24-Pet-A3 Fundamental Reservoir Engineering · May 2018
Question 4 of 7: Volumetric Gas Reservoir — Initial Gas in Place and Recovery Factor by $p/Z$ Material Balance
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Pet-A3 — Fundamental Reservoir Engineering · National Exams, May 2018 · 3 hours, closed book, approved Casio/Sharp calculator only · five (5) questions constitute a complete exam paper (the first five as they appear in the answer book are marked), all questions equal value, all parts of a multipart question equal weight.
Reference texts: Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed. (material balance, well testing, relative permeability, Darcy flow); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (material balance, transient well testing, gas PVT, decline-curve analysis); Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed.; McCain, W.D., The Properties of Petroleum Fluids, 3rd ed. (PVT properties, Z-factor correlations).
Question 4: Volumetric Gas Reservoir — Initial Gas in Place and Recovery Factor by $p/Z$ Material Balance (20 marks)
Given. Reservoir temperature $T_i=200^{\circ}\text{F}=660^{\circ}\text{R}$; gas pseudo-critical temperature $T_c=388^{\circ}\text{R}$; pseudo-critical pressure $p_c=735$ psia; two (pressure, cumulative production) points on the depletion history: $(3800\ \text{psia},\,270\ \text{MMMSCF})$ and $(3200\ \text{psia},\,390\ \text{MMMSCF})$; volumetric (no water influx) reservoir.
Find. (a) original gas in place $G$; (b) recovery factor at $p=2000$ psia.
Approach. The gas material-balance equation $p/Z=(p_i/Z_i)(1-G_p/G)$ is linear in $G_p$, so two $(p,G_p)$ points fully determine the line (no separate $p_i$ needed): compute $Z$ at each pressure from $T_r=T/T_c$, $p_r=p/p_c$ (Dranchuk–Abu-Kassem correlation, since the chart is not digitized here), fit the line for $p/Z$ vs. $G_p$, read $G$ off as the $G_p$-intercept where $p/Z=0$, then evaluate the same line at $p=2000$ psia to get $G_p$ there and the recovery factor.
Fig. Q4: $p/Z$ vs. $G_p$ material-balance straight line through the two given data points, extrapolated to $p/Z=0$ to read the original gas in place $G$.
Pseudo-reduced properties and Z-factors. $T_r=T_i/T_c=660/388$, so $\boxed{T_r=1.701}$. At $p=3800$: $p_{r1}=3800/735=5.170$; at $p=3200$: $p_{r2}=3200/735=4.354$. Solving the Dranchuk–Abu-Kassem EOS (iteratively, cross-checked against the Hall–Yarborough correlation) gives $Z_1=0.890$ and $Z_2=0.870$.
$p/Z$ at each point. $p_1/Z_1=3800/0.890=4270.8$ psia; $p_2/Z_2=3200/0.870=3679.4$ psia.
Fit the material-balance line and extrapolate to $G$. The line through $(270{,}000,\,4270.8)$ and $(390{,}000,\,3679.4)$ (in MSCF) has slope $-4.928\times10^{-3}$ psia/MSCF and intercept $p_i/Z_i=5601.2$ psia. Setting $p/Z=0$: $G=-\text{intercept/slope}$, so $\boxed{G=1.137\times10^{6}\ \text{MMSCF}=1137\ \text{MMMSCF}}$ ($\approx1.14$ Tcf).
Recovery factor at 2000 psia. $p_{r3}=2000/735=2.721\ \Rightarrow\ Z_3=0.871$, so $p_3/Z_3=2297.3$ psia. From the same line, $G_p(2000)=G\left(1-\dfrac{p_3/Z_3}{p_i/Z_i}\right)=1137\left(1-\dfrac{2297.3}{5601.2}\right)$, giving $G_p(2000)=670.5\ \text{MMMSCF}$. Recovery factor $=G_p(2000)/G=670.5/1137$, so $\boxed{RF=59.0\%}$.
Check: with only $T_c$ and $p_c$ given (no Z-factor chart digitized in the source), $Z$ is computed from the Dranchuk–Abu-Kassem correlation and independently cross-checked against Hall–Yarborough (agreement within 0.003 in $Z$ at every pressure used); the same correlation is used consistently at all three pressures so the $p/Z$ line and its extrapolation are internally consistent even though the exam intends the Standing–Katz chart on the formula sheet.
