Question 6 of 7: Combination Gas-Cap/Oil-Zone Material Balance

Question 6: Combination Gas-Cap/Oil-Zone 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. Saturated reservoir with an existing gas cap; $m=V_{\text{gas cap}}/V_{\text{oil zone}}=1/3$ (oil zone is 3× the gas cap); $A=400$ acres, $\phi=0.17$, $S_{gi}=0$ (oil zone), $S_{wi}=0.20$ (both zones), $p_i=3400$ psia, $T=160^\circ\text{F}=620^\circ\text{R}$, $B_{oi}=1.33$ bbl/STB, $B_{o}=1.25$ bbl/STB at 2800 psia, $W_e=1$ MMbbl, $N_p=2$ MMSTB, $G_p=2.6$ MMMSCF, $Z=1-0.0001p$, $R_{so}=0.2p$; no water production or rock/water-compressibility data given (both taken as negligible).

Find. Original gas in place $G$ and thickness of the oil zone $h_{\text{oil}}$.

Approach. Because the reservoir is saturated with an existing gas cap, $B_{ti}=B_{oi}$ and $R_{soi}=0.2p_i$ by the given correlation. Solve the combined gas-cap/oil-zone material balance for $N$ (dropping the negligible rock/water-expansion term), get the gas cap's reservoir volume from $mNB_{ti}$ and hence $G$ from $B_{gi}$, then get the oil-zone net thickness from the standard volumetric equation using $N$ and $B_{oi}$.

1. Solution GOR and gas properties at each pressure. $R_{soi}=0.2(3400)=680$, $R_{so}=0.2(2800)=560\ \text{SCF/STB}$. $Z_i=1-0.0001(3400)=0.66$, $Z=1-0.0001(2800)=0.72$. $B_g=0.02827\,ZT/p$ (ft³/SCF), converted to bbl/SCF ($\div5.615$): $B_{gi}=0.02827(0.66)(620)/3400/5.615$, so $\boxed{B_{gi}=6.06\times10^{-4}\ \text{bbl/SCF}}$; $B_g=0.02827(0.72)(620)/2800/5.615$, so $\boxed{B_g=8.03\times10^{-4}\ \text{bbl/SCF}}$.
2. Two-phase FVF. $B_{ti}=B_{oi}=1.33$ bbl/STB (saturated at $t=0$, no free gas has yet evolved from the oil column). $B_t=B_o+B_g(R_{soi}-R_{so})=1.25+8.03\times10^{-4}(680-560)$, so $\boxed{B_t=1.3463\ \text{bbl/STB}}$.
3. Cumulative producing GOR. $R_p=G_p/N_p=2.6\times10^9/2\times10^6$, so $\boxed{R_p=1300\ \text{SCF/STB}}$.
4. Oil in place, $N$. With the rock/water-expansion term dropped (no $c_w,c_f$ given), the combination material balance reduces to $N(B_t-B_{ti})+Nm\dfrac{B_{ti}}{B_{gi}}(B_g-B_{gi})+W_e=N_p[B_t+B_g(R_p-R_{soi})]$. Substituting $m=1/3$: coefficient of $N$ is $(1.3463-1.33)+\tfrac13\left(\tfrac{1.33}{6.06\times10^{-4}}\right)(8.03\times10^{-4}-6.06\times10^{-4})=0.1603$; RHS $=2\times10^6[1.3463+8.03\times10^{-4}(1300-680)]-1\times10^6=2.688\times10^6$. So $N=2.688\times10^6/0.1603$, giving $\boxed{N=16.77\times10^{6}\ \text{STB}\ (16.77\ \text{MMSTB})}$.
5. Original gas in place. Gas-cap reservoir volume $=mNB_{ti}=\tfrac13(16.77\times10^6)(1.33)=7.44\times10^6$ bbl. $G=\dfrac{mNB_{ti}}{B_{gi}}=\dfrac{7.44\times10^6}{6.06\times10^{-4}}$, so $\boxed{G=1.227\times10^{10}\ \text{SCF}\ (12{,}271\ \text{MMSCF})}$.
6. Oil-zone net thickness. $N=\dfrac{7758\,A\,h\,\phi(1-S_{wi})}{B_{oi}}\ \Rightarrow\ h=\dfrac{NB_{oi}}{7758A\phi(1-S_{wi})}=\dfrac{16.77\times10^6(1.33)}{7758(400)(0.17)(0.80)}$, giving $\boxed{h_{\text{oil}}=52.9\ \text{ft}}$.