Cementing · casing · field units
Cement volume calculation: slurry, sacks, mix water and displacement
To calculate cement for a casing job, work out the slurry volume (annulus × excess + shoe track + rathole), convert it to cubic feet, and divide by the slurry yield to get the number of sacks. Mix water is sacks × water per sack, and the displacement volume is the casing capacity from surface to the float collar. In field units: annular capacity = (Dh² − OD²) ÷ 1029.4 bbl/ft, casing capacity = ID² ÷ 1029.4 bbl/ft, ft³ = bbl × 5.6146 and sacks = ft³ ÷ yield. This page explains each step, gives yields for neat API Class A, G and H cement, and works one 9⅝ in casing job end to end.
Reference material only. Slurry design, excess, top of cement and pumping schedule come from the operator's cementing programme and the cementing service company's lab-tested recipe. Worked examples are illustrative arithmetic, not a job design.
1. The cement calculation formulas
Annular capacity
Ca (bbl/ft) = (Dh² − OD²) ÷ 1029.4
Casing capacity
Cc (bbl/ft) = ID² ÷ 1029.4
Slurry volume
V = Ca × L × (1 + excess) + shoe track + rathole
Barrels to cubic feet
ft³ = bbl × 5.6146
Number of sacks
sacks = slurry ft³ ÷ yield (ft³/sk)
Displacement
Vdisp (bbl) = Cc × depth to float collar
Dh is the hole diameter, OD the casing outside diameter and ID the casing inside diameter, all in inches; L is the cemented length in feet. The 1029.4 constant turns diameter in inches squared into barrels per foot; it is the same constant used for annular capacity and drill pipe capacity. Where the annulus is inside a previous casing string, use that casing's ID as Dh and normally no excess.
2. Slurry volume: annulus, excess, shoe track and rathole
Cement has to fill four spaces, and each is calculated separately:
- Open-hole annulus. Annular capacity × length from the shoe up to the planned top of cement (or to the next interval with a different slurry). Bit size understates the real hole, so an excess factor is applied. The best basis is a caliper log volume; without one, larger percentages are used. The figure is always set by the cementing programme.
- Cased-hole annulus. If cement is brought back inside the previous casing, use the previous casing ID instead of the bit size, with little or no excess.
- Shoe track. The joints between the float collar and the float shoe stay full of cement after the top plug lands, so they are added at casing capacity × shoe track length.
- Rathole. Any open hole drilled below the casing shoe, at open-hole capacity (Dh² ÷ 1029.4) × length.
Jobs with a lighter lead slurry above a denser tail slurry are calculated the same way, one interval at a time, because each slurry has its own yield.
3. Slurry yield, density and water for API cement
Slurry yield is the volume of slurry made by one sack of cement. For API cement sold in 94 lb sacks, the cement itself occupies about 3.59 gal (94 lb ÷ (3.14 specific gravity × 8.33 lb/gal)). Add the mix water and any additives, then divide by 7.4805 gal/ft³. Slurry density is total mass ÷ total volume. The table uses the API Spec 10A water ratios for neat slurries.
| Neat API cement | Water per 94 lb sack | Yield and density |
|---|---|---|
| Class A | 5.2 gal (46%) | 1.18 ft³/sk · 15.6 ppg |
| Class G | 5.0 gal (44%) | 1.15 ft³/sk · 15.8 ppg |
| Class H | 4.3 gal (38%) | 1.06 ft³/sk · 16.4 ppg |
Calculated as (3.59 gal + water) ÷ 7.4805 and (94 + 8.33 × water) ÷ (3.59 + water). Extended, light-weight and additive-loaded slurries have very different yields; always use the yield on the service company's lab report for the actual blend.
4. Worked example: 9⅝ in casing in 12¼ in hole
9⅝ in 47 lb/ft casing (ID 8.681 in) is run to 8,000 ft in 12¼ in hole. The float collar is 80 ft above the shoe, and 10 ft of rathole is left below it. One tail slurry of neat Class G at 15.8 ppg (1.15 ft³/sk, 5.0 gal/sk water) covers 1,500 ft of open hole, from 8,000 ft up to 6,500 ft, with 30% excess. Mud is 10.5 ppg. All figures are illustrative.
| Step | Working | Answer |
|---|---|---|
| Annular capacity | (12.25² − 9.625²) ÷ 1029.4 | 0.05578 bbl/ft |
| Annulus, gauge hole | 0.05578 × 1,500 | 83.7 bbl |
| Annulus with 30% excess | 83.7 × 1.30 | 108.8 bbl |
| Shoe track | (8.681² ÷ 1029.4) × 80 | 5.9 bbl |
| Rathole | (12.25² ÷ 1029.4) × 10 | 1.5 bbl |
| Total slurry | 108.8 + 5.9 + 1.5 | 116.1 bbl |
| Slurry in ft³ | 116.1 × 5.6146 | 651.8 ft³ |
| Sacks of cement | 651.8 ÷ 1.15 | 567 sacks |
| Mix water | 567 × 5.0 ÷ 42 | 67.5 bbl |
| Dry cement weight | 567 × 94 lb | 53,298 lb (≈ 24.2 t) |
Real jobs order more dry cement and water than the calculated figure to cover surface lines, mixing losses and tank bottoms; the allowance is a programme decision.
5. Displacement volume and strokes to bump the plug
After the slurry, the top plug is pumped down with mud (or another displacement fluid) until it lands on the float collar. The displacement volume is the casing capacity from surface to the float collar:
| Step | Working | Answer |
|---|---|---|
| Casing capacity | 8.681² ÷ 1029.4 | 0.07321 bbl/ft |
| Displacement volume | 0.07321 × 7,920 | ≈ 580 bbl |
| Pump output | 0.000243 × 6.5² × 12 × 0.97 | 0.1195 bbl/stk |
| Strokes to bump | 580 ÷ 0.1195 | ≈ 4,850 strokes |
The pump output is for a triplex pump with a 6½ in liner, 12 in stroke and 97% volumetric efficiency; see the triplex pump output formula. Because volumetric efficiency is never known exactly, crews compare the calculated strokes with the actual pressure response and agree in advance how far past the calculated volume they will pump if the plug does not bump. Over-displacing leaves a wet shoe; under-displacing leaves cement inside the casing.
Near the end of displacement the heavier cement in the annulus pushes back against the lighter mud inside the casing. With 1,420 ft of 15.8 ppg slurry in the annulus above the float collar and 10.5 ppg mud inside, the hydrostatic difference is 0.052 × (15.8 − 10.5) × 1,420 ≈ 391 psi, before any friction. This is the same hydrostatic pressure formula used everywhere else on the rig, and the slurry's added hydrostatic plus circulating friction is checked against fracture limits in the same way as ECD.
6. Checks before the job
- Units. Mixing up gallons, barrels and cubic feet is the classic error. Keep everything in barrels until the last step, then convert once with 5.6146.
- Casing weight. ID changes with weight and grade; 9⅝ in casing ID runs from about 8.4 in to 9.0 in across common weights. Use the ID from the tally, not a nominal value.
- Float collar depth. Take it from the casing tally, not from the shoe depth minus a standard length.
- Top of cement. Confirm the required top of cement and any regulatory or programme overlap with the previous shoe before calculating lengths.
- Returns. Losses while cementing mean the planned top of cement may not be reached; record volumes pumped and returns observed in the daily drilling report.
7. In DrillX
DrillX brings daily drilling reports with IADC-style time logs, a rig fleet view, NPT insights and engineering calculators into one tool. The calculators cover hydrostatic pressure, ECD, annular pressure loss, triplex pump output, annular capacity and velocity, buoyancy and the kill sheet. This page explains the cement volume arithmetic; it does not describe a DrillX cementing calculator. DrillX is in final testing; email info@drillx.app for a demo or early access.
Questions
How do you calculate cement volume for a casing job?
Add up every space the slurry must fill. Annular volume is (hole diameter² − casing OD²) ÷ 1029.4 bbl/ft times the cemented length, multiplied by an excess factor for open hole. Add the shoe track (casing ID² ÷ 1029.4 × length from float collar to shoe) and any rathole below the shoe. In the worked example 1,500 ft of 9⅝ in casing in 12¼ in hole with 30% excess needs about 116 bbl of slurry.
How many sacks of cement do I need?
Convert the slurry volume to cubic feet (bbl × 5.6146) and divide by the slurry yield in ft³ per sack. 116.1 bbl is 651.8 ft³; at a Class G neat yield of 1.15 ft³/sk that is 566.8, rounded up to 567 sacks of 94 lb.
What is cement slurry yield?
Yield is the volume of slurry one sack of cement makes, in ft³/sk. It is the absolute volume of the cement plus the mix water and any additives, divided by 7.4805 gal/ft³. A 94 lb sack of API cement has an absolute volume of about 3.59 gal, so Class G neat with 5.0 gal of water per sack yields (3.59 + 5.0) ÷ 7.4805 ≈ 1.15 ft³/sk at about 15.8 ppg.
How much mix water is needed for a cement job?
Multiply the number of sacks by the water requirement per sack and divide by 42 to get barrels. Class G neat uses about 5.0 gal/sk (44% water by weight of cement), so 567 sacks need 2,835 gal, or 67.5 bbl, before any allowance for surface lines and tank bottoms.
How do you calculate cement displacement volume?
Displacement is the internal capacity of the casing from surface to the float collar: casing ID² ÷ 1029.4 × depth to the float collar. For 9⅝ in 47 lb/ft casing (ID 8.681 in) with the float collar at 7,920 ft that is about 580 bbl. Divide by the pump output per stroke to get strokes to bump the plug; at 0.1195 bbl/stk it is about 4,850 strokes.
How much excess cement should be used?
Excess covers hole enlargement that the bit size does not show. It is set by the operator's cementing programme, ideally from a caliper log. When a caliper volume is available, the programme often adds a smaller percentage on top of it; without one, larger open-hole excess is typical. Cased-hole sections normally use little or no excess. The 30% in this page's example is illustrative only.
Sources and further reading
- E. B. Nelson and D. Guillot (eds.), Well Cementing, 2nd ed. (Schlumberger, 2006): slurry design, yield and density, excess and displacement practice.
- N. J. Lapeyrouse, Formulas and Calculations for Drilling, Production and Workover, 2nd ed. (Gulf Professional Publishing, 2002): cementing calculations in field units (1029.4, 5.6146, sacks from yield, mix water, displacement strokes).
- API Specification 10A, Cements and Materials for Well Cementing: cement classes and the water ratios used for neat slurries (46% Class A, 44% Class G, 38% Class H).
- A. T. Bourgoyne Jr. et al., Applied Drilling Engineering, SPE Textbook Series Vol. 2 (1986), chapter 3: cementing, slurry density and yield calculations.
Reference summary only. Worked examples are illustrative arithmetic. Slurry recipes, yields, excess and displacement volumes are set by your operator's cementing programme and the cementing service company.