Question 8 of 8

Question 8 (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. A volumetric, initially-undersaturated reservoir depleting from the bubble point (1,700 psia) to 1,600 psia by solution-gas drive.

Find. $N_p$, $G_p$, $R_p$, $S_o$, and the instantaneous GOR at $p=1{,}600$ psia.

Approach. Compute OOIP volumetrically at $p_i$, then apply Tarner's iterative solution-gas-drive method for the single 1,700→1,600 psia depletion step: 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.

1. OOIP (volumetric). $$N=\frac{7{,}758\,Ah\phi(1-S_{wc})}{B_{oi}}=\frac{7{,}758(1{,}000)(60)(0.20)(0.80)}{1.4500}$$ $$\boxed{N\approx 51.36\text{ MMSTB}}$$
2. 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.
3. Oil and gas saturation as a function of the trial $N_p$. $$S_o=(1-S_{wc})\left(1-\frac{N_p}{N}\right)\frac{B_o}{B_{oi}},\qquad S_g=1-S_{wc}-S_o$$
4. 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 in 5 steps to $$S_o\approx71.6\%,\quad S_g\approx8.42\%,\quad k_{rg}/k_{ro}=0.005e^{10(0.0842)}\approx0.0116$$ $$\text{GOR(1,600 psia, instantaneous)}=800+0.0116(10.7)\frac{1.4404}{0.0017}$$ $$\boxed{\text{GOR}\approx905.2\text{ scf/STB}}$$ $$R_p=\text{average producing GOR over the step}=\tfrac12(900+905.2)$$ $$\boxed{R_p\approx902.6\text{ scf/STB}}$$
5. Converged $N_p$ and $G_p$. $$N_p=\frac{N\big[(B_o-B_{oi})+(R_{si}-R_s)B_g\big]}{B_o+(R_p-R_s)B_g}$$ $$\boxed{N_p\approx5.10\text{ MMSTB}\ (9.93\%\text{ of OOIP})}$$ $$G_p=N_p\times R_p=5.10\times902.6$$ $$\boxed{G_p\approx4{,}605\text{ MMscf}}$$