Isothermal Gas Compressibility from the Z-Factor Chart: 24-Pet-A2 Petroleum Reservoir Fluids, December 2018, Q5

Question 5: Isothermal Gas Compressibility from the Z-Factor Chart (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.

Quantity	Value
Pressure, $p$	1400 psia
Temperature, $T$	200°F
Gas specific gravity, $\gamma_g$	0.6
Pseudo-critical pressure, $p_c$	670 psia
Pseudo reduced pressure, $p_{pr}$	2.0
Pseudo reduced temperature, $T_{pr}$	1.8

Find. Real-gas isothermal compressibility $c_g$ from the $Z$-chart slope, and the ideal-gas $c_g$ for comparison.

Approach. Use the formula-sheet relation for real-gas compressibility in terms of the chart slope, $c_g=\dfrac{1}{p}-\dfrac{1}{Z}\left(\dfrac{1}{p_c}\dfrac{dZ}{dp_{pr}}\right)_{T_{pr}}$, reading $Z$ and its local slope with respect to $p_{pr}$ off the Standing–Katz chart at the given $T_{pr}=1.8$ (reproduced numerically here with the Dranchuk–Abou-Kassem fit, since the chart cannot be read to more than 2–3 significant figures by eye); the ideal-gas case simply drops the $Z$-derivative term.

Check: the formula sheet transcribes this relation as $c_g=-\dfrac{1}{p}-\dfrac{1}{Z}(\ldots)$ (a leading minus sign on the $1/p$ term); that sign is inconsistent with the standard definition of isothermal compressibility, $c_g\equiv-\dfrac{1}{V}\left(\dfrac{\partial V}{\partial p}\right)_T$, and with the well-known result $c_g\to1/p$ as $Z\to1$ (ideal gas). It is treated here as a misprint and the standard form $c_g=\dfrac{1}{p}-\dfrac{1}{Z}\left(\dfrac{dZ}{dp}\right)_T$ is used.

Z-factor and its slope at the given reduced conditions. At $T_{pr}=1.8$, $p_{pr}=2.0$ the Standing–Katz chart gives $\boxed{Z\approx0.914}$. Reading the local slope of the same $T_{pr}=1.8$ curve around $p_{pr}=2.0$ gives $\left(\dfrac{dZ}{dp_{pr}}\right)_{T_{pr}}\approx-0.0282$ (the curve is still descending at this reduced pressure, below its eventual minimum).
Real-gas isothermal compressibility. $c_g=\dfrac{1}{p}-\dfrac{1}{Z\,p_c}\left(\dfrac{dZ}{dp_{pr}}\right)_{T_{pr}}=\dfrac{1}{1400}-\dfrac{1}{0.914\times670}\times(-0.0282)$, so $\boxed{c_g\approx7.60\times10^{-4}\ \text{psi}^{-1}}$.
Ideal-gas isothermal compressibility. For an ideal gas $Z\equiv1$ and $dZ/dp=0$, so the relation collapses to $c_{g,ideal}=\dfrac{1}{p}=\dfrac{1}{1400}$, giving $\boxed{c_{g,ideal}\approx7.14\times10^{-4}\ \text{psi}^{-1}}$.

Quantity	Value
Real-gas compressibility, $c_g$	$7.60\times10^{-4}$ psi$^{-1}$
Ideal-gas compressibility, $c_{g,ideal}$	$7.14\times10^{-4}$ psi$^{-1}$

Topic: Isothermal gas compressibility and its real-gas correction via the $Z$-factor chart.

Key relations: $c_g=\dfrac{1}{p}-\dfrac{1}{Z}\left(\dfrac{dZ}{dp}\right)_T=\dfrac{1}{p}-\dfrac{1}{Zp_c}\left(\dfrac{dZ}{dp_{pr}}\right)_{T_{pr}}$; ideal-gas limit $c_g\to1/p$.

Why this works: $c_g$ is defined from $V\propto Z/p$ at fixed $T,n$, so differentiating $\ln V=\ln Z-\ln p+\text{const}$ with respect to $p$ splits $c_g$ into the familiar $1/p$ ideal-gas term plus a correction carrying exactly the local slope of the $Z$-factor curve; where $Z$ decreases with $p$ (as here, below the chart's eventual minimum for this $T_{pr}$), the correction adds to $1/p$ and the real gas is MORE compressible than an ideal gas at the same $p,T$.

Common pitfall: Assuming a real gas is always LESS compressible than an ideal gas; that is only true once $dZ/dp$ turns positive (higher reduced pressures), which has not yet happened at $p_{pr}=2$ on the $T_{pr}=1.8$ curve.

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 2–3 (gas compressibility, Z-factor); Lyons, Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed.

Question 5 of 7: Isothermal Gas Compressibility from the Z-Factor Chart

Question 5: Isothermal Gas Compressibility from the Z-Factor Chart (20 marks)