2. Basic Calculations

Table of Content

1. Volumes and Strokes
2. Slug Calculations
3. Accumulator Capacity — Usable Volume Per Bottle
4. Bulk Density of Cuttings (Using Mud Balance)
5. Drill String Design (Limitations)
6. Ton-Mile (TM) Calculations
7. Cementing Calculations
8. Weighted Cement Calculations
9. Calculations for the Number of Sacks of Cement Required
10. Calculations for the Number of Feet to Be Cemented
11. Setting a Balanced Cement Plug
12. Differential Hydrostatic Pressure Between Cement in the Annulus and Mud Inside the Casing
13. Hydraulicing Casing
14. Depth of a Washout
15. Lost Returns — Loss of Overbalance
16. Stuck Pipe Calculations
17. Calculations Required for Spotting Pills
18. Pressure Required to Break Circulation

2.7 Cementing Calculations

Cement additive calculations

a. Weight of additive per sack of cement:

Weight, lb = percent of additive × 94 lb/sk

b. Total water requirement, gal/sk, of cement:

Water, gal/sk = Cement water requirement, gal/sk + Additive water requirement, gal/sk

c. Volume of slurry, gal/sk:

$\mathrm{Vol\; gal/sk}=\frac{\mathrm{94\; lb}}{\mathrm{SG\; of\; cement}\times \mathrm{8.33\; lb/gal}}+\frac{\mathrm{weight\; of\; additive,\; lb}}{\mathrm{SG\; of\; additive}\times \mathrm{8.33\; lb/gal}}+\mathrm{water\; volume,\; gal}$

d. Slurry yield, ft3/sk:

$\mathrm{Yield,\; ft3/sk}=\frac{\mathrm{vol.\; of\; slurry,\; gal/sk}}{{\mathrm{7.48\; gal/ft}}^3}$

e. Slurry density, lb/gal:

$\mathrm{Density,\; lb/gal}=\frac{94+\mathrm{wt\; of\; additive}+(8.33\times \mathrm{vol.\; of\; water/sk})}{\mathrm{vol.\; of\; slurry,\; gal/sk}}$

Example: Class A cement plus 4% bentonite using normal mixing water:
Determine the following:
Amount of bentonite to add
Total water requirements Slurry yield Slurry weight

a. Weight of additive:

Weight, lb/sk = 0.04 × 94 lb/sk
Weight = 3.76 lb/sk

b. Total water requirement:

Water = 5.1 (cement) + 2.6 (bentonite)
Water = 7.7 gal/sk of cement

c. Volume of slurry:

$\mathrm{Vol\; gal/sk}=\frac{94}{3.14\times 8.33}+\frac{3.76}{2.65\times 8.33}+7.7$
Vol. gallsk = 3.5938 + 0.1703 + 7.7
Vol. = 11.46 gal/sk

d. Slurry yield, ft³/sk:

Yield, ft³/sk = 11.46 gal/sk ÷ 7.48 gal/ft³
Yield = 1.53 ft³/sk

e. Slurry density, lb/gal:

$\mathrm{Density,\; lb/gal}=\frac{94+3.76+(8.33\times 7.7)}{11.46}$
$\mathrm{Density,\; lb/gal}=\frac{61.90}{11.46}$
Density = 14.13 lb/gal

Water requirements

a. Weight of materials, lb/sk:

Weight, lb/sk = 94 + (8.33 × vol of water, gal) + (% of additive × 94)

b. Volume of slurry, gal/sk:

$\mathrm{Vol\; gal/sk}=\frac{\mathrm{94\; lb/sk}}{\mathrm{SG}\times 8.33}+\frac{\mathrm{weight\; of\; additive,\; lb/sk}}{\mathrm{SG}\times 8.33}+\mathrm{water\; volume,\; gal}$

c. Water requirement using material balance equation:

D₁V₁ = D₂V₂

Example: Class H cement plus 6% bentonite to be mixed at 14.0 lb/gal. Specific gravity of bentonite = 2.65.
Determine the following:
Bentonite requirement, lb/sk Water requirement, gallsk
Slurry yield, ft3/sk Check slurry weight, lb/gal

a. Weight of materials, lb/sk:

Weight, lb/sk = 94 + (0.06 × 94) + (8.33 × “y”)
Weight, lb/sk = 94 + 5.64 + 8.33 “y”
Weight = 99.64 + 8.33”y”

b. Volume of slurry, gal/sk:

$\mathrm{Vol\; gal/sk}=\frac{94}{3.14\times 8.33}+\frac{5.64}{3.14\times 8.33}+\mathrm{“y”}$
Vol, gal/sk = 3.6 + 0.26 + “y”
Vol, gal/sk = 3.86

c. Water requirements using material balance equation

99.64 + 8.33”y” = (3.86 + ”y”) × 14.0
99.64 + 8.33”y” = 54.04 + 14.0 “y”
99.64 - 54.04 = 14.0”y” - 8.33”y”
45.6 = 5.67”y”
45.6 : 5.67 = “y”
8.0 = ”y” Thus , water required = 8.0 gal/sk of cement

d. Slurry yield, ft³/sk:

${\mathrm{Yield,\; ft}}^3\text{/sk}=\frac{3.6+0.26+8.0}{7.48}$
${\mathrm{Yield,\; ft}}^3\text{/sk}=\frac{11.86}{7.48}$
Yield = 1.59 ft³/sk

e. Check slurry density, lb/gal:

$\mathrm{Density,\; lb/gal}=\frac{94+5.64+(8.338.0)}{11.86}$
$\mathrm{Density,\; lb/gal}=\frac{166.28}{11.86}$
Density = 14.0 lb/gal

Field cement additive calculations

When bentonite is to be pre-hydrated, the amount of bentonite added is calculated based on the total amount of mixing water used.

Cement program: 240 sk cement; slurry density = 13.8 ppg; 8.6 gal/sk mixing water; 1.5% bentonite to be pre-hydrated:

a. Volume of mixing water, gal:

Volume = 240 sk × 8.6 gal/sk
Volume = 2064 gal

b. Total weight, lb, of mixing water:

Weight = 2064 gal × 8.33 lb/gal
Weight = 17,193 lb

c. Bentonite requirement, Lb:

Bentonite = 17,193 lb × 0.015%
Bentonite = 257.89 lb

Other additives are calculated based on the weight of the cement:

Cement program: 240 sk cement; 0.5% Halad; 0.40% CFR-2:

a. Weight of cement:

Weight = 240 sk × 94 lb/sk
Weight = 22,560 lb

b. Halad = 0.5%

Halad = 22,560 lb × 0.005
Halad = 112.8 lb

c. CFR-2 = 0.40%

CFR-2 = 22,560 lb × 0.004
CFR-2 = 90.24 lb

Table 2-1 Water Requirements and Specific Gravity of Common Cement Additives

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