1. Basic Formulas

Table of Content

1. Pressure Gradient
2. Hydrostatic Pressure
3. Converting Pressure into Mud Weight
4. Specific Gravity
5. Equivalent Circulation Density
6. Maximum Allowable Mud Weight
7. Pump Output
8. Annular Velocity
9. Capacity Formulas
10. Control Drilling
11. Buoyancy Factor
12. Hydrostatic Pressure Decrease POOH
13. Loss of Overbalance Due to Falling Mud Level
14. Formation Temperature
15. Hydraulic Horsepower
16. Drill Pipe, Drill Collar Calculation
17. Pump Pressure, Pump Strokes
18. Cost Per Foot
19. Temperature Conversion Formulas

1.9 Capacity Formulas

Annular capacity between casing or hole and drill pipe, tubing, or casing

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

Example: hole size = 12-1/4 in, and, pipe OD = 5.0 in.
$\mathrm{Annular\; capacity,\; bbl/ft}=\frac{{12.25}^2-5^2}{1029.4}$
Annular capacity = 0.12149 bbl/ft

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

Example: hole size = 12-1/4 in, and, pipe OD = 5.0 in.
$\mathrm{Annular\; capacity,\; ft/bbl}=\frac{1029.4}{{12.25}^2-5^2}$
Annular capacity = 8.23 ft/bbl

$\mathrm{Annular\; capacity,\; gal/ft}=\frac{{\mathrm{Dh}}^2-{\mathrm{Dp}}^2}{24.51}$

Example: hole size = 12-1/4 in, and, pipe OD = 5.0 in.
$\mathrm{Annular\; capacity,\; gal/ft}=\frac{{12.25}^2-5^2}{24.51}$
Annular capacity = 5.1 gal/ft

$\mathrm{Annular\; capacity,\; ft/gal}=\frac{24.51}{{\mathrm{Dh}}^2-{\mathrm{Dp}}^2}$

Example: hole size = 12-1/4 in, and, pipe OD = 5.0 in.
$\mathrm{Annular\; capacity,\; ft/gal}=\frac{24.51}{{12.25}^2-5^2}$
Annular capacity, ft/gal = 0.19598 ft/gal

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

Example: hole size = 12-1/4 in, and, pipe OD = 5.0 in.
${\mathrm{Annular\; capacity,\; ft}}^3\mathrm{/Iinft}=\frac{{12.25}^2-5^2}{183.35}$
Annular capacity = 0.682097 ft³/linft

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

Example: hole size = 12-1/4 in, and, pipe OD = 5.0 in.
${\mathrm{Annular\; capacity,\; linft/ft}}^3=\frac{183.35}{{12.25}^2-5^2}$
Annular capacity = 1.466 linft/ft³

Annular capacity between casing and multiple strings of tubing

$\text{Annular capacity, bbl/ft}=\frac{{\mathrm{Dh}}^2-({T_1}^2+{T_2}^2)}{1029.4}$

Example: Using two strings of tubing of same size:
Dh = casing — 7.0 in. — 29 lb/ft - ID = 6.184 in.
T1 = tubing No. 1 — 2-3/8 in. - OD = 2.375 in.
T2 = tubing No. 2 — 2-3/8 in. - OD = 2.375 in.
$\text{Annular capacity, bbl/ft}=\frac{{6.184}^2-({2.375}^2+{2.375}^2)}{1029.4}$
$\text{Annular capacity, bbl/ft}=\frac{38.24-11.28}{1029.4}$
Annular capacity = 0.02619 bbl/ft

$\text{Annular capacity, ft/bbl}=\frac{1029.4}{{\mathrm{Dh}}^2-({T_1}^2+{T_2}^2)}$

Example: Using two strings of tubing of same size:
Dh = casing — 7.0 in. — 29 lb/ft - ID = 6.184 in.
T1 = tubing No. 1 — 2-3/8 in. - OD = 2.375 in.
T2 = tubing No. 2 — 2-3/8 in. - OD = 2.375 in.
$\text{Annular capacity, bbl/ft}=\frac{1029.4}{{6.184}^2-({2.375}^2+{2.375}^2)}$
$\text{Annular capacity, bbl/ft}=\frac{1029.4}{38.24-11.28}$
Annular capacity = 38.1816 ft/bbl

$\text{Annular capacity, gal/ft}=\frac{{\mathrm{Dh}}^2-({T_1}^2+{T_2}^2)}{24.51}$

Example: Using two strings of tubing of same size:
Dh = casing — 7.0 in. — 29 lb/ft - ID = 6.184 in.
T1 = tubing No. 1 — 2-3/8 in. - OD = 2.375 in.
T2 = tubing No. 2 — 3-1/2 in. - OD = 3.5 in.
$\text{Annular capacity, gal/ft}=\frac{{6.184}^2-({2.375}^2+{3.5}^2)}{24.51}$
$\text{Annular capacity, gal/ft}=\frac{38.24-17.89}{24.51}$
Annular capacity = 0.8302733 gal/ft

$\text{Annular capacity, ft/gal}=\frac{24.51}{{\mathrm{Dh}}^2-({T_1}^2+{T_2}^2)}$

Example: Using two strings of tubing of same size:
Dh = casing — 7.0 in. — 29 lb/ft - ID = 6.184 in.
T1 = tubing No. 1 — 2-3/8 in. - OD = 2.375 in.
T2 = tubing No. 2 — 3-1/2 in. - OD = 3.5 in.
$\text{Annular capacity, ft/gal}=\frac{24.51}{{6.184}^2-({2.375}^2+{3.5}^2)}$
$\text{Annular capacity, ft/gal}=\frac{24.51}{38.24-17.89}$
Annular capacity = 1.2044226 ft/gal

${\mathrm{Annular\; capacity,\; ft}}^3\mathrm{/Iinft}=\frac{{\mathrm{Dh}}^2-({T_1}^2+{T_2}^2+{T_3}^2)}{183.35}$

Example: Using two strings of tubing of same size:
Dh = casing — 9-5/8 in. — 47 lb/ft ID = 8.681 in.
T1 = tubing No. 1 — 3-1/2 in. — OD = 3.5 in.
T2 = tubing No. 2 — 3-1/2 in. — OD = 3.5 in.
T3 = tubing No. 3 — 3-1/2 in. — OD = 3.5 in.
${\mathrm{Annular\; capacity,\; ft}}^3\mathrm{/Iinft}=\frac{{8.681}^2-({3.5}^2+{3.5}^2+{3.5}^2)}{183.35}$
${\mathrm{Annular\; capacity,\; ft}}^3\mathrm{/Iinft}=\frac{75.359-36.75}{183.35}$
Annular capacity = 0.2105795 ft³/linft

${\mathrm{Annular\; capacity,\; linft/ft}}^3=\frac{183.35}{{\mathrm{Dh}}^2-({T_1}^2+{T_2}^2+{T_3}^2)}$

Example: Using two strings of tubing of same size:
Dh = casing — 9-5/8 in. — 47 lb/ft ID = 8.681 in.
T1 = tubing No. 1 — 3-1/2 in. — OD = 3.5 in.
T2 = tubing No. 2 — 3-1/2 in. — OD = 3.5 in.
T3 = tubing No. 3 — 3-1/2 in. — OD = 3.5 in.
${\mathrm{Annular\; capacity,\; linft/ft}}^3=\frac{183.35}{{8.681}^2-({3.5}^2+{3.5}^2+{3.5}^2)}$
${\mathrm{Annular\; capacity,\; linft/ft}}^3=\frac{183.35}{75.359-36.75}$
Annular capacity = 4.7487993 linft/ft³

Capacity of tubulars and open hole: drill pipe, drill collars, tubing, casing, hole, and any cylindrical object

$\text{Capacity, bbl/ft}=\frac{{\mathrm{ID,\; in}}^2}{1029.4}$

Example: Determine the capacity, bbl/ft, of a 12-1/4 in. hole:
$\text{Capacity, bbl/ft}=\frac{{12.25}^2}{1029.4}$
Capacity = 0.1457766 bbl/ft

$\text{Capacity, ft/bbl}=\frac{1029.4}{{\mathrm{ID,\; in}}^2}$

Example: Determine the capacity, ft/bbl, of 12-1/4 in. hole:
$\text{Capacity, ft/bbl}=\frac{1029.4}{{12.25}^2}$
Capacity = 6.8598 ft/bbl

$\text{Capacity, gal/ft}=\frac{{\mathrm{ID,\; in}}^2}{24.51}$

Example: Determine the capacity, gal/ft, of 8-1/2 in. hole:
$\text{Capacity, gal/ft}=\frac{{8.5}^2}{24.51}$
Capacity = 2.9477764 gal/ft

$\text{Capacity, ft/gal}=\frac{24.51}{{\mathrm{ID,\; in}}^2}$

Example: Determine the capacity, ft/gal, of 8-1/2 in. hole:
$\text{Capacity, ft/bbl}=\frac{24.51}{{8.5}^2}$
Capacity = 0.3392 ft/gal

${\mathrm{Capacity,\; ft}}^3\mathrm{/Iinft}=\frac{{\mathrm{ID,\; in}}^2}{183.35}$

Example: Determine the capacity, ft³/linft, for a 6.0 in. hole:
${\mathrm{Capacity,\; ft}}^3\mathrm{/Iinft}=\frac{{6.0}^2}{183.35}$
Capacity = 0.1963 ft³/linft

${\mathrm{Capacity,\; Iinft/ft}}^3=\frac{183.35}{{\mathrm{ID,\; in}}^2}$

Example: Determine the capacity, linft/ft³, for a 6.0 in. hole:
${\mathrm{Capacity,\; Iinft/ft}}^3=\frac{183.35}{{6.0}^2}$
Capacity = 5.09305 linft/ft³

Amount of cuttings drilled per foot of hole drilled

a. BARRELS of cuttings drilled per foot of hole drilled:

$\mathrm{Barrels}=\frac{{\mathrm{Dh}}^2}{1029.4}\times (1-\mathrm{\%\; porosity})$

Example: Determine the number of barrels of cuttings drilled for one foot of 12-1/4 in. -hole drilled with 20% (0.20) porosity:
$\mathrm{Barrels}=\frac{{12.25}^2}{1029.4}\times (1-0.20)$
Barrels = 0.1457766 × 0.80
Barrels = 0.1166213

b. CUBIC FEET of cuttings drilled per foot of hole drilled:

$\mathrm{Cubic\; feet}=\frac{{\mathrm{Dh}}^2}{144}\times 0.7854\times (1-\mathrm{\%\; porosity})$

Example: Determine the number of barrels of cuttings drilled for one foot of 12-1/4 in. -hole drilled with 20% (0.20) porosity:
$\mathrm{Cubic\; feet}=\frac{{12.25}^2}{144}\times 0.7854\times (1-0.20)$
$\mathrm{Cubic\; feet}=\frac{150.0626}{144}\times 0.7854\times 0.80$

c. Total solids generated:

Wcg = 350 × Ch × L × (1 − P) × SG

where:
Wcg = solids generated, pounds
Ch = capacity of hole, bbl/ft
L = footage drilled, ft
SG = specific gravity of cuttings
P = porosity, %

Example: Determine the total pounds of solids generated in drilling 100 ft of a 12-1/4 in. hole (0.1458 bbl/ft). Specific gravity of cuttings = 2.40 gm/cc. Porosity = 20%:
Wcg = 350 × 0.1458 × 100 × (1 − 0.2) × 2.4
Wcg = 9797.26 pounds

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