cálculos de cementación en una etapa

Single Stage Calculations

2 Single Stage Cementing

Engineer’s Handbook

1 foot

1. OD2. LBS/FT3. CUFT/FT4. Displacement BBL/FT

1 foot

1. OD outside casing2. LBS/FT outside casing3. OD inside casing4. CUFT/FT

1 foot

1. OD outside casing2. LBS/FT outside casing3. OD inside casing4. CUFT/FT

1. OD casing2. OH diameter3. CUFT/FT

Volumes to be Calculated : Step A

Open Hole

PreviousCasing

CollarShoe

11. Casing in casing

Open Hole

PreviousCasing

CollarShoe

11. Casing in casing

2. Casing in open hole

Open Hole

PreviousCasing

CollarShoe

1. Casing in casing

2. Casing in open hole

3. Inside casing(between collar and shoe)

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/FtShoe : 4600’Collar : 4510’Previous casing : 9 5/8 - 47 LB/FtShoe : 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

OnlyHere

Excess in only calculated on AnnularVolume in THE OPEN HOLE

Example: Problem 1

OH 50% Excess

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

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 calledYIELD

+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 calledYIELDYIELD

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 calledYIELDYIELD

WATER =

The unit of slurry yield isThe unit of slurry yield is cuftcuft/sack/sack

+CEMENTCEMENT SLURRYSLURRY

1 sack of cementof 94 lbs = 1= 1 cuftcuft

4.97 gals of water = 0.66= 0.66 cuftcuft

1 sack of cement1 sack of cementof 94 lbsof 94 lbs = 1= 1 cuftcuft

4.97 gals of water4.97 gals of water = 0.66= 0.66 cuftcuft

11 cuftcuft + 0.66+ 0.66 cuftcuft = 1.66= 1.66 cuftcuft

1 sack of cement1 sack of cementof 94 lbsof 94 lbs = 1= 1 cuftcuft

But the But the Handbook states:Handbook states:

4.97 gals of water4.97 gals of water 1.151.15 cuftcuft/sack/sack= 0.66= 0.66 cuftcuft

WHY ?WHY ?11 cuftcuft + 0.66+ 0.66 cuftcuft = 1.66= 1.66 cuftcuft

Bulk volume vs Absolute volume

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

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

Bulk volume vs Absolute volume

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

Pour waterPour waterinto cement

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

into cement

Air in pore spaces willbe displaced by water

Note………

Yield varies with densityBasically, 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

BentoniteAPI States:

Ideal Mix (15.8 ppg) Lightweight Slurry (12.8) ppg

Cement Water

1 cu/ft 4.97 gals

Bentonite

12% ExtraBWOC Water

1 cu/ft 4.97 gals

Cement Water

1.15 cuft/sack

BentoniteAPI States:

Ideal Mix (15.8 ppg) Lightweight Slurry (12.8) ppg

Cement Water

1 cu/ft 4.97 gals

Bentonite 7.18 gals

12% ExtraBWOC Water

1 cu/ft 4.97 gals

Cement Water

1.15 cuft/sack

Only 4.97 gals water is required for the cement.A chemical has to be added to react with the extra water.

BentoniteBentonite is used in large quantities in oilfield slurries.As much as 25% BWOC is possibleOffshore this presents a problem due to limited storage capacities and dry blending facilities.If Bentonite is mixed with fresh mix water instead of being dry blended only 1/4 of the amount is needed

This is called: PREHYDRATED BENTONITEExample : Dry In Mix Water

2%BWOC 0.5%BWOC4%BWOC 1% BWOC

10%BWOC 2.5%BWOC

Handbook Gives:Ref

Class “G” Neat - Density 15.8 lbs/gals Yield = 1.15 cuft/sack 700.06

Water 4.97 gals/sack

Class “G” + 6% D20 - Density 13.9 lbs/gals Yield = 1.655 cuft/sack 700.013

Water 8.56 gal/sx

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 cuft385.35 cuft

Example: Problem 1 (Continued)Previous Problem:Lead volume [1] : 240.2 cuft Tail volume [2 + 3] :

367.05 cuft+ 18.2 cuft385.35 cuft

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

Cement volume Lead:

Example: Problem 1 (Continued)Previous Problem:Lead volume [1] : 240.2 cuft Tail volume [2 + 3] :

367.05 cuft+ 18.2 cuft385.35 cuft

Use: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:

Example: Problem 1 (Continued)Previous Problem:Lead volume [1] : 240.2 cuft Tail volume [2 + 3] : 367.05 cuft

+ 18.2 cuft385.35 cuft

Use:Lead Slurry : 6% D20 at 13.9 ppgTail Slurry Neat : 15.8 ppgCement volume Lead:

240.2 cuft = 145.2 sacks1.655 cuft/sack

385.25 cuft = 335.0 sacks1.15 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 multiplyMix 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

Example: Problem 1 (Continued)Lead mix water 145sx x 8.56 gal/sx

Example: Problem 1 (Continued)Lead mix water 145sx x 8.56 gal/sx = 1241.2 gal

= 29.55 bbls

42 gal/bbl

Example: Problem 1 (Continued)

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

Example: Problem 1 (Continued)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

Example: Problem 1 (Continued)Lead mix water 145sx x 8.56 gal/sx = 1241.2 gal = 29.55 bbls

42 gal/bbl

Total mix water 69.19 bbls

42 gal/bbl

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

42 gal/bbl

No water ahead, behind, CW, spacer 0 bbls

42 gal/bbl

Priming, washing up 20 bbls

42 gal/bbl

Priming, washing up 20 bbls

Total water needed 103 bbls

Additives - Step ESolid AdditivesCalculated as a % by weight of dry cement.– Except: Salt– Small amounts will not effect calculation

Liquid AdditivesCalculated as gallons of additives per sack of bulk cementSubtract total amount of liquid additives from amount of mix fluid to have real amount of mix water.

Example: Problem 1 (Continued)Lead 6% D20 BWOC (145 sks)Tail 0.2 gal/sack of D80 (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 bbl42 gal/bbl

DisplacementTo FLOAT COLLAR unless otherwise stated

Exceptions1. 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 statedExceptions1. 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 GMIXING TIME

+DISPLACEMENT TIME

+10 MINUTES (DROP PLUGS ...)

= ESTIMATION OF JOB TIME

1 1/2 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 11minsTail 68.6 : 8bpm 8.5 mins 9 mins

Displacement: 4510ft x 0.0361 bbl/ft = 162.8 bbl: 12 bpm13.6 mins 14 mins

Dropping Plugs 10 minsTotal 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, DisplacementJob 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 ppg12% BWOC Bentonite0.3% BWOC D13 Tail: 3 bpm 9800’ - 9300’Class “G” Neat @ 15.8 ppg0.05% BWOC D280.5% BWOC D65

Displacement: 8 bpm

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

Mix Water, Additives, DisplacementJob 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 ppg12% BWOC Bentonite0.3% BWOC D13 Tail: 3 bpm 9800’ - 9300’Class “G” Neat @ 15.8 ppg0.05% BWOC D280.5% 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 1Capacities– 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

Solution - Example 1• Capacities

CAS/CAS :0.3354 CUFT/FTCAS OH :0.3132 CUFT/FTCAS : 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

• 2. Yield

Handbook 700.013 2.165 cuft/sk (lead)

700.006 1.15 cuft/sk (tail)

Solution - Example 13. CementLead : 3477 cuft = 1606 sacks

2.165 cuft/sk

Tail : 211.7 cuft = 184 sacks1.15 cuft/sk

4. Mix FluidLead : 1606sk x 12.15 gals/sk = 464.5 bbls

42 gal/bbl

Tail : 184sk x 4.97 gals/sk = 21.8 bbls42 gal/bbl

5. AdditivesLead : D20 : 1606sk x 94lbs/sk x 0.12 = 18116 lbs

D13 : 1606sk x 94lbs/sk x 0.003 = 453 lbsTail : D28 : 184sk x 94lbs/sk x 0.0005 = 8.6 lbs

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

Solution - Example 16. Displacement

9760ft x 0.0708 bbl/ft = 691 bbls

7. Job Thickening TimeLead 3477 cuft = 124 min

5.6146 cuft/bbl x 5 bpmTail : 211.7 cuft = 13 min

5.6146 cuft/bbl x 3 bpmDisplacement: 691 bbls = 86 min

8 bpmDrop Plugs = 10 minSafety = 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 VolumeMix water AdditivesDisplacement @ 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 2Capacities

– 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. VolumesTail : 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 22. Cement sacks

Lead : Handbook : 13.5 ppgClass"G"8% D20Yield = 1.825 cuft/Sk

1516.1 cuft = 830.7 = 831 sacks1.825 cuft/sack

Tail : 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 24. Additives

Lead: D20 : 831 sks x 94 x 0.08 = 6249 lbsD109 : 831 sks x 0.05 gals/sk = 41.5 galsD80 : 831 sks x 0.1 gals/sk = 83.1 galsTotal fluid = 124.6 gals

Tail:D66 : 250 sks x 94 x 0.3 = 7050 lbsD81 : 250 sks x 0.1 gals/sk = 25 galsD603 : 250 sks x 0.5 gals/sk = 125 galsTotal 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 26. 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

Job Pressures

9 5/8", 53.5 lbs/ft

13 3/8" , 68 lbs/ft

12 1/4" OH

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 HydrostaticMud Hydrostatic

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

Annular HydrostaticTail 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

Long String Cement Calcs

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’

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).

Long String Cement Calcs - Solution

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

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/ftDisplacement 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/ftHeight of Chemical wash in Csg/Csg

Ann= 50 / 0.0564= 886 ft

Height of Tail = 9980 - 9480 = 500 ftHeight of Lead = 9480 - 2100 = 7380 ftHeight 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 psiDifferential pressure at float collar :

= psi outside csg - psi inside csg= 6423 - 5883= 540 psi

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