hence the derrick capacity must be approximately 750,000 lb to allow fo r extra loading such as wind , pipe setback load etc.

TOTAL POWER REQUIREMENTS

8.

The total power requirement of a rig is the sum of the power requirement of: 1. 2. 3. 4. 5.

Drawworks mud pumps Rotary system Auxiliary power requirements for lighting etc. life support system

The above total power may not be required in a continuous but in an intermittent mode. The actual power required will depend on the drilling job being carried out. The maximum power used is during hoisting and circulation. The least power use d is during wireline operations. The majority of rigs in current use require between 1000 – 3000 horsepower. The power on modern rigs is most commonly generated by diesel-electric power units. The power produced is AC current which is then converted to DC current by the use of SCR (Silicon Controlled Rectifier) . The current is delivered by cables to electric motors attached directly to the equipment involved such as mud pumps, rotary table, Drawworks etc.


Rig Sizing - Text

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