Undersaturated-to-Saturated Oil Material Balance: 24-Pet-A3 Fundamental Reservoir Engineering, May 2014, Q5

Question 5: Undersaturated-to-Saturated Oil 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.

Given.

Pressure (psia)	$B_o$ (bbl/STB)	$R_{so}$ (SCF/STB)	Z	$B_g$ (bbl/SCF)
4000 ($p_i$)	1.18	1000	0.850	0.000766
3500	1.20	1000	0.785	0.000706
3000	1.10	800	0.765	0.000860

$S_{wi}=20\%$; no initial gas cap, no water influx (volumetric, closed reservoir).

Find. (a) $N_p/N$ at bubble-point pressure; (b) current gas saturation at 3000 psia if $N_p/N=10\%$.

Approach. $R_{so}$ stays at 1000 SCF/STB from 4000 down to 3500 psia and only falls (to 800) by 3000 psia — solution gas has not yet begun evolving at 3500 psia, so the bubble point is $p_b=3500$ psia. Above $p_b$ recovery is by fluid expansion only ($N_p/N=(B_{ob}-B_{oi})/B_{ob}$); below $p_b$, use a saturation balance to convert produced oil (at the current $B_o$) into liberated gas saturation, holding connate water constant.

Part (a) — recovery at the bubble point. Above $p_b$, with no free gas yet released, $N_p\,B_{ob}=N(B_{ob}-B_{oi})$ (fluid-expansion drive only), so: $$\frac{N_p}{N}=\frac{B_{ob}-B_{oi}}{B_{ob}}=\frac{1.20-1.18}{1.20}$$ giving $\boxed{N_p/N=1.67\%}$ — the small recovery typical of depletion above the bubble point, driven only by fluid expansion.
Part (b) — oil saturation remaining at 3000 psia. With $N_p/N=10\%$, the remaining oil occupies a reservoir volume of $(1-N_p/N)\,N\,B_o$ against an original hydrocarbon pore volume of $N B_{oi}/(1-S_{wi})$ (holding total pore volume and $S_{wi}$ fixed, no water influx): $$S_o=(1-N_p/N)\frac{B_o}{B_{oi}}(1-S_{wi})=(0.90)\left(\frac{1.10}{1.18}\right)(0.80)=0.6712$$
Current gas saturation. $S_g=1-S_{wi}-S_o=1-0.20-0.6712$, so $\boxed{S_g=12.9\%}$.

Check: assumes connate water saturation stays fixed at 20% (no water influx/production) and that no secondary gas cap has segregated updip of the well/log location by 3000 psia, so the liberated gas simply fills the pore space vacated by expanding/produced oil in place — the standard simplifying assumption for a volumetric solution-gas-drive material balance absent explicit rock/water-compressibility data.

Quantity	Value
Bubble-point pressure (identified)	3500 psia
(a) $N_p/N$ at bubble point	1.67%
(b) Oil saturation at 3000 psia ($N_p/N=10\%$)	67.1%
(b) Gas saturation at 3000 psia	12.9%

Topic: Oil material balance across the bubble point — fluid-expansion drive above $p_b$, solution-gas-drive saturation balance below it.

Key relations: Above $p_b$: $N_p/N=(B_{ob}-B_{oi})/B_{ob}$. Below $p_b$: $S_o=(1-N_p/N)(B_o/B_{oi})(1-S_{wi})$, then $S_g=1-S_{wi}-S_o$.

Why this works: Identifying $p_b$ from where $R_{so}$ stops tracking its initial value is the standard PVT-table diagnostic; once below $p_b$, converting stock-tank oil produced into a reservoir-volume saturation change (rather than tracking absolute gas volumes) sidesteps needing the free-gas material-balance term entirely for saturation-only questions.

Common pitfall: Using $B_{oi}$ (the pressure-$p_i$ value, 1.18) instead of $B_{ob}$ (the bubble-point value, 1.20) in part (a)'s expansion-drive formula, or vice-versa in part (b) — the anchor for expansion-drive recovery is always the bubble-point property, while the anchor for the original-oil-in-place volume is the true initial ($p_i$) property.

Source: Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 2 (oil material balance); Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 5.

Question 5 of 7: Undersaturated-to-Saturated Oil Material Balance

Question 5: Undersaturated-to-Saturated Oil Material Balance (20 marks)