chapter one basic formulas

C H A P T E R O N E

BASIC FORMULAS

Pressure Gradient

Pressure gradient, psilft, using mud weight, ppg

psi/ft = m u d weight , p p g x 0.052

Example: 12 .0ppg fluid

psi/ft = 12 .0ppg x 0.052 psi/ft = 0.624

Pressure gradient, psi/ft, using mud weight, lblft 3

psi/fl = m u d weight , lb/ft 3 • 0 .006944

Example: 100 lb/f l 3 fluid

psi/ft = 1001b/ft 3 • 0 .006944 psi/ft = 0 .6944

O R

psi/fl = m u d weight , lb/fl3 + 144

Example." 100 lb/f t 3 fluid

psi/ft = 1001b/ft3 + 144 psi/ft = 0 .6944

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

psi/ft = m u d weight , S G • 0.433

Example: 1.0 S G fluid

psi/ft = 1 . 0 S G x 0.433 psi/ft = 0.433

2 Formulas and Calculations

Metric calculations

Pressure gradient , ba r /m = dri l l ing fluid densi ty kg/l x 0.0981 Pressure gradient , ba r /10m = dri l l ing fluid densi ty kg/1 x 0.981

S.I. units calculations

Pre s su re grad ien t , kPa /m = dri l l ing fluid densi ty, kghn 3 + 102

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

ppg = pressure gradient , psi/ft + 0.052

Example: 0.4992 psi/ft

ppg - 0.4992psi /f t + 0.052 ppg - 9.6

Convert pressure gradient, psffft, to mud weight, lblft a

lb/ft 3 = pressure gradient , psi/ft + 0.006944


lb[ft 3 = 0.6944psi/ft + 0.006944 lb/ft 3 = 100

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

S G = pressure gradient , psi/fl + 0.433


S G = 0.433psi/ft + 0.433 S G = 1.0

Metric calculations

Dri l l ing fluid density, kg/1 = pressure gradient , ba r /m - 0.0981 Dri l l ing fluid density, kg/1 - pressure gradient , ba r /10m - 0.981

Basic Formulas 3


Dri l l ing fluid density, kg/m 3 = pressure gradient , kP a / m • 102

Hydrostatic Pressure (HP)

Hydrostatic pressure using ppg and feet as the units of measure

H P = m u d weight, ppg x 0.052 x t rue vert ical dep th (TVD), ft

Example." m u d weight = 13.5 ppg t rue vert ical dep th = 12,000 ft

H P = 13.5ppg x 0.052 x 12,000ft

H P = 8424psi

Hydrostatic pressure, psi, using pressure gradient, psilft

H P = psi/ft x t rue vert ical depth, fl

Example." pressure gradient = 0. 624 psi/ft t rue vert ical dep th = 850Oft

H P = 0. 624 psi/ft x 8500ft

H P = 5304psi

Hydrostatic pressure, psi, using mud weight, lbfft 3

H P = m u d weight, lb/ft 3 x 0.006944 x TVD, ft

Example." m u d weight = 90 lb/ft 3 t rue vert ical dep th = 7500 ft

H P = 901b/ft 3 x 0.006944 x 7500ft

H P = 4687psi

Hydrostatic pressure, psi, using meters as unit of depth

H P = m u d weight, ppg x 0.052 x T V D , m x 3.281

Example." m u d weight = 12 .2ppg true vert ical dep th = 3700 mete rs


H P = 12 .2ppg x 0.052 • 3700 • 3.281

H P = 7701psi

Metric calculations

H y d r o s t a t i c _ dri l l ing fluid - x 0.0981 x

pressure , ba r densi ty, kg/1

t rue vert ical

depth , m


H y d r o s t a t i c _ dri l l ing fluid density, kg/m 3 B

pressure, k P a 102 x t rue vert ical dep th , m

Converting Pressure into Mud Weight

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

M u d weight , ppg = pressure , psi + 0.052 + T V D , ft

Example." pressure = 2600psi t rue vert ical dep th = 5000f t

Mud , ppg = 2600psi + 0.052 + 5000ft

M u d = 10.0ppg

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

M u d weight, ppg = pressure , psi + 0.052 + T V D , m + 3.281

Example: pressure = 3583 psi t rue vert ical dep th = 2000 mete r s

M u d wt, ppg = 3583psi + 0.052 + 2 0 0 0 m + 3.281

M u d wt = 10.5ppg

Basic Formulas 5

Metric calculations

E q u i v a l e n t dr i l l ing pressure , = - 0.0981 +

fluid densi ty , kg/1 b a r

t rue ver t ica l

dep th , m


E q u i v a l e n t dr i l l ing _ pressure , x 102 + fluid densi ty , kg/m3 - k P a

t rue ver t ica l

dep th , m

Specific Gravity (SG)

Specific gravity using mud weight, ppg

S G = m u d weight , p p g + 8.33

Example: 15.0 p p g fluid

S G = 15 .0ppg + 8.33

S G = 1.8

Specific gravity using pressure gradient, psilft

S G = p ressure g rad ien t , psi/ft + 0.433

Example: pressure g r a d i e n t = 0.624 psi /f t

S G = 0. 624 psi/ft + 0.433

S G = 1.44

Specific gravity using mud weight, lblft 3

S G = m u d weight , lb/ft 3 - 62.4

Example." m u d we igh t = 1201b/ft 3

S G = 1201b/ft 3 + 62.4

S G = 1.92


Convert specific gravity to mud weight, ppg

M u d weight, ppg = specific gravi ty x 8.33

Example." specific gravity = 1.80

M u d wt, ppg = 1.80 x 8.33 M u d wt = 15.0ppg

Convert specific gravity to pressure gradient, psi/ft

psi/ft = specific gravi ty x 0.433


psi/ft = 1.44 x 0.433 psi/ft = 0.624

Convert specific gravity to mud weight, lblft 3

lb/ft 3 = specific gravi ty x 62.4


lb/ft 3 = 1.92 x 62.4

lb/ft 3 = 120

Equivalent Circulating Density (ECD), ppg

ECD, ppg = a n n u l a r pressure / loss, p s i )

m u d weight , ] + 0.052 + TVD, ft + \ i n use, ppg )

Example." annu l a r pressure loss = 200 psi t rue vert ical dep th = 10,000 ft m u d weight = 9 .6ppg

ECD, ppg = 200psi + 0.052 + 10,000ft + 9 . 6 p p g E C D = 10.0ppg

Basic Formulas 7

Metric calculation

Equivalent drilling _ annular pressure fluid density, kg/1 loss, bar

+ 0.0981 + TVD, m + mud wt, kg/1


Equivalent circulating _ annular pressure loss, kPa x 102 density, kg/1 - TVD, m

+ mud density, kg/m

Maximum Allowable Mud Weight from Leak-off Test Data

=( leak-of f / + 0 . 0 5 2 (casing shoe'] (mud weight,) PPg \pressure, psi - / , T V D , ft J + \ ppg

Example." leak-off test pressure = 1140 psi casing shoe TVD = 4000 ft mud weight = 10.0 ppg

ppg = l l40psi + 0.052 + 4000ft + 10.0ppg ppg = 15.48

Pump Output (PO)

Triplex Pump

Formula I

(liner )2 PO, bbl/stk = 0.000243 x k, diameter, in. (stroke in.)

x \length,

Example: Determine the pump output, bbl/stk, at 100% efficiency for a 7-in. by 12-in. triplex pump:

PO @ 100% = 0.000243 x 72 x 12 PO @ 100% = 0.142884bbl/stk

Adjust the pump output for 95% efficiency:


Decimal equivalent = 95 + 100 - 0.95

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

Formula 2

PO, gpm = [3 (D 2 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. triplex pump at 80 strokes per minute:

PO, gpm = [3 (72 x 0.7854)1210.00411 x 80 PO, gpm = 1385.4456 x 0.00411 x 80 PO = 455.5gpm

Duplex Pump

Formula 1

( l iner )2 ( s t r o k e ) 0.000324 • (d iameter , in. • ( length, in. =

stro e ) -0.000162 x / ,d iamete r , in. x / , length , in. =

p u m p output @ 100% eft =

bbl/stk

bbl/stk

bbl/stk

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

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

Pump output @ 100% eft = 0.128142 bbl/stk

Adjust p u m p ou tpu t for 85% efficiency:

Decimal equivalent = 85 + 100 = 0.85

Basic Formulas 9

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

Formula 2

PO, bbl/stk : 0.000162 x S [2(D) 2 - d 2]

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. duplex p u m p @ 100% efficiency. Rod d iameter = 2.0in.:

PO @ 100% : 0.000162 x 14 x [2 (5.5) 2 - 22]

PO @ 100% = 0.000162 x 14 x 56.5 PO @ 100% = 0.128142 bbl/stk

Adjust p u m p ou tpu t for 85% efficiency:

P 0 @ 8 5 % = O.128142bbl/stk x 0.85 PO @ 85% = O. 10892 bbl/stk

Metric calculation

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

S.I. units calculation

Pum p output , m3/min = pump output , liter/stk x p u m p speed, spm

Annular Velocity (AV)

Annular velocity (AV), ftimin

Formula 1

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


Example: p u m p o u t p u t = 12 .6bbl /min a n n u l a r capaci ty = 0.1261 bbl/ft

A V = 12.6bbl /min + 0.1261bbl/f t

A V = 99.92 ft /min

Formula 2

AV, ft /min = 24.5 x Q

D h 2 _ D p 2

where Q = c i rcula t ion rate, g p m D h = inside d i ame te r o f casing or hole size, in. D p = outs ide d i ame te r o f pipe, tub ing o r collars, in.

Example." p u m p o u t p u t = 5 3 0 g p m hole size = 12-1/4in. p ipe O D = 4-1/2 in.

24.5 • 530 A V =

12.252 - 4.52

A V = 12,985

129.8125

A V = 100ft /min

Formula 3

AV, ft /min = PO, bbl /min x 1029.4

D h 2 _ D p 2

Example." p u m p o u t p u t = 12 .6bbl /min hole size = 12-1/4 in. p ipe O D = 4-1/2in.

12 .6bbl /min • 1029.4 A V =

12.252 - 4.52

A V = 12970.44

129.8125

A V = 99.92 f t /min

Basic Formulas 11

Annular velocity (AV), ft/sec

AV, ft/sec = 17.16 • PO, bbl/min

Dh 2 _ Dp 2

Example." p u m p ou tpu t = 12.6bbl/min hole size = 12-1/4 in. pipe O D = 4-1/2 in.

17.16 • 12.6bbl/min A V =

12.252 - 4.52

A V = 216.216

129.8125

AV = 1.6656 ft/sec

Metric calculations

Annular velocity, m/min = pump output , liter/rnin + annular volume, lkn

Annular velocity, m/sec = pump output , liter/min + 60 + annular volume, l/m


Annula r velocity, m/min = p u m p ou tpu t , m3/min + annu la r volume, m3/m

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

Pump output , gpm = AV, ft/min (Dh 2 - D P 2)

24.5

where AV = desired annular velocity, f t /min Dh = inside diameter of casing or hole size, in. Dp = outside d iameter of pipe, tubing or collars, in.

Example." desired annular velocity = 120 ft/min hole size = 12-1/4in. pipe O D = 4-1/2in.

P O = 120 (12.252 - 4.52)

24.5


P O = 120 • 129.8125

24.5

P O = 15,577.5

24.5

PO = 635 .8gpm

Strokes per minute (SPM) required for a given annular velocity

SPM = annu la r velocity, f t /min x annu l a r capacity, bbl/ft

p u m p output , bbl/stk

Example." annu la r velocity = 120 f t /min annu la r capaci ty = 0.1261 bbl/ft D h = 12-1/4in. D p = 4-1/2 in. p u m p o u t p u t = 0.136 bbl/stk

S P M = 120ft/min x 0.1261bbl/ft

0 .136bbl/s tk

S P M = 15.132

0.136

SPM = 111.3

Capacity Formulas

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

a) A n n u l a r capacity, bbl/ft = D h 2 _ D p 2

1029.4

Example." Hole size (Dh) = 12-1/4in. Dril l pipe O D (Dp) = 5.0in.

12.252 - 5.02 A n n u l a r capacity, bbl/ft =

1029.4

A n n u l a r capaci ty = 0.12149 bbl/ft

Basic Formulas 13

1029.4 b) Annula r capacity, fffbbl = (1.)11"'--2 _ 1_) ) " p 2 "

Example." Hole size (Dh) = 12-1/4in. Drill pipe O D (Dp) = 5.0in.

1029.4 Annula r capacity, ft/bbl =

(12.252 - 5.02)

Annula r capacity - 8.23 ft/bbl

c) Annula r capacity, gal/fl = Dh 2 - D p 2

24.51

Example." Hole size (Dh) = 12-1/4 in. Drill pipe O D (Dp) = 5.0in.

12.252 - 5.02 Annula r capacity, gal/ft =

24.51

Annula r capacity = 5.1 gal/ft

d) Annula r capacity, ft/gal = 24.51

(Dh 2 - D p 2)


24.51 Annula r capacity, ft/gal =

(12.252 - 5.02)

Annula r capacity = 0.19598 ft/gal

e) Annula r capacity, fl3/linfl = Dh 2 _ Dp 2

183.35


12.252 - 5.02 Annula r capacity, ft3/linft =

183.35

Annula r capacity = 0.682097 ft3/linft


f) Annula r capacity, linft/fl 3 = 183.35

(Dh 2 - Dp 2)


183.35 Annu la r capacity, linft/ft 3 =

(12.25 - 5.o )

Annula r capaci ty = 1.4661inft/ft 3

Annular capacity between casing and multiple strings of tubing

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

Annu la r capacity, bbl/ft = 1029.4

Example." Using two strings of tubing of same size: Dh = cas ing- -7 .0 in . - -29 lb/ft ID = 6.184 in. T~ = tubing No. 1--2-3/8in. O D = 2.375in. T2 = tubing No. 2--2-3/8 in. O D = 2.375 in.

6.1842 - (2.3752 + 2.3752) Annula r capacity, bbl/ft =

1029.4

Annula r capacity, bbl/fl = 38.24 - 11.28

1029.4

Annula r capaci ty = 0.02619 bbl/ft

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

Annu la r capacity, ft/bbl = 1029.4

Example: Using two strings of tubing of same size: D h = cas ing--7 .0 in . - -29 lb/ft ID = 6.184 in. T~ = tubing No. l h 2 - 3 / 8 i n . O D = 2.375in. T2 = tubing No. 2 h 2 - 3 / 8 in. O D = 2.375 in.

Annu la r capacity, ft/bbl = 1029.4

6.1842 - (2.3752 + 2.3752)

Basic Formulas 15

Annula r capacity, ft/bbl = 1029.4

38.24 - 11.28

Annula r capacity = 38.1816ft/bbl

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

Annula r capacity, gal/ft = Oh - +

24.51

Example: Using two tubing strings of different size: Dh = cas ingm7.0 in . - -29 lb/ft ID = 6.184 in. T~ = tubing No. l h 2 - 3 / 8 i n . O D = 2.375in. T2 = tubing No. 2- -3-1/2 in . O D = 3.5in.

Annula r capacity, gal/fl = 6.1842 - (2.3752 + 3.52)

24.51

Annula r capacity, gal/ft = 38.24 - 17.89

24.51

Annular capacity = 0.8302733gal/ft

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

Annula r capacity, ft/gal = 24.51

Dh - +

Example." Using two tubing strings of different sizes: Dh = cas ing--7 .0 in . - -29 lb/ft ID = 6.184 in. T~ = tubing No. l h 2 - 3 / 8 i n . O D = 2.375in. T2 = tubing No. 2- -3-1/2 in . O D = 3.5in.

Annular capacity, ft/gal = 24.51

6.1842 - (2.3752 + 3.52)

Annula r capacity, ft/gal = 24.51

38.24 - 17.89

Annula r capacity = 1.2044226ft/gal


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

Annular capacity, ft3/linft = Dh 2 - [ (T1) 2 +(T2) 21

183.35

Example." Using three strings of tubing: D h = casing--9-5/8 in . - -47 lb/ft ID = 8.681 in. T~ = tubing No. 1--3-112in. O D = 3.5 in. T2 = tubing No. 2--3-112in. O D = 3.5 in. T3 = tubing No. 3--3-112in. O D = 3.5 in.

8.6812 - (3.52 + 3.52 + 3.52) Annula r capacity =

183.35

Annular capacity, fl3/linfl = 183.35

7 5 . 3 5 9 - 36.75

Annular capacity = 0.2105795 ft3/linft

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

183.35 Annular capacity, linft/ft 3 =

Dh 2 _ [(T1)2 + (T 2)2]

Example." Using three strings tubing of same size: D h = casing--9-5/8 in . - -47 lb/ft ID = 8.681 in. TI = tubing No. 1--3-1/2in. O D = 3.5in. T2 = tubing No. 2--3-1/2in. O D = 3.5 in. T3 = tubing No. 3--3-1/2in. O D = 3.5 in.

183.35 Annula r capacity =

8.6812 - (3.52 + 3.52 + 3.52)

183.35 Annular capacity, linft/ft 3 =

7 5 . 3 5 9 - 36.75

Annular capacity = 4.74879931inft/ft 3

Uapacity of tubulars and open hole: drill pipe, drill collars, tubing, :asing, hole, and any cylindrical object

~) Capacity, bbl/ft = ID, in. 2 1029.4

Basic Formulas 17

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

Capacity, bbl/ft =

12.252

1029.4

Capacity = 0.1457766 bbl/ft

b) Capacity, ft/bbl = 1029.4 Dh 2

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

1029.4 Capacity, ft/bbl =

12.252

Capacity = 6.8598 ft/bbl

c) Capacity, gal/ft = ID, in. 2

24.51

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

Capacity, gal/ft = 8.5 2

24.51

Capacity = 2.9477764gal/ft

d) Capacity, ft/gal = 24.51

ID, in. 2

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

Capacity, ft/gal = 24.51 8.52

Capacity = 0.3392 ft/gal

e) Capacity, ft3/linft = ID 2

183.35

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

Capacity, ft3/linft = 6.02

183.35


Capac i ty = 0.1963 ft3/linft

f) Capaci ty , linft/ft 3 = 183.35 ID, in. 2

Example." D e t e r m i n e the capacity, l inft/ft 3, for a 6.0in. hole"

Capaci ty , linft/ft 3 = 183.35 6.02

Capac i ty = 5.09305 linft/ft 3

Amount of cuttings drilled per foot of hole drilled

a) B A R R E L S of cut t ings dri l led per foot of hole drilled:

D h 2 Barrels - ~ (1 - % poros i ty)

1029.4

Example: D e t e r m i n e the n u m b e r o f barre ls o f cut t ings dri l led for one foot o f 12-1/4in.-hole dri l led with 20% (0.20) poros i ty :

12.252 Barrels = ~ (1 - 0.20)

1029.4

Barrels = 0.1457766 x 0.80

Barrels = 0.1166213

b) C U B I C F E E T of cut t ings dri l led per foot o f hole drilled:

D h 2 Cubic feet =

144 x 0.7854 (1 - % poros i ty )

Example: D e t e r m i n e the cubic feet o f cut t ings dri l led for one foot of 12-1/4in. hole with 20% (0.20) porosi ty:

Cubic feet = 12.252

144 x 0.7854 (1 - 0.20)

Cubic feet = 150.0626

144 • 0.7854 • 0.80

Cubic feet = 0.6547727

Basic Formulas 19

c) Total solids generated:

Wcg -- 350 Ch x L (1- P) SG

where Wcg -~ solids generated, pounds Ch = capacity of hole, bbl/ft L = footage drilled, ft SG = specific gravity of cuttings P = porosity, %

Example." Determine the total pounds of solids generated in drilling 100ft of a 12-1/4in. hole (0.1458bbl/ft). Specific gravity of cuttings = 2.40grn/cc. Porosity = 20%.

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

Wcg = 9797.26 pounds

Control Drilling

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

MDR, ft/hr =

67 • (mud wt _ mud wt~ • (circulation~ \out , ppg in, ppg j \ ra te , gpm j

Dh 2

Example." Determine the MDR, ft/hr, necessary to keep the mud weight coming out at 9.7 ppg at the flow line"

Data: Mud weight in - 9.0ppg Circulation rate - 530gpm Hole size = 17-1/2 in.

67 (9.7 - 9.0) 530 MDR, ft/hr -

17.52

67 x 0.7 x 530 MDR, ft/hr =

306.25

MDR, ft/hr - 24,857 306.25

M D R = 81.16ft/hr


Buoyancy Factor (BF)

Buoyancy factor using mud weight, ppg

B F = 65.5 - mud weight, ppg

65.5

Example." Determine the buoyancy factor for a 15.0ppg fluid"

B F = 65.5 - 15.0

65.5

BF = 0.77099

Buoyancy factor using mud weight, lb/ft 3

B F = 489 - mud weight, lb/ft 3

489

Example." Determine the buoyancy factor for a 120 lb/ft 3 fluid:

B F = 4 8 9 - 120

489

BF = 0.7546

Hydrostatic Pressure ( H P )

Decrease When Pulling Pipe out of the Hole

When pulling DRY pipe

Step 1

Barrels displaced

number average pipe = of stands x length per x displacement

pulled stand, ft bbl/ft

Step 2

HP, psi _ n

decrease barrels displaced

iasing pipe a p a c l t y , - displacement,] bl/ft bbl/ft )

x 0.052 x mud weight, ppg

Basic Formulas 21

Example." Determine the hydrostatic pressure decrease when pulling DRY pipe out of the hole:

Number of stands pulled = 5 Average length per stand = 92 ft Pipe displacement = 0.0075 bbl/ft Casing capacity = 0.0773 bbl/ft Mud weight = 11.5 ppg

Step 1

Barrels displaced

Barrels displaced

= 5 stands x 92ft/std x 0.0075bbl/ft

= 3.45

Step 2

HP, psi _ 3.45 barrels decrease (0.0773 - 0.0075"~

~bbgft bbgft J

x 0.052 x 11.5ppg

HP, psi _ 3.45 barrels E

decrease 0.0698 x 0.052 x 11.5ppg

HP = 29.56 psi decrease

When pulling WET pipe

Step 1

Barrels displaced

number average /pipe disp., bbl/ft] = of stands x length per x +

pulled stand, ft \pipe cap., bbl/ft )

Step 2

barrels displaced

HP' psi = ( c a s i n g ) /pipe disp" b b l / f t ~ ~ , [ c a p a c i t Y , b b l / f t - ~ pipe cap.,+ bbl/ft

x 0.052 x mud weight, ppg


Example." Determine the hydrostatic pressure decrease when pulling W E T pipe out of the hole"

Number of stands pulled = 5 Average length per stand = 92ft Pipe displacement = 0.0075 bbl/ft Pipe capacity = 0.01776 bbl/ft Casing capacity = 0.0773 bbl/ft Mud weight = 11.5 ppg

Step 1

Barrels displaced

/~ ~176 / = 5 stands x 92ft/std x .01776bbl/ft)

Barrels = 11.6196

displaced

Step 2

HP, psi _ decrease

11.6196 barrels

/o l0 ~176 / bbl/ft ~,0.01776 bbl/ft)

HP, psi _ 11.6196 m

decrease 0.05204 x 0.052 x 11.5ppg

x 0.052 • 11.5ppg

HP = 133.52psi

decrease

Loss of Overbalance Due to Falling Mud Level

Feet of pipe pulled DRY to lost overbalance

Feet = overbalance, psi (casing c a p . - pipe disp., bbl/ft)

mud wt., ppg x 0.052 x pipe disp., bbl/fl

Example." Determine the F E E T of DRY pipe that must be pulled to lose the overbalance using the following data:

Basic Formulas 23

Amount of overbalance = 150psi Casing capacity = 0.0773 bbl/ft Pipe displacement = 0.0075 bbl/ft Mud weight = 11.5 ppg

150psi (0.0773 - 0.0075) F t =

11.5ppg x 0.052 x 0.0075

10.47 F t =

0.004485

Ft = 2334

Feet of pipe pulled WET to lose overbalance

overbalance, psi • (casing c a p . - pipe c a p . - pipe disp.) Feet =

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 lose the overbalance using the following data:

Amount of overbalance = 150psi Casing capacity = 0.0773 bbl/ft Pipe capacity = 0.01776 bbl/ft Pipe displacement = 0.0075bbl/ft Mud weight = 11.5 ppg

Feet = 150psi x (0.0073 - 0.01776 - 0.0075bbl/ft)

l l .5ppg x 0.052 (0.01776 + 0.0075bbl/ft)

Feet = 150 psi x 0.05204

11.5 ppg x 0.052 x 0.02526

Feet = 7.806

0.0151054

Feet = 516.8

Metric calculations

Pressure drop per meter tripping dry pipe, bar/m

drilling fluid density, kg/l

metal displacement, x 0.0981 1/m

casing capacity, 1/m

metal displacement, l/m


Pressure drop per meter tripping dry pipe, bar/m

drilling fluid density, bar/m

metal displacement, x 0.0981 I/m

casing capacity, l/m

metal displacement, l/m

Pressure drop per meter tripping wet pipe, bar/m

Pressure drop per meter tripping wet pipe, bar/m

drilling fluid density, kg/1

metal di~p., 1/m ) x 0.0981

pipe capacity, l/m annular capacity, l/m

drilling fluid density, bar/m

metal di~p., 1/m

pipe capacity, l/m) annular capacity, l/m

Level drop for POOH drill collars

length of drill collars, m x metal disp., 1/m casing capacity, 1/m


Pressure drop per meter tripping dry pipe, kPa/m

drilling fluid density, kg/m 3

metal disp., m3/m

casing capacity, m3/m

metal disp., x 102 m3/m

Pressure drop per meter tripping wet pipe, kPa/m

drilling fluid density, kg/m 3

metal dis~, m3/m

pipe capacity, m3/111) annular capacity, • 102 m3/m

Level drop for POOH drill collars, m

length of drill collars, m x metal disp., m3/m casing capacity, m3/m

Formation Temperature (FT)

FT, ~ = (ambient / surface temperature, ~

(temperature ) + k, increase ~ per ft of depth x TVD, ft

Basic Formulas 25

Example." If the temperature increase in a specific area is 0 .012~ of depth and the ambient surface temperature is 70~ determine the estimated formation temperature at a T V D of 15,000 ft"

FT, ~ = 70~ + (0.012~ x 15,000fl) FT, ~ = 70~ + 180~ FT = 250~ (estimated format ion temperature)

Hydraulic Horsepower (HHP)

H H P = P x Q

1714

where H H P = hydraulic horsepower P = circulating pressure, psi Q = circulating rate, gpm

Example." circulating pressure = 2950 psi circulating rate = 520 gpm

2950 • 520 H H P =

1714

H H P = 1,534,000 1714

H H P = 894.98

Drill Pipe/Drill Collar Calculations

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

Capacity, bbl/fl =

ID, in. 2

1029.4

Displacement, bbl/ft = OD, in} - ID, in. 2 1029.4

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


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

Drill collar OD = 8.0in. Drill collar ID = 2-13/16 in.

Convert 13/16 to decimal equivalent:

13 - 16 = 0.8125

2.81252 a) Capacity, bbl/ft =

1029.4

Capaci ty = 0.007684bbl/ft

8.02 _ 2.81252 b) Displacement, bbl/ft =

1029.4

56.089844 Displacement, bbl/ft =

1029.4

Displacement = 0.0544879 bbl/ft

c) Weight, lb/ft = 0.0544879bbl/ft x 2747 lb/bbl Weight = 149.678 lb/ft

Rule of thumb formulas

Weight, lb/ft, for R E G U L A R D R I L L C O L L A R S can be approximated using the following formula:

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

Example: Regular drill collars

Drill collar O D = 8.0in. Drill collar ID = 2-13/16in. Decimal equivalent = 2.8125 in.

Weight, lb/ft = (8.02 - 2.81252) 2.66 Weight, lb/ft = 56.089844 x 2.66 Weight = 149.19898 lb/ft

Weight, lb/ft, for SPIRAL D R I L L C O L L A R S can be approximated using the following formula:

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

Basic Formulas 27

Example: Spiral drill collars

Drill collar OD - 8.0in. Drill collar ID - 2-13/16 in. Decimal equivalent = 2.8125 in.

Weight, lb/ft = (8.02 - 2.81252) 2.56 Weight, lb/ft = 56.089844 x 2.56 Weight = 143.59 lb/ft

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

Basic formula

present New circulating = circulating pressure, psi pressure, psi

new pump rate, spm/2 •

old pump rate, spm

Example." Determine the new circulating pressure, psi using the following data:

Present circulating pressure = 1800psi Old pump rate - 60spm New pump rate - 30 spm

New circulating = 1800psi ( 3 0 s p a / 2 pressure, psi 60spm

New circulating _ 1800psi x 0.25 pressure, psi

New circulating _ 450psi pressure

Determination of exact factor in above equation

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

Fac tor = log (pressure 1 - pressure 2)

log (pump rate 1 + pump rate 2)


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

Factor = log (2500psi + 450psi) log (315gpm + 120gpm)

Factor = log (5.5555556) log (2.625)

Factor = 1.7768

Example." Same example as above but with correct factor:

New circula.ting = 1800psi (30spm)l7768 pressure, psi 60spm

New circulating = 1800psi x 0.2918299 pressure, psi

New circulating = 525psi pressure

Metric calculation

new pump pressure with new pump strokes, bar

current = X

pressure, bar new S P M )2

old SPM

S.I. units calculation

new pump pressure with new pump strokes, kPa

2 current (new SPM) pressure, kPa x old SPM

Cost per Foot

C T = B + CR (t + T)

Example." Determine the drilling cost (Cx), dollars per foot, using the following data:

Basic Formulas 29

Bit cost (B) = $2500 Rig cost (CR) = $900/hour Rota t ing t ime (T) = 65 hours R o u n d trip t ime (T) = 6 hours (for d e p t h m l 0 , 0 0 0 ft) Foo tage per bit (F) = 1300ft

2500 + 900 (65 + 6) C'r = 1300

66, 400 C T =

1300

C T = $51.08 per foot

Temperature Conversion Formulas

Convert temperature, ~ (F) to ~ or Celsius (C)

o c = (~ - 32) 5 O R ~ = ~ - 32 x 0.5556

Example." Conver t 95 ~ to ~

~ (95 - 32) 5

O R ~ = 95 - 32 • 0.5556

~ = 35 ~ = 35

Convert temperature, ~ or Celsius (C) to ~

OF = (~ x 9)

5 + 3 2 O R ~ 1 7 6 2 1 5 1.8 + 3 2

Example." Conver t 24 ~ to ~

OF = (24 • 9) 5

+ 3 2 O R ~ 2 1 5

~ = 75.2 ~ = 75.2

Convert temperature, ~ Celsius (C) to ~ (K)

~ = ~ + 273.16


Example: C o n v e r t 35 ~ to ~

~ = 35 + 273 .16

~ = 308 .16

Convert temperature, ~ (F) to ~ (R)

~ = ~ + 459 .69

Example." C o n v e r t 2 6 0 ~ to ~

~ = 260 + 459 .69

~ = 719.69

Rule of thumb formulas for temperature conversion

a) C o n v e r t ~ to ~

~ = ~ - 30 - 2

Example: C o n v e r t 95 ~ to ~

~ = 95 - 3 0 + 2

~ = 32.5

b) C o n v e r t ~ to ~

~ = ~ + ~ + 30

Example." C o n v e r t 24 ~ to ~

~ = 24 + 24 + 30 OF = 78

chapter one basic formulas

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