Primary and Remedial Calculations

Calculations

Schlumberger Private

Primary and Remedial Primary and Remedial Calculations

Cementing Calculations We want to calculate: • • • • •

2

Initials

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

Important Rule Cement 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 3 Initials 9/8/2006

Slurry Yield When water is added to dry cement the resulting Slurry normally has more volume than the original Sack 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 4

Initials

Bulk and Absolute Volumes Bulk Volume : The volume occupied by a certain weight of dry material including void spaces between solid particles.

CEMENT

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

Absolute Volume : The volume occupied by the same weight of material, less the void spaces between particles.

5 Initials 9/8/2006

B

Bulk and Absolute Volumes Cement 1 drum = 1 cu.ft = 7.48 gal

A

Absolute Volume of Cement:

A

7.48 gal – 3.89 gal = 3.59 gal

Air in pore spaces will be displaced by water

Water

B

A 3.89 gal

6

Initials

B

Slurry Yield 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)

Class G API mix 1 Sack 1 cu.ft

4.97 Gal 0.66 cu.ft

CEMENT + AIR

+

WATER

1.15 cu.ft

=

Slurry Yield = 1.15 cu.ft / sk 7 Initials 9/8/2006

SLURRY

Mix Water Requirement Definition :

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

Unit:

gal/sack Class G API mix 1 Sack 1 cu.ft

CEMENT + AIR

4.97 Gal 0.66 cu.ft

+

WATER

1.15 cu.ft

=

Water Required = 4.97 gps 8 Initials 9/8/2006

SLURRY

Slurry Density Definition :

The weight of 1 gal of slurry

Unit:

lb/gal Class G API mix 1 Sack 1 cu.ft

CEMENT + AIR

4.97 Gal 0.66 cu.ft

+

WATER

1.15 cu.ft

=

SLURRY

1 gal of slurry will weight 15.8 pounds

Slurry Density = 15.80 ppg 9 Initials 9/8/2006

Calculations - Example 1 All 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

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

94

*

0.0382

= 3.59

H20 (44%) 41.36

*

1/8.33

= 4.97

Total

*

135.36

1. Density =

135.36 lb/sk 8.56 gal/sk

= 8.56 = 15.81 lb/gal 2. Yield =

8.56 gal/sk 7.48 gal/cu.ft

3. Water required = 4.97 gal/sk (from the table) 10 Initials 9/8/2006

= 1.144 cu.ft/sk

Calculations - Example 2 Class G, mix @ 15.5 ppg Material Weight (lb) * Absolute Volume (gal/lb) = Volume (gal) Class G H20 Total

94 8.33X

*

0.0382

= 3.59

*

1/8.33

=X

94 + 8.33X *

Density = 15.5 ppg =

94 + 8.33X

X = Water required = 5.35 gal/sk

3.59 + X Yield =

11 Initials 9/8/2006

= 3.59 + X

3.59 + 5.35 7.48

= 1.195 cu.ft/sk

Calculations - Example 3 Class G, mix with 5.05 gps of water requirement Material Weight (lb) * Absolute Volume (gal/lb) = Volume (gal) Class G

94

*

0.0382

= 3.59

H20

42.07

*

1/8.33

= 5.05

Total

136.07

*

Density =

136.07 8.64

= 15.75 gal/sk

= 8.64

Yield =

8.64 7.48

Water required = 5.05 gal/sk (from the table) 12 Initials 9/8/2006

= 1.16 cu.ft/sk

Calculations - Example 4 Class G, Given slurry yield – 1.06 cu.ft/sk Material Weight (lb) * Absolute Volume (gal/lb) = Volume (gal) Class G H20 Total

94 8.33X

*

0.0382

= 3.59

*

1/8.33

=X

94 + 8.33X *

Yield = 1.06 cu.ft/sk =

3.59 + X

Density = 13 Initials 9/8/2006

= 3.59 + X

7.48

X = Water required = 4.34 gal/sk

94 + 8.33 * 4.34 3.59 + 4.34

= 16.41 ppg

Calculations - Example 5 Class H, 3% S001. Mix by API Material Weight (lb) * Absolute Volume (gal/lb) = Volume (gal) Class H

*

0.0382

= 3.59

2.82

*

0.0687

= 0.194

H20

94 (0.38)

*

1/8.33

= 4.288

Total

132.54

S 001

Density =

94

= 8.072

132.54 8.072

= 16.42 ppg

Yield =

Water required = 4.288 gal/sk 14 Initials 9/8/2006

8.072 7.48

= 1.079 cu.ft/sk

Additives Requiring Additional Water D020, 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, Barite „

0.024 gal additional water for each 1 lb D31 added.

D042, Kolite „

1 gal additional water for each 25 lb D42 added.

D066, Silica Flour 15 Initials 9/8/2006

„

0.343% (BWOC) additional water for each 1 % D66 added;

„

therefore 12% for 35% D66.

Calculations - Example 6 Class A, D020 – 2% BWOC. Mix by API Material Weight (lb)

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

Class A

*

0.0382

= 3.59

*

0.0454

= 0.085

H20

94[0.46+2(0.053)] *

1/8.33

= 6.384

Total

149.08

D 020

Density =

94 1.88

149.08 10.059

= 10.059

= 14.82 ppg

Yield =

Water required = 6.384 gal/sk 16 Initials 9/8/2006

10.059 7.48

= 1.345 cu.ft/sk

Calculations - Example 7 Class 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

3.76

*

0.0454

= 0.171

H20

8.33X

*

1/8.33

= X

Total

110.26 + 8.33X

D 020

Density = 13.8 ppg =

110.26 + 8.33X 4.917 + X Yield =

17 Initials 9/8/2006

= 4.917 + x

X = Water required = 7.75 gal/sk

4.917 + 7.75 7.48

= 1.69 cu.ft/sk

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

*

Abs. Volume (gal/lb)

Class H

*

0.0382

= 3.59

*

0.0454

= 0.0854

*

0.0456

= 1.5002

*

1/8.33

= 10.2012

D 020 D 030

94 1.88 32.9

H20

94[0.38+8(0.053)+0.1]

Total

213.756

Density =

= 15.3768

213.756 15.3768

= 13.90 ppg

Yield =

Water required = 10.20 gal/sk 18 Initials 9/8/2006

= Volume (gal)

15.3768 7.48

= 2.056 cu.ft/sk

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

= 0.2

H20

8.33X

*

1/8.33

Total

116.17 +8.33X

Density = 16.5 ppg =

= 6.09 + X

116.17 + 8.33X 6.09 + X Yield =

19 Initials 9/8/2006

=X

6.09 + 1.92 7.48

X = Water required = 1.92 gal/sk

= 1.071 cu.ft/sk

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. 20 Initials

Draw a diagram

Csg/Csg Annulus Displacement Volume

13⅜ inch 68ppf

5000 ft OH/Csg Annulus

9⅝ inch 47ppf

Shoetrack

8500 ft

Slurry Volume Calculations (1) Vol 1 (Csg/Csg Ann) 5000 x 0.3354 =

1677 ft3

Vol 2 (OH/Csg Ann) (8500 – 5000) x 0.3131 x 1.214 =

1330.4 ft3

Vol 3 (Shoetrack) 80 x 0.4110 =

32.9

Total Volume 3040 ft3

ft3

Vol 4 (Displ Vol) (8500 – 80) x 0.0732 =

616.3 bbls

Vol 5 (Sacks cement) 21 Initials

3040 ÷ 1.144 =

2657 sx

22 Initials

Slurry Volume Calculations (2 & 3) (2)

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.

23 Initials

well requires the 20 inch 94ppf (3) Acasing 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.

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

70.7 ft3

Vol 2 (OH/Csg Ann)

237.1 ft3

Vol 3 (Shoetrack)

17.7 ft3 Total Volume 326.1 ft3

Vol 4 (Displ Vol) 69.73 +177

= 260.9 bbls

Vol 5 (Sacks cement) 24 Initials

326.1 ÷ 1.38 =

236 sx

Homework Calculate: Slurry Volumes, Cement Volumes, Mix Water, Additives, Displacement Job 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’ - surface Class “G” @ 12.8 ppg 12% BWOC Bentonite 0.3% BWOC D13 12.15 gps H20 Yield: 2.165 cu.ft/sk

25 Initials

Tail: 3 bpm 9800’ - 9300’ Class “G” Neat @ 15.8 ppg 0.05% BWOC D28 0.5% BWOC D65 4.97 gps H20 Yield: 1.15 cu.ft/sk Displacement: 8 bpm

13 3/8" , 68 lbs/ft

1

2900' 9 5/8", 53.5 lbs/ft

2

2

9300'

3 4

12 1/4" OH 9760'

3 9800'

Solution - Homework

Capacities

„ CAS/CAS

: 0.3354 cuft/ft

„ CAS OH

: 0.3132 cuft/ft

„ CAS

: 0.3973 cuft/ft , 0.0708 bbl/ft

1. Volumes Lead (1) 2900ft x 0.3354 cuft/ft (2) 9300ft - 2900ft x 0.3132

= 972.7 cuft = 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

Total = 211.7 cuft

z 2. Yield Handbook 26 Initials

700.013 700.006

= 39.2 cuft

2.165 cuft/sk (lead) 1.15 cuft/sk (tail)

Solution - Example 1 3. Cement Lead : Tail :

3477 cuft 2.165 cuft/sk 211.7 cuft 1.15 cuft/sk

= 1606 sacks = 184 sacks

4. Mix Fluid Lead : Tail : 27 Initials

1606sk x 12.15 gals/sk = 464.5 bbls 42 gal/bbl 184sk x 4.97 gals/sk = 21.8 bbls 42 gal/bbl

Solution - Example 1 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

28 Initials

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

8 bpm Drop Plugs Safety

= 10 min = 120 = 353 min

Minimum TT for LS Minimum TT for TS 29 Initials

is is

5 hrs 53 min 3 hrs 49 min