Question 7 of 7: Dry Gas Reservoir — Depletion and Gas Produced

Question 7: Dry Gas Reservoir — Depletion and Gas Produced (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.

Find. (i) The average reservoir pressure at 50% recovery ($G_p/G=0.5$); (ii) the volume of gas produced (SCF) when the reservoir depletes from $p_i$ to 500 psia, given the reservoir originally held 1 MMft$^3$ of reservoir-condition gas.

Approach. Build the pseudo-critical properties and $Z_i$ at $p_i$, then use the volumetric $p/Z=(p_i/Z_i)(1-G_p/G)$ relation: for (i), solve for the pressure at which $G_p/G=0.5$ (iteratively, since $Z$ itself depends on $p$); for (ii), convert the given 1 MMft$^3$ reservoir volume to gas-in-place $G$ (SCF) via $B_{gi}$, then apply the same relation at $p_f=500$ psia to get $G_p$.

1. Pseudo-criticals and $Z_i$. $T_{pc}=168+325(0.65)-12.5(0.65)^2=374.0\,{}^{\circ}\text{R}$; $p_{pc}=677+15.0(0.65)-37.5(0.65)^2=670.9$ psia; $T_r=620/374.0=1.658$. At $p_i=6000$: $p_r=8.943$, Standing–Katz gives $\boxed{Z_i=1.064}$, so $p_i/Z_i=6000/1.064=5637$ psia.
2. Pressure at 50% recovery. Require $p/Z=0.5(5637)=2818.5$ psia. Solving $p/Z(p)=2818.5$ iteratively (with $Z$ from the same correlation at each trial $p_r$) gives $\boxed{p=2385\ \text{psia}}$ (at which $Z=0.846$).
3. Gas in place from reservoir volume. $B_{gi}=0.02827\dfrac{Z_iT}{p_i}=0.02827\dfrac{(1.064)(620)}{6000}=0.003109$ ft$^3$/SCF. $\boxed{G=\dfrac{1{,}000{,}000}{0.003109}=321.6\ \text{MMSCF}}$.
4. Gas produced at $p_f=500$ psia. $p_r=500/670.9=0.745$, $Z_f=0.949$, $p_f/Z_f=526.9$ psia. $G_p=G\left(1-\dfrac{p_f/Z_f}{p_i/Z_i}\right)=321.6\left(1-\dfrac{526.9}{5637}\right)=321.6(0.9065)$, so $\boxed{G_p=291.5\ \text{MMSCF}}$.