Undersaturated Oil Reservoir — Volumetric Material Balance: 24-Pet-A3 Fundamental Reservoir Engineering, May 2016, Q6

Question 6: Undersaturated Oil Reservoir — Volumetric Material Balance (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: at both 5000 and 3900 psia the table's solution GOR is unchanged (700 SCF/STB), confirming both pressures lie above the stated bubble point (3500 psia) and the oil remains undersaturated (single phase) throughout — consistent with $p_i=5000$ psia $>p_b=3500$ psia as given.

Given.

Porosity, $\phi$	0.20
Reservoir area, $A$	1000 acres
Formation thickness, $h$	70 ft
Initial water saturation, $S_{wi}$	0.25
Initial pressure, $p_i$ / $B_{oi}$	5000 psia / 1.315 bbl/STB
Pressure at 3900 psia, $B_o$	1.320 bbl/STB
Bubble-point pressure, $p_b$	3500 psia (both pressures above it)

Find. Cumulative oil production, $N_p$, when reservoir pressure drops to 3900 psia (undersaturated, rock and water compressibility ignored).

Approach. Compute the original oil-in-place $N$ volumetrically from the reservoir's rock properties, then apply the simplified undersaturated material balance: with $R_s$ unchanged (no free gas evolved) and rock/water compressibility neglected as instructed, the only drive mechanism is oil expansion, $N_p B_o=N(B_o-B_{oi})$.

Original oil in place. $N=\dfrac{7758\,A\,h\,\phi\,(1-S_{wi})}{B_{oi}}=\dfrac{7758(1000)(70)(0.20)(1-0.25)}{1.315}=\dfrac{7758(1000)(70)(0.20)(0.75)}{1.315}$, so $\boxed{N=61.95\times10^6\ \text{STB}}$.
Undersaturated material balance (oil expansion only). With $c_f,c_w$ neglected and $R_s$ constant (no free gas), $N_pB_o=N(B_o-B_{oi})$: $N_p=N\dfrac{B_o-B_{oi}}{B_o}=61.95\times10^6\times\dfrac{1.320-1.315}{1.320}=61.95\times10^6\times0.003788$, so $\boxed{N_p=2.35\times10^5\ \text{STB}\ (0.235\ \text{MMSTB})}$.

Quantity	Value
Original oil in place, $N$	61.95 MMSTB
Cumulative oil production at 3900 psia, $N_p$	0.235 MMSTB (235,000 STB)
Recovery factor, $N_p/N$	0.38%

Topic: Volumetric oil-in-place and the simplified undersaturated (single-phase liquid expansion) material balance.

Key relations: $N=\dfrac{7758\,A\,h\,\phi\,(1-S_{wi})}{B_{oi}}$ (STB, $A$ in acres, $h$ in ft); undersaturated MB with $c_f,c_w$ neglected, $N_pB_o=N(B_o-B_{oi})$.

Why this works: Above the bubble point no free gas evolves, so the reservoir simply loses volume equal to the produced stock-tank oil's reservoir-barrel equivalent, $N_pB_o$; that lost volume is exactly supplied by the small expansion of the remaining oil in place, $N(B_o-B_{oi})$, once rock and connate-water expansion are (as instructed) neglected.

Common pitfall: Because liquid oil is only weakly compressible, $(B_o-B_{oi})/B_o$ is tiny (here $\approx0.38\%$) — a very small recovery factor is the physically correct answer for pure undersaturated liquid expansion over a modest pressure drop, not a sign of an arithmetic error.

Source: Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 1, 5 (volumetric OOIP, undersaturated material balance); Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 5.

Question 6 of 7: Undersaturated Oil Reservoir — Volumetric Material Balance

Question 6: Undersaturated Oil Reservoir — Volumetric Material Balance (20 marks)