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

2.2 Slug Calculations

Barrels of slug required for a desired length of dry pipe

Step 1 - Hydrostatic pressure required to give desired drop inside drill pipe:

HP, psi = mud wt, ppg × 0.052 × ft of dry pipe

Step 2 - Difference in pressure gradient between slug weight and mud weight:

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

Step 3 - Length of slug in drill pipe:

Slug length, ft = pressure, psi ÷ difference in pressure gradient, psi/ft

Step 4 - Volume of slug, barrels:

Slug vol., bbl = slug length, ft × drill pipe capacity, bbl/ft

Example: Example: Determine the barrels of slug required for the following:
Desired length of dry pipe (2 stands) = 184 ft, Mud weight = 12.2 ppg
Drill pipe capacity 4-1/2 in. — 16.6 lb/ft = 0.01422 bbl/ft, Slug weight = 13.2 ppg
Step 1 Hydrostatic pressure required:
HP, psi = 12.2 ppg × 0.052 × 184 ft
HP = 117 psi
Step 2 Difference in pressure gradient, psi/ft:
psi/ft = (13.2 ppg − 12.2 ppg) × 0.052
psi/ft = 0.052
Step 3 Length of slug in drill pipe, ft:
Slug length, ft = 117 psi ÷ 0.052
Slug length = 2250 ft
Step 4 Volume of slug, bbl:
Slug vol., bbl = 2250 ft × 0.01422 bbl/ft
Slug vol. = 32.0 bbl

Weight of slug required for a desired length of dry pipe with a set volume of slug

Step 1 - Length of slug in drill pipe, ft:

Slug length, ft = slug vol., bbl ÷ drill pipe capacity, bbl/ft

Step 2 - Hydrostatic pressure required to give desired drop inside drill pipe:

HP, psi = mud wt, ppg × 0.052 × ft of dry pipe

Step 3 - Weight of slug, ppg:

Slug wt, ppg = HP, psi ÷ 0.052 ÷ slug length, ft + mud wt, ppg

Example: Determine the weight of slug required for the following:
Desired length of dry pipe (2 stands) = 184 ft, Mud weight = 12.2 ppg
Drill pipe capacity 4-1/2 in. — 16.6 lb/ft = 0.0 1422 bbl/ft, Volume of slug = 25 bbl
Step 1 Length of slug in drill pipe, ft:
Slug length, ft = 25 bbl ÷ 0.01422 bbl/ft
Slug length = 1758 ft
Step 2 Hydrostatic pressure required:
HP, Psi = 12.2 ppg × 0.052 × 184 ft
HP, Psi = 117psi
Step 3 Weight of slug, ppg:
Slug wt, ppg = 117 psi ÷ 0.052 ÷ 1758 ft + 12.2 ppg
Slug wt, ppg = 1.3 ppg + 12.2 ppg
Slug wt = 13.5 ppg

Volume, height, and pressure gained because of slug:

Volume gained in mud pits after slug is pumped, due to U-tubing:

Vol., bbl = ft of dry pipe × drill pipe capacity, bbl/ft

Height, ft, that the slug would occupy in annulus:

Height, ft = annulus vol., ft/bbl × slug vol., bbl

Hydrostatic pressure gained in annulus because of slug:

HP, psi = height of slug in annulus, ft × difference in gradient, psi/ft between slug wt and mud wt

Example: Feet of dry pipe (2 stands) = 184 ft
Slug volume = 32.4 bbl
Slug weight = 13.2 ppg
Mud weight = 12.2 ppg
Drill pipe capacity 4-1/2 in. 16.6 lb/ft = 0.01422 bbl/ft
Annulus volume (8-1/2 in. by 4-1/2 in.) = 19.8 ft/bbl
Volume gained in mud pits after slug is pumped due to U-tubing:
Vol., bbl = 184 ft × 0.01422 bbl/ft
Vol. = 2.62 bbl
Height, ft, that the slug would occupy in the annulus:
Height, ft = 19.8 ft/bbl × 32.4 bbl
Height = 641.5 ft
Hydrostatic pressure gained in annulus because of slug:
HP, psi = 641.5 ft (13.2 &minus 12.2) × 0.052
HP, psi = 641.5 ft × 0.052
HP = 33.4 psi

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