Given. A volumetric, initially-undersaturated reservoir depleting from the true initial condition (3,100 psia) through the bubble point (1,700 psia) to 1,600 psia by solution-gas drive.
Given data
$A$
1,000 ac
$h$
60 ft
$\phi$
0.20
$S_{wc}$
0.20
$p_i$
3,100 psia
$p_b$
1,700 psia
$B_{oi}$ (at $p_i$)
1.4500 rb/STB
$R_{si}$
900 scf/STB
Find. $N_p$, $G_p$, $R_p$, $S_o$, and the instantaneous GOR at $p=1{,}600$ psia.
Approach. Compute OOIP volumetrically using $B_o$ at the TRUE initial pressure (3,100 psia, not the bubble point), then apply Tarner's iterative solution-gas-drive method for the 1,700→1,600 psia depletion step, keeping every term in the generalized material balance referenced to that same true initial condition (which automatically absorbs the undersaturated 3,100→1,700 psia interval since $R_s$ is unchanged there): guess $N_p$, get the implied oil saturation and hence $S_g$, evaluate the instantaneous GOR from the given $k_{rg}/k_{ro}$ correlation, average it with the start-of-step GOR, and re-solve the material balance for $N_p$ until the two agree.
OOIP (volumetric, referenced to the true initial pressure).
$$N=\frac{7{,}758\,Ah\phi(1-S_{wc})}{B_{oi}}=\frac{7{,}758(1{,}000)(60)(0.20)(0.80)}{1.4500}$$
$$\boxed{N\approx 51.36\text{ MMSTB}}$$
Material balance for the 1,700→1,600 psia step (no gas cap, no water influx).
$$N\big[(B_o-B_{oi})+(R_{si}-R_s)B_g\big]=N_p\big[B_o+(R_p-R_s)B_g\big]$$
At 1,600 psia: $N\big[(1.4404-1.4500)+(900-800)(0.0017)\big]=N(0.1604)=8.24$ MMres bbl of total voidage to be matched by production – this is a large per-STB voidage because 100 scf/STB of gas coming out of solution occupies substantial reservoir volume at $B_g=0.0017$ res bbl/scf.
Oil and gas saturation as a function of the trial $N_p$.
$$S_o=(1-S_{wc})\left(1-\frac{N_p}{N}\right)\frac{B_o}{B_{oi}},\qquad S_g=1-S_{wc}-S_o$$
Tarner iteration. Starting the interval at $p_b$ (where $S_g=0$, so GOR$_1=R_{si}=900$), iterate: (i) solve the material balance for $N_p$ using the current average producing GOR estimate; (ii) get $S_o$, $S_g$ from step 3; (iii) compute the instantaneous GOR from $\text{GOR}=R_s+\dfrac{k_{rg}}{k_{ro}}\dfrac{\mu_o}{\mu_g}\dfrac{B_o}{B_g}$; (iv) average with GOR$_1$ and repeat. The iteration converges to
$$S_o\approx71.6\%,\quad S_g\approx8.42\%,\quad k_{rg}/k_{ro}=0.005e^{10(0.0842)}\approx0.0116$$
$$\text{GOR(1,600 psia, instantaneous)}=800+0.0116(10.7)\frac{1.4404}{0.0017}$$
$$\boxed{\text{GOR}\approx905.2\text{ scf/STB}}$$
$$R_p=\text{average producing GOR over the step}=\tfrac12(900+905.2)$$
$$\boxed{R_p\approx902.6\text{ scf/STB}}$$
Converged $N_p$ and $G_p$.
$$N_p=\frac{N\big[(B_o-B_{oi})+(R_{si}-R_s)B_g\big]}{B_o+(R_p-R_s)B_g}$$
$$\boxed{N_p\approx5.10\text{ MMSTB}\ (9.93\%\text{ of OOIP})}$$
$$G_p=N_p\times R_p=5.10\times902.6$$
$$\boxed{G_p\approx4{,}605\text{ MMscf}}$$
Question 5 – final results (at $p=1{,}600$ psia)
Quantity
Value
OOIP, $N$
51.36 MMSTB
Cumulative oil, $N_p$
5.10 MMSTB (9.93% of OOIP)
Cumulative gas, $G_p$
4,605 MMscf
Cumulative (average) producing GOR, $R_p$
902.6 scf/STB
Instantaneous producing GOR
905.2 scf/STB
Remaining oil saturation, $S_o$
71.6%
Free gas saturation, $S_g$
8.42%
Check – the small rock/water-compressibility contribution above the bubble point (3,100→1,700 psia) is neglected in this single-step Tarner treatment, consistent with how negligible it is next to the 0.1604 res bbl/STB of gas-evolution expansion between 1,700 and 1,600 psia (about 65× larger); it would shift $N_p$ by well under 1%.
