Undersaturated Oil Reservoir — Material Balance and Gas Reinjection: 24-Pet-A3 Fundamental Reservoir Engineering, December 2014, Q6

Question 6: Undersaturated Oil Reservoir — Material Balance and Gas Reinjection (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.

Check: the paper does not restate the initial reservoir pressure elsewhere. Since $R_s$ is unchanged (950 SCF/STB) between 4500 and 4000 psia and only drops at 3500 psia, both 4500 and 4000 psia lie at or above the bubble point; consistent with "initially undersaturated," the highest tabulated pressure, 4500 psia, is taken as $p_i$, so $B_{ti}=B_{oi}=1.31$ bbl/STB.

Given. $R_p=3300$ SCF/STB, $N_p=1.5$ MMSTB (both at 3500 psia); PVT table above; $R_{soi}=950$ SCF/STB, $B_{ti}=1.31$ bbl/STB (at $p_i=4500$ psia, per the callout); volumetric reservoir, no initial gas cap ($m=0$). Formula sheet: $N=\dfrac{N_p[B_t+B_g(R_p-R_{soi})]}{(B_t-B_{ti})}$ (with $m=0$); $B_t=B_o+B_g(R_{soi}-R_{so})$; 1 bbl $=5.615$ ft$^3$.

Find. (a) Fractional recovery $N_p/N$ at 3500 psia as produced; (b) fractional recovery at 3500 psia if two-thirds of the produced gas is returned (reinjected).

Approach. Compute the two-phase formation volume factor $B_t$ at 3500 psia and at initial conditions, then apply the $m=0$ oil material balance once with the actual (gross) producing GOR for part (a), and again with the reduced net producing GOR (only the un-reinjected third actually leaves the reservoir) for part (b).

Two-phase FVF. Converting $B_g$ to bbl/SCF ($\div5.615$): $B_g(3500)=0.0041198/5.615=0.0007337$ bbl/SCF. $B_t(3500)=B_o+B_g(R_{soi}-R_{so})=1.31+0.0007337(950-900)=\boxed{1.3467\ \text{bbl/STB}}$; $B_{ti}=1.31$ bbl/STB (no correction needed, since $R_{so}=R_{soi}$ at $p_i$).
Part (a) — as produced. $N=\dfrac{N_p[B_t+B_g(R_p-R_{soi})]}{B_t-B_{ti}}=\dfrac{1.5[1.3467+0.0007337(3300-950)]}{1.3467-1.31}=\dfrac{1.5(1.3467+1.7244)}{0.0367}=\dfrac{1.5(3.0711)}{0.0367}$, so $\boxed{N=125.6\ \text{MMSTB}}$ and $\boxed{N_p/N=1.5/125.6=1.19\%}$.
Part (b) — two-thirds of produced gas reinjected. Only the un-reinjected third of the produced gas actually leaves the reservoir, so the net producing GOR is $R_p'=\tfrac13(3300)=1100$ SCF/STB. $N'=\dfrac{1.5[1.3467+0.0007337(1100-950)]}{0.0367}=\dfrac{1.5(1.3467+0.1101)}{0.0367}=\dfrac{1.5(1.4568)}{0.0367}$, so $\boxed{N'=59.6\ \text{MMSTB}}$ and $\boxed{N_p/N'=1.5/59.6=2.52\%}$.

Quantity	Value
$B_t$ at 3500 psia	1.3467 bbl/STB
(a) $N$ (as produced)	125.6 MMSTB
(a) Fractional recovery	1.19%
(b) $N$ (2/3 gas reinjected)	59.6 MMSTB
(b) Fractional recovery	2.52%

Topic: Undersaturated-oil material balance ($m=0$) and the effect of gas reinjection on apparent OOIP.

Key relations: $N=\dfrac{N_p[B_t+B_g(R_p-R_{soi})]}{B_t-B_{ti}}$; $B_t=B_o+B_g(R_{soi}-R_{so})$.

Why this works: This early in depletion (only 1000 psi below $p_i$, and $p_i$ itself only slightly above the bubble point), the denominator $B_t-B_{ti}$ is very small, so the material-balance $N$ is extremely sensitive to the free-gas term $B_g(R_p-R_{soi})$ — reinjecting gas reduces the net producing GOR, shrinks that term, and roughly halves the computed $N$, which is why the two parts give such different answers for the same $N_p$.

Common pitfall: Forgetting to convert $B_g$ from ft$^3$/SCF to bbl/SCF before combining it with $B_o$ (bbl/STB) and $R_s$ (SCF/STB) — the units must match for $B_t$ to come out in bbl/STB.

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 5 (generalized material balance, gas reinjection); Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 5.

Question 6 of 7: Undersaturated Oil Reservoir — Material Balance and Gas Reinjection

Question 6: Undersaturated Oil Reservoir — Material Balance and Gas Reinjection (20 marks)