Volumetric Gas Reservoir — p/Z Material Balance: 24-Pet-A3 Fundamental Reservoir Engineering, December 2014, Q4

Question 4: Volumetric 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.

Check: the paper labels cumulative production "MMMSCF" (triple-M); this reads as a misprint of the standard "MMSCF" (10$^6$ SCF) unit, but since parts (a) and (b) below both reduce to a ratio $G_p/G$, the numeric answer is unaffected by which reading is intended — the unit label is carried through as printed.

Given. $\gamma_g=0.67$; $T=237\,{}^{\circ}\text{F}=697\,{}^{\circ}\text{R}$ (constant); two ($p$, $G_p$) data pairs above. Formula sheet: $T_{pc}=168+325\gamma_g-12.5\gamma_g^2$, $p_{pc}=677+15.0\gamma_g-37.5\gamma_g^2$; $p/Z=(p_i/Z_i)(1-G_p/G)$.

Find. (a) Initial gas in place, $G$; (b) recovery factor at $p=500$ psia.

Fig. 3 — The two ($G_p$, $p/Z$) points define a straight line; its $G_p$-intercept is the initial gas in place $G$.

Approach. Build pseudo-critical properties from $\gamma_g$, form $Z$ at each tabulated pressure via the Standing–Katz correlation, plot $p/Z$ against $G_p$, and fit the straight line $p/Z=(p_i/Z_i)-(p_i/Z_i)/G\cdot G_p$ through the two points to solve simultaneously for the intercept and $G$; apply the same relation at 500 psia for part (b).

Pseudo-critical properties. $T_{pc}=168+325(0.67)-12.5(0.67)^2=380.1\,{}^{\circ}\text{R}$; $p_{pc}=677+15.0(0.67)-37.5(0.67)^2=670.2$ psia; $T_r=697/380.1=\boxed{1.834}$ (constant for this isothermal reservoir).
$Z$ and $p/Z$ at the two data points. At $p=5000$: $p_r=7.460$, Standing–Katz gives $Z=1.011$, so $p/Z=4946$ psia. At $p=4000$: $p_r=5.968$, $Z=0.952$, $p/Z=4203$ psia.
Fit the line and solve for $G$. Slope $=\dfrac{4203-4946}{520-300}=-3.376$ psia/MMMSCF; intercept $p_i/Z_i=4946-(-3.376)(300)=5959$ psia. $\boxed{G=5959/3.376=1765\ \text{MMMSCF}}$.
Recovery factor at 500 psia. $p_r=500/670.2=0.746$, $Z=0.965$, $p/Z=518.3$ psia. $G_p(500)=(5959-518.3)/3.376=1611.5$ MMMSCF. $\boxed{\text{RF}=1611.5/1765=91.3\%}$.

Quantity	Value
$T_r$ (constant)	1.834
(a) Initial gas in place, $G$	1765 MMMSCF
(b) $G_p$ at 500 psia	1611.5 MMMSCF
(b) Recovery factor at 500 psia	91.3%

Topic: $p/Z$ material balance for a volumetric (no-water-drive) gas reservoir.

Key relations: $p/Z=(p_i/Z_i)(1-G_p/G)$ — a straight line in $(G_p,\,p/Z)$ space whose $G_p$-axis intercept is $G$; Standing pseudo-criticals; Standing–Katz $Z$-factor correlation.

Why this works: For a closed (volumetric) gas reservoir, the real-gas law applied at initial and current conditions collapses to a line in $p/Z$ vs. $G_p$; any two (or more) production/pressure pairs pin down that line, and extrapolating it to $p/Z=0$ (i.e. $G_p=G$) gives the reservoir's ultimate gas-in-place without needing volumetric (porosity/area/thickness) data at all.

Common pitfall: Using raw pressure instead of $p/Z$ — the material balance line is only straight in $p/Z$ because $Z$ itself varies materially with pressure at these high $p_r$ values (here $Z$ ranges from about 0.95 to 1.01 across the tabulated range).

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 5 (gas material balance); Lyons (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed. (Standing–Katz $Z$-factor correlation).

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

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