Oilwell Cementing Calculation

transcript

Single Stage Calculations

2 Single Stage Cementing

1. Casing in casing

Open Hole

PreviousCasing

Collar

Volumes to be Calculated : Step A

1. Casing in casing

2. Casing in open hole

Open Hole

PreviousCasing

Collar

1. Casing in casing

2. Casing in open hole

3. Inside casing (between collar and shoe)

Open Hole

PreviousCasing

Collar

Cement Calculations

Volume of open hole below the casing shoe is not calculated.

Cement circulates out of the shoe and up the annulus.

Any cement displacing the mud below by gravity is considered negligible.

Problem 1

Casing : 7” - 32 LB/Ft

Shoe : 4600’

Collar : 4510’

Previous casing : 9 5/8 - 47 LB/Ft

Shoe : 2670’

Open hole : 8.5”

Top of cement : 1000’

Calculate Total Amount of Slurry Required in CuFt

SolutionCapacities

1. Cas-Cas - 0.1438 cuft/ft

2. Cas - OH - 0.1268 cuft/ft

3. Cas - 0.2025 cuft/ft, 0.0361 bbl/ft

Volumes

1000ft

2670ft

4510ft

4600ft

SolutionCapacities

2. Cas - OH - 0.1268 cuft/ft

3. Cas - 0.2025 cuft/ft, 0.0361 bbl/ft

Volumes

1. (2670ft - 1000ft) x 0.1438 cuft/ft = 240.2 cuft

1000ft

2670ft

4510ft

4600ft

SolutionCapacities

2. Cas - OH - 0.1268 cuft/ft

3. Cas - 0.2025 cuft/ft, 0.0361 bbl/ft

Volumes

1000ft

2670ft

4510ft

4600ft

SolutionCapacities

2. Cas - OH - 0.1268 cuft/ft

3. Cas - 0.2025 cuft/ft, 0.0361 bbl/ft

Volumes

1000ft

2670ft

4510ft

4600ft

SolutionCapacities

2. Cas - OH - 0.1268 cuft/ft

3. Cas - 0.2025 cuft/ft, 0.0361 bbl/ft

Volumes

TOTAL VOLUME = 503.1 CUFT

1000ft

2670ft

4510ft

4600ft

Excess

Excess in only calculated on AnnularVolume in THE OPEN HOLE

OnlyHere

Example: Problem 1

OH 50% Excess

OH volume (2) of the previous problem:244.7 cuft

Excess Volume: 244.7 cuft x 50/100 = 122.35 cuft

Total OH Volume: 367.05 cuft

OR 244.7 cuft x 1.50 = 367.05 cuft

Slurry Yield - Step B

The volume of slurry produced when one sack of cement ismixed with water is called

Slurry Yield - Step BThe volume of slurry produced when one sack of cement is

mixed with water is called YIELD

+CEMENTCEMENT

Slurry Yield - Step BThe volume of slurry produced when one sack of cement isThe volume of slurry produced when one sack of cement is

mixed with water is calledmixed with water is called YIELDYIELD

WATER =+CEMENTCEMENT

Slurry Yield - Step BThe volume of slurry produced when one sack of cement isThe volume of slurry produced when one sack of cement is

mixed with water is calledmixed with water is called YIELDYIELD

WATER =

The unit of slurry yield is cuft/sackThe unit of slurry yield is cuft/sack

+ SLURRYSLURRYCEMENTCEMENT

1 sack of cement of 94 lbs

4.97 gals of water

= 1 cuft= 1 cuft

= 0.66 cuft= 0.66 cuft

1 sack of cement1 sack of cement of 94 lbsof 94 lbs

4.97 gals of water4.97 gals of water

1 cuft + 0.66 cuft1 cuft + 0.66 cuft

= 1 cuft= 1 cuft

= 0.66 cuft= 0.66 cuft

= 1.66 cuft= 1.66 cuft

1 sack of cement1 sack of cement of 94 lbsof 94 lbs

4.97 gals of water4.97 gals of water

1 cuft + 0.66 cuft1 cuft + 0.66 cuft

= 1 cuft= 1 cuft

= 0.66 cuft= 0.66 cuft

= 1.66 cuft= 1.66 cuft

1.15 cuft/sack1.15 cuft/sack

But the But the Handbook states:Handbook states:

WHY ?WHY ?

Bulk vs Absolute

One drum full One drum full of dry cementof dry cement(1 drum = 1 cuft)(1 drum = 1 cuft)

Same size drumSame size drum0.66 of water0.66 of water

Bulk vs Absolute

One drum full One drum full of dry cementof dry cement(1 drum = 1 cuft)(1 drum = 1 cuft)

Same size drumSame size drum0.66 of water0.66 of water

Air in pore spaces willbe displaced by water

Pour waterPour waterinto cementinto cement

Yield varies with density

Basically, when you change the density you alter the

Water / Cement ratio

....this changes the yield

Normally ........

as you increase the water content the yield increases

Step C - Cement Calculations

If you know what volume of slurry a sack of cement [YIELD] produces at a certain weight, you can calculate the amount of sacks required when you know a volume.

Step C - Cement Calculations

If you know what volume of slurry a sack of cement [YIELD] produces at a certain weight, you can calculate the amount of sacks required when you know a volume.

VOLUME[cuft]YIELD [cuft/sacks]

= Sacks

Example: Problem 1 (Continued)

Previous Problem:Lead volume [1] : 240.2 cuft Tail volume [2 + 3] : 367.05 cuft + 18.2 cuft 385.35 cuft

Example: Problem 1 (Continued)Previous Problem:

Lead volume [1] : 240.2 cuft Tail volume [2 + 3] : 367.05 cuft

+ 18.2 cuft 385.35 cuft

Use:Lead Slurry : Class G + 6% D20 at 13.9 ppgTail Slurry Neat : Class G at 15.8 ppg

Cement volume Lead:

+ 18.2 cuft 385.35 cuftUse:Lead Slurry : 6% D20 at 13.9 ppgTail Slurry Neat : 15.8 ppgCement volume Lead: 240.2 cuft = 145.2 sacks 1.655 cuft/sack

Cement volume Tail:

+ 18.2 cuft 385.35 cuftUse:Lead Slurry : 6% D20 at 13.9 ppgTail Slurry Neat : 15.8 ppgCement volume Lead: 240.2 cuft = 145.2 sacks 1.655 cuft/sack

385.25 cuft = 335.0 sacks 1.15 cuft/sack

Cement volume Tail:

Mix Water - Step D

If you know the number of sacks then you find the gal/sk of mix water required and multiply

Mix Fluid = gal/sack x number of sacks

CONVERT TO BBLS

(Your displacement tanks are in bbls)

Mix Water Quality

In the laboratory, clean location water is used for cement test. Test results are only obtainable in the field when same water (= clean) is used.

Causes for contamination:– Dirty rig tanks– Left over chemicals/mud in lines and pumps– Dirty displacement tanks

Total Water Available 1. Calculated mix water for cement + 20%

safety

2. Water (ahead, behind, CW, spacer)

20 bbls minimum (priming and washing up)

= TOTAL WATER NEEDED

Lead mix water 145sx x 8.56 gal/sx

Lead mix water 145sx x 8.56 gal/sx = 1241.2 gal = 29.55 bbls

42 gal/bbl

Tail mix water 335sx x 4.97 ga/sx

42 gal/bbl

Tail mix water 335sx x 4.97 ga/sx = 1664.95 gal = 39.64 bbls

42 gal/bbl

Lead mix water 145sx x 8.56 gal/sx = 1241.2 gal

= 29.55 bbls

42 gal/bbl

Tail mix water 335sx x 4.97 ga/sx = 1664.95 gal

= 39.64 bbls

42 gal/bbl

Total mix water 69.19 bbls

= 29.55 bbls

42 gal/bbl

Tail mix water 335sx x 4.97 ga/sx = 1664.95

gal = 39.64 bbls

42 gal/bbl

20% safety margin : 66.19 bbl x 20/100 = 13.84 bbls

= 29.55 bbls

42 gal/bbl

Tail mix water 335sx x 4.97 ga/sx = 1664.95

gal = 39.64 bbls

42 gal/bbl

No water ahead, behind, CW, spacer 0 bbls

Lead mix water 145sx x 8.56 gal/sx = 1241.2 gal =

29.55 bbls

42 gal/bbl

Tail mix water 335sx x 4.97 ga/sx = 1664.95 gal =

39.64 bbls

42 gal/bbl

Priming, washing up 20 bbls

Lead mix water 145sx x 8.56 gal/sx = 1241.2 gal = 29.55

42 gal/bbl

Tail mix water 335sx x 4.97 ga/sx = 1664.95 gal =

39.64 bbls

42 gal/bbl

Priming, washing up 20 bbls

Total water needed 103 bbls

Additives - Step E

Solid Additives Calculated as a % by weight of dry cement.

– Except: Salt– Small amounts will not effect calculation

Liquid Additives Calculated as gallons of additives per sack of bulk

cement Subtract total amount of liquid additives from

amount of mix fluid to have real amount of mix water.

Example: Problem 1 (Continued)Lead 6% Bentonite BWOC (145 sks)

Tail 0.2 gal/sack of Dispersant (335 sks)Solid Additives

145 sacks x 94 lbs/sack x 6/100 = 817.8 lbsLiquid Additives

335 sacks x 0.2 gal/sack = 67 galNote : Mix water of tail slurry will change

Mix fluid: 335sx x 4.97 gal/sk =1664.95 galMix water = Mix fluid - liquid additives 1664.95 gal - 67 gal = 38.1 bbl 42 gal/bbl

Displacement

To FLOAT COLLAR unless otherwise stated

Exceptions

1. STAB-IN CEMENTING:– Displacement to 10-15’ before stab-in shoe– Check returns - OK - pull out and circulate

2. GUIDE SHOE:– Displace to 20-30’ before shoe

NEVER OVERDISPLACE

Displacement

To FLOAT COLLAR unless otherwise stated

Exceptions

1. STAB-IN CEMENTING:– Displacement to 10-15’ before stab-in shoe– Check returns - OK - pull out and circulate

2. GUIDE SHOE:– Displace to 20-30’ before shoe

NEVER OVERDISPLACE

Estimated Job Time - Step G MIXING TIME + DISPLACEMENT TIME + 10 MINUTES (DROP PLUGS ...) = ESTIMATION OF JOB TIME 1 HOUR (SAFETY)

= MINIMUM THICKENING TIME

Estimated Job Time - Problem 1 (Continued)

Mixing time + Displacement time + Plugs mixing time:

Lead 42.8 : 4 bpm 10.5 mins 11mins Tail 68.6 : 8bpm 8.5 mins 9 mins

Displacement: 4510ft x 0.0361 bbl/ft = 162.8 bbl : 12 bpm 13.6 mins 14 mins Dropping Plugs 10 mins Total 44 mins

To allow a safe margin to do this job we expect at least 1 hr extra for the cement to be still pumpable after it is put in place.

ie, 1hrs 44 mins

Examples - Example 1Calculate : Slurry Volumes, Cement Volumes,

Mix Water, Additives, Displacement

Job Time

Well Data: 9 5/8” 53.5 lb/ft casing, shoe at 9800’,

collar at 9760’, OH 12 1/4” + 25% excess,

previous casing 13 3/8” 68 lb/ft at 2900’

Slurries: Lead: 5 bpm 9300’ - surface

Class “G” @ 12.8 ppg

12% BWOC Bentonite

0.3% BWOC D13

Tail: 3 bpm 9800’ - 9300’

Class “G” Neat @ 15.8 ppg

0.05% BWOC D28

0.5% BWOC D65

Displacement: 8 bpm

Examples - Example 1Calculate : Slurry Volumes, Cement Volumes,

Mix Water, Additives, Displacement

Job Time

Well Data: 9 5/8” 53.5 lb/ft casing, shoe at 9800’,

collar at 9760’, OH 12 1/4” + 25% excess,

previous casing 13 3/8” 68 lb/ft at 2900’

Sluuries: Lead: 5 bpm 9300’ - surface

Class “G” @ 12.8 ppg

12% BWOC Bentonite

0.3% BWOC D13

Tail: 3 bpm 9800’ - 9300’

Class “G” Neat @ 15.8 ppg

0.05% BWOC D28

0.5% BWOC D65

Displacement: 8 bpm

9 5/8", 53.5 lbs/ft

13 3/8" , 68 lbs/ft

12 1/4" OH

Solution - Example 1

Capacities– CAS/CAS : 0.3354 cuft/ft– CAS OH : 0.3132 cuft/ft– CAS : 0.3973 cuft/ft , 0.0708 bbl/ft

1. Volumes

Solution - Example 1 Capacities

– CAS/CAS : 0.3354 cuft/ft– CAS OH : 0.3132 cuft/ft– CAS : 0.3973 cuft/ft , 0.0708 bbl/ft

1. Volumes

Tail (3) 9800ft - 9300ft x 0.3132 cuft/ft = 156.cuft (4) 40ft x 0.3973 cuft/ft = 15.9 cuft Excess 156.6 cuft x 0.25 = 39.2 cuft Total = 211.7 cuft

Lead (1) 2900ft x 0.3354 cuft/ft = 972.7 cuft (2) 9300ft - 2900ft x 0.3132 = 2004 cuft Excess = 2004 cuft x 0.25 = 501 cuft Total = 3477 cuft

Tail (3) 9800ft - 9300ft x 0.3132 cuft/ft = 156.cuft (4) 40ft x 0.3973 cuft/ft = 15.9 cuft Excess 156.6 cuft x 0.25 = 39.2 cuft Total = 211.7 cuft

Lead (1) 2900ft x 0.3354 cuft/ft = 972.7 cuft (2) 9300ft - 2900ft x 0.3132 = 2004 cuft Excess = 2004 cuft x 0.25 = 501 cuft Total = 3477 cuft

• CapacitiesCAS/CAS :0.3354 CUFT/FTCAS OH :0.3132 CUFT/FTCAS : 0.3973 CUFT/FT, 0.0708 BBL/FT

• 1. Volumes

• 2. Yield

Handbook 700.013 2.165 cuft/sk (lead)

700.006 1.15 cuft/sk (tail)

Solution - Example 1 3. Cement

Lead : 3477 cuft = 1606 sacks 2.165 cuft/sk

Tail : 211.7 cuft = 184 sacks 1.15 cuft/sk

4. Mix Fluid

Lead : 1606sk x 12.15 gals/sk = 464.5 bbls 42 gal/bbl

Tail : 184sk x 4.97 gals/sk = 21.8 bbls 42 gal/bbl

5. Additives

Lead : D20 : 1606sk x 94lbs/sk x 0.12 = 18116 lbs

D13 : 1606sk x 94lbs/sk x 0.003 = 453 lbs

Tail : D28 : 184sk x 94lbs/sk x 0.0005 = 8.6 lbs

D65 : 184sk x 94lbs/sk x 0.005 = 86 lbs

Solution - Example 1 6. Displacement

9760ft x 0.0708 bbl/ft = 691 bbls

7. Job Thickening Time

Lead 3477 cuft = 124 min

5.6146 cuft/bbl x 5 bpm

Tail : 211.7 cuft = 13 min

5.6146 cuft/bbl x 3 bpm

Displacement: 691 bbls = 86 min

Drop Plugs = 10 min

Safety = 60 min

298 min

Minimum TT is 4 hrs 53 min

Example 2Data :Casing : 9 5/8”, 53.5 lbs/ft, 8600’ 7”, 29 lbs/ft, 14300’ Float collar - 14230’Open hole : 8 1/2” Section from 10200’ - 10800’ : 9”Cement : Lead : - Top surface @ 7 bpm - “G” at 13.5 ppg - 0.05 gals/sk D109 - 0.1 gals/sk D80

- 8% BWOC - D20 Tail : - 250 sacks @ 5 bpm - Class “G” @ 15.7 ppg - 30% BWOC D66 - 0.1 gals/sk D81 - 0.5 gals/sk D603

Calculate : Slurry Volume Cement Volume Mix water Additives Displacement @ 7 bpm Job Time

210200'

7", 29 lbs/ft

9 5/8" , 53.5 lbs/ft

8600'8.5" OH

14230'

14300'

10800'

11507'

Solution - Example 2 Capacities– CAS/CAS : 0.1301 cuft/ft CAS OH : 0.1268 cuft/ft– CAS/OHX : 0.1745 cuft/ft CAS : 0.2086 cuft/ft , 0.0371

bbl/ft 1. Volumes

Tail : Yield is 1.475 cuft/sk 250 sacks

---> 250 x 1.475 = 368.75 cuftVolume (5) + (6) = 368.75 cuft

(6) = 70 x 0.2086 = 14.6 cuft (5) = 368.75 - 14.6 = 354.15 cuft

Height in annulus 354.15 cuft = 2793’ 0.1268 cuft/ft 14300 - 2793' = 11, 507'Lead: Surface - 11, 507' (1) 8600 x 0.1301 = 1118.86 cuft (2) (10200 - 8600) x 0.1268 = 202.88 cuft (3) (10800 - 10200) x 0.1745 = 104.70 cuft (4) (11507 - 10800) x 0.1268 = 89.64 cuft Total Lead = 1516.1 cu ft

Solution - Example 2 2. Cement sacks

Lead : Handbook : 13.5 ppg Class"G" 8% D20 Yield = 1.825 cuft/Sk 1516.1 cuft = 830.7 = 831 sacks

1.825 cuft/sackTail : 250 sacks

3. Mix fluidLead : 831sk x 9.76 gal/sk = 193.1 bbls 42 gal/bblTail : 250 sks x 6.15 gals/sk = 36.6 bbls

42 gal/bbl

Solution - Example 2 4. Additives

Lead: D20 : 831 sks x 94 x 0.08 = 6249 lbs D109 : 831 sks x 0.05 gals/sk =

41.5 gals D80 : 831 sks x 0.1 gals/sk =

83.1 gals Total fluid = 124.6 galsTail: D66 : 250 sks x 94 x 0.3 = 7050 lbs D81 : 250 sks x 0.1 gals/sk = 25 gals D603 : 250 sks x 0.5 gals/sk = 125

gals Total fluid = 150 gals

5. Mix waterLead: 193.bbls - 124.6gal = 190.1 bbls 42gal/bblTail: 36.6bbls - 150gal = 33.0 bbls 42gal/bbl

Solution - Example 2 6. Displacement

– 14230ft' x 0.0371 bbl/ft = 527.9 bbl

7. Job Thickening Time

Lead : 270 bbls : 7 bpm = 38.5 min

Tail : 65.7 bbls : 5 bpm = 13.1 min

Plugs : = 10 min

Displacement 527.9 bbls : 7 bpm = 75 min

Safety Margin : = 60 min

Min TT is 3 hrs 16 min

9 5/8", 53.5 lbs/ft

13 3/8" , 68 lbs/ft

12 1/4" OH

Job Pressures

Example No 1

Tail Slurry at 15.8 ppg

– Top of Tail at 9300 Ft

– Lead slurry at 12.8 ppg to surface

Displace with 11.5 ppg Mud

Find differential pressure at end of Job

Hydrostatic Pressure Calcs

Tubing Hydrostatic Mud Hydrostatic

– Pmud = 11.5 x 9760 x 0.052– Pmud = 5836 psi

Annular Hydrostatic Tail Slurry Hydrostatic

– PTail = 15.8 x(9760 - 9300)x 0.052– PTail = 378 psi

Lead Slurry Hydrostatic– PLead= 12.8 x 9300 x 0.052– PLead = 6190 psi

Differential pressure – PDiff= Annular Hyd. - Tubing Hyd.

= (6190 +378) - 5836 – PDiff = 732 psi = Surface pump

pressure

9 5/8", 53.5 lbs/ft

13 3/8" , 68 lbs/ft

12 1/4" OH

9 5/8", 40 lbs/ft

13 3/8" , 77 lbs/ft

12 1/4" OH

Mud @ 11.3 ppg

Chemical wash @ 8.4 ppg

Lead Slurry @ 12.8 ppg

Tail Slurry @ 15.8 ppg

2100’

Long String Cement Calcs

Primary Cement Job– 9 5/8” Csg , 40 lbs/ft– Shoe @ 9980’ - Collar @ 9900’– Tail slurry @ 15.8 ppg to 9480’– Lead slurry @ 12.8 ppg to 2100’– 50 Bbls Chemical Wash ahead.

Previous Casing– 13 3/8“ Csg ,77 lbs/ft– Shoe @ 3150’

12 1/4” OH - Assume 10% Excess.– Find Lead, Tail and

Displacement Volumes required.

– Calculate Differential pressure at the end of the job (ignore friction).

9 5/8", 40 lbs/ft

13 3/8" , 77 lbs/ft

12 1/4" OH

Mud @ 11.3 ppg

2100’

Long String Cement Calcs - Solution

Capacities– Csg/OH Ann = 0.0558 bbls/ft– Csg/Csg Ann = 0.0564 bbls/ft– 9 5/8” Csg cap (40 #/ft) = 0.0758

bbls/ft Tail volume = Vol of Shoe track

+ Vol outside csg (with XS) = 0.0758 x (9980 - 9900) + 0.0558 x (9980 - 9480) x 1.10 = 6.06 + 30.69 = 36.75 = 37 Bbls= 206.3 cuft

Lead volume = Vol in Csg/OH Ann (with XS) + Vol in Csg/Csg Ann = 0.0558 x (9480 - 3150) x 1.10 + 0.0564 x (3150 - 2100) = 388.53 + 59.22 = 447.75 = 448 Bbls = 2514.04 cuft

9 5/8", 40 lbs/ft

13 3/8" , 77 lbs/ft

12 1/4" OH

Mud @ 11.3 ppg

2100’

Capacities– Csg/OH Ann = 0.0558 bbls/ft– Csg/Csg Ann = 0.0564 bbls/ft– 9 5/8” Csg cap (40 #/ft) =

0.0758 bbls/ft Displacement volume = Csg Vol

to float collar = 0.0758 x 9900 = 750 Bbls

Differential pressure at end of Job := Hydrostatic of Lead and tail cement columns + Hydrostatic of wash + Hydrostatic of remaining mud in annulus LESS Hydrostatic of displacing fluid + Hydrostatic of Tail cement in shoe joint

9 5/8", 40 lbs/ft

13 3/8" , 77 lbs/ft

12 1/4" OH

Mud @ 11.3 ppg

2100’

Capacities– Csg/OH Ann = 0.0558 bbls/ft– Csg/Csg Ann = 0.0564 bbls/ft– 9 5/8” Csg cap (40 #/ft) =

0.0758 bbls/ft Height of Chemical wash in

Csg/Csg Ann= 50 / 0.0564= 886 ft

Height of Tail = 9980 - 9480 = 500 ft

Height of Lead = 9480 - 2100 = 7380 ft

Height of Chem Wash = 886 ft Remainder is Mud behind casing

= 9900 - (420 + 7380 + 886) = 1214 ft

9 5/8", 40 lbs/ft

13 3/8" , 77 lbs/ft

12 1/4" OH

Mud @ 11.3 ppg

2100’

Annulus hydrostatic is total of Cement slurries + Chem Wash + Mud := 15.8 x 500 x 0.052 ( =Tail slurry)+ 12.8 x 7380 x 0.052 ( = Lead slurry)+ 8.4 x 886 x 0.052 (= Chem Wash)+ 11.3 x 1214 x 0.052 ( Mud in Ann)

= 410 + 4912 + 387 + 713= 6423 psi

Hydrostatic inside casing := 11.3 x 9900 x 0.052 (=Mud, disp.)

+ 15.8 x 80 x 0.052 (=Tail slurry)= 5817 + 65.7 = 5883 psi

Differential pressure at float collar := psi outside csg - psi inside csg= 6423 - 5883= 540 psi

Oilwell Cementing Calculation

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