This article gives an example of calculation for a typical balanced cement plug job.

Parameters to be Calculated

The following parameters shall be calculated for the setting of a balanced plug.

• 150 % of drillpipe/stinger contents.
• required slurry volume taking all excess requirements into account.
• tonnage of cement.
• mixwater requirements.
• spacer volumes.
• theoretical TOC in balance.
• TOC inside drillpipe when under-displaced by required amount between 250 - 500 ft.
• top of postflush inside drillpipe when cement is under-displaced by required amount.
• volume of mud to under-displace cement and afterflush by required amount

Hole Data

12.1/4in hole was drilled to 7 153 ft. The hole was logged and RFT measurements done.  All logs showed a good hole, in gauge, with an average hole size of 12.1/2in and a hole capacity of 0.1518 bbls/ft.

The RFT measurement showed the following:

It was decided to plug back to 6 660 ft.  Minimum required overbalance is 102 psi above 4 922 ft and 152 psi deeper down.

The plug will be set with the balanced plug method. A 591 ft 3.1/2in tubing stinger will be run on 5in DP.

Hole Size:                    12.1/4in

Average size:               12.1/2in

Capacity:                     0.1518 bbls/ft

Content:                       mud

Gradient:                      0.4686 psi/ft

Surface lines                2 bbls

Tubular Data

Tubing Weight:                                    9.2 lbs/ft

Capacity:                                 0.00863 bbls/ft

Displacement:                          (open end) 0.00339 bbls/ft

(closed end) 0.01202 bbls/ft

Drillpipe weight:                      19.5 lbs/ft

Capacity:                                 0.01776 bbls/ft

Displacement:                          0.02492 bbls/ft

Cement Recipe

Cement Type:                          Class “G”

Water requirements:                 5 gals/sk

Slurry Yield:                            1.14 cu.ft/sk

Slurry Density:                         0.8223 psi/ft

Additives

D603:                                      0.0125 gal/sk

D080:                                      0.0287 gal.sk

D047:                                      0.0018 gal/sk

Calculations Pre-Flush and Post-Flush Volume

The formation pressure, EMG 0.442 psi/ft, needs to be overcome, with a safety margin of 152 psi, in all circumstances.  The depth at which the hydrostatic head of the mud column is just enough to provide this safety margin is:

D x 0.4686 = D x 0.4421 + 152 gives D = 5 742 ft.

This means that, assuming pressure communication does exist from 6480 ft to
5 742 ft, any preflush/spacers is not allowed to rise higher than 5 742 ft. With a planned TOC at 6 660 ft yields a maximum spacer length of 919 ft. If however, the actual TOC, due to excess cement being pumped, would be higher than 6 660 ft a spacer of exactly 919 ft could rise beyond 5 742 ft. The spacer is therefore limited to 656 ft annular length.

656 ft annular length occupies a volume of : 98 ft stinger:

398 ft x 558 ft x (0.1518 - 0.01202) bbls/ft = 13.8 bbls.

558 ft DP : 558 ft x (0.1518 0 - 0.0249) = 70.8 bbls

Total Pre-Flush volume = 84.5 bbls

To balance the plug the height of afterflush in the DP has to be 656 ft as well:

98 ft stinger :   98 ft x 0.0086 bbls/ft  = 0.843 bbls

558 ft DP:        558 ft x 0.0178 bbls/ft = 9.9 bbls

Total Post Flush Volume = 10.8 bbls

Material Requirements

Slurry Volume

Slurry volume  = plug back length x average hole size

= 493 ft x 0.1518

= 74.8 bbls (logs show 75.5 bbls)

Allowing for an excess of 10 %.

Total slurry volume = 1.1 x 75.5 bbls = 83 bbls = 466 cu.ft

An excess of 10 % means an additional 49 ft plug length, which is in agreement with the reduced spacer length.

Dry Cement

Cement needed

= slurry volume/yield

= 466 cu.ft/1.15 cu.ft/sk

Total cement = 405 sk cement

Mixwater

Mixwater required:

for 405 sk cement:  405 sk x 5 gal/sk  = 2025 gal = 48.2 bbls

excess 10 %:   0.10 x 48.2 bbls  = 4.8 bbls

dead volume tanks:      18.9 bbls

Total Mixwater Required : 71.9 bbls.

Additives

All additives are liquids and added to the mixwater. The use of the additives is calculated back to a concentration mixed in the mixwater:

D603   =          0.707 gal/bbl

D080   =          1.615 gal/bbl

D047   =          0.101 gal/bbl

Thus to mix in 71.9 bbls of site water:

D603:  71.9 bbls x 0.707 gal/bbl = 50.83 gal in drums of 15 gal each; 3.39 drums

D080:  71.9 bbls x 1.615 gal/bbl = 111.8 gal in drums of 15 gal each; 7.74 drums

D047:  71.9 bbls x 0.101 gal/bbl = 7.26 gal in drums of 5 gal each; 1.42 drums

To obtain whole numbers of drums D080 to be used, the volume of mixwater is adjusted to 74.5 bbls, by increasing the percentage excess to 12 %.  The numbers then become 3.5 drums of D603, 8 drums of D080 and 1.6 drums of D047, to be mixed in 74.5 bbls site water.

The total mix fluid to be held by the mix tank on the rig is:

74.5 bbls + (8 drums x 0.358 bbls/drum) + (3.5 drums x 0.358 bbls/drum) + (1.6 drums x 0.113 bbls/drum) = 78.8 bbls

Displacement Volumes

To calculate the volume to displace accurately, the TOC with the stinger in the cement, has to be calculated.

String + annulus OH - tubing total capacity
= 0.1518 bbls/ft - 0.0034 bbls/ft = 0.1484 bbls/ft.

String + annulus OH - DP total capacity
= 0.1518 bbls/ft - 0.00783 bbls/ft = 0.144 bbls/ft.

83 bbls slurry occupies 83 bbls / 0.1484 bbls/ft = 559 ft annular height (i.e. 559 ft along the tubular).

The highest estimate of TOC gives the maximum displacement volume allowed:

TOC = 7 153 ft -561 ft = 6 592 ft

Top of water, afterflush in DP, will be at 6 592 ft - 656 ft = 5 936 ft

Total volume to displace          = 5 936 ft + 3 ft overstand of DP + vol. Surf. Lines

= 5 939 ft x 0.0178 bbls/ft  + 2 bbls

= 107.7 bbls

Allowing for under displacement of 1 stand of DP (2 bbls):

Displacement Volume = 105.7 bbls.

Displacement Rates

When the cement reaches the end of the stinger the pump rate has to be slowed down to meet plug flow condition (< 30 ft/min annular velocity) during placement of the slurry.

The total string volume

= ( 6 562 ft x 0.0178 bbls/ft) + (591 ft x 0.00863 bbls/ft)

= 122 bbls

Thus pumping should be slowed down after string volume minus slurry volume minus afterflush has been displaced:

= 122 bbls - 83 bbls - 10.8 bbls

= 28.2 bbls

Accepting an annular velocity of 30 ft/min yields (not taking any U-tubing effects into account) a maximum pumprate Q of:

Q (max) = (0.1518 - 0.01202) bbls/ft  x 30 ft/min = 4.2 bbls/min

Duration of the Job

Pump preflush:                        84.6 bbls @ 9.44 bbls/min       »            9.0 min

Mix/pump slurry:         90 bbls   @ 6.29 bbls/min        »          14.3 min

Pump after-flush:         11 bbls   @ 9.44 bbls/min        »            1.2 min

Chase:                          105 bbls @ 3.9 bbls/min          »          27.0 min

TOTAL                                                                                    52.0 min

Total Job Time = 1 hr