p/Z Material Balance for a Volumetric Dry-Gas Reservoir: 24-Pet-A3 Fundamental Reservoir Engineering, December 2018, Q2

Question 2: p/Z Material Balance for a Volumetric Dry-Gas Reservoir (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. Volumetric (no water drive) dry-gas reservoir; at $p_i$, $Z_i=0.85$, $G_p=0$; at 2500 psia, $Z=0.80$, $G_p=500$ MMMSCF; at 1500 psia, $Z=0.75$, $G_p=1000$ MMMSCF; at 1000 psia, $Z=0.70$.

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

Approach. The gas material balance $p/Z=(p_i/Z_i)(1-G_p/G)$ is a straight line in $p/Z$ vs. $G_p$. Two known $(p,Z,G_p)$ points fix that line; its $G_p=0$ intercept gives $p_i/Z_i$ (hence $p_i$, since $Z_i$ is given) and its $p/Z=0$ intercept gives $G$.

Line through the two known data points. $p/Z=2500/0.80=3125$ psia at $G_p=500$; $p/Z=1500/0.75=2000$ psia at $G_p=1000$. Slope $m=\dfrac{2000-3125}{1000-500}=-2.25\ \text{psia/MMMSCF}$; intercept $b=3125-(-2.25)(500)$, so $\boxed{p/Z\big|_{G_p=0}=4250\ \text{psia}}$.
(a) Initial pressure. $p_i/Z_i=4250$ and $Z_i=0.85\Rightarrow p_i=4250(0.85)$, so $\boxed{p_i=3612.5\ \text{psia}}$.
(b) Original gas in place. $G$ is the $G_p$-axis intercept of the same line ($p/Z=0$): $G=-b/m=4250/2.25$, so $\boxed{G=1888.9\ \text{MMMSCF}}$ ($1.889\times10^{12}$ SCF).
(c) Cumulative production at 1000 psia. At 1000 psia, $p/Z=1000/0.70=1428.6$ psia. From the line, $G_p=\dfrac{4250-1428.6}{2.25}$, so $\boxed{G_p=1254.0\ \text{MMMSCF}}$.
(d) Recovery factor. $RF=G_p/G=1254.0/1888.9$, so $\boxed{RF=0.664\ (66.4\%)}$.

Quantity	Value
(a) Initial reservoir pressure, $p_i$	3612.5 psia
(b) Original gas in place, $G$	1888.9 MMMSCF ($1.889\times10^{12}$ SCF)
(c) Cumulative production at 1000 psia, $G_p$	1254.0 MMMSCF
(d) Recovery factor at 1000 psia	66.4%

Topic: $p/Z$ material balance for a volumetric dry-gas reservoir.

Key relations: $\dfrac{p}{Z}=\dfrac{p_i}{Z_i}\left(1-\dfrac{G_p}{G}\right)$ — linear in $G_p$ with intercept $p_i/Z_i$ and $G_p$-intercept $G$.

Why this works: for a volumetric gas reservoir the pore volume is fixed, so the real-gas law applied at initial and current conditions gives $G_pB_g=(G-G_i)B_{gi}\ldots$, which rearranges directly into a straight line in $p/Z$ vs. $G_p$ — the single most-used gas-reservoir engineering tool because it needs no separate volumetric (area×thickness×porosity) estimate once at least two pressure/production points are known.

Common pitfall: using the given $Z_i=0.85$ directly with the raw (unknown) $p_i$ instead of first solving for the $p_i/Z_i$ ratio from the straight line — the line only ever gives the ratio, so $Z_i$ must be supplied separately to split it into $p_i$.

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 5 (gas material balance); Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 2.

Question 2 of 7: p/Z Material Balance for a Volumetric Dry-Gas Reservoir

Question 2: p/Z Material Balance for a Volumetric Dry-Gas Reservoir (20 marks)