Weighted Cement Calculations

Amount of high density additive required per sack of cement to achieve a required cement slurry density

(wt x 11.207983, .. .„ ,, „„n - + (wt x CW) - 94 - (8.33 x CW)

where x = additive required, pounds per sack of cement Wt = required slurry density, lb/gal SGc = specific gravity of cement CW = water requirement of cement AW = water requirement of additive SGa = specific gravity of additive

Water Requirement Specific

Additive gal/94 lb/sk Gravity

Hematite 0.34 5.02

Ilmenite 0 4.67

Barite 2.5 4.23

Sand 0 2.63 API Cements

Example: Determine how much hematite, lb/sk of cement would be required to increase the density of Class H cement to 17.51b/gal:

Water requirement of cement = 4.3gal/sk

Water requirement of additive (hematite) = 0.34gal/sk Specific gravity of cement =3.14

Specific gravity of additive (hematite) = 5.02

_ 62.4649 + 75.25 - 94 - 35.819 X ~ 1.0034 - 0.418494 - 0.0595

7.8959

0.525406

x = 15.11b of hematite per sk of cement used

Calculations for the Number of Sacks of Cement Required

If the number of feet to be cemented is known, use the following:

Step 1

Determine the following capacities:

a) Annular capacity, ft3/ft:

y J 183.35

b) Casing capacity, ft3/ft:

c) Casing capacity, bbl/ft:

Casing capacity, bbl/ft =-

Step 2

Determine the number of sacks of LEAD or FILLER cement required:

required cemented ft4 LEAD cement

Step 3

Determine the number of sacks of TAIL or NEAT cement required:

Sacks feet annular . . , „ 3/ , required = to be x capacity, x excess -s- ^^ ^s annulus cemented ft3/ft cement

Sacks no. of feet casing required = between float x capacity, + * . '

>1 o i_ r, -t TAIL cement casing collar & shoe ft3/ft

Total Sacks of TAIL cement required:

Sacks = sacks required in annulus + sacks required in casing

Step 4

Determine the casing capacity down to the float collar:

Casing . uu1/f, feet of casing

Z. ,,,= casing capacity, bbl/ft x , _ 6„ capacity, bbl 6 F to the float collar

Step 5

Determine the number of strokes required to bump the plug: Strokes = casing capacity, bbl pump output, bbl/stk

Example: From the data listed below determine the following:

  1. How many sacks of LEAD cement will be required?
  2. How many sacks of TAIL cement will be required?
  3. How many barrels of mud will be required to bump the plug?
  4. How many strokes will be required to bump the top plug?

Data: Casing setting depth = 3000 ft

Casing ID = 12.615in.

Float collar (number of feet above shoe) = 44 ft Pump (5-1/2in. by 14in. duplex @ 90%efT) = 0.112bbl/stk

Cement program: LEAD cement (13.8 lb/gal) = 2000 ft slurry yield = 1.59 ft3/sk

TAIL cement (15.8 lb/gal) = 1000ft slurry yield = 1.15ft7sk

Excess volume = 50%

Step 1

Determine the following capacities:

a) Annular capacity, ftVft:

183.35

, 127.35938 Annular capacity, ft /ft = 35

Annular capacity = 0.6946ft3/ft b) Casing capacity, ft3/ft:

Casing capacity, ft7ft = -

159.13823 Casing capacity, ft /ft = ^ 35

Casing capacity = 0.8679ft3/ft c) Casing capacity, bbl/ft:

Casing capacity, bbl/ft = ]Q29 4

159.13823 Casing capacity, bbl/ft = l0294

Casing capacity = 0.1545 bbl/ft Step 2

Determine the number of sacks of LEAD or FILLER cement required: Sacks required = 2000ft x 0.6946ft3/ft x 1.50 - 1.59ft3/sk Sacks required =1311

Step 3

Determine the number of sacks of TAIL or NEAT cement required:

Sacks required annulus = 1000ft x 0.6946ft3/ft x 1.50 + 1.15ft3/sk Sacks required annulus = 906

Sacks required casing = 44ft x 0.8679ft3/ft + 1.15ft3/sk Sacks required casing =33

Total sacks of TAIL cement required:

Step 4

Determine the barrels of mud required to bump the top plug:

Casing capacity, bbl = (3000 ft - 44 ft) x 0.1545 bbl/ft Casing capacity = 456.7 bbl

Step 5

Determine the number of strokes required to bump the top plug:

Strokes = 456.7bbl h- 0.112bbl/stk Strokes = 4078

__Calculations for the Number of Feet to Be Cemented

If the number of sacks of cement is known, use the following: Step 1

Determine the following capacities:

a) Annular capacity, ft3/ft:

b) Casing capacity, ft3/ft:

Casing capacity, ft3/ft =

183.35

Step 2

Determine the slurry volume, ft3

Slurry _ number of sacks of slurry yield, vol, ft3 cement to be used ft3/sk

Step 3

Determine the amount of cement, ft3, to be left in casing: Cement in _ f feet of setting depth of "i (casing casing, ft3 leasing cementing tool, ft J ^capacity, ft3/ft Step 4

Determine the height of cement in the annulus—feet of cement:

'slurry cement (annular '

Feet = vol, - remaining in + capacity, excess Jt3 casing, ft3 J ftVft

Step 5

Determine the depth of the top of the cement in the annulus:

„ ^ casing setting ft of cement Depth, ft = , ,fc „ 6 - .

depth, ft in annulus

Step 6

Determine the number of barrels of mud required to displace the cement:

Barrels =

ft of drill pipe capacity, drill pipe bbl/ft Step 7

Determine the number of strokes required to displace the cement:

Strokes - ^ re1uired to ^ pump output, displace cement ' bbl/stk

Example: From the data listed below, determine the following:

  1. Height, ft, of the cement in the annulus
  2. Amount, ft3, of the cement in the casing
  3. Depth, ft, of the top of the cement in the annulus
  4. Number of barrels of mud required to displace the cement
  5. Number of strokes required to displace the cement

Data: Casing setting depth = 3000ft

Casing ID = 12.615in.

Pump (7in. by 12in. triplex @ 95% eff.) = 0.136bbl/stk Cementing tool (number of feet above shoe) = 100 ft

Cementing program: NEAT cement = 500sk

Slurry yield = 1.15ft3/sk Excess volume = 50%

Step 1

Determine the following capacities:

  1. Annular capacity between casing and hole, ft3/ft:
  2. 52 - 13.3752

Annular capacity, ft3/ft = Annular capacity, ft'/ft

183.35 127.35938

183.35

Annular capacity = 0.6946 ft3/ft b) Casing capacity, ft3/ft:

12.6152

Casing capacity, ft3/ft = Casing capacity, ft3/ft =

183.35

159.13823 183.35

Casing capacity = 0.8679 ft3/ft

Step 2

Determine the slurry volume, ft3:

Slurry vol, ft3 = 500 sk x 1.15 ft3/sk Slurry vol = 575 ft3

Step 3

Determine the amount of cement, ft3, to be left in the casing:

Cement in casing, ft3 = (3000 ft - 2900 ft) x 0.8679 ft3/ft Cement in casing, ft3 = 86.79 ft3

Step 4

Determine the height of the cement in the annulus—feet of cement:

Feet = (575ft3 - 86.79ft3) h- 0.6946ft3/ft - 1.50 Feet = 468.58

Step 5

Determine the depth of the top of the cement in the annulus:

Step 6

Determine the number of barrels of mud required to displace the cement:

Barrels = 2900ft x 0.01776bbl/ft Barrels = 51.5

Step 7

Determine the number of strokes required to displace the cement:

Strokes = 51.5 bbl - 0.136bbl/stk Strokes = 379

Step 1

Determine the following capacities:

a) Annular capacity, ft3/ft, between pipe or tubing and hole or casing:

cmc Dh, in.2 - Dp, in.2 Annular capacity, ft /ft = -

183.35

b) Annular capacity, ft/bbl, between pipe or tubing and hole or casing:

1029.4

Annular capacity, ft/bbl =-r-7

c) Hole or casing capacity, ft3/ft: Hole or capacity, ft3/ft =

183.35

d) Drill pipe or tubing capacity, ft3/ft: Drill pipe or tubing capacity, ft3/ft =

183.35

e) Drill pipe or tubing capacity, bbl/ft:

Drill pipe or tubing capacity, bbl/ft = 94

Step 2

Determine the number of SACKS of cement required for a given length of plug, OR determine the FEET of plug for a given number of sacks of cement:

a) Determine the number of SACKS of cement required for a given length of plug:

Sacks plug ■ slurry

of = length, x x excess + yield, cement ft ftS/ft" ^ ft3/sk

NOTE: If no excess is to be used, omit the excess step.

b) Determine the number of FEET of plug for a given number of sacks of cement:

, hole or sacks slurry casi

NOTE: If no excess is to be used, omit the excess step. Step 3

Determine the spacer volume (usually water), bbl, to be pumped behind the slurry to balance the plug:

Spacer .. spacer tubing i lli = capacity, excess x ' , , x .

vol, bbl vol ahead, bbl capacity, ft/bbl bbl/ft NOTE: if no excess is to be used, omit the excess step. Step 4

Determine the plug length, ft, before the pipe is withdrawn:

Plug sacks slurry annular length, = of x yield, + capacity, x excess + .

ft cement ft3/sk ft3/ft ft?ft

NOTE: If no excess is to be used, omit the excess step. Step 5

Determine the fluid volume, bbl, required to spot the plug:

length plug or spacer vol

Vol, bbl = of P'Pe - length, x tubln? - behind slurry or tubing, . capacity, ...

ft bbl/ft

Example I: A 300 ft plug is to be placed at a depth of 5000ft. The open hole size is 8-1/2in. and the drill pipe is 3-1/2in.—13.3lb/ft; ID—2.764in. Ten barrels of water are to be pumped ahead of the slurry. Use a slurry yield of 1.15 ft3/sk. Use 25% as excess slurry volume:

Determine the following:

  1. Number of sacks of cement required
  2. Volume of water to be pumped behind the slurry to balance the plug
  3. Plug length before the pipe is withdrawn
  4. Amount of mud required to spot the plug plus the spacer behind the plug

Step 1

Determined the following capacities:

a) Annular capacity between drill pipe and hole, ft3/ft:

183.35

Annular capacity = 0.3272 ft7ft b) Annular capacity between drill pipe and hole, ft/bbl:

Annular capacity = 17.1569 ft/bbl c) Hole capacity, ft'/ft:

8 52

F 3 183.35

Hole capacity = 0.3941 ft3/ft d) Drill pipe capacity, bbl/ft:

2.7642

Drill pipe capacity, bbl/ft =

1029.4

Drill pipe capacity = 0.00742 bbl/ft e) Drill pipe capacity, ft3/ft Drill pipe capacity, ft3/ft Drill pipe capacity

Step 2

Determine the number of sacks of cement required:

Sacks of cement = 300ft x 0.3941 ft3/ft x 1.25 -s- 1.15ft3/sk Sacks of cement =129

= 0.0417ft3/ft

Step 3

Determine the spacer volume (water), bbl, to be pumped behind the slurry to balance the plug:

Spacer vol, bbl = 17.1569ft/bbl + 1.25 x lObbl x 0.00742bbl/ft Spacer vol = 1.018 bbl

Step 4

Determine the plug length, ft, before the pipe is withdrawn:

r.1 . .u ft fl29 1.15 ^ f0.3272 , ^ 0.04 m Plug length, ft = xft3/skj.^3/ft x 1.25 + ft3/ft j

Plug length = 329 ft

Step 5

Determine the fluid volume, bbl, required to spot the plug: Vol, bbl = [(5000 ft - 329 ft) x 0.00742 bbl/ft] - 1.0 bbl Vol, bbl = 34.66bbl - l.Obbl Volume = 33.6 bbl

Example 2: Determine the number of FEET of plug for a given number of SACKS of cement:

A cement plug with lOOsk of cement is to be used in an 81/2 in, hole. Use 1.15 ft3/sk for the cement slurry yield. The capacity of 8-1/2in. hole = 0.3941 ftVft. Use 50% as excess slurry volume:

Feet = lOOsk x 1.15ftVsk - 0.3941 ft3/ft - 1.50 Feet = 194.5

Differential Hydrostatic Pressure Between Cement in __the Annulus and Mud Inside the Casing_

  1. Determine the hydrostatic pressure exerted by the cement and any mud remaining in the annulus.
  2. Determine the hydrostatic pressure exerted by the mud and cement remaining in the casing.
  3. Determine the differential pressure.

Example: 9-5/8in. casing — 43.5lb/ft in 12-l/4in. hole:

Well depth = 8000 ft

Cementing program:

LEAD slurry 2000ft = 13.8lb/gal

TAIL slurry 1000ft = 15.8lb/gal

Mud weight = lO.Olb/gal

Float collar (No. of feet above shoe) = 44 ft

Determine the total hydrostatic pressure of cement and mud in the annulus a) Hydrostatic pressure of mud in annulus:

HP, psi = lO.Olb/gal x 0.052 x 5000ft HP = 2600 psi b) Hydrostatic pressure of LEAD cement:

HP. psi = 13.8 lb/gal x 0.052 x 2000 ft HP = 1435 psi c) Hydrostatic pressure of TAIL cement:

HP, psi = 15.8 lb/gal x 0.052 x 1000ft HP = 822 psi d) Total hydrostatic pressure in annulus:

Determine the total pressure inside the casing a) Pressure exerted by the mud:

HP, psi = 10.0 lb/gal x 0.052 x (8000 ft - 44 ft) HP = 4137psi b) Pressure exerted by the cement:

HP, psi = 15.8 lb/gal x 0.052 x 44ft HP = 36 psi c) Total pressure inside the casing:

Differential pressure

_Hydraulicing Casing_

These calculations will determine if the casing will hydraulic out (move upward) when cementing

Determine the difference in pressure gradient, psi/ft, between the cement and the mud psi/ft = (cement wt, ppg - mud wt, ppg) x 0.052

Determine the differential pressure (DP) between the cement and the mud

DP, psi = .. ... x casing length, ft pressure gradients, psi/ft

Determine the area, sq in., below the shoe

Determine the Upward Force (F), lb. This is the weight, total force, acting at the bottom of the shoe

,, differential pressure rorce, lb = area, sq in. x , ^ . . .

between cement and mud, psi

Determine the Downward Force (W), lb. This is the weight of the casing

Weight, lb = casing wt, lb/ft x length, ft x buoyancy factor Determine the difference in force, lb

Differential force, lb = upward force, lb - downward force, lb

Pressure required to balance the forces so that the casing will not hydraulic out (move upward)

Mud weight increase to balance pressure

Mud wt, ppg = Prc*ssure required ^ 0.052 -s- casing length, ft to balance forces, psi

New mud weight, ppg

Check the forces with the new mud weight a) psi/ft = (cement wt, ppg - mud wt, ppg) x 0.052

  1. psi = difference in pressure gradients, psi/ft x casing length, ft c) Upward force, lb = pressure, psi x area, sq in.
  2. Difference in , r ,, , , r ., r „ = upward force, lb - downward force, lb force, lb

Example: Casing size = 13 3/8 in. 54 lb/ft Cement weight = 15.8ppg Mud weight = 8.8ppg Buoyancy factor = 0.8656 Well depth = 164 ft (50 m)

Determine the difference in pressure gradient, psi/ft, between the cement and the mud psi/ft = (15.8 - 8.8) x 0.052 psi/ft = 0.364

Determine the differential pressure between the cement and the mud psi = 0.364psi/ft x 164ft psi = 60

Determine the area, sq in., below the shoe area, sq in. = 13.3752 x 0.78 54 area, = 140.5 sq in.

Determine the upward force. This is the total force acting at the bottom of the shoe

Determine the downward force. This is the weight of casing

Weight, lb = 54.5 lb/ft x 164ft x 0.8656 Weight = 77371b

Determine the difference in force, lb

Differential force, lb = downward force, lb - upward force, lb Differential force, lb = 77371b - 84301b Differential force = -693 lb

Therefore: Unless the casing is tied down or stuck, it could hydraulic out (move upward).

Pressure required to balance the forces so that the casing will not hydraulic out (move upward)

Mud weight increase to balance pressure

Mud wt, ppg = 4.9psi -h 0.052 h- 164ft Mud wt = 0.57ppg

New mud weight, ppg

New mud wt, ppg = 8.8ppg + 0.6ppg New mud wt = 9.4ppg

Check the forces with the new mud weight a) psi/ft = (15.8 - 9.4) x 0.052 psi/ft = 0.3328

  1. psi = 0.3328psi/ft x 164ft psi = 54.58
  2. Upward force, lb = 54.58psi x 140.5 sq in. Upward force = 76681b d) Differential = downward force - upward force force, lb 77371b 76681b

Differential = +69 lb force

Depth of a Washout

Method 1

Pump soft line or other plugging material down the drill pipe and note how many strokes are required before the pump pressure increases.

Depth of strokes pUtmpt driUpipe

, „ = , x output, capacity, washout, ft required ,.. , .

bbl/stk bbl/ft

Example: Drill pipe = 3-1/2in.—13.3 lb/ft capacity = 0.00742 bbl/ft Pump output = 0.112bbl/stk (5-l/2in. by 14in. duplex @ 90% efficiency)

NOTE: A pressure increase was noted after 360 strokes.

Depth of = 36Qstk x Q ] 12bbl/stk ■*■ 0.00742bbl/ft washout, ft

Depth of =5434ft washout

Method 2

Pump some material that will go through the washout, up the annulus and over the shale shaker. This material must be of the type that can be easily observed as it comes across the shaker. Examples: carbide, corn starch, glass beads, brightly colored paint, etc.

Depth of washout, ft strokes required pump x output, bbl/stk drill pipe capacity, bbl/ft +

annular capacity, bbl/ft

Example: Drill pipe = 3-1/2in. 13.31b/ft capacity = 0.00742 bbl/ft Pump output = 0.112bbl/stk (5-l/2in. x 14in. duplex @ 90% efficiency)

Annulus hole size = 8-l/2in.

NOTE: The material pumped down the drill pipe came over the shaker after 2680 strokes.

Drill pipe capacity plus annular capacity:

Depth of = 2680stk x o.l 12bbl/stk + 0.0657 bbl/ft washout, ft

Depth of =4569ft washout

Lost Returns—Loss of Overbalance

Number of feet of water in annulus

Feet = water added, bbl + annular capacity, bbl/ft

Bottomhole (BHP) pressure reduction

BHP (mud wt, wt of A n fftof j ■ = x 0-052 x , , , decrease, psi vppg water, ppg,.' V water added

Equivalent mud weight at TD

EMW, ppg = mud wt, ppg - (BHP decrease, psi h- 0.052 + TVD, ft)

Example: Mud weight = 12.5ppg Weight of water = 8.33 ppg TVD = 10,000 ft

Annular capacity = 0.1279bbl/ft (12-1/4 x 5.0in.) Water added = 150 bbl required to fill annulus

Number of feet of water in annulus

Bottomhole pressure decrease

BHP decrease, psi = (12.5ppg - 8.33ppg) x 0.052 x 1173ft BHP decrease = 254 psi

Equivalent mud weight at TD

EMW, ppg = 12.5 - (254psi + 0.052 + 10,000ft) EMW = 12.0ppg

_Stuck Pipe Calculations_

Determine the feet of free pipe and the free point constant Method 1

The depth at which the pipe is stuck and the number of feet of free pipe can be estimated by the drill pipe stretch table below and the following formula.

Table 2-2 Drill Pipe Stretch Table

Stretch

Table 2-2 Drill Pipe Stretch Table

Stretch

Nominal

Wall

Constant

Free

ID,

Weight,

ID,

Area,

in/1000 lb

Point

in.

lb/ft

in.

sqin.

/1000ft

constant

2-3/8

4.85

1.995

1.304

0.30675

3260.0

6.65

1.815

1.843

0.21704

4607.7

2-7/8

6.85

2.241

1.812

0.22075

4530.0

10.40

2.151

2.858

0.13996

7145.0

3-1/2

9.50

2.992

2.590

0.15444

6475.0

13.30

2.764

3.621

0.11047

9052.5

15.50

2.602

4.304

0.09294

10760.0

4.0

11.85

3.476

3.077

0.13000

7692.5

14.00

3.340

3.805

0.10512

9512.5

4-1/2

13.75

3.958

3.600

0.11111

9000.0

16.60

3.826

4.407

0.09076

11017.5

18.10

3.754

4.836

0.08271

12090.0

20.00

3.640

5.498

0.07275

13745.0

5.0

16.25

4.408

4.374

0.09145

10935.0

19.50

4.276

5.275

0.07583

13187.5

5-1/2

21.90

4.778

5.828

0.06863

14570.0

24.70

4.670

6.630

0.06033

16575.0

6-5/8

25.20

5.965

6.526

0.06129

16315.0

Feet of _ stretch, in. x free point constant free pipe pull force in thousands of pounds

Example: 3-l/2in. 13.30lb/ft drill pipe

20in. of stretch with 35,0001b of pull force

From drill pipe stretch table:

Free point constant = 9052.5 for 3-l/2in. drill pipe 13.301b/ft

fFeetof = 5173ft tree pipe

Determine free point constant (FPC)

The free point constant can be determined for any type of steel drill pipe if the outside diameter, in., and inside diameter, in., are known:

where As = pipe wall cross sectional area, sq in.

Example 1: From the drill pipe stretch table:

FPC = (4.52 - 3.8262 x 0.7854) x 2500 FPC = 4.407 x 2500 FPC = 11,017.5

Example 2: Determine the free point constant and the depth the pipe is stuck using the following data:

2-3/8 in. tubing—6.5 lb/ft—ID = 2.441 in. 25 in. of stretch with 20,0001b of pull force a) Determine free point constant (FPC):

FPC = (2.8752 - 2.4412 x 0.7854) x 2500 FPC = 1.820 x 2500 FPC = 4530

b) Determine the depth of stuck pipe:

feel0f = 5663ft tree pipe

Method 2

differential pull, lb where e = pipe stretch, in.

Wdp = drill pipe weight, lb/ft (plain end)

Plain end weight, lb/ft, is the weight of drill pipe excluding tool joints:

Weight, lb/ft = 2.67 x pipe OD, in.2 - pipe; ID, in.2

Example: Determine the feet of free pipe using the following data: 5.0in. drill pipe; ID^t.276in.; 19.5lb/ft Differential stretch of pipe = 24in. Differential pull to obtain stretch = 30,0001b

Weight, lb/ft = 2.67 x (5.02 - 4.2762) Weight = 17.93 lb/ft

F F 30,000

Free pipe = 10,547 ft

Determine the height, ft, of unweighted spotting fluid that will balance formation pressure in the annulus:

a) Determine the difference in pressure gradient, psi/ft, between the mud weight and the spotting fluid:

psi/ft = (mud wt, ppg - spotting fluid wt, ppg) x 0.052

b) Determine the height, ft, of unweighted spotting fluid that will balance formation pressure in the annulus:

Hei ht ft - amount ^ difference in overbalance, psi ' pressure gradient, psi/ft

Example: Use the following data to determine the height, ft, of spotting fluid that will balance formation pressure in the annulus:

Data: Mud weight = 11.2ppg

Weight of spotting fluid = 7.0ppg Amount of overbalance = 225.0psi a) Difference in pressure gradient, psi/ft:

psi/ft = (11.2ppg - 7.0ppg) x 0.052 psi/ft = 0.2184

b) Determine the height, ft, of unweighted spotting fluid that will balance formation pressure in the annulus:

Height, ft = 225 psi + 0.2184psi/ft Height = 1030 ft

Therefore: Less than 1030 ft of spotting fluid should be used to maintain a safety factor that will prevent a kick or blowout.

______Calculations Required for Spotting Pills_____

The following will be determined:

a) Barrels of spotting fluid (pill) required b) Pump strokes required to spot the pill

Step 1

Determine the annular capacity, bbl/ft, for drill pipe and drill collars in the annulus:

1029.4

Step 2

Determine the volume of pill required in the annulus: Vol, bbl = annular cap., bbl/ft x section length, ft x washout factor

Step 3

Determine total volume, bbl, of spotting fluid (pill) required:

Barrels = Barrels required in annulus plus barrels to be left in drill string

Step 4

Determine drill string capacity, bbl: Barrels = drill pipe/drill collar capacity, bbl/ft x length, ft

Step 5

Determine strokes required to pump pill: Strokes = vol of pill, bbl + pump output, bbl/stk

Step 6

Determine number of barrels required to chase pill:

Barrels - slr'nS _ vo' left in drill vol, bbl string, bbl

Step 7

Determine strokes required to chase pill:

. , pump strokes required , bbl required . . . , Strokes = t jjase -|| output, + to displace bbl/stk surface system

Step 8

Total strokes required to spot the pill:

^ , , , , strokes required strokes required Total strokes = .„ + . ?..

to pump pill to chase pill

Example: Drill collars are differentially stuck. Use the following data to spot an oil-based pill around the drill collars plus 200 ft (optional) above the collars. Leave 24 bbl in the drill string:

Data: well depth = 10,000 ft

Washout factor = 20%

Drill pipe = 5.0in.—19.5lb/ft capacity = 0.01776 bbl/ft length = 9400 ft

capacity = 0.0061 bbl/ft length = 600 ft

Pump output =0.117 bbl/stk

Strokes required to displace surface system from suction tank to the drill pipe = 80stk.

Step 1

Annular capacity around drill pipe and drill collars:

a) Annular capacity around drill collars:

Annular capacity, bbl/ft =-

Annular capacity = 0.02914 bbl/ft b) Annular capacity around drill pipe:

Annular capacity, bbl/ft =-

Annular capacity = 0.0459 bbl/ft

Step 2

Determine total volume of pill required in annulus:

a) Volume opposite drill collars:

Vol, bbl = 0.02914bbl/ft x 600ft x 1.20 Vol = 21.0 bbl b) Volume opposite drill pipe:

Vol, bbl = 0.0459bbl/ft x 200ft x 1.20 Vol = 11.0 bbl c) Total volume, bbl, required in annulus:

Step 3

Total bbl of spotting fluid (pill) required:

Barrels = 32.0 bbl (annulus) + 24.0 bbl (drill pipe) Barrels = 56.0 bbl

Step 4

Determine drill string capacity:

a) Drill collar capacity, bbl:

Capacity, bbl = 0.0062 bbl/ft x 600 ft Capacity = 3.72 bbl b) Drill pipe capacity, bbl:

Capacity, bbl = 0.01776 bbl/ft x 9400 ft Capacity = 166.94bbl c) Total drill string capacity, bbl:

Capacity, bbl = 3.72bbl + 166.94bbl Capacity = 170.6bbl

Step 5

Determine strokes required to pump pill:

Strokes = 56 bbl + 0.117bbl/stk Strokes = 479

Step 6

Determine bbl required to chase pill:

Barrels = 170.6bbl - 24bbl Barrels = 146.6

Step 7

Determine strokes required to chase pill:

Strokes = 146.6bbl + 0.117 bbl/stk + 80stk Strokes = 1333

Step 8

Determine strokes required to spot the pill:

Total strokes = 479 +1333 Total strokes =1812

___Pressure Required to Break Circulation

Pressure required to overcome the mud's gel strength inside the drill string

where Pgs = pressure required to break gel strength, psi y = lOmin gel strength of drilling fluid, lb/100 sq ft d = inside diameter of drill pipe, in. L = length of drill string, ft

Example:

Pgs = (10 - 300 + 4.276) 12,000 ft Pgs = 0.007795 x 12,000ft Pgs = 93.5 psi

Therefore, approximately 94 psi would be required to break circulation.

Pressure required to overcome the mud's gel strength in the annulus

where Pgs = pressure required to break gel strength, psi L = length of drill string, ft y = lOmin. gel strength of drilling fluid, lb/100 sq ft Dh = hole diameter, in. Dp = pipe diameter, in.

Example: L = 12,000 ft y = 10 lb/100 sq ft Dh = 12-l/4in. Dp = 5.0in.

Pgs =10-5- [300 x (12.25 - 5.0)] x 12,000 ft Pgs = 10 + 2175 x 12,000ft Pgs = 55.2 psi

Therefore, approximately 55 psi would be required to break circulation.

References

API Specification for Oil-Well Cements and Cement Additives, American Petroleum Institute, New York, N.Y., 1972.

Chenevert, Martin E. and Reuven Hollo, 77-39 Drilling Engineering Manual, PennWell Publishing Company, Tulsa, 1981.

Crammer Jr., John L., Basic Drilling Engineering Manual, PennWell Publishing Company, Tulsa, 1983.

Drilling Manual, International Association of Drilling Contractors, Houston, Texas, 1982.

Murchison, Bill, Murchison Drilling Schools Operations Drilling Technology and Well Control Manual, Albuquerque, New Mexico.

Oil-Well Cements and Cement Additives, API Specification 10A, December 1979.

CHAPTER THREE

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