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Primary and Remedial Calculations
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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
4 Initials
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
5 Initials9/8/2006
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
6 Initials
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
7 Initials9/8/2006
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
8 Initials9/8/2006
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
WATER =+ SLURRYSLURRYCEMENT + AIRCEMENT + AIR
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
9 Initials9/8/2006
Definition : The weight of 1 gal of slurry
Unit: lb/gal
Slurry Density
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 Density = 15.80 ppg
1 gal of slurry will weight 15.8 pounds
10 Initials9/8/2006
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)
11 Initials9/8/2006
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
Material Weight (lb) * Absolute Volume (gal/lb) = Volume (gal)
Class G 94 * 0.0382 = 3.59H20 8.33X * 1/8.33 = X
Total 94 + 8.33X * = 3.59 + X
12 Initials9/8/2006
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)
Material Weight (lb) * Absolute Volume (gal/lb) = Volume (gal)
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 =
13 Initials9/8/2006
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
Material Weight (lb) * Absolute Volume (gal/lb) = Volume (gal)
Class G 94 * 0.0382 = 3.59H20 8.33X * 1/8.33 = X
Total 94 + 8.33X * = 3.59 + X
14 Initials9/8/2006
Calculations - Example 5Class H, 3% S001. Mix by API
Water required = 4.288 gal/sk
Material Weight (lb) * Absolute Volume (gal/lb) = Volume (gal)
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
15 Initials9/8/2006
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.
16 Initials9/8/2006
Calculations - Example 6Class A, D020 – 2% BWOC. Mix by API
Water required = 6.384 gal/sk
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
17 Initials9/8/2006
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 =
18 Initials9/8/2006
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 =
Water required = 10.20 gal/sk
19 Initials9/8/2006
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 =
20 Initials
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
21 Initials
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
22 Initials
23 Initials
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)
24 Initials
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
25 Initials
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
26 Initials
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)
27 Initials
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
28 Initials
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
29 Initials
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