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CEMENTING CALCULATIONS Page: 1 of 9
Cementing calculations are an essential part of the designing stage of a cement job. Here, accepted methods and examples are given for:
- Slurry related calculation, weight, yield, and water requirements.
- Fill-up calculations.
- Balancing a cement plug by the "balanced-plug" method.
- Pressure-related calculations, hydrostatic pressure, frictional Pressure, and Horsepower needs.
- Turbulent flow rate.
Slurry calculations
Used in conjunction with fill-up calculations, the slurry weight, slurry volume, and water requirement calculations determine the proper amount of dry-blended cement and water needed for a particular job.
Calculation aids. The following will help in making calculation.
All additive concentrations except salt are based on the weight of cement. When using blended cement systems, the additives are based on the weight of the mixture of cements. Salt percentages are based on the weight of the water. Additives used in low concentrations (less than S%) do not appreciably affect calculations and can generally be ignored. Among such additives are retarders, metasilicate, Diacel A, salt, dispersant, Diacel LWL, KC1, borax and CaC12.
Additives used in larger concentrations are included in the calculations. Such additives are barite, Diacel D, silica sand, Gilsonite, Thixad, and salt.
Class A, C, and H cement weigh 94 lb/sk (sk= sack) and have an absolute volume of 3.60 gal/sk (absolute volume is the volume occupied only by the solids and does not include air. See Table 1). Commercial lightweight cement weight 75 lb/sk and have an absolute volume of 3.22 gal/sk. Weight and volume of blended pozzolan or talc cement are calculated from the ratio of each component. For example, 50:50 Pozment Class H consists of 1/2 bulk cu.ft. (37 lb) of Pozment and 1/2 bulk cu. ft. (47 lb) of Class H. Water weighs at 8.33 lb/gal at room temperature. Set up a three-column table once the slurry composition is known. (See example calculations).
Material Wt, lb Vol, gal
Place the names of the cement and additives in the "material" column. Unless the weight is known in lb, this amount is determined by multiplying the additive percentage time the weight of the cement. Place the volume of the additives in gal. in the "Volume: column. This volume is found by multiplying the additive's weight by the absolute volume column in Table 1. Add up the weight column (lb). Add up the volume column (gal). Divide the weight by the volume to obtain the slurry weight in lb per gal.
Example calculations
Example 1: Calculations
Example 1: Calculate the weight, and yield and the water needed to mix a slurry
consisting of Class H, 35 % sand, and 0.5 % retarder, 46% water.
Solution: Since the retarder appears in low concentration, we can neglect it during
the calculations without introducing a serious error. As a basis for our calculation we will use a sack of Class H cement. Thus:
Wt Wt
Contribution contribution
Material lb gal
1 sk Class H 94.0 3. 35 % sand 32.9 1. 46 % water 43. 2 5. Total 170.1 lb 10.29 gal
Slurry weight = 170.1 lb / 10.29 gal = 16.5 lb/gal Slurry yield = (10.29 gal/sk of cement) (7.48 gal/cu ft) = 1.376 cu ft/sk Mixing water = 5.19 gal/sk of cement
Example 3: Calculate the pounds of hematite needed to obtain an 18.5 lb/gal slurry if
the basic composition is Class H, 35 % sand, hematite and water.
Solution: Here we have, at least on paper, an infinite number of possible solutions
since the amount of hematite will change with concentration of water. If we decide on a concentration of water, we can then calculate the amount of hematite needed. Let's say that we would like to consider 46% water. Call X the lb of hematite per sk of cement. Thus;
Material Wt, lb Vol, gal
1 sk of Class H 94.0 3. 35 % sand 32.9 1. 40 % water 43. 2 5. Xlb hematite X 0.0239X
Total 170.1 + X 10.29 + 0.0239 X
Slurry weight = 18.5 lb/gal (170.1 + X ) (10.29 + 0.0239 X ) Solving for X we get:
X = 36.3 lb hematite/sk of cement
Slurry yield = (10.29 + 0.0239 X) / 7.48 = 1.492 cu ft/sk) Mixing water = 5.19 gal/sk.
P-WAH Slurry
% Water lb/sk yield
Well conditions, slurry cost, and actual lab testing of the slurry will determine which one to use for the job.
Example 4: Calculate the weight, yield, and water requirement for a slurry of Class H,
18 % salt, and 46 % water.
Solution: Here we have to remember that the salt is based on the weight of the
water. Table 2 will be used to aid in the calculations.
Material Wt,lb Vol, gal
1 sk of Class H 94.0 3. 46 % water 43. 2 5. 18 % salt 7. 8 0. 3 Total 145.0 9.
TABLE II
Percent salt
% Salt by wt of water Absolute vol. (dissolved), gal/lb
Slurry density = 145.0 / 9.11 = 15.9 lb/.gal. Slurry yield = 9.11 / 7.48 = 1.22 cu ft/sk of cement Mixing water = 5.19 gal. sk of cement.
Example 6: Six hundred ft of 13 3/8 in surface casing (68 lb/ft) is to be cemented in a 16" hole
using a 116 pcf slurry and 75 % excess.
Solution: From cementer's handbook, the capacity of the annulus to be cemented
0.4206 cu ft/ft.
Slurry volume required = (600 ft) (0.4206 cu ft/ft) = 252.4 cu ft. Sacks of cement needed = 252.4 cu ft/1.18 cu ft/sk = 214 sk. Total number of sacks needed including the excess = (214) (1.75) = 375 sk. Volume of water needed for the slurry = (375 sk) (5.2 gal/sk) (lbbl/42 gal) = 46.4 bbl.
From the cementer's handbook we can also obtain the capacity of the pipe 0.1497 bbl/ft. Therefore, displacement volume needed is:
(600 ft) (0.1497 bbl/ft) = 90 bbl
Balancing a plug
A cement plug must be balanced to help insure that it will be the proper length and in the proper place. If the plug is not balanced, it may migrate down the hole, become contaminated, or fail. The quantities which must be calculated when designing a cementing plug by the balanced plug method include the length of plug; volumes of spacer needed before and after the cement to balance the plug properly; the height of the plug before the pipe is withdrawn; and the volume of displacement fluid needed to balance plug. These examples illustrate the calculations.
Example 7: A 600-ft plug is to be place at a depth of 8,000 ft. The open hole size is 6
1/2 in. Tubing size is 2 3/8 in OD (4.6 lb/ft). Ten bbl of water is to be pumped ahead of the slurry. Slurry yield is 1.18 cu ft/sk. Calculate the number of sacks needed for the job.
Solution: The following equation calculates the number of sacks. N=(L) (Ch)/Y(1)
Where: N = Number of sk of cement L = Plug length, ft Ch = Hole capacity, cu ft/ft Y = Slurry yield, cu ft/sk
The hole capacity (Ch) can be obtained from a cementer's handbook.
Back to Equation 1: N = (600) (0. 2304) / 1.18 = 117. 2 sk
Example 8: If 150 sk are mixed (rather than 117) what would be the length of the plug?
Solution: Equation 1 above can be rearranged to solve for L, the length of the plug:
L = (N) (Y)/Ch (2) and L=(150) (1.18)/0.2304 = 768ft
Example 9: Calculate the volume of water to be pumped behind the slurry to balance the plug.
Solution: The following equation will be used.
Vb = (Cp) (Va)/Ca (3)
where:
Va = Volume of spacer ahead of the slurry, bbl Vb = Volume of spacer behind the slurry, bbl Ca=Capacity of the annulus, cu ft/ft Cp = Capacity of the pipe, cu ft/ft
Capacities may be obtained from a handbook
Ca = 0.1997 cu ft/ft Cp = 0.02171 cu ft/ft
Finally: Vb = (0.02171) (10)/0.1997 = 1.09 bbl
Example 10: Assume that you want to pump a 2-bbl spacer behind the slurry. What is the
volume of spacer needed ahead of the slurry to balance the plug?
Solution: Equation 3 can be solved for Va giving:
Va= (Ca) (Vb)/Cp (4) Va= (0.1997) (2) / 0.02171 = 1.85 bbl
