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

2.11 Setting a Balanced Cement Plug

Step 1 Determine the following capacities:

a. Annular capacity, ft3/ft, between pipe or tubing and hole or casing:

${\mathrm{Annular\; capacity,\; ft}}^3\text{/ft}=\frac{{\mathrm{Dh,\; in.}}^2-{\mathrm{Dp,\; in.}}^2}{183.35}$

b. Annular capacity, ft/bbl between pipe or tubing and hole or casing:

$\mathrm{Annular\; capacity,\; ft}\text{/bbl}=\frac{1029.4}{{\mathrm{Dh,\; in.}}^2-{\mathrm{Dp,\; in.}}^2}$

c. Hole or casing capacity, ft3/ft:

${\mathrm{Hole\; or\; casing\; capacity,\; ft}}^3\text{/ft}=\frac{{\mathrm{ID,\; in.}}^2}{183.35}$

d. Drill pipe or tubing capacity, ft3/ft:

${\mathrm{Drill\; pipe\; or\; tubing\; capacity,\; ft}}^3\text{/ft}=\frac{{\mathrm{ID,\; in.}}^2}{183.35}$

e. Drill pipe or tubing capacity, bbl/ft:

$\mathrm{Drill\; pipe\; or\; tubing\; capacity,\; bbl/ft}=\frac{{\mathrm{ID,\; in.}}^2}{1029.4}$

Step 2 Determine the number of SACKS of cement required for a given length of plug, OR determine the FEET of plug for a given number of sacks of cement:

a. Determine the number of SACKS of cement required for a given length of plug:

Sacks of cement = plug length, ft × hole or casing capacity, ft³/ft × excess ÷ slurry yield, ft³/sk

NOTE: If no excess is to be used, simply omit the excess step.

OR

b. Determine the FEET of plug for a given number of SACKS of cement:

Feet = sacks of cement × slurry yield, ft³/sk ÷ hole or casing capacity, ft³/ft ÷ excess

NOTE: If no excess is to be used, simply omit the excess step.

Step 3 Determine the spacer volume (usually water), bbl, to be pumped behind the slurry to balance the plug:

Spacer vol, bbl = annular capacity, ft/bbl ÷ excess × spacer volume ahead, bbl × pipe or tubing capacity, bbl/ft

NOTE: If no excess is to be used, simply omit the excess step.

Step 4 Determine the plug length, ft, before the pipe is withdrawn:

Plug length, ft = sacks of cement × slurry yield, ft³/sk ÷ annular capacity, ft³/ft × excess + pipe or tubing capacity, ft³/ft

NOTE: If no excess is to be used, simply omit the excess step.

Step 5 Determine the fluid volume, bbl, required to spot the plug:

Vol, bbl = (length of pipe or tubing, ft − plug length, ft) × pipe or tubing capacity, bbl/ft − spacer vol behind slurry, bbl

Example 1:
A 300 ft plug is to be placed at a depth of 5000 ft. The open hole size is 8-1/2 in. and the drill pipe is 3-1/2 in. — 13.3 lb/ft; ID — 2.764 in. Ten barrels of water are to be pumped ahead of the slurry. Use a slurry yield of 1.15 ft3/sk. Use 25% as excess slurry volume:
Determine the following:
1. Number of sacks of cement required
2. Volume of water to be pumped behind the slurry to balance the plug
3. Plug length before the pipe is withdrawn
4. Amount of mud required to spot the plug plus the spacer behind the plug

Step 1 Determined the following capacities:

a. Annular capacity between drill pipe and hole, ft3/ft:

${\mathrm{Annular\; capacity,\; ft}}^3\text{/ft}=\frac{{8.5}^2-{3.5}^2}{183.35}$
Annular capacity = 0.3272 ft³/ft

b. Annular capacity between drill pipe and hole, ft/bbl:

$\mathrm{Annular\; capacity,\; ft}\text{/bbl}=\frac{1029.4}{{8.5}^2-{3.5}^2}$
Annular capacity = 17.1569 ft/bbl

c. Hole capacity, ft3/ft:

${\mathrm{Hole\; capacity,\; ft}}^3\text{/ft}=\frac{{8.5}^2}{183.35}$
Hole capacity = 0.3941 ft³/ft

d. Drill pipe capacity, bbl/ft:

$\mathrm{Drill\; pipe\; capacity,\; bbl/ft}=\frac{{2.764}^2}{1029.4}$
Drill pipe capacity = 0.00742 bbl/ft

e. Drill pipe capacity, ft3/ft:

${\mathrm{Drill\; pipe\; capacity,\; ft}}^3\text{/ft}=\frac{{2.764}^2}{183.35}$
Drill pipe capacity = 0.0417 ft³/ft

Step 2 Determine the number of sacks of cement required:

Sacks of cement = 300 ft × 0.3941 ft³/ft × 1.25 ÷ 1.15 ft³/sk
Sacks of cement = 129

Step 3 Determine the spacer volume (water), bbl, to be pumped behind the slurry to balance the plug:

Spacer vol, bbl = 17.1569 ft/bbl ÷ 1.25 × 10 bbl × 0.00742 bbl/ft
Spacer vol, bbl = 1.018 bbl

Step 4 Determine the plug length, ft, before the pipe is withdrawn:

Plug length, ft = (129 sk × 1.15 ft³/sk) ÷ (0.3272 ft³/ft × 1.25 + 0.0417 ft³/ft)
Plug length, ft = 148.35 ft³ ÷ 0.4507 ft³/ft
Plug length, ft = 329 ft

Step 5 Determine the fluid volume, bbl, required to spot the plug:

Vol, bbl = [(5000 ft − 329 ft) × 0.00742 bbl/ft] − 1.0 bbl
Vol, bbl = 33.6 bbl

Example 2:
Determine the number of FEET of plug for a given number of SACKS of cement:
A cement plug with 100 sk of cement is to be used in an 8-1/2 in, hole. Use 1.15 ft3/sk for the cement slurry yield. The capacity of 8-1/2 in. hole = 0.3941 ft3/ft. Use 50% as excess slurry volume:

Feet = 100 sk × 1.15 ft3/sk ÷ 0.3941 ft3/ft ÷ 1.50
Feet = 194.5

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