Volumetric Dry Gas Reservoir — p/Z Material Balance: 24-Pet-A3 Fundamental Reservoir Engineering, May 2016, Q4

Question 4: 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. Two production-history points and the target-pressure $Z$ above (table).

Find. (a) Initial gas in place, $G$ (MMMSCF); (b) cumulative gas production, $G_p$, when $p$ has declined to 1000 psia.

Approach. For a volumetric dry-gas reservoir, $p/Z$ falls on a straight line in $G_p$: $p/Z=(p_i/Z_i)-\left[(p_i/Z_i)/G\right]G_p$. Fit that line through the two given ($p/Z$, $G_p$) points to get its intercept ($p_i/Z_i$) and slope, read $G$ off the intercept/slope, then use the same line to convert the given $p/Z$ at 1000 psia into $G_p$.

$p/Z$ at the two history points. At 0.5 yr: $p/Z=1680/0.870=1931.03$ psia. At 2.0 yr: $p/Z=1335/0.900=1483.33$ psia.
Fit the material-balance line. Slope $b=\dfrac{1483.33-1931.03}{3.92-0.96}=\dfrac{-447.70}{2.96}=-151.25\ \text{psia/MMMSCF}$; intercept (at $G_p=0$) $a=p_i/Z_i=1931.03-(-151.25)(0.96)=2076.23$ psia. Setting $p/Z=0$ (fully depleted) gives $\boxed{G=-a/b=2076.23/151.25=13.73\ \text{MMMSCF}}$.
Cumulative production at 1000 psia. $p/Z=1000/0.92=1086.96$ psia. From the same line, $G_p=(a-p/Z)/(-b)=(2076.23-1086.96)/151.25$, so $\boxed{G_p=6.54\ \text{MMMSCF}}$ (recovery factor $G_p/G=6.54/13.73=47.6\%$).

Quantity	Value
$p_i/Z_i$ (line intercept)	2076.2 psia
(a) Initial gas in place, $G$	13.73 MMMSCF
(b) $G_p$ at $p=1000$ psia	6.54 MMMSCF
Recovery factor at 1000 psia	47.6%

Both given production-history points already sit almost exactly on the fitted $p/Z$ line, so the intercept ($p_i/Z_i$) and slope come from the data directly, without needing an independent gas-gravity/pseudo-critical $Z$-correlation to backfill missing points. A recovery factor of 47.6% at 1000 psia (roughly a third of the initial pressure) is a physically reasonable depletion fraction for a volumetric dry-gas reservoir with no water influx to support it — well above what an equivalent undersaturated-oil reservoir would give over a comparable relative pressure drop, since gas itself provides essentially all of its own drive energy through expansion.

Topic: p/Z straight-line 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 $x$-intercept $G$.

Why this works: For a volumetric (no water influx) dry-gas reservoir, the real-gas law applied at reservoir conditions before and after production shows $p/Z$ declines exactly linearly with cumulative gas withdrawn; because the two given production-history points already lie on this line, no gas-gravity/pseudo-critical $Z$-correlation is needed here — the line can be fit directly from the tabulated $Z$ values.

Common pitfall: Fitting a straight line through raw pressure vs. $G_p$ instead of $p/Z$ vs. $G_p$ — only the $p/Z$ form is linear once $Z$ itself varies with pressure.

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 5 (gas material balance); Lyons (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed.

Question 4 of 7: Volumetric Dry Gas Reservoir — p/Z Material Balance

Question 4: Volumetric Dry Gas Reservoir — p/Z Material Balance (20 marks)