Dry-Gas Reservoir Material Balance (p/Z Depletion): 24-Pet-A2 Petroleum Reservoir Fluids, December 2015, Q6

Question 6: Dry-Gas Reservoir Material Balance (p/Z Depletion) (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. $p_1=2500$ psia; $T=190\,{}^{\circ}\text{F}=650\,{}^{\circ}\text{R}$ constant; volumetric reservoir ($V$ constant); composition table above; $n_2=n_1/2$ after withdrawal. Formula sheet: real gas law $n=pV/(ZRT)$; Standing pseudo-criticals; $M_{av}=\sum y_iM_i$.

Find. The reservoir pressure $p_2$ once half the original moles of gas have been produced.

Approach. Compute the apparent molecular weight and specific gravity of the mixture from composition, get pseudo-critical properties (Standing) and hence $Z_1$ at the initial state; since $V,T$ are constant, moles are proportional to $p/Z$, so $p_2/Z_2=\tfrac12\,p_1/Z_1$; solve this implicit equation for $p_2$ (iterating $Z_2$ with the same correlation).

Apparent molecular weight and gravity. $M_{av}=\sum y_iM_i=0.9132(16.04)+0.0443(30.07)+0.0212(44.10)+0.0136(58.12)+0.0042(72.15)+0.0015(86.18)+0.0020(104.00)=18.35\ \text{lb}_m/\text{lb-mol}$ (mole % sums to 100.00, confirming the basis). $\gamma_g=M_{av}/28.97=18.35/28.97=0.6333$.
Pseudo-critical properties and initial state. $T_{pc}=168+325(0.6333)-12.5(0.6333)^2=368.8\,{}^{\circ}\text{R}$; $p_{pc}=677+15.0(0.6333)-37.5(0.6333)^2=671.5$ psia. $T_r=650/368.8=1.763$ (fixed for both states, isothermal). At $p_1=2500$ psia: $p_{r1}=2500/671.5=3.723$; solving the $Z$-correlation at $(T_r,p_{r1})$ gives $\boxed{Z_1=0.883}$.
Material balance ($p/Z$ proportional to moles). Since $V,T$ constant, $n\propto p/Z$, so $n_1/n_2=(p_1/Z_1)/(p_2/Z_2)=2$. Thus $\dfrac{p_2}{Z_2}=\dfrac{1}{2}\times\dfrac{p_1}{Z_1}=\dfrac{1}{2}\times\dfrac{2500}{0.883}=\dfrac{1}{2}\times2831=1416\ \text{psia}$ (target $p/Z$ at state 2).
Solve for $p_2$ (iterative, since $Z_2$ itself depends on $p_2$). Using the same correlation with $T_r=1.763$, search $p_2$ until $p_2/Z(p_2,T_r)=1416$: this converges to $p_2\approx1287$ psia with $Z_2\approx0.909$ (check: $1287/0.909=1416$ ✓). So $\boxed{p_2=1287\ \text{psia}}$.

Quantity	Value
$M_{av}$, $\gamma_g$	18.35 lb$_m$/lb-mol, 0.6333
$T_{pc}$, $p_{pc}$	368.8°R, 671.5 psia
$Z_1$ at 2500 psia	0.883
Target $p_2/Z_2$	1416 psia
$p_2$ (half the gas withdrawn)	1287 psia ($Z_2\approx0.909$)

Topic: Volumetric dry-gas reservoir material balance via the $p/Z$ relationship.

Key relations: $n=pV/(ZRT)$; for constant $V,T$, $p/Z\propto n$, so the familiar straight-line $p/Z$-vs-cumulative-production reserves plot follows directly.

Why this works: A volumetric gas reservoir neither gains nor loses pore volume and stays at reservoir temperature throughout depletion, so the real gas law reduces the moles-remaining ratio to a simple ratio of $p/Z$ at the two states — converting any stated fractional withdrawal directly into a solvable equation for the new pressure once $Z$ is re-evaluated self-consistently at that new pressure.

Common pitfall: Using the initial-state $Z_1$ unchanged when solving for $p_2$, instead of iterating $Z_2=Z(T_r,p_{r2})$ at the NEW pressure — because $Z$ itself varies significantly over this pressure range (0.883 to 0.909), a one-shot linear guess for $p_2$ understates the true pressure.

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 5 ($p/Z$ material balance for volumetric gas reservoirs); Craft & Hawkins, 3rd ed., Ch. 2.

Question 6 of 7: Dry-Gas Reservoir Material Balance (p/Z Depletion)

Question 6: Dry-Gas Reservoir Material Balance (p/Z Depletion) (20 marks)