Example Calculation 9.5/8in Casing Cementation

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299 casing-cement-calculationThis article presents an example of calculation for a typical 9 5/8" casing cementation (cement volumes, displacement volumes, pressure at bump, etc.).

Relevant Data

  • Vertical exploration well.
  • Previous casing: 13 3/8in, 72 lbs/ft, N‑80, BTC, set at 3724 ft BDF.
  • 12.1/4in hole drilled to 5102 ft BDF.
  • 9.5/8in casing, 47 lbs/ft, N‑80, VAM, run to 5 053 ft BDF and float collar at 4954 ft BDF (2 joints shoetrack).
  • Casing to be cemented with 15.8 ppg slurry.
  • Mixwater recipe: industrial water + 0.4 % w/w fluid loss control additive + 0.2 % w/w friction reducer (dispersant) + 0.03 % v/v defoamer.
  • Surface lines capacity = 1.0 bbl.
  • Mud in use: CMS (CMC/Bentonite) mud, 10.4ppg.

Cement Volume Calculations

a) The relevant capacities are given in the following table:

SECTION

CAPACITY(bbls/ft)

9.5/8in casing

0.0732

12.1/4in (gauge) hole / 9.5/8in casing annulus

0.0558

13.3/8in / 9.5/8in casing annulus

0.0581

b) The relevant cement slurry yields and water/cement ratios are given in the following table:

SLURRY GRADIENT (psi/ft)

YIELD
(cu.ft/sk)

WATER / CEMENT RATIO (gal/sk)

0.822

1.15

5

c) If the open hole section is gauge, the annular capacity would be:

(5 053 ft - 3 724 ft) x 0.0558bbls/ft = 74 bbls.

Assume an SHDT calliper log has been run, giving an integrated cement volume of 89 bbls. No excess will be used over this volume.

d) The volume of the shoetrack is (approx.):

(5 053 ft - 4954 ft) x 0.0732bbls/ft = 7.2 bbls.

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

89 bbls + 7.2 bbls = 96 bbls = 539 cu.ft

f) Yield = 1.15 cu.ft/sk, so:

539 cu.ft »          469 sx cement required
1.15 cu.ft/sk

Water / cement ratio = 5 gal/sk, so:

g) 469 sk x 5 gal/sk = 2 345 gal = 56 bbls mixwater required.

Displacement Calculation

9.5/8in casing capacity from cementing head (assumed to be at drillfloor level) to float collar is:

4 954 ft x 0.0732 bbls/ft = 363 bbls

After the top plug has been dropped, 5 bbls of water behind will be pumped.

So the total displacement volume, to be pumped with the rig pumps is:

363 bbls - 5 bbls = 358 bbls

Using 0.2013 bbls/stroke for 100 % efficiency of the mud pumps, this equals:

1 778 strokes at 100 % efficiency

1  823 strokes at 97.5 % efficiency

1867 strokes at 95 % efficiency

Shoetrack volume is 7.2 bbls = 36 strokes (at 100 % efficiency).

If 97.5 % is the calibrated efficiency of the mud pumps, not more than 1 823 strokes should be pumped (plug should bump after 1 823 strokes). This is done to avoid over-displacement of the cement in the case of leaking plugs; 1 840 strokes equals the theoretical displacement volume at 97.5 % volumetric efficiency of the mud pumps plus half the volume of the shoetrack.

Mixwater Preparation

A mixwater volume of 56 bbls is required. Prepare some excess (to allow for suction loss of the mixwater tank, eventualities, etc.).

56 bbls = 0.119 bbls/sk

469 sk

The water / cement ratio = 0.119 bbls/sk, so 15 bbls mixwater is enough for excess

   71 bbls        =  597 sk

0.119 bbls/sk

 

0.4 % w/w fluid loss additive =

0.4 x 597 sk x 94 lbs/sk =    224 lbs

             100

 

0.2 % w/w friction reducer =

 

0.2 x 597 sk x 94 lbs/sk =          112 lbs

           100

 

The defoamer should be added to the industrial water before the other additives.

150 bbls mixwater

0.03 % v/v defoamer =  0 03 x 71 bbls =          0.02 bbls = 0.85 gal » 1 gal

Spacers

In this case, we assume the programme specified the use of the standard spacers for production casing: a 500 ft industrial water spacer followed by 500 ft mudflush and 500 ft scavenger slurry.

Calculate the volumes required for 500 ft spacers in the average (washed out) hole ‑ 9.5/8in casing annulus.

The total integrated cement volume was 89 bbls over (5 053 ft ‑ 3 724 ft) = 1 329 ft.

So the average annulus capacity is:

 

89 bbls =          0.07 bbls/ft

1 329 ft

 

500 ft spacer will require 500 ft x 0.07 bbls/ft = 35 bbls = 196 cu.ft

For a 0.522 psi/ft scavenger slurry, a yield of 5.1 cu.ft/sk and a water / cement ratio of 34.4 gal/sk can be assumed, So for 196 cu.ft, we need:

 

196 bbls =          38.4 sk

5.1cu.ft/sk

 

and,

38.4 sk x 34.4 gal/sk = 1322 gal = 31.51 bbls

After the top plug has been dropped, 5 bbls water behind water will be pumped.

Reduction in Overbalance

It has to be checked if the minimum required overbalance is held on the formations during all stages of the cementation. If we assume 639 ft column of spacers (“worst case”, in gauge hole), the resulting maximum temporary reduction in overpressure can be calculated as follows:

FLUID

GRADIENT (psi/ft)

Mud

0.539

Preflush

0.460

Scavenger

0.522

639 ft x (0.539 - 0.460) psi/ft           =          51 psi

639 ft x (0.539 - 0.522) psi/ft           =          11. psi

Total reduction in overpressure =                 122 psi

Hydrostatic Bump Pressure and Estimated TOC

An estimation of the position of the TOC can be obtained by observing the pump pressure just before bumping the plug at a predetermined, slow rate and comparing this to the pump pressure observed when circulating mud around at the same rate.

Hydrostatic bump pressure is defined as the standpipe pressure that would be seen at surface, just before bumping, at no rate. It can be calculated before the job to predict what sort of bump pressure can be expected.

In our case (TOC at 3 215 ft - 500 ft inside 13.3/8in shoe):

a) outside the casing:

cement : (5 053 ft ‑ 3 215 ft) x 0.822 psi/ft = 1 510 psi

spacers: all spacers will be inside the 13.3/8in - 9.5/8in annulus at the moment of bumping.

31.5 bbls spacer means   

     

31.5 bbls =          5 421 ft length spacer

0.0581 bbls/ft

 

5 421 ft x 0.460 psi/ft (mudflush)               =          249 psi

5 421 ft x 0.522 psi/ft (scavenger)               =          283 psi

Total                                                            =          532 psi

mud: above the spacers is 0.539 psi/ft  mud. Prior to bumping, the top of the seawater spacer will be at 1 589 ft.

1 589 ft x 0.539 psi/ft              =          857 psi

So the total hydrostatic pressure at shoe depth, due to fluids outside the casing will be:

1 510 psi + 532 psi + 857 psi = 2 899 psi

b) inside the casing:

shoetrack: (5 053 ft ‑ 4 954 ft) x 0.822 psi = 81 psi

mud: prior to bumping, the top of the 31.5 bbls seawater spacer will be at 1 589 ft.

(4 954 ft ) x 0.541 psi/ft = 2 680 psi

So the total hydrostatic pressure at shoe depth, due to fluids inside the casing will be:

81 psi + 2 680 psi = 2 761 psi

c) pressure differential and estimation of TOC.

 

The predicted hydrostatic bump pressure will be:

2 899 psi - 2 761psi = 138 psi

Imagine that the observed pump pressure when circulating mud at 30 strokes per minute was 200 psi, while the observed bump pressure at this rate was 638 psi.

638 psi means a hydrostatic Bump Pressure of

638 psi - 200 psi = 438 psi

This is 438 psi - 138 psi = 300 psi  higher than the predicted pressure.

Every foot of extra cement column will cause an additional pressure of (0.822 psi/ft - 0.541 psi/ft = 0.281 psi/ft.

 

300 psi =          1 068 ft

0.281 psi/ft

 

So the estimated TOC will be 1 068 ft higher than planned:

 

3 215 ft - 1 068 ft -  = 2 147 ft BDF.

 

The expected differential pressure for various TOC’s can be plotted on a graph, to get an idea of the position of the cement slurry at any time during the displacement.

 

 

Comments  

#6 Ayodele hope 2017-07-08 22:40
Nice one,thanks
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#5 زهرا 2017-02-21 12:07
How do we calculate what proportion of materials used?
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#4 aziz moussa 2016-06-22 15:38
How to calculate the force downward & upward, including the differential pressure in the above mentioned example.
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#3 Tamerziton 2016-06-15 16:50
Thanks.
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#2 ramawork 2015-08-06 06:45
good job

Do you have the calculations in Metric??
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#1 mmonteiro2014 2013-12-28 16:55
very good........!!!
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