Cement Volume Calculator

Overview

Cementing calculations determine the volumes, weights, and mix ratios needed for primary and remedial cementing operations. Primary cementing seals the annulus between casing and formation, providing zonal isolation, casing support, and corrosion protection. Accurate slurry design ensures proper fill-up, preventing channels and ensuring long-term well integrity.

Theory

Cement slurry is a mixture of cement powder, water, and additives. The key parameters are slurry density, slurry yield, water requirement, and mix water ratio. API classes (A, C, G, H) define cement properties for different depth/temperature ranges. Class H and G are the most commonly used in oil and gas operations.

Formulas

Annular Volume

V_annular = (Dh^2 - OD_casing^2) / 1029.4 * Length  (bbl)

Convert to cubic feet: V_ft3 = V_bbl * 5.615

Convert to sacks: sacks = V_ft3 / slurry_yield

Slurry Yield

Yield (ft³/sack) = (94 + Ww + Wa) / (ρ_slurry * 7.48)

where 94 lb = weight of 1 sack of cement, Ww = weight of water per sack, Wa = weight of additives per sack.

Mix Water Ratio

Water (gal/sack) = (water_ratio * 94) / 8.33

For Class H neat cement: water ratio = 0.38 (by weight), water = 4.29 gal/sack.

Slurry Density

ρ_slurry = (94 + Ww + Wa) / (V_cement + V_water + V_additives)  (lb/gal)

Absolute volume of cement: 0.0382 gal/lb (Portland).

Absolute volume of water: 0.1200 gal/lb.

Hydrostatic Pressure of Cement Column

P_cement = 0.052 * ρ_slurry * h_cement

Displacement Volume (Inside Casing)

V_displacement = ID_casing^2 / 1029.4 * (MD - shoe_depth + shoe_track)

Excess Volume Factor

V_total = V_annular * (1 + excess_factor)

Typical excess: 50–100% in open hole, 0–10% in cased hole.

Worked Example

Given: 7" casing (OD = 7.0", ID = 6.276") in 8.75" hole, cement interval = 2,000 ft, Class H neat cement (ρ = 16.4 ppg, yield = 1.18 ft³/sack).

Step 1: Annular volume:

V_annular = (8.75^2 - 7.0^2) / 1029.4 * 2,000
           = (76.5625 - 49.0) / 1029.4 * 2,000
           = 27.5625 / 1029.4 * 2,000
           = 0.02677 * 2,000
           = 53.5 bbl = 300.5 ft³

Step 2: With 50% excess:

V_total = 300.5 * 1.50 = 450.8 ft³

Step 3: Cement sacks:

Sacks = 450.8 / 1.18 = 382 sacks

Step 4: Mix water:

Water = 382 * 4.29 = 1,639 gal = 39.0 bbl

Step 5: Displacement volume (well TD = 8,000 ft, shoe at 6,000 ft):

V_disp = 6.276^2 / 1029.4 * (8,000 - 2,000) = 0.03827 * 6,000 = 229.6 bbl

Valid Ranges

References

1. Nelson, E.B. & Guillot, D. (2006). Well Cementing. Schlumberger.
2. API Specification 10A — Cements and Materials for Well Cementing.
3. Smith, D.K. (1990). Cementing. SPE Monograph, Vol. 4.
4. PetroWiki — Cementing: https://petrowiki.spe.org/Cementing