This article gives an example of calculation for a typical 20" casing cemented with stinger (cement volumes, displacement volumes, collapse and floating).

Relevant Data

Derrick floor elevation: 100 ft (above Mean Sea Level  (MSL)).

Water level: 100ft

30in conductor driven to 300 ft BDF.

26in hole drilled to 600 ft BDF (vertical, gauge hole).

Cement: class G with 1.5% CaCl₂.

20in casing run to 580 ft BDF (K55, BTC, 94 lbs/ft).

Surface lines capacity: 1 bbl.

Mud Weight: 8.9 ppg (sea water)

The casing will be cemented using a 14.8 ppg (or less) slurry (2 % bentonite as mixwater). 5in drillpipe will be used as a stinger.

As a stinger is used, pumping of 14.8 ppg slurry will be continued until good cement returns to seabed, unless serious losses are encountered (in that case, 200 % excess over gauge open hole volume (or more) may be pumped). The cement volume calculated in this example is just to get an idea of how much is approximately needed.

If no calliper log has been run, a best guess of the average hole size should be made. In this case, we shall assume 100 % excess over the 26in (gauge) hole / 20”-30”in conductor casing annulus. The cement will be under displaced in such a way that approx. 2 bbl of cement is left inside the drillpipe. It will be assumed that cement enters the pocket

Calculation cement volume

a. Capacities:

b. Slurry yields and water-cement ratios:

c. Annular capacity of 26in (gauge) hole / 20in  casing:

(580 - 300) ft x  0.2681 bbls/ft = 75 bbls (including 100 % excess » 150 bbls)

d. Annular capacity 30in conductor / 20in casing:

(300-200) ft x  0.428 bbls/ft = 42.8 bbls

pocket capacity: 20 ft x 0.6567 bbls/ft = 13 bbls (including 200 % excess  =39 bbls)

e. Add (c) and (d) for the total 14.8ppg slurry volume required:

150 bbls + 42.8 bbls + 39 bbls » 232 bbls » 1302 cu.ft

f. 14.8 ppg slurry = 1302 cu.ft slurry required. Yield = 1.34 cu.ft/sk, so

1302 cu.ft     =  971.6 sx cement required
1.34 cu.ft/sk

Water / cement ratio = 6.3 gal/sk, therefore mixwater required:

972 sx x 6.3gal/sk = 6123.6 gal = 145.8 bbls » 146 bbls

g. Total wt: of cement = 972 sxs x 94 lbs/sx = 91,368 lbs

Wt: of CaCl2 (1.5% BWOC) = 0.015 x 91368 = 1370 lbs

Calculate Displacement Volume

The capacity of the drillpipe from derrick floor to 50 ft above the shoe is:

(580 ft - 50 ft) x 0.0176bbls/ft = 9.3  bbls

Surface lines capacity = 1 bbl

So displace with 9.3 bbls + 1 bbl = 10.3 bbls mud.

Mixwater Preparation Calculations

From the calculations, it follows that 146 bbls of mixwater is required for a slurry (14.8 ppg) volume that allows for 100 % excess over open hole volume.

Preferably some excess + dead volume of mixwater should be prepared, as cement will be pumped until it reaches sea bed (if no losses). The amount of excess mixwater should be determined by the cost of the additives required and reasoned estimates of what is required.

If dry Bentonite is added, it should be done well in advance of the job, to allow full hydration.

Premix can be diluted instead of using dry bentonite. Determine the bentonite content of the premix, using the Methylene blue test.

A defoamer should be added to the premix before diluting or to the industrial water before adding the dry bentonite in a concentration 0.03 % v/v of the mixwater

= 4.4gal.

Check for Collapse and Float of the Casing

The mud gradient is 0.4628 psi/ft (8.9ppg).

Check for Collapse:

a) pressure at shoe depth outside the casing:

580 ft x 14.8ppg x 0.052 = 453 psi

b) pressure at shoe depth inside the casing:

580 ft x 0.4628psi/ft = 268 psi

So, collapse pressure is 453 psi - 268 psi = 185 psi.

This is smaller than the collapse resistance of this casing (522 psi).

Check for Float

a) Downward forces

The total (dry) weight of the casing is: 580 ft x 94 lbs/ft = 54 520 lbs 

The internal diameter of this casing is 19.124in

Cross‑sectional area:

Pi/4 x (19.124)² = 287in²

Downward force due to mud and cement inside the casing is:

pressure inside the casing at shoe depth (see a above) x cross - sectional area

= 268 psi x 287in² = 76 916 lbs

This added to the weight of the casing gives the total downward force:

76 916 lbs + 54 520 lbs = 131 436 lbs 

b) Upward forces

The external diameter of this casing is 20in.

Cross‑sectional area:

(PI/4) x (20)² = 314in²

Upward force due to cement outside the casing is equal to

pressure outside the casing at shoe depth x cross - sectional area

= 453 psi x 314in² = 142 242 lbs

c) Resultant force:

downward:   131 436 lbs

upward:        142 242  lbs

Net Force:    10 806 lbs upwards

In this case the 20in casing will tend to float. Depending on the actual well data, a heavier mud inside the casing or a lighter slurry should be used.