Determine the following:
Amount of bentonite to add
Total water requirements
Slurry yield
Slurry weight
1) Weight of additive:
Weight, lb/sk = 0.04 x 94 lb/sk
Weight = 3.76 lb/sk
2) Total water requirement:
Water = 5.1 (cement) + 2.6 (bentonite)
Water = 7.7 gal/sk of cement
3) Volume of slurry:
Vol,gal/sk = (94 / (3.14 x 8.33)) + (3.76 /
(2.65 x 8.33)) +7.7
Vol, gal/sk = 3.5938 + 0.1703 + 7.7
Vol = 11.46 gal/sk.
4) Slurry yield, ft’/sk:
Yield, ft3/sk = 11.46gal/sk + 7.48gal/ft³
Yield = 1.53ft³/sk
5 ) Slurry density. lb/gal:
Density, lb/gal = (94 + 3.76 + (8.33 x 7.7))
/ 11.46
Density, Ib/gal = 161.90 / 11.46
Density = 14.13 lb/gal
Water requirements
a) Weight of materials, Ib/sk:
Weight, Ib/sk = 94 + (8.33 x vol of water, gal) + (%of additive x
94)
b) Volume of slurry, gal/sk:
Vo1,gal/sk = (94 Ib/sk / SG x 8.33) + (wt of additive,
Ib/sk / SG x 8.33) + water vol, gal
c) Water requirement using material balance equation:
D1V1 = D2V2
Example:
Class H cement plus 6% bentonite to be mixed at 14.0lblgal.
Specific gravity of bentonite = 2.65.
Determine the following:
Bentonite requirement, Ib/sk
Water requirement, gal/sk
Slurry yield, ft³/sk
Check slurry weight, lb/gal
1) Weight of materials, lb/sk:
Weight, lb/sk = 94 + (0.06 x 94) +
(8.33 x “y”)
Weight, lb/sk = 94 + 5.64 + 8.33“y”
Weight = 99.64 + 8.33“y”
2) Volume of slurry, gaVsk:
Vol, gal/sk = (94 / 3.14 x 8.33)+ (5.64
/ 2.65 x 8.33) + “Y”
Vol, gal/sk = 3.6 + 0.26 + “y”
Vol, gal/sk = 3.86 + ‘‘y”
3) Water requirement using material balance
equation:
99.64 + 8.33“y”= (3.86
+ “y”) x 14.0
99.64 + 8.33“y”= 54.04+14.0“y”
99.64  54.04 = l4.0“y” 8.33“y”
45.6 = 5.67“y”
45.6 ÷ 5.67 = “y”
8.0 = “y” Thus, water requirement = 8.0gal/sk of cement
4) Slurry yield, ft³/sk:
Yield, ft³/sk = 3.6 + 0.26 + 8.0 / 7.48
Yield, ft3/sk = 11.86 / 7.48
Yield = 1.59ft³/sk
5) Check slurry density, lb/gal:
Density, lb/gal = (94+5.64+(8.33x 8.0)) / 11.86
Density, lb/gal = 166.28 / 11.86
Density = 14.0lb/gal
Field cement additive calculations
When bentonite is to be prehydrated, the amount of
bentonite added is calculated based on the total amount of mixing
water used.
Cement program:
240sk cement; slurry density = 13.8 ppg.
8.6gal/sk mixing water; 1.5% bentonite to be prehydrated.
a) Volume of mixing water, gal:
Volume = 240sk x 8.6gal/sk
Volume = 2064 gal.
b) Total weight, lb, of mixing water:
Weight = 2064 gal x 8.33 lb/gal
Weight = 17,193 lb
c) Bentonite requirement, lb:
Bentonite = 17,193 lb x 0.015'%,
Bentonite = 257.89 lb
Other additives are calculated based on the weight of the cement:
Cement program: 240sk cement; 0.5% Halad;
0.40% CFR2:
a) Weight of cement:
Weight = 240sk x 94 lb/sk
Weight = 22,560 lb
b) Halad = 0.5%
Halad = 22,560 lb x 0.005
Halad = 112.8 lb
C)CFR2 = 0.40%
CFR2 = 22,560 lb x 0.004
CFR2 = 90.24 lb
Note:
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