Primary and Remedial Calculations

Primary and Remedial Calculations

Schlumberger Private

Primary and Remedial Calculations

2 Initials

Cementing Calculations

We want to calculate:

• Slurry Volumes• Sacks of cement required• Displacement Volume• Estimated Job time• Correct Plug bumping Pressure

3 Initials9/8/2006

Important RuleCement slurries should always have density specified by API.

Density can only be changed by using the appropriate additive.

If water/solids ratio is not correct, may get :

High viscosity / unpumpable slurry.

Excessive free water.

If the cement composition and one of the properties are known, other two properties can be calculated

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Slurry YieldWhen water is added to dry cement the resultingSlurry normally has more volume than the originalSack of 94lbs based on a material balance calculation.

1 sack of cement = 94lbs = 1 cubic foot

Dry Cement absolute volume = 0.0382 gal/lb

1 sack of cement = 3.59 gal

Class G cement slurry @ 15.8 ppg (1.9 SG) uses 44% mix water or 4.97 gal/sx

7.48 gallons = 1 cubic foot

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Bulk and Absolute VolumesBulk Volume : The volume occupied by a certain weight of dry material

including void spaces between solid particles.

Absolute Volume : The volume occupied by the same weight of material,

less the void spaces between particles.

CEMENTCEMENT 1 Sack = 1 cubic foot (cu.ft) = 94 pounds

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A

B

Bulk and Absolute Volumes

A

Cement 1 drum = 1 cu.ft = 7.48 gal

3.89 gal

B

Water

BA

Air in pore spaces willbe displaced by water

Absolute Volume of Cement:

7.48 gal – 3.89 gal = 3.59 gal

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Definition : The volume of slurry produced when 1 sack of dry cement (and additives) are mixed with water

Unit: cubic foot/sack (cu.ft/sk)

Slurry Yield

WATER =+ SLURRYSLURRYCEMENT + AIRCEMENT + AIR

1 Sack 1 cu.ft

4.97 Gal 0.66 cu.ft 1.15

cu.ft

Class G API mix

Slurry Yield = 1.15 cu.ft / sk

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Definition : The amount of water needed to hydrate 1 sack of dry cement (and additives) to create a pumpable liquid

Unit: gal/sack

Mix Water Requirement


1 Sack 1 cu.ft

4.97 Gal 0.66 cu.ft 1.15

cu.ft

Class G API mix

Water Required = 4.97 gps

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Definition : The weight of 1 gal of slurry

Unit: lb/gal

Slurry Density


1 Sack 1 cu.ft

4.97 Gal 0.66 cu.ft 1.15

cu.ft

Class G API mix

Slurry Density = 15.80 ppg

1 gal of slurry will weight 15.8 pounds

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Calculations - Example 1All calculations based on one sack of cement Note: Absolute volumes from Field Data Handbook, Page:700.005

Example: Class G cement mixed by API specifications

8.56 gal/sk

135.36 lb/sk= 15.81 lb/gal1. Density =

Material Weight (lb) * Absolute Volume (gal/lb) = Volume (gal)

Class G 94 * 0.0382 = 3.59H20 (44%) 41.36 * 1/8.33 = 4.97

Total 135.36 * = 8.56

7.48 gal/cu.ft

8.56 gal/sk= 1.144 cu.ft/sk2. Yield =

3. Water required = 4.97 gal/sk (from the table)

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Calculations - Example 2Class G, mix @ 15.5 ppg

3.59 + X94 + 8.33XDensity = 15.5 ppg =

7.483.59 + 5.35 = 1.195 cu.ft/skYield =

X = Water required = 5.35 gal/sk


Class G 94 * 0.0382 = 3.59H20 8.33X * 1/8.33 = X

Total 94 + 8.33X * = 3.59 + X

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Calculations - Example 3Class G, mix with 5.05 gps of water requirement

7.488.64

= 1.16 cu.ft/skYield =

Water required = 5.05 gal/sk (from the table)


Class G 94 * 0.0382 = 3.59H20 42.07 * 1/8.33 = 5.05

Total 136.07 * = 8.64

8.64136.07

= 15.75 gal/skDensity =

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Calculations - Example 4Class G, Given slurry yield – 1.06 cu.ft/sk

7.483.59 + XYield = 1.06 cu.ft/sk =

3.59 + 4.3494 + 8.33 * 4.34 = 16.41 ppgDensity =

X = Water required = 4.34 gal/sk


Class G 94 * 0.0382 = 3.59H20 8.33X * 1/8.33 = X

Total 94 + 8.33X * = 3.59 + X

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Calculations - Example 5Class H, 3% S001. Mix by API

Water required = 4.288 gal/sk


Class H 94 * 0.0382 = 3.59

S 001 2.82 * 0.0687 = 0.194H20 94 (0.38) * 1/8.33 = 4.288

Total 132.54 = 8.072

8.072132.54 = 16.42 ppgDensity =

7.488.072Yield = = 1.079 cu.ft/sk

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Additives Requiring Additional WaterD020, Bentonite

5.3% (BWOC) additional water for each 1% D20 added.D024, Gilsonite

1 gal additional water for each 25 lb D24 added.D030, Silica Sand

0.286% (BWOC) additional water for each 1 % D30 added; therefore 10% for 35% D30.

D031, Barite0.024 gal additional water for each 1 lb D31 added.

D042, Kolite1 gal additional water for each 25 lb D42 added.

D066, Silica Flour 0.343% (BWOC) additional water for each 1 % D66 added; therefore 12% for 35% D66.

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Calculations - Example 6Class A, D020 – 2% BWOC. Mix by API


10.059149.08 = 14.82 ppgDensity =

7.4810.059Yield = = 1.345 cu.ft/sk

Material Weight (lb) * Abs. Volume (gal/lb) = Volume (gal)

Class A 94 * 0.0382 = 3.59

D 020 1.88 * 0.0454 = 0.085H20 94[0.46+2(0.053)] * 1/8.33 = 6.384

Total 149.08 = 10.059

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Calculations - Example 7Class G, D042 - 12.5 lb/sk, D020 – 4% BWOC. Mix @ 13.8 ppg

Material Weight (lb) * Abs. Volume (gal/lb) = Volume (gal)

Class G 94 * 0.0382 = 3.59

D 042 12.5 * 0.0925 = 1.156

D 020 3.76 * 0.0454 = 0.171H20 8.33X * 1/8.33 = X

Total 110.26 + 8.33X = 4.917 + x

4.917 + X110.26 + 8.33XDensity = 13.8 ppg = X = Water required = 7.75 gal/sk

7.484.917 + 7.75 = 1.69 cu.ft/skYield =

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Calculations - Example 8Class H, D020 – 2% BWOC (Pre-hydrated). D030 – 35% BWOC. Mix by

APIMaterial Weight (lb) * Abs. Volume (gal/lb) = Volume (gal)

Class H 94 * 0.0382 = 3.59

D 020 1.88 * 0.0454 = 0.0854

D 030 32.9 * 0.0456 = 1.5002H20 94[0.38+8(0.053)+0.1] * 1/8.33 = 10.2012

Total 213.756 = 15.3768

7.4815.3768 = 2.056 cu.ft/skYield =

15.3768213.756 = 13.90 ppgDensity =


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Calculations - Example 9Class H, D600 – 2.0 gps. D080 – 0.3 gps. D801 – 0.2gps. Mix @ 16.5

ppgMaterial Weight (lb) * Abs. Volume (gal/lb) = Volume (gal)

Class H 94 * 0.0382 = 3.59

D 600 17.09 * 0.117 = 2

D 080 3.08 * 0.0973 = 0.3

D 801 2 * 0.1 = 0.2H20 8.33X * 1/8.33 = X

Total 116.17 +8.33X = 6.09 + X

6.09 + X116.17 + 8.33XDensity = 16.5 ppg = X = Water required = 1.92 gal/sk

7.486.09 + 1.92 = 1.071 cu.ft/skYield =

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Slurry Volume Calculations (1)A well requires the 9⅝ inch 47ppf casing at 8500 feet cemented to surface with neat Class G cement. Previous casing is 13 ⅜ inch 68ppf set at 5000 feet. There are two joints of casing between the Float Collar and Float Shoe and the open hole requires an excess of 21.4%. Bit size is 12¼ inch.

5000 ft

8500 ft

13⅜inch 68ppf

9⅝inch 47ppf

Csg/CsgAnnulus

OH/CsgAnnulus

Shoetrack

DisplacementVolume

Draw a diagram

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Slurry Volume Calculations (1)Vol 1 (Csg/Csg Ann)

Vol 2 (OH/Csg Ann)

Vol 3 (Shoetrack)

Vol 4 (Displ Vol)

Vol 5 (Sacks cement)

5000 x 0.3354 = 1677 ft3

(8500 – 5000) x 0.3131 x 1.214 = 1330.4 ft3

80 x 0.4110 = 32.9 ft3Total Volume

3040 ft3

3040 ÷ 1.144 = 2657 sx

(8500 – 80) x 0.0732 = 616.3 bbls

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Slurry Volume Calculations (2 & 3)A well requires the 7 inch 23ppf liner cemented at 12,200 feet with an overlap inside the 9⅝ inch 47ppf of 150m using Class G cement + 35% Silica Flour at 16.55ppg. Previous casing is set at 10,500 feet. There are two joints of casing between the Float Collar and Float Shoe and the open hole requires an excess of 10%. Bit size is 8½ inch. Running tool to be used is 5” DP, 19.5 ppf.

A well requires the 20 inch 94ppf casing cemented at 1500 feet using an inner cement stinger made up from 5 inch 19.5ppf DP. The previous casing was a 30 inch, 1 inch wall conductor which was driven to 300 feet. There is no float collar only a float shoe and the hole seems large so a guestimate at volumes to bring the cement to surface is 150% on OH size. Slurry is Neat Class G.

(2) (3)

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Slurry Volume Calculations (2)Vol 1 (Csg/Csg Ann)

Vol 2 (OH/Csg Ann)

Vol 3 (Shoetrack)

Vol 4 (Displ Vol)

Vol 5 (Sacks cement)

70.7 ft3

237.1 ft3

17.7 ft3

Total Volume326.1 ft3

326.1 ÷ 1.38 = 236 sx

69.73 +177 = 260.9 bbls

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HomeworkCalculate: Slurry Volumes, Cement Volumes,

Mix Water, Additives, DisplacementJob Time, Surface Pressure @ end of job

Well Data: 9 5/8” 53.5 lb/ft casing, shoe at 9800’,collar at 9760’, OH 12.25” + 25% excess,previous casing 13 3/8” 68 lb/ft at 2900’Mud 10ppg

Slurries: Lead: 5 bpm 9300’ - surfaceClass “G” @ 12.8 ppg12% BWOC Bentonite0.3% BWOC D1312.15 gps H20Yield: 2.165 cu.ft/sk

Tail: 3 bpm 9800’ - 9300’Class “G” Neat @ 15.8 ppg0.05% BWOC D280.5% BWOC D65

4.97 gps H20Yield: 1.15 cu.ft/sk

Displacement: 8 bpm

2

34

113 3/8" , 68 lbs/ft

2900'

12 1/4" OH

9760'

9800'

9300'

3

2 9 5/8", 53.5 lbs/ft

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Solution - HomeworkCapacities

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

1. VolumesTail(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. YieldHandbook 700.013 2.165 cuft/sk (lead)

700.006 1.15 cuft/sk (tail)

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Solution - Example 1

3. CementLead : 3477 cuft = 1606 sacks

2.165 cuft/skTail : 211.7 cuft = 184 sacks

1.15 cuft/sk4. Mix Fluid

Lead : 1606sk x 12.15 gals/sk = 464.5 bbls42 gal/bbl

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

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Solution - Example 15. Additives

Lead : D20 : 1606sk x 94lbs/sk x 0.12 =18116 lbsD13 : 1606sk x 94lbs/sk x 0.003= 453 lbs

Tail : D28 : 184sk x 94lbs/sk x 0.0005 = 8.6 lbsD65 : 184sk x 94lbs/sk x 0.005 = 86 lbs

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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 = 120

= 353 minMinimum TT for LS is 5 hrs 53 minMinimum TT for TS is 3 hrs 49 min

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Primary and Remedial Calculations