Question 2 of 7: Volumetric Dry Gas Reservoir — $p/Z$ Material Balance

Question 2: Volumetric Dry Gas Reservoir — $p/Z$ 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. A volumetric (no water influx), dry-gas reservoir depletion history:

Find. (a) $p_i$; (b) original gas in place $G$; (c) $G_p$ when $p=1000$ psia; (d) recovery factor at 1000 psia.

Approach. For a volumetric dry-gas reservoir the $p/Z$ material-balance equation, $p/Z=(p_i/Z_i)(1-G_p/G)$, plots as a straight line in $G_p$. Fit that line through the two rows where both $p$ and $G_p$ are known, then use the fitted line to read off everything else.

1. Compute $p/Z$ at the two known points. At 2500 psia: $p/Z=2500/0.80=3125$ psia. At 1500 psia: $p/Z=1500/0.75=2000$ psia.
2. Fit the material-balance line. The line has the form $p/Z=b+mG_p$. Slope $m=\dfrac{2000-3125}{1000-500}=\dfrac{-1125}{500}=-2.25$ psia/MMMSCF. Intercept (value at $G_p=0$, i.e. $p_i/Z_i$): $b=3125-m(500)=3125+1125=4250$ psia.
3. Part (a) — initial pressure. Since $b=p_i/Z_i=4250$ and $Z_i=0.85$: $p_i=4250\times0.85$, so $\boxed{p_i=3612.5\text{ psia}}$.
4. Part (b) — original gas in place. The slope of the $p/Z$ line is $m=-b/G$, so $G=-b/m=4250/2.25$, giving $\boxed{G=1888.9\text{ MMMSCF}\ (\approx1.889\times10^{12}\text{ SCF})}$.
5. Part (c) — cumulative production at 1000 psia. $p/Z=1000/0.70=1428.6$ psia. Solving the line for $G_p$: $G_p=(1428.6-4250)/(-2.25)$, so $\boxed{G_{p}=1254.0\text{ MMMSCF}}$.
6. Part (d) — recovery factor at 1000 psia. $RF=G_p/G=1254.0/1888.9$, so $\boxed{RF=66.4\%}$.