Dry Gas Material Balance and Isothermal Gas Compressibility: 24-Pet-A2 Petroleum Reservoir Fluids, December 2014, Q6

Question 6: Dry Gas Material Balance and Isothermal Gas Compressibility (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 as printed asks for “the gas compressibility factor at 1000 psia”, but the data fix $Z$ only at the state reached after half the moles are produced, which the same question gives as 1125 psia, and the next sentence asks for $c_g$ at 1125 psia. The 1000 psia is therefore read as a typo for 1125 psia, and $Z$ is reported there (Note 1 on the cover invites a stated assumption). If 1000 psia was really meant, extending the same $p$–$Z$ secant from Step 3 gives $Z(1000)\approx 0.900+(-1.143\times10^{-4})(1000-1125)\approx 0.914$. That is an extrapolation, not a value the given data determine.

Given. $p_1=2000$ psia, $Z_1=0.80$; $T=120\,{}^{\circ}\text{F}$ constant (volumetric reservoir, $V$ constant); $p_2=1125$ psia after half the initial moles have been produced ($n_2=n_1/2$). Formula sheet: $c_g=\dfrac{1}{p}-\dfrac{1}{Z}\left(\dfrac{dZ}{dP}\right)_T$.

Find. $Z_2$ at 1125 psia and $c_g$ at 1125 psia.

Approach. Apply the real-gas law $n=\dfrac{pV}{ZRT}$ at both states with $V,T$ constant to relate $n_1,n_2$ to $p_1/Z_1$ and $p_2/Z_2$; solve for $Z_2$; then estimate $dZ/dP$ from the two known $(p,Z)$ states and evaluate $c_g$ at state 2.

Relate moles to p/Z. Since $V$ and $T$ are constant, $n=\dfrac{pV}{ZRT}\propto \dfrac{p}{Z}$, so $\dfrac{n_1}{n_2}=\dfrac{p_1/Z_1}{p_2/Z_2}$. With $n_2=n_1/2$, $\dfrac{n_1}{n_2}=2$.
Solve for $Z_2$. $2=\dfrac{p_1 Z_2}{p_2 Z_1} \Rightarrow Z_2=\dfrac{2\,p_2\,Z_1}{p_1}=\dfrac{2(1125)(0.80)}{2000}=\dfrac{1800}{2000}$. So $\boxed{Z_2 = 0.900}$ at 1125 psia.
Estimate $dZ/dP$. With only the two states available, use a straight-line (secant) estimate over the interval: $\left(\dfrac{dZ}{dP}\right)_T \approx \dfrac{Z_2-Z_1}{p_2-p_1}=\dfrac{0.900-0.800}{1125-2000}=\dfrac{0.100}{-875}=-1.143\times10^{-4}\ \text{psi}^{-1}$.
Evaluate $c_g$ at 1125 psia. $c_g=\dfrac{1}{p_2}-\dfrac{1}{Z_2}\left(\dfrac{dZ}{dP}\right)_T=\dfrac{1}{1125}-\dfrac{1}{0.900}\times(-1.143\times10^{-4})=8.889\times10^{-4}+1.270\times10^{-4}$. So $\boxed{c_g = 1.016\times10^{-3}\ \text{psi}^{-1}}$ at 1125 psia.

Quantity	Value
$Z_2$ at 1125 psia	0.900
$dZ/dP$ (secant estimate)	$-1.143\times10^{-4}\ \text{psi}^{-1}$
$c_g$ at 1125 psia	$1.016\times10^{-3}\ \text{psi}^{-1}$

Topic: Volumetric dry-gas material balance and isothermal gas compressibility.

Key relations: $n=\dfrac{pV}{ZRT}$; $p/Z$ material balance ($p/Z$ vs. cumulative production is linear for a volumetric gas reservoir); $c_g=\dfrac{1}{p}-\dfrac{1}{Z}\left(\dfrac{dZ}{dP}\right)_T$.

Why this works: For a volumetric (constant pore volume) gas reservoir at constant temperature, moles present are directly proportional to $p/Z$, so any stated fraction of the original gas produced converts immediately into a ratio of $p/Z$ values — this is exactly the physical basis of the standard $p/Z$ straight-line reserves plot used field-wide.

Common pitfall: Using $c_g=1/p$ alone (the ideal-gas approximation) instead of including the $-\frac{1}{Z}(dZ/dP)_T$ correction — at these pressures, where $Z$ is changing appreciably, that correction materially changes the result, as it does here (about $+14\%$).

Source: Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 2 (Volumetric gas reservoirs, $p/Z$ material balance); Lyons (ed.), Standard Handbook, Ch. 1.

Question 6 of 7: Dry Gas Material Balance and Isothermal Gas Compressibility

Question 6: Dry Gas Material Balance and Isothermal Gas Compressibility (20 marks)