1 basic formulas

CHAPTER ONE

BASIC FORMULAS

Pressure Gradient

Pressure gradient, psi/ft, using mud weight, ppg

psi/ft = mud weight, ppg x 0.052

Example: 12.0 ppg fluidpsi/ft = 12.0ppg x 0.052psi/ft = 0.624

Pressure gradient, psi/ft, using mud weight, lb/ft3

psi/ft = mud weight, lb/ft3 x 0.006944

Example: 100 lb/ft3 fluidpsi/ft = 1001b/ft3 x 0.006944psi/ft = 0.6944

OR

psi/ft = mud weight, lb/ft3 -s- 144

Example: 100 lb/ft3 fluidpsi/ft = 1001b/ft3 + 144psi/ft = 0.6944

Pressure gradient, psi/ft, using mud weight, specific gravity (SG)

psi/ft = mud weight, SG x 0.433

Example: 1.0 SG fluidpsi/ft = 1.0SG x 0.433psi/ft = 0.433

Metric calculations

Pressure gradient, bar/m = drilling fluid density kg/1 x 0.0981Pressure gradient, bar/1 Om = drilling fluid density kg/1 x 0.981

S.I. units calculations

Pressure gradient, kPa/m = drilling fluid density, kg/m3 -5-102

Convert pressure gradient, psi/ft, to mud weight, ppg

ppg = pressure gradient, psi/ft + 0.052

Example: 0.4992 psi/ftppg = 0.4992psi/ft -s- 0.052ppg = 9.6

Convert pressure gradient, psi/ft, to mud weight, lb/ft3

lb/ft3 = pressure gradient, psi/ft + 0.006944

Example: 0.6944psi/ftlb/ft3 = 0.6944psi/ft + 0.006944lb/ft3 = 100

Convert pressure gradient, psi/ft, to mud weight, SG

SG = pressure gradient, psi/ft -s- 0.433

Example: 0.433 psi/ftSG = 0.433psi/ft + 0.433SG = 1.0

Metric calculations

Drilling fluid density, kgA = pressure gradient, bar/m -s- 0.0981Drilling fluid density, kg/1 = pressure gradient, bar/lOm *- 0.981


Drilling fluid density, kg/m3 = pressure gradient, kPa/m x 102

Hydrostatic Pressure (HP)

Hydrostatic pressure using ppg and feet as the units of measure

HP = mud weight, ppg x 0.052 x true vertical depth (TVD), ft

Example: mud weight = 13.5ppgtrue vertical depth = 12,000 ft

HP = 13.5ppg x 0.052 x 12,000ftHP = 8424psi

Hydrostatic pressure, psi, using pressure gradient, psi/ft

HP = psi/ft x true vertical depth, ft

Example: pressure gradient = 0.624 psi/fttrue vertical depth = 8500ft

HP = 0.624psi/ft x 8500ftHP = 5304psi

Hydrostatic pressure, psi, using mud weight, Ib/ft3

HP = mud weight, lb/ft3 x 0.006944 x TVD, ft

Example: mud weight = 90 lb/ft3

true vertical depth = 7500 ftHP = 901b/ft3 x 0.006944 x 7500ftHP = 4687psi

Hydrostatic pressure, psi, using meters as unit of depth

HP = mud weight, ppg x 0.052 x TVD, m x 3.281

Example: mud weight = 12.2 ppgtrue vertical depth = 3700 meters

HP = 12.2ppg x 0.052 x 3700 x 3.281HP = 7701psi

Metric calculations

Hydrostatic _ drilling fluid w n n o 0 1 ^ true vertical

pressure, bar density, kg/I depth, m


Hydrostatic _ drilling fluid density, kg/m3pressure, kPa " m X t r u e v e r t i c a l d e p t h ' m

Converting Pressure into Mud Weight

Convert pressure, psi, into mud weight, ppg, using feet asthe unit of measure

Mud weight, ppg = pressure, psi -s- 0.052 -s- TVD, ft

Example: pressure = 2600 psitrue vertical depth = 5000 ft

Mud, ppg = 2600psi - 0.052 - 5000ftMud = lO.Oppg

Convert pressure, psi, into mud weight, ppg, using meters as the unitof measure

Mud weight, ppg = pressure, psi -s- 0.052 + TVD, m -s- 3.281

Example: pressure = 3583 psitrue vertical depth = 2000 meters

Mud wt, ppg = 3583psi - 0.052 -* 2000 m - 3.281Mud wt = 10.5 ppg

Metric calculations

Equivalent drilling _ pressure, ^ ^ ^ 1 ^ true verticalfluid density, kg/1 bar ' depth, m


Equivalent drilling _ pressure, ^ t r u e verticalfluid density, kg/m3 ~ kPa ' depth, m

Specific Gravity (SG)

Specific gravity using mud weight, ppg

SG = mud weight, ppg H- 8.33

Example: 15.0ppg fluidSG = 15.0ppg H- 8.33SG = 1.8

Specific gravity using pressure gradient, psi/ft

SG = pressure gradient, psi/ft * 0.433

Example: pressure gradient = 0.624 psi/ftSG = 0.624psi/ft + 0.433SG = 1.44

Specific gravity using mud weight, Ib/ft3

SG = mud weight, lb/ft3 + 62.4

Example: mud weight = 120 lb/ft3

SG = 1201b/ft3 H- 62.4SG = 1.92

Convert specific gravity to mud weight, ppg

Mud weight, ppg = specific gravity x 8.33

Example: specific gravity = 1.80Mud wt, ppg = 1.80 x 8.33Mud wt = 15.0ppg

Convert specific gravity to pressure gradient, psi/ft

psi/ft = specific gravity x 0.433

Example: specific gravity = 1.44psi/ft = 1.44 x 0.433psi/ft = 0.624

Convert specific gravity to mud weight, lb/ft3

lb/ft3 = specific gravity x 62.4

Example: specific gravity = 1.92lb/ft3 = 1.92 x 62.4lb/ft3 = 120

Equivalent Circulating Density (ECD), ppg

(annular "\ , * . - >.ECD, ppg = pressure

+ 0.052 + TVD, ft + fmud w e i g h Mf Vm use, ppg JUoss, psij v F F & y

Example: annular pressure loss = 200 psitrue vertical depth = 10,000 ftmud weight = 9.6ppg

ECD, ppg = 200psi + 0.052 + 10,000ft + 9.6ppgECD = lO.Oppg

Metric calculation

Equivalent drilling = annular pressure + + m + m u d

fluid density, kg/1 loss, bar


Equivalent circulating annular pressure loss, kPa x 102 . ,density, kg/1 = T V D ^ + m u d d e n S l t y ' ^

Maximum Allowable Mud Weight from Leak-off Test Data

_ Heak-off ^ _ n n

Decimal equivalent = 95 - 100 = 0.95

PO@95% = 0.142884bbl/stk x 0.95PO@95% = 0.13574bbl/stk

Formula 2

PO, gpm = [3(D2 x 0.7854) S] 0.00411 x SPM

where D = liner diameter, in.S = stroke length, in.SPM = strokes per minute

Example: Determine the pump output, gpm, for a 7-in. by 12-in. triplexpump at 80 strokes per minute:

PO, gpm = [3(72 x 0.7854) 12] 0.00411 x 80PO, gpm = 1385.4456 x 0.00411 x 80PO = 455.5 gpm

Duplex Pump

Formula 1

0.000324 x {!!ner t . f x (f\t . ) = bbl/stk

^diameter, m.J v length, m.J-0.000162 x frd

t . f x f f 0 ^ . ) = bbl/stkV diameter, m.J [length, m.J

pump output @ 100% eff = bbl/stk

Example: Determine the output, bbl/stk, of a 5-1/2 in. by 14-in. duplexpump at 100% efficiency. Rod diameter = 2.0in.:

0.000324 x 5.52 x 14 = 0.137214bbl/stk-0.000162 x 2.02 x 14 = 0.009072bbl/stk

Pump output 100% eff = 0.128142 bbl/stk

Adjust pump output for 85% efficiency:Decimal equivalent = 85 + 100 = 0.85

PO@85% = 0.128142bbl/stk x 0.85PO@85% = 0.10892 bbl/stk

Formula 2

PO, bbl/stk = 0.000162 x S [2(D)2 - d2]

where S = stroke length, in.D = liner diameter, in.d = rod diameter, in.

Example: Determine the output, bbl/stk, of a 5-1/2-in. by 14-in. duplexpump @ 100% efficiency. Rod diameter = 2.0in.:

PO @ 100% = 0.000162 x 14 x [2 (5.5)2 - 22]PO @ 100% = 0.000162 x 14 x 56.5PO @ 100% = 0.128142bbl/stk

Adjust pump output for 85% efficiency:PO@85% = 0.128142bbl/stk x 0.85PO@85% = 0.10892 bbl/stk

Metric calculation

Pump output, liter/min = pump output, liter/stk x pump speed, spm

S.I. units calculation

Pump output, mVmin = pump output, liter/stk x pump speed, spm

Annular Velocity (AV)

Annular velocity (AV), ft/min

Formula 1

AV = pump output, bbl/min * annular capacity, bbl/ft

Example: pump output = 12.6bbl/minannular capacity = 0.1261 bbl/ft

AV = 12.6bbl/min -s- 0.1261 bbl/ftAV = 99.92ft/min

Formula 2

where Q = circulation rate, gpmDh = inside diameter of casing or hole size, in.Dp = outside diameter of pipe, tubing or collars, in.

Example: pump output = 530 gpmhole size = 12-l/4in.pipe OD = 4-l/2in.

AV = lOOft/min

Formula 3

Example: pump output = 12.6bbl/minhole size = 12-1/4 in.pipe OD = 4-l/2in.

AV = 99.92 ft/min

Annular velocity (AV), ft/sec

Example: pump output = 12.6bbl/minhole size = 12-l/4in.pipe OD = 4-1/2 in.

AV = 1.6656 ft/sec

Metric calculations

Annular velocity, m/min = pump output, liter/min * annular volume, IAn

Annular velocity, m/sec = pump output, liter/min -s- 60 -s- annular volume, 1/m


Annular velocity, m/min = pump output, m3/min -* annular volume, m3/m

Pump output, gpm, required for a desired annular velocity, ft/min

Pump output, gpm =

where AV = desired annular velocity, ft/minDh = inside diameter of casing or hole size, in.Dp = outside diameter of pipe, tubing or collars, in.

Example: desired annular velocity = 120 ft/minhole size = 12-1/4 in.pipe OD = 4-1/2 in.

PO = 635.8gpm

Strokes per minute (SPM) required for a given annular velocity_ annular velocity, ft/min x annular capacity, bbl/ft

pump output, bbl/stk

Example: annular velocity = 120 ft/minannular capacity = 0.1261 bbl/ftDh = 12-1/4 in.Dp =4-1/2 in.pump output = 0.136 bbl/stk

_ 120ft/min x 0.1261bbl/ft0.136bbl/stk

SPM = 111.3

Capacity Formulas

Annular capacity between casing or hole and drill pipe, tubing,or casing

Dh2 - Dp2a) Annular capacity, bbl/ft =

Example: Hole size (Dh) = 12-l/4in.Drill pipe OD (Dp) = 5.0in.

12 252 - 5 O2Annular capacity, bbl/ft = : '-

Annular capacity = 0.12149bbl/ft

1029.4b) Annular capacity, ft/bbl = -^"2\

Example: Hole size (Dh) = 12-l/4in.Drill pipe OD (Dp) = 5.0in.

1029 4Annular capacity, ft/bbl = ^2 _' 5 02\

Annular capacity = 8.23ft/bbl

c) Annular capacity, gal/ft = 24.51

Example: Hole size (Dh) = 12-1/4 in.Drill pipe OD (Dp) = 5.0in.

A i uft 1 2 - 2 5 2 - 5 - 2Annular capacity, gal/ft =Z*r.J 1

Annular capacity =5.1 gal/ft

24 51d) Annular capacity, ft/gal =

2 ^ p


Annular capacity, ft/gal = ^2 '_ 5 Q2X

Annular capacity = 0.19598 ft/gal

Dh2 - Dp2e) Annular capacity, ft3/linft =

183.35Example: Hole size (Dh) = 12-1/4 in.

Drill pipe OD (Dp) = 5.0 in.12 252 5 O2Annular capacity, ft3/linft = ' '

Annular capacity = 0.682097ft3/linft

f) Annular capacity, linft/ft3 = --: ^-


Annular capacity, linft/ft3 = '_ Q

Annular capacity = 1.466 linft/ft3

Annular capacity between casing and multiple strings of tubing

a) Annular capacity between casing and multiple strings of tubing, bbl/ft:

Dh 2 - [(T1)2 + (T2)2IAnnular capacity, bbl/ft = -

Example: Using two strings of tubing of same size:Dh = casing7.0 in.29 lb/ft ID = 6.184 in.T1 = tubing No. 12-3/8 in. OD = 2.375 in.T2 = tubing No. 22-3/8 in. OD = 2.375 in.

. l i r 6.1842 - (2.3752 + 2.3752)Annular capacity, bbl/ft = -

A T uuiift 38 .24-11 .28Annular capacity, bbl/ft =

Annular capacity = 0.02619 bbl/ft

b) Annular capacity between casing and multiple strings of tubing, ft/bbl:

1029.4Annular capacity, ft/bbl = = ^ Dh 2 - [(T1)2 + (T2)2]

Example: Using two strings of tubing of same size:Dh = casing7.0 in.29 lb/ft ID = 6.184 in.T1 = tubing No. 12-3/8 in. OD = 2.375in.T2 = tubing No. 22-3/8 in. OD = 2.375 in.

1029 4Annular capacity, ft/bbl = ^ _ ^ 3 7 5 2 + ^37fi)

1029 4Annular capacity, ft/bbl = :

F y 38.24-11.28

Annular capacity = 38.1816ft/bbl

c) Annular capacity between casing and multiple strings of tubing,gal/ft:

Dh 2 - [(T1)2 + (T2)2IAnnular capacity, gal/ft = -

Example: Using two tubing strings of different size:Dh = casing7.0 in.29 lb/ft ID = 6.184 in.T1 = tubing No. 12-3/8 in. OD = 2.375 in.T2 = tubing No. 23-1/2in. OD = 3.5in.

A 1 U I/ft 6 ' 1 8 4 2 " I 2 ' 3 7 5 ' + 3 - 5 ' )Annular capacity, gal/ft = - -

A i ,/* 3 8 . 2 4 - 1 7 . 8 9Annular capacity, gal/ft =

24.51Annular capacity = 0.8302733gal/ft

d) Annular capacity between casing and multiple strings of tubing,ft/gal:

Annular capacity, ft/gal = =' rrDh 2 - [(T1)2 + (T2)2]

Example: Using two tubing strings of different sizes:Dh = casing7.0 in.29 lb/ft ID = 6.184 in.T1 = tubing No. 12-3/8 in. OD = 2.375in.T2 = tubing No. 23-1/2in. OD = 3.5in.

Annular capacity, ft/gal = ^ 2 _ ^ 5 2 + 3 ^

Annular capacity, ft/gal = 3 8 2 4 '_ 1 ? 8 9

Annular capacity = 1.2044226 ft/gal

e) Annular capacity between casing and multiple strings of tubing, ft3/linft:

Dh 2 - [(T1)2 + (T2)2IAnnular capacity, fr/linft = -

183.35Example: Using three strings of tubing:

Dh = casing9-5/8 in.471b/ft ID = 8.681 in.T1 = tubing No. 13-1/2in. OD = 3.5in.T2 = tubing No. 23-1/2in. OD = 3.5 in.T3 = tubing No. 33-1/2in. OD = 3.5in.

A , . 8.6812 - (3.52 + 3.52 + 3.52)Annular capacity = -

183.35

Annular capacity, ft3/linft = '

Annular capacity = 0.2105795 ft3/linftf) Annular capacity between casing and multiple strings of tubing, linft/ft3:

Annular capacity, linft/ft3 = '- Dh 2 - [(T1)2 + (T2)2]

Example: Using three strings tubing of same size:Dh = casing9-5/8 in.471b/ft ID = 8.681 in.T1 = tubing No. 13-1/2in. OD = 3.5in.T2 = tubing No. 23-1/2in. OD = 3.5in.T3 = tubing No. 33-1/2in. OD = 3.5in.

Annular capacity = 8.6812 _ ( 3 . " " 3.52 + 3.52)

Annular capacity, linft/ft3 = 7 5 3 5 9 1 3 6 7 5

Annular capacity = 4.7487993 linft/ft3

Capacity of tubulars and open hole: drill pipe, drill collars, tubing,casing, hole, and any cylindrical object

a) Capacity, bbl/ft = 1^j

Example: Determine the capacity, bbl/ft, of a 12-1/4 in. hole:

Capacity, bbl/ft = ^^~F y 1029.4

Capacity = 0.1457766 bbl/ft

b) Capacity, ft/bbl = 19A

Example: Determine the capacity, ft/bbl, of 12-1/4 in. hole:1029 4

Capacity, ft/bbl = ~ r

Capacity = 6.8598ft/bbl

c) Capacity, gal/ft = ^

Example: Determine the capacity, gal/ft, of 8-1/2 in. hole:8 52Capacity, gal/ft = - ^ j -

Capacity = 2.9477764 gal/ft

d) Capacity, ft/gal = ~ - ^

Example: Determine the capacity, ft/gal, of 8-1/2 in. hole:

Capacity, ft/gal = '8.5

Capacity = 0.3392 ft/gal

ID2e) Capacity, ftVlinft = - ^ ^

Example: Determine the capacity, ftVlinft, for a 6.0 in. hole:

Capacity, ft3/linft = ^ ^

Capacity =0.1963 ft 3/linft

183 35f) Capacity, linft/ft3 = | -^

Example: Determine the capacity, linft/ft3, for a 6.0 in. hole:

Capacity, linft/ft3 = ^ f ^

Capacity = 5.09305 linft/ft3

Amount of cuttings drilled per foot of hole drilled

a) BARRELS of cuttings drilled per foot of hole drilled:

Barrels = (1 - % porosity)

Example: Determine the number of barrels of cuttings drilled for onefoot of 12-l/4in.-hole drilled with 20% (0.20) porosity:

Barrels = ! ^ ( 1 - - 2 0 )

Barrels = 0.1457766 x 0.80Barrels = 0.1166213

b) CUBIC FEET of cuttings drilled per foot of hole drilled:

Dh2Cubic feet = x 0.7854 (1 - % porosity)

144Example: Determine the cubic feet of cuttings drilled for one foot of

12-1/4 in. hole with 20% (0.20) porosity:Cubic feet = )2^L

x 0.7854(1 - 0.20)144

Cubic feet = 1 5 0 6 2 6 x 0.7854 x 0.80144

Cubic feet = 0.6547727

c) Total solids generated:W c g -350ChXL(I -P)SG

where Wcg = solids generated, poundsCh = capacity of hole, bbl/ftL = footage drilled, ftSG = specific gravity of cuttingsP = porosity, %

Example: Determine the total pounds of solids generated in drilling100ft of a 12-l/4in. hole (0.1458bbl/ft). Specific gravity ofcuttings = 2.40gm/cc. Porosity = 20%.

Wcg = 350 x 0.1458 x 100(1 - 0.20) x 2.4Wcg = 9797.26 pounds

Control Drilling

Maximum drilling rate (MDR), ft/hr, when drilling large diameterholes (14-3/4 in. and larger)

rn fmudwt _ mud wt^ j (circulation^MDR ft/hr = lout, ppg in, ppg J {rate, gpm J

Dh2

Example: Determine the MDR, ft/hr, necessary to keep the mud weightcoming out at 9.7 ppg at the flow line:

Data: Mud weight in = 9.0ppgCirculation rate = 530 gpmHole size = 17-1/2 in.

MDR, ft/hr = 6 7 C 9 - 7 - 9 - 0 ) 5 3 0

17.52

> ft/u 6 7 x 0 . 7 x 5 3 0MDR, ft/hr =

306.25

MDR = 81.16ft/hr

Buoyancy Factor (BF)

Buoyancy factor using mud weight, ppg

B F = 65.5 - mud weight, ppg65.5

Example: Determine the buoyancy factor for a 15.0ppg fluid:

BF = 65-5 - 15-65.5

BF = 0.77099

Buoyancy factor using mud weight, Ib/ft3

_ 489 - mud weight, lb/ft3B F =

489

Example: Determine the buoyancy factor for a 120 lb/ft3 fluid:

OI7 489 - 120

BF = 0.7546

Hydrostatic Pressure (HP)Decrease When Pulling Pipe out of the Hole

When pulling DRY pipe

Step 1

Barrels n u m b e r average pipe,. . i = of stands x length per x displacement

displaced A x , jf , , , , , ,

^ pulled stand, ft bbl/ftStep 2

HP, psi = barrels displaced x 0 052 x m u d

decrease /casing pipe "\ ' weight, ppgcapacity, - displacement,

[bbl/ft bbl/ft J

Example: Determine the hydrostatic pressure decrease when pulling DRYpipe out of the hole:

Number of stands pulled = 5Average length per stand = 92 ftPipe displacement = 0.0075 bbl/ftCasing capacity = 0.0773 bbl/ftMud weight = 11.5ppg

Stepl

Barrels 5 s t a n d s x 9 2 f t / s t d x 0.0075 bbl/ft

displacedBarrels ^ 3 4 5displaced

Step 2

HP,psi 3.45 barrelsdecrease = ^.0773 - 0.0075^ X -0 5 2 X 1 L 5 p p g

Ubl/ft bbl/ft JHP,psi 3.45 barrelsA

F = x 0.052 x 11.5ppgdecrease 0.0698

1 P = 29.56psidecrease

When pulling WET pipe

Step 1

R . number average fpipe disp., bbl/ft ^barrels

o f s t a n d g x l e n g t h p e r x +aispiacea

p u l l e d ^ ^ ft ^ p i p e c a p ? b b ] / f t j

Step 2

_ barrels displaced o n s ? mud

'P " ^casing A f pipe disp., bbl/ft^ weight,ppg

capacity, - +(,bbl/ft J l^ pipe cap., bbl/ft J

Example: Determine the hydrostatic pressure decrease when pullingWET pipe out of the hole:

Number of stands pulled = 5Average length per stand = 92 ftPipe displacement = 0.0075 bbl/ftPipe capacity =0.01776bbl/ftCasing capacity = 0.0773 bbl/ftMud weight = 11.5ppg

Stepl

f 0.0075 bbl/ft i^Barrels

5 s t a n d s x 9 2 f t / s t d x +d l s P l a c e d [o.O1776bbl/ftJBarrels

= H^6 1 9 6displaced

Step 2

HP, psi 11.6196 barrels

Ibbl/ft J [0.01776bbl/ftJHP, psi 1L6196, ^ = x 0.052 x 11.5ppg

decrease 0.05204TTp^

1 = 133.52psi

decrease

Loss of Overbalance Due to Falling Mud Level

Feet of pipe pulled DRY to lost overbalance_ overbalance, psi (casing cap. - pipe disp., bbl/ft)

mud wt., ppg x 0.052 x pipe disp., bbl/ftExample: Determine the FEET of DRY pipe that must be pulled to lose

the overbalance using the following data:

Amount of overbalance = 150psiCasing capacity = 0.0773 bbl/ftPipe displacement = 0.0075 bbl/ftMud weight =11.5 ppg

Feet of pipe pulled WET to lose overbalance

_ overbalance, psi x (casing cap. - pipe cap. - pipe disp.)mud wt., ppg x 0.052 x (pipe cap. + pipe disp., bbl/ft)

Example: Determine the feet of WET pipe that must be pulled to losethe overbalance using the following data:

Amount of overbalance = 150 psiCasing capacity = 0.0773 bbl/ftPipe capacity =0.01776 bbl/ftPipe displacement = 0.0075 bbl/ftMud weight =11.5 ppg

Metric calculations

_ j drilling fluid ^ metal displacement, ^ n nQQ1Pressure drop per , ^ ,

n x ^ x 0.0981A . . ^

F density, kg/1 1/m

meter tripping = r - : r~r;dry pipe, bar/m ^ m g C a p a C l t y ' - t a l d l s P l a c e m e n t '

, drilling fluid ^ metal displacement, v n nQQ1Pressure drop per , ..

u , x ^ x 0.098I. .

F F density, bar/m 1/m

meter tripping = : TT.:

, , , casing capacity, metal displacement,dry pipe, bar/m ^ -

y m

f metal disp., 1/m Adrilling fhaid

x + I 0 0 9 8 1Pressure drop per density, kg/1 I j

c k ^ Imeter tripping = ^ - F .

u J

-

^ . f T annular capacity, 1/mwet pipe, bar/m v J'

n . , /^ metal disp., 1/m ^drilling fluid I f^

Pressure drop per density, bar/m I 1 {i ymjmeter tripping = , v r-^ ^

. . u , annular capacity,

wet pipe, bar/m .. ^ J9Level drop for _ length of drill collars, m x metal disp., 1/mPOOH drill collars ~ casing capacity, 1/m


drilling fluid metal disp.,Pressure drop per density, kg/m* X mVmmeter tripping =

: ^f -rr.i . , j^ , casing capacity, metal disp., . . .dry pipe, kPa/m & ^ J - x 102m3/m m3/mA -Ii a -A /"metal disp., m3/m "\drilling fluid *_

Pressure drop per density, kg/m3 I p i p e c a p a c i t y ? m 3 / m

meter tripping = ; ^ :

wet pipe, kPa/m anfular c aP a c i ty' x 102nr/m

Level drop for POOH _ length of drill collars, m x metal disp., m3/mdrill collars, m casing capacity, m3/m

Formation Temperature (FT)

fambient ^ ,. x

o _ /temperature |N }

, a C e

+ OA ^increase 0F per ft of depth x TVD, ftJVtemperature, 0Fy v v v y

Example: If the temperature increase in a specific area is 0.012F/ft ofdepth and the ambient surface temperature is 700F, determinethe estimated formation temperature at a TVD of 15,000ft:

FT, 0F = 700F + (0.012F/ft x 15,000ft)FT, 0F = 700F + 1800FFT = 2500F (estimated formation temperature)

Hydraulic Horsepower (HHP)

HHP = i ^ r

1714where HHP = hydraulic horsepower

P = circulating pressure, psiQ = circulating rate, gpm

Example: circulating pressure = 2950 psicirculating rate = 520 gpm

HHP = 2 9 5 X 5 2 1714

HHP = 1^5 3 4 '0 0 01714

HHP = 894.98

Drill Pipe/Drill Collar Calculations

Capacities, bbl/ft, displacement, bbl/ft, and weight, lb/ft, can becalculated from the following formulas:

Capacity, bbl/ft = 1^-

T^ 1 UUI/ft O D > i n * 2 ~ I D > m ' 2

Displacement, bbl/ft =F 1029.4

Weight, lb/ft = displacement, bbl/ft x 27471b/bbl

Example: Determine the capacity, bbl/ft, displacement, bbl/ft, and weight,lb/ft, for the following:

Drill collar OD = 8.0in.Drill collar ID = 2-13/16in.Convert 13/16 to decimal equivalent:13 + 16 = 0.8125

2 81252a) Capacity, bbl/ft = ^ 2 9 ^

Capacity = 0.007684 bbl/ft8 O2 - 2 81252

b) Displacement, bbl/ft = :

T .^ 1 m i r 56.089844Displacement, bbl/ft =

y 1029.4

Displacement = 0.0544879 bbl/ftc) Weight, lb/ft = 0.0544879bbl/ftx27471b/bbl

Weight = 149.6781b/ft

Rule of thumb formulas

Weight, lb/ft, for REGULAR DRILL COLLARS can be approximatedusing the following formula:

Weight, lb/ft = (OD, in.2 - ID, in.2) 2.66

Example: Regular drill collarsDrill collar OD = 8.0in.Drill collar ID = 2-13/16in.Decimal equivalent = 2.8125 in.Weight, lb/ft - (8.02 - 2.81252) 2.66Weight, lb/ft = 56.089844 x 2.66Weight = 149.19898 lb/ft

Weight, lb/ft, for SPIRAL DRILL COLLARS can be approximated usingthe following formula:

Weight, lb/ft = (OD, in.2 - ID, in.2) 2.56

Example: Spiral drill collarsDrill collar OD = 8.0in.Drill collar ID = 2-13/16 in.Decimal equivalent = 2.8125 in.Weight, lb/ft = (8.02 - 2.81252) 2.56Weight, lb/ft = 56.089844 x 2.56Weight = 143.59 lb/ft

Pump Pressure/Pump Stroke Relationship(Also Called the Roughneck's Formula)

Basic formula

New circulating =

V^i{ x (^ new pump rate, spm V

pressure, psi pressure, psi ^ o l d P u mP r a t e ' s P m '

Example: Determine the new circulating pressure, psi using the follow-ing data:

Present circulating pressure = 1800 psiOld pump rate = 60spmNew pump rate = 30spm

New circulating = 1 8 0 0 j p O s p m f

pressure, psi p V60spmJ

Newcirculating = 1 8 0 s i x Q 2 5

pressure, psiNew circulating

= .

pressure

Determination of exact factor in above equation

The above formula is an approximation because the factor 2 is a rounded-off number. To determine the exact factor, obtain two pressure readings atdifferent pump rates and use the following formula:

_ log (pressure 1 -*- pressure 2)log (pump rate 1 * pump rate 2)

Example: Pressure 1 = 2500 psi @ 315 gpmPressure 2 = 450 psi @ 120 gpm

F a c t o r = log (2500psi + 450psi)log (315gpm + 120gpm)

Factor = to ^ 5 5 5 5 5 5 6 >log (2.625)

Factor = 1.7768

Example: Same example as above but with correct factor:

New circulating _ IOQQ pOspmV'7768pressure, psi " pS1 V60spmJ

New circulating = 18OOpsi x 0.2918299

pressure, psiNew circulating _ ^ c pressure

Metric calculation

new pump pressure with _ current (new SPM^jnew pump strokes, bar ~ pressure, bar ^

old SPM J

S.I. units calculation

new pump pressure with _ current (new SPM^jnew pump strokes, kPa ~ pressure, kPa ^

old SPM J

Cost per Foot

Example: Determine the drilling cost (Cx), dollars per foot, using thefollowing data:

Bit cost (B) = $2500Rig cost (CR) = $900/hourRotating time (T) = 65 hoursRound trip time (T) = 6 hours(for depth10,000 ft)Footage per bit (F) = 1300 ft

_ 2500 + 900(65 + 6)T " 1300

= 66,400T ~ 1300

CT = $51.08 per foot

Temperature Conversion Formulas

Convert temperature, Fahrenheit (F) to Centigrade or Celsius (C)

C0F - 32) 50C = ^ - -^- OR 0C = 0F - 32 x 0.5556

Example: Convert 95 0F to 0C:

0C = ( 9 5 ~ 3 2 ) 5 OR 0C = 95 - 32 x 0.5556

0C = 35 0C = 35

Convert temperature, Centigrade or Celsius (C) to Fahrenheit

0F = ( C X 9^ + 32 OR 0F = 0C x 1.8 + 32

Example: Convert 24 0C to 0F

oF = (24>^9) + 3 2 Q R oF = 2 4 x L 8 + 32

0F = 75.2 0F = 75.2

Convert temperature, Centigrade, Celsius (C) to Kelvin (K)0K = 0C H- 273.16

Example: Convert 35 0C to 0K:0K = 35 + 273.160K = 308.16

Convert temperature, Fahrenheit (F) to Rankine (R)0R = 0F + 459.69

Example: Convert 2600F to 0R:0R = 260 + 459.690R = 719.69

Rule of thumb formulas for temperature conversion

a) Convert 0F to 0C0C = 0F - 30 * 2

Example: Convert 95 0F to 0C:0C = 95 - 30 + 20C = 32.5

b) Convert 0C to 0F0F = 0C + 0C + 30

Example: Convert 24 0C to 0F:0F = 24 + 24 + 300F = 78

Front MatterTable of Contents1. Basic FormulasPressure GradientHydrostatic Pressure (HP)Converting Pressure into Mud WeightSpecific Gravity (SG)Equivalent Circulating Density (ECD)Maximum Allowable Mud Weight from Leak-off Test DataPump Output (PO)Annular Velocity (AV)Capacity FormulasControl DrillingBuoyancy Factor (BF)Hydrostatic Pressure (HP) Decrease When Pulling Pipe Out of the HoleLoss of Overbalance Due to Falling Mud LevelFormation Temperature (FT)Hydraulic Horsepower (HHP)Drill Pipe/Drill Collar CalculationsPump Pressure/Pump Stroke Relationship (Also Called the Roughneck's Formula)Cost per FootTemperature Conversion Formulas

Index

1 basic formulas

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