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

2.13 Hydraulicing Casing

These calculations will determine if the casing will hydraulic out (move upward) when cementing

Determine the difference in pressure gradient, psi/ft, between the cement and the mud

psi/ft = (cement wt, ppg − mud wt, ppg) × 0.052

Determine the differential pressure (DP) between the cement and the mud

DP, psi = difference in pressure gradients, psi/ft × casing length, ft

Determine the area, sq in., below the shoe

Area, sq in. = casing diameter, in.2 × 0.7854

Determine the Upward Force (F), lb. This is the weight, total force, acting at the bottom of the shoe

Force, lb = area, sq in. × differential pressure between cement and mud, psi

Determine the Downward Force (W), lb. This is the weight of the casing

Weight, lb = casing wt, lb/ft × length, ft × buoyancy factor

Determine the difference in force, lb

Differential force, lb = upward force, lb − downward force, lb

Pressure required to balance the forces so that the casing will not hydraulic out (move upward)

psi = force, lb − area, sq in.

Mud weight increase to balance pressure

Mud wt, ppg = pressure required ÷ 0.052 ÷ casing length, ft to balance forces, psi

New mud weight, ppg

Mud wt, ppg = mud wt increase, ppg + mud wt, ppg

Check the forces with the new mud weight

a. psi/ft = (cement wt, ppg — mud wt, ppg) x 0.052
b. psi = difference in pressure gradients, psi/ft x casing length, ft
c. Upward force, lb = pressure, psi x area, sq in.
d. Difference in = upward force, lb — downward force, lb force, lb

Example:
Casing size = 13 3/8 in. 54 lb/ft
Cement weight = 15.8 ppg
Mud weight = 8.8 ppg
Buoyancy factor = 0.8656
Well depth = 164 ft (50 m)

Determine the difference in pressure gradient, psi/ft, between the cement and the mud

psi/ft = (15.8 ppg − 8.8 ppg) × 0.052
psi/ft = 0.364

Determine the differential pressure between the cement and the mud

DP, psi = 0.364 psi/ft × 164 ft
DP = 60 psi

Determine the area, sq in., below the shoe

Area, sq in. = 13.375 in.2 × 0.7854
Area = 140.5 sq in.

Determine the upward force. This is the total force acting at the bottom of the shoe

Force, lb = 140.5 sq in. × 60 psi
Force = 8430 lb

Determine the downward force. This is the weight of the casing

Weight, lb = 54. lb/ft × 164 ft × 0.8656
Weight = 7737 lb

Determine the difference in force, lb

Differential force, lb = 8430 lb − 7737 lb
Differential force = 693 lb
Therefore: Unless the casing is tied down or stuck, it could possibly hydraulic out (move upward).

Pressure required to balance the forces so that the casing will not hydraulic out (move upward)

psi = 693 lb ÷ 140.5 sq in.
psi = 4.9

Mud weight increase to balance pressure

Mud wt, ppg = 4.9 psi ÷ 0.052 ÷ 164 ft
Mud wt increase = 0.57 ppg

New mud weight, ppg

New mud wt, ppg = 8.8 ppg + 0.57 ppg
New mud wt = 9.4 ppg

Check the forces with the new mud weight

a. psi/ft = (15.8 ppg − 9.4 ppg) × 0.052 = 0.3328 psi/ft
b. psi = 0.3328 psi/ft × 164 ft = 54.58 psi
c. Upward force, lb = 54.58 psi × 140.5 sq in. = 7668 lb
d. Differential force, lb = downward force − upward force
Difference in force, lb = 7737 lb − 7668 lb = +69 lb

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