Given. Casing/hole geometry, a two-slurry cement program (500 ft high-strength tail, 2000 ft low-density lead, together spanning the full 2500 ft to surface), and each slurry's mix-water ratio and additive percentages.
Casing OD / ID
13.375 in / 12.415 in
Hole size
17 in
Setting depth / shoe joint
2500 ft / 40 ft (inside casing)
High-strength column (bottom)
500 ft (incl. the 40 ft shoe joint)
Low-density column (upper annulus)
2000 ft
Annulus excess factor
1.75
HS slurry: Class A + 2% $CaCl_2$, water ratio
5.2 gal/sack
LD slurry: Class A + 16% bentonite + 5% NaCl, water ratio
13 gal/sack
Specific gravities: cement / bentonite / $CaCl_2$ / NaCl
3.14 / 2.65 / 1.96 / 2.16
Find. The slurry volume and number of sacks required for each of the two slurries.
Check: additive percentages (2% $CaCl_2$, 16% bentonite, 5% NaCl) are taken by weight of cement (bwoc), the standard convention when a source states percentages without naming a different basis; the excess factor is applied to the full annular interval for both slurries (none inside the casing/shoe joint, which is a closed pipe with no washout to excess for).
Approach. Get the annulus and casing-ID capacities per foot, split the two slurry columns into their casing-ID and annulus portions, then run each slurry's mix through the standard cement-yield calculation (absolute volumes of cement, water and additives per sack) to get its yield and density, and divide each required volume by its own yield to get the sack count.
Capacities. Annulus (hole–casing OD), no excess: $\dfrac{\pi}{4}\dfrac{17^2-13.375^2}{144}=0.6005$ ft³/ft; with the 1.75 excess factor, $\boxed{0.6005\times1.75=1.051\ \text{ft}^3/\text{ft}}$. Casing ID (shoe joint, no excess — a closed pipe): $\dfrac{\pi}{4}\dfrac{12.415^2}{144}$. $\boxed{0.8407\ \text{ft}^3/\text{ft}}$.
High-strength slurry volume. 40 ft of shoe joint (casing-ID capacity) plus the remaining $500-40=460$ ft of annulus (excess capacity): $V_{HS}=40(0.8407)+460(1.051)=33.6+483.5$. $\boxed{V_{HS}=517\ \text{ft}^3}$.
Low-density slurry volume. The full 2000 ft is annulus only: $V_{LD}=2000\times1.051$. $\boxed{V_{LD}=2102\ \text{ft}^3}$.
High-strength slurry yield and density. Per sack (94 lb cement): $V_{cement}=94/(3.14\times8.33)=3.593$ gal, $w_{CaCl_2}=0.02\times94=1.88$ lb $\Rightarrow V_{CaCl_2}=1.88/(1.96\times8.33)=0.115$ gal, $V_{water}=5.2$ gal. Total volume $=3.593+0.115+5.2=8.909$ gal/sack $=8.909/7.4805$. $\boxed{\text{yield}=1.191\ \text{ft}^3/\text{sack}}$; total weight $=94+1.88+5.2(8.33)=139.2$ lb/sack, density $=139.2/8.909$. $\boxed{15.62\ \text{ppg}\ (116.9\ \text{lb/ft}^3)}$.
Low-density slurry yield and density. $w_{bentonite}=0.16\times94=15.04$ lb $\Rightarrow V_{bentonite}=15.04/(2.65\times8.33)=0.681$ gal; $w_{NaCl}=0.05\times94=4.70$ lb $\Rightarrow V_{NaCl}=4.70/(2.16\times8.33)=0.261$ gal; $V_{water}=13$ gal; $V_{cement}=3.593$ gal. Total $=3.593+0.681+0.261+13=17.54$ gal/sack $=17.54/7.4805$. $\boxed{\text{yield}=2.344\ \text{ft}^3/\text{sack}}$; total weight $=94+15.04+4.70+13(8.33)=222.0$ lb/sack, density $=222.0/17.54$. $\boxed{12.66\ \text{ppg}\ (94.7\ \text{lb/ft}^3)}$ — correctly lighter than the high-strength mix, confirming the "low density" label.
Sack counts. $n_{HS}=V_{HS}/\text{yield}_{HS}=517/1.191=434.2\Rightarrow\boxed{435\ \text{sacks}}$ (round up to whole sacks). $n_{LD}=V_{LD}/\text{yield}_{LD}=2102/2.344=896.6\Rightarrow\boxed{897\ \text{sacks}}$.
Fig. 6 — Two-slurry cement job: low-density lead slurry over the upper 2000 ft of annulus, high-strength tail slurry over the bottom 500 ft (460 ft of annulus plus the 40 ft casing-ID shoe joint), all above the float collar and guide shoe.
