rig sizing - text

8/17/2019 Rig Sizing - Text

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1 Network of Excellence in Training

Rig Sizing

1. Fundamentals .................................................................................................................................... 21.1. Types of Rigs ............................................................................................................................ 2

2. Hoisting System................................................................................................................................ 32.1. Hoisting Design Considerations ............................................................................................... 52.2. Power Requirements of the Drawworks................................................................................... 7

3. Drilling Line Design Considerations .............................................................................................. 12

3.1. Ton-Miles (Mega joules) Of A Drilling Line ......................................................................... 123.2. Evaluation Of Total Service And Cut-Off Practice................................................................ 15

4. ROTATING EQUIPMENT............................................................................................................ 165. Circulating System.......................................................................................................................... 17

5.1. Volumetric Efficiency ............................................................................................................ 18

5.2. HORSEPOWER..................................................................................................................... 185.3. Pump Output ........................................................................................................................... 19

5.4. Pump Factors .......................................................................................................................... 195.5. Centrifugal Pumps .................................................................................................................. 205.6. Mud Handling Equipment ...................................................................................................... 20

6. Pressure Cont rol Equipment ........................................................................................................... 217. Derrick Capacity and Substructure................................................................................................. 23

8. Total Power Requirements ............................................................................................................. 27


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Rig Sizing

1. FFUUNNDDAAMMEENNTTAALLSS

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 well2. 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

♦ Drillships2. 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. Hoisting2. Rotating Equipment

3. Circulating System4. Tubular Goods


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5. Pressure Control6. Derrick Capacity And Substructure7. Power Requirements

2. HHOOIISSTTIINNGG SS Y YSSTTEEMM

The hoisting system consists of: (Figure 1)

♦ Drawwork : this is an assembly of a rotating drum, a series of shafts, clutches, chains and gears forchanging speed and for reversing, figure 2. It also contains the main brake for stopping the drillingline. The drilling line is wound a number of times around the drum, the end of the line then passeson the crown and travelling blocks.

♦ Crown Block : A block located at the top of the derrick. It contains a number of sheaves on whichis wound the dril ling line . The crown block provides a means of taking the drilling line from thehoisting drum to the travelling block. The crown block is stationary and is firmly fastened to thetop 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 goingto an anchoring clamp called “ DEAD LINE ANCHOR”. The static line is called the deadline. Theline section connecting the drum with the crown block is called the fastline.

Figure 1 Schematic Of The Hoisting System

Crown Block

Fixed sheaves

Deadline

Anchor Hook

W

Drawworks

Drilling

Lines

FastlineW/4

W/4W/4

W/4 W/4W/4

Travelling Block

Deadline

Crown Block

Fixed sheaves

Deadline

Anchor Hook

W

Drawworks

Drilling

Lines

FastlineW/4

W/4W/4

W/4 W/4W/4

Travelling Block

Deadline


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Hence during hoisting operations, if there are 10 lines between the crown and travelling block, thefastline line travels 10 times faster than the travelling block in order to spool or unspool drilling linefrom the hoisting drum.

The DEAD LINE ANCHOR anchors the last line coming from the crown block and also storesdrilling 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. Theworn parts are regularly cut and removed, Slip and Cut Practice. Slipping the line, then cutting it offhelps to increase the lifetime of the drilling line.

♦ Travelling Block: a diamond-shaped block containing a number of sheaves which is always lessthan 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 entiredrilling 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 constructionwith Independent Wire Rope Core (IWRC). This construction implies that there are 6 strands and eachstrand containing 19 filler wires.

The size of the drilling line varies from ½ “ to 2 “.

Figure 2 Drawworks, Courtesy of National Oilwell


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2.1. Hoisting Design Considerations

The procedure for carrying out hoisting design calculations are as follows:

1. Determine the deepest hole to be drilled2. Determine the worst drilling loads or casing loads3. 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 staticconditions:

Fast- line load (FL) = Hook load /4

Dead-line load (DL) = Hook load /4

HL HL

HL Hl

SDL2

3

44=++= Equation 1

Figure 3 Travelling Block and Hook, Courtesy of National Oilwell


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

where

N = number of lines strung to travelling blockHL = hook load

2. Efficiency Of The Hoisting Systems

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

NxEF

HL FL =

where

K = sheave and line efficiency per sheave

Deadline-load is given by:

HL x K N DL = ––––––––––

N x EF

If the breaking strength of the drilling line is known, then a design factor, DF, may becalculated as follows:

nominal strength of wire rope (lb)DF = –––––––––––––––––––––––––––––––

fast-line load (lb)

B. Lowering Operations

During lowering of pipe, the efficiency factor and fast- line load are given by

HL N

N )2(+

=

)1(

)1(

K N

K K EF

n

−

−=

Equation 2

Equation 3

Equation 4

Equation 5

Equation 6


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

(1-K N)

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

(1-K N)

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. Assumethe value of K to be 0.9615.

Solution

K x (1 - K N)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 Efficiency factor

6 0.8748 0.842

10 0.81112 0.782

2.2. Power Requirements of the Drawworks

As a rule of thumb, the drawwork should have 1 HP for every 10 ft to be drilled.

Equation 7

Equation 8


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

a) Velocity of fast- line load (Vf )

Vf = N x VL

where

VL = velocity of travelling block

N = number of lines strung

b) Power output at drum = FL x Vf

(HL) x VL P = ---------

EF

In the Imperial system, power is quoted in horse-power and the above equation becomes:

HL x VL

Drum output = ----------- - horsepowerEF x 33,000

Example 2: Hook Loads

The following data refer to a 1.5 in block line with 10 lines of extra improved ploughsteel wire rope strung to the travelling block.

hole depth = 10,000 ftdrillpipe = 5 in OD/4.276 in ID, 19.5 lb/ftdrill collars = 500 ft, 8 in/2,825 in, 150 lb/ftmud weight = 10 ppgline and sheave efficiency coefficient = 0.9615

Calculate:

(1) weight of drill string in air and in mud;(2) hook load, assuming weight of travelling block and hook to be 23,500 lb;(3) deadline and fast-line loads, assuming an efficiency factor of 0.81;(4) dynamic crown load;(5) wireline design factor during drilling if breaking strength of wire is 228,000 lb(6) design factor when running 7 in casing of 29 lb/ft.

Solution

(1) Weight of string in air

Equation 9

Equation 10

Equation 11


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= 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

= 220,432 lb

(2) Hook load

= weight of string in mud

+ weight of travelling block, etc

= 220,432 + 23,500

= 243,932 lb

(3) Deadline load

HL K 10 243,932 x 0.961510 = –––––-––– = –––––––––––––––––

N EF 10 x 0.81

= 20,336 lb

HL 243,932Fast-line load = ––––––– = ––––––––––

N x EF 10 x 0.81

= 30,115 lb

(4) Dynamic crown load


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= DL + FL + HL

= 20,336 + 30,115 + 243,932

= 294,383 lb

breaking strength(5) Design factor = –––––––––––––––

fast-line load

228,000

= –––––––– = 7.630,115

(6) Weight of casing in mud

= 10,000 x 29 x BF

= 245,630 lb

HL = weight of casing in mud


= 245,630 + 23,500

= 269,130 lb

HL 269,130FL = –––––– = ––––––––––– = 33,226 lb

N x EF 10 x 0.81

228,000DF = –––––––– = 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 lbfhook load speed = 120 ft/minHoisting drum diameter = 32"


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Mechanical efficiency of draw works = 0.88

Calculate

(i) The power at the drawwork(ii) The motor power required(iii) The fastline(iv) 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

= 500,000 x 120 x 10.81 x 33000

= 2245 HP

Power at drum = motor power x mechanicalefficiency

2245 = 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

)000,33

1( x

EF

HLxV w=


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Motor speedGear ratio = ––––––––––

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

1200

Gear ratio = ––––– = 2.5477.5

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

3. DDRRIILLLLIINNGG LLIINNEE DDEESSIIGGNN CCOONNSSIIDDEERRAATTIIOONNSS

3.1. Ton-Miles (Mega joules) Of A Drilling Line

The drilling line, like any other drilling equipment, does work at any time it isinvolved 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 lineis not replaced it will eventually break.

The amount of work done need to be calculated to determine when to change thedrilling line. The following gives equations for calculating the work done on thedrilling 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

where

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

Equation 12


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Wdc = weight of drill collars in airWdp = weight of drill pipe in airL = length of drill collars

b) In drilling a length of section from d1 to depth d2 the work done is given by

Td = 3(T2 - T1)

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

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 D (Ls+D) x Wcs MDTs = ––– [ –––––––––––– + ––––––––– ]

2 10,560,000 2,640,000

where

Wcs = effective weight per unit length of casing in mudLs = length of casing jointM = mass of travelling assembly (lb)D = 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 ifone 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 10,000to 10,180 ft.

Solution

(a) From Equation (12):

Equation 13

Equation 14

Equation 15


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D (LS+D) We D (M+C/2)Tr = ––––––––––– + –––––––––––– ton-miles

10,560,000 2,640,000

M = 23.500 lbC = (L x Wdc - L x Wdp) BF

= (500 x 150 - 500 x 19.5) x 0.847= 55,267

D = 10,000 ftLs = 93 ft

We =19.5 x BF = 167.52 lb/ft

Therefore,

10,000 x (93 + 10,000) x 16.52

Tr = –––––––––––––––––––––––––10,560,000

10,000 x (23,500 + 55,267/2)+ –––––––––––––––––––––––––––––

2,640,000

= 157.9 + 193.7

= 351.6 ton-miles

(b)1 D x (Ls + D) x Wcs D x M

Ts = –– [–––––––––––––––– + –––––]2 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

1 10,000 x (40 +10,000) 10,000 x 23,500Ts = –– [––––––––––––––– + ––––––––––––––––]

2 10,560,000 2,640,000


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= 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 10,000ft, where coring started. Therefore,

10,180 x (93 + 10,180) x 16.52T2 = ––––––––––––––––––––––––––––

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


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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. RROOTTAATTIINNGG EEQQUUIIPPMMEENNTT

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

Equation 16

Figure 4 Rotary Table, Courtesy Of National Oilwell


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5. CCIIRRCCUULLAATTIINNGG SS Y YSSTTEEMM

The heart of the circulating system is the mud pumps. There are two types of pumps used in the oilindustry: 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 fluidon both the forward and backward stokes (Duplex). Figure 7 shows a triplex mud pump.

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

Figure 7 Triplex Mud Pump, Courtesy of National Oilwell


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Pump Liners: Pump liners fit inside the pump cavity. These affect the pressure rating and flow rate fromthe pump. For a given pump, a liner has the same OD but with different Ids. The smaller liner (smallID) 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 propertiesand size of nozzles used. A full hydraulics program needs to be calculated to determine the pressurerequirement 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 ashorter 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 (psi)1713.6

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

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

Hydraulic horsepower = 0.02448 (bbl/min) x pressure (psi)

Hydraulic horsepower = brake horsepower x efficiency of powertrain to pump x pump efficiency

[Power in kW = 0.01667 Pressure (Kpa) x Flow rate (m3/min)]

Equation 17

Equation 18

Equation 19


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5.3. Pump Output

Double acting duplex pump

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

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

Single acting triplex pump

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

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

5.4. 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 - dr

2) x Ev

F p = –––––––––––––––––––––––42 x 294

For triplex

1s x dl2 x Ev

F p = –––––––––––42 x 98.03

where

Nc = number of cylinders

1s = length of stroke, inch

dl = liner diameter, inch

dr = rod diameter, inch=

Ev = volumetric efficiency, fraction

Equation 20

Equation 21

Equation 22

Equation 23

Equation 24

Equation 25


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F p = pump factor, bbl/stroke

Example 5: Horse Power Requirement of a Mud Pump

Calculate the power requirement for the following pump:

Flow rate = 1200 gpmPressure = 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 insidea cylinder. Centrifugal pumps are used to supercharge mud pumps and providing fluid tosolids 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 motionof the shaker. Deep holes require more than the customary three shakers.


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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 ftlong 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 degasser4. Centrifuges and mud cleaners5. Desanders and desilters

The selection of the above equipment determines the loading on the derrick.

6. PPRREESSSSUURREE CCOONNTTRROOLL EEQQUUIIPPMMEENNTT

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 thewell. 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 preventer3. Ram preventers (determine minimum size of rams required to suit the drillstring)4. Blind or Shear rams5. Choke manifold6. HCR valves7. Choke and Kill lines8. 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.


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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

BOPs are rated by API as 3M (3000 psi), 5M, 10 M and 15 M. For HPHT, BOPS are either 15 Mor 20 M.

All the above equipment must be rated to the highest pressure to be expected at the well during a kickor 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-upsystem 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.

Figure 8 A Basic BOP Configuration


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7. DDEERRRRIICCKK CCAAPPAACCIITT Y Y AANNDD SSUUBBSSTTRRUUCCTTUURREE

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 andwind 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 withstandextra dynamic stresses due to rolling, pitching and heaving of the support and due to stresses fromwinds. 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 theuse of drawers after completing the well and the rig is ready to move to another location. themast 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 thevertical by the drawworks.The derrick consists of four legs connected by horizontal structural members described as girts. thederrick is further strengthened by bracing members connecting the girts.

The derrick sits on a substructure on which drilling equipment is mounted. The substructure iscomposed of derrick supports and rotary supports.The derrick supports consist of four posts and exterior bracing between the supports. the rotarysupports 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 andthe 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

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

HL HL

HL Hl

2

3

44=++=


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where

N = number of lines strung to travelling blockHL = 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 ploughsteel wire rope strung to the travelling block.

hole depth = 10,000 ft

drillpipe = 5 in OD/4.276 in ID, 19.5 lb/ftdrill collars = 500 ft, 8 in/2,825 in, 150 lb/ftmud weight = 10 ppgline and sheave efficiency coefficient = 0.9615

Calculate

(1) weight of drill string in air and in mud;(2) hook load, assuming weight of travelling block and hook to be 23,500 lb;(3) deadline and fast-line loads, assuming an efficiency factor of 0.81;(4) dynamic crown load;

(5) wireline design factor during drilling if breaking strength of wire is 228,000 lb (1,010 kN);(6) design factor when running 9 5/8 in casing of 53.5 lb/ft .(7) dynamic derrick load when running the 9 5/8” casing

Solution

(1) Weight of string in air

= weight of drillpipe + weight of drill collars

HL N

N )2( +=


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= (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

= 220,432 lb

(2) Hook load

= weight of string in mud


= 220,432 + 23,500

HL = 243,932 lb

(3) Deadline load

HL K 10 243,932 x 0.961510 = ––– x –––– = ––––––––––––––- N EF 10 x 0.81

= 20,336 lb

HL 243,932Fast-line load = ––––– = ––––––– N x EF 10 x 0.81

= 30,115 lb

(4) Dynamic derrick Loading during drilling

= DL + FL + HL


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= 20,336 + 30,115 + 243,932

= 294,383 lb

breaking strength(5) Design factor = –––––––––––––––

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

HL 476.645FL = ––––– = –––––– = 58,845 lb

N x EF 10 x 0.81

228,000DF = ------- = 3.9

58,845(7) Dynamic derrick loading during running casing = FLL + HL + DLL

Deadline load

HL K 10 476,645 x 0.961510 = –––––––––- = –––––––––––––––

N EF 10 x 0.81

= 39,738

Dynamic derrick loading = 58,845+ 476,645 + 39,737

= 575,228 lb


27/27

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

8. TTOOTTAALL PPOOWWEERR RREEQQUUIIRREEMMEENNTTSS

The total power requirement of a rig is the sum of the power requirement of:

1. Drawworks2. mud pumps3. Rotary system4. Auxiliary power requirements for lighting etc.5. 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 usedis during hoisting and circulation. The least power used 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 ControlledRectifier) . The current is delivered by cables to electric motors attached directly to the equipmentinvolved such as mud pumps, rotary table, Drawworks etc.

rig sizing - text

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