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

2.16 Stuck Pipe Calculations

Determine the feet of free pipe and the free point constant

Method 1

The depth at which the pipe is stuck and the number of feet of free pipe can be estimated by the drill pipe stretch table below and the following formula.

Table 2-2 Drill Pipe Stretch Table

Feet of free pipe = stretch, in. × free point constant free pipe ÷ pull force in thousands of pounds

Example:
3-1/2 in. 13.30 lb/ft drill pipe 20 in. of stretch with 35,000 lb of pull force

From drill pipe stretch table:
Free point constant = 9052.5 for 3-1/2 in. drill pipe 13.30 lb/ft
$\mathrm{Feet\; of\; free\; pipe}=\frac{\mathrm{20,\; in.}\times 9052.5}{35}$
Feet of free pipe = 5173 ft

Determine free point constant (FPC)

The free point constant can be determined for any type of steel drill pipe if the outside diameter, in., and inside diameter, in., are known:

FPC = As × 2500

where: As = pipe wall cross sectional area, sq in.

Example:
Determine the free point constant and the depth the pipe is stuck using the following data:
2-3/8 in. tubing — 6.5 lb/ft — ID = 2.441 in.
25 in. of stretch with 20,000 lb of pull force

a. Determine free point constant (FPC):

FPC = [(2.875² − 2.441²) × 0.7854] × 2500
FPC = 1.820 × 2500
FPC = 4530

b. Determine the depth of stuck pipe:

$\mathrm{Feet\; of\; free\; pipe}=\frac{\mathrm{25,\; in.}\times 4530}{20}$
Feet of free pipe = 5663 ft

Method 2

$\mathrm{Free\; pipe,\; ft}=\frac{\mathrm{735,294}\times e\times \mathrm{Wdp}}{\text{differential pull, lb}}$

where e = pipe stretch, in.
Wdp = drill pipe weight, lb/ft (plain end)

Plain end weight, lb/ft, is the weight of drill pipe excluding tool joints:
Weight, lb/ft = 2.67 × (pipe OD, in.² − pipe; ID, in.²)

Example:
Determine the feet of free pipe using the following data:
5.0 in. drill pipe; ID — 4.276 in.; 19.5 lb/ft
Differential stretch of pipe = 24 in.
Differential pull to obtain stretch = 30,000 lb

Solution: Weight, lb/ft = 2.67 × (5.0² − 4.276²)
Weight, lb/ft = 17.93 lb/ft
$\mathrm{Free\; pipe,\; ft}=\frac{\mathrm{735,294}\times 24\times 17.93}{\mathrm{30,000}}$
Free pipe = 10,547 ft

Determine the height, ft of unweighted spotting fluid that will balance formation pressure in the annulus:

a. Determine the difference in pressure gradient, psi/ft, between the mud weight and the spotting fluid:

psi/ft = (mud wt, ppg &minus spotting fluid wt, ppg) × 0.052

b. Determine the height, ft, of unweighted spotting fluid that will balance formation pressure in the annulus:

Height ft = amount of overbalance, psi ÷ difference in pressure gradient, psi/ft

Example:
Use the following data to determine the height, ft, of spotting fluid that will balance formation pressure in the annulus:
Data:
Mud weight = 11.2 ppg
Weight of spotting fluid = 7.0 ppg
Amount of overbalance = 225.0 psi

a. Difference in pressure gradient, psi/ft:

psi/ft = (11.2 ppg − 7.0 ppg) × 0.052
psi/ft = 0.2184

b. Determine the height, ft. of unweighted spotting fluid that will balance formation pressure in the annulus:

Height, ft = 225.0 psi ÷ 0.2184 psi/ft
Height = 1030 ft

Therefore: Less than 1030 ft of spotting fluid should be used to maintain a safety factor to prevent a kick or blow-out.

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