Question 1 of 6: Combination Material Balance — Undersaturated-to-Saturated Solution-Gas-Drive Reservoir

Question 1: Combination Material Balance — Undersaturated-to-Saturated Solution-Gas-Drive 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. $p_i=2500$ psia (above $p_b=2300$ psia, so the reservoir starts undersaturated), $T=140^\circ\text{F}$, no water drive/production, $N_p=600$ MMSTB, $G_p=800$ MMMSCF as pressure falls from 2500 to 1500 psia; the PVT table above; $S_{wc}=25\%$, $\phi=15\%$.

Find. (a) initial oil in place $N$; (b) free (liberated) gas volume in the reservoir at 1500 psia; (c) reservoir bulk volume.

Approach. Because $p_i$ is above the bubble point, $R_{soi}$ and $B_{ti}$ are read at $p_i$ (no free gas exists yet); apply the general oil-material-balance equation ($m=0$, no aquifer, no water production) using the two-phase (total) formation volume factor $B_t=B_o+(R_{soi}-R_{so})B_g$ to solve for $N$, then close a separate gas balance for the free-gas volume, then get the bulk volume from the initial oil pore volume.

1. Cumulative GOR and initial solution-gas ratio. $R_p=G_p/N_p=\dfrac{800\times10^9}{600\times10^6}$, so $\boxed{R_p=1333.3\ \text{SCF/STB}}$. Since $p_i=2500 > p_b=2300$, $R_{so}$ has not yet changed from its bubble-point value: $R_{soi}=800\ \text{SCF/STB}$, and $B_{ti}=B_{oi}=1.40\ \text{bbl/STB}$ (no free gas above the bubble point).
2. Two-phase FVF at 1500 psia. $B_t=B_o+(R_{soi}-R_{so})B_g=1.35+(800-600)(0.0022)=1.35+0.44$, so $\boxed{B_t=1.79\ \text{bbl/STB}}$ at 1500 psia.
3. Initial oil in place. With $m=0$, $W_e=0$, $W_p=0$, the material balance reduces to $N_p[B_t+(R_p-R_{soi})B_g]=N(B_t-B_{ti})$. Substituting, $N=\dfrac{600\times10^6\left[1.79+(1333.3-800)(0.0022)\right]}{1.79-1.40}=\dfrac{600\times10^6(1.79+1.173)}{0.39}$, so $\boxed{N=4.56\times10^9\ \text{STB}\ (4559\ \text{MMSTB})}$.
4. Free gas within the reservoir at 1500 psia. Total solution gas originally in place is $N R_{soi}=4.56\times10^9(800)=3.647\times10^{12}\ \text{SCF}$. Gas still dissolved in the remaining oil is $(N-N_p)R_{so}=(4.56\times10^9-0.6\times10^9)(600)=2.375\times10^{12}\ \text{SCF}$. The balance $N R_{soi}-G_p-(N-N_p)R_{so}$ is the gas that has come out of solution and remains in the reservoir as a free-gas phase: $3.647\times10^{12}-0.8\times10^{12}-2.375\times10^{12}$, so $\boxed{V_{\text{free gas}}=4.72\times10^{11}\ \text{SCF}\ (471.8\ \text{MMMSCF})}$.
5. Reservoir bulk volume. Before any gas evolves, the oil alone fills the non-water pore space at $p_i$: $N B_{oi}=V_b\,\phi\,(1-S_{wc})$. So $V_b=\dfrac{N B_{oi}}{\phi(1-S_{wc})}=\dfrac{4.56\times10^9(1.40)}{0.15(0.75)}$, giving $\boxed{V_b=5.67\times10^{10}\ \text{bbl}}$ ($\approx 7.31\times10^6$ acre-ft).