Question 5 of 7: Drive-Mechanism Indices for a Combination-Drive Oil Reservoir

Question 5: Drive-Mechanism Indices for a Combination-Drive Oil Reservoir (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $N=67$ MMSTB (initial oil in place); gas cap $G_{fgi}=38$ MMMSCF at discovery; $c_f=c_w=0$; no water production ($W_p=0$); PVT/production table above (4100 psia = initial condition, 3950 psia = current condition).

Find. Depletion drive index (DDI), gas-cap drive index (SDI/GCDI), and water drive index (WDI) at 3950 psia.

Approach. Compute the two-phase FVF $B_t$ at both conditions, the gas-cap size ratio $m$, and the total underground withdrawal $F$; get DDI and SDI directly from their standard material-balance terms, then use the fact that the three indices must sum to 1 (and $c_f=c_w=0$ removes the rock/water-expansion term entirely) to obtain WDI by difference, since no water-influx volume $W_e$ is given directly.

1. Two-phase FVF, $B_t$. At $p_i=4100$ psia (bubble point/initial, $R_{so}=R_{soi}$): $B_{ti}=B_{oi}=1.41$ bbl/STB. At $p=3950$: $B_t=B_o+B_g(R_{soi}-R_{so})=1.34+0.000602(950-800)$, so $\boxed{B_t=1.4303\ \text{bbl/STB}}$.
2. Gas-cap ratio, $m$. Gas-cap reservoir volume $=G_{fgi}B_{gi}=38\times10^{9}(0.000591)=2.246\times10^{7}$ bbl. Oil-zone reservoir volume $=NB_{ti}=67\times10^{6}(1.41)=9.447\times10^{7}$ bbl. $m=\dfrac{G_{fgi}B_{gi}}{NB_{ti}}=\dfrac{2.246\times10^{7}}{9.447\times10^{7}}$, so $\boxed{m=0.2377}$.
3. Underground withdrawal, $F$. Cumulative GOR $R_p=G_p/N_p=1140/1.314=867.6$ SCF/STB. $F=N_p\left[B_t+B_g(R_p-R_{soi})\right]=1.314\times10^{6}\left[1.4303+0.000602(867.6-950)\right]$, so $\boxed{F=1.814\times10^{6}\ \text{bbl}}$.
4. Depletion drive index. $\text{DDI}=\dfrac{N(B_t-B_{ti})}{F}=\dfrac{67\times10^{6}(1.4303-1.41)}{1.814\times10^{6}}$, so $\boxed{\text{DDI}=0.750\ (75.0\%)}$.
5. Gas-cap drive index. $\text{SDI}=\dfrac{Nm\frac{B_{ti}}{B_{gi}}(B_g-B_{gi})}{F}=\dfrac{67\times10^{6}(0.2377)\frac{1.41}{0.000591}(0.000602-0.000591)}{1.814\times10^{6}}$, so $\boxed{\text{SDI}=0.230\ (23.0\%)}$.
6. Water drive index, by difference. With $c_f=c_w=0$, the rock-and-water expansion term vanishes and, since no $W_e$ (water influx volume) is given, the three drive indices are found to sum to 1: $\text{WDI}=1-\text{DDI}-\text{SDI}=1-0.750-0.230$, so $\boxed{\text{WDI}=0.020\ (2.0\%)}$.