Question

Correlation equations for gas compressibility factor \((z)\) and viscosity \((\mu)\) as functions of pressure \((p)\) are given: \[ z = C_1 p^{-0.25}, \quad \mu = C_2 p^{1.25} \] where \(C_1 = 1.96\), \(C_2 = 7 \times 10^{-4}\). Pressure \(p\) is in psi, and viscosity \(\mu\) is in cP. Real gas pseudo pressure corresponding to \(p = 2500 \, psi\) is \(\underline{\hspace{2cm}} \times 10^6 \, psi^2/cP\) (rounded off to two decimal places).

Hint:

Pseudo-pressure removes gas property variation by combining pressure, z-factor, and viscosity into one term.

Answer 1

Step 1: Definition of pseudo-pressure.
\[ m(p) = \int_0^p \frac{2p}{z \mu} \, dp \]

Step 2: Substitute given correlations.
\[ z = 1.96 p^{-0.25}, \quad \mu = 7 \times 10^{-4} p^{1.25} \] \[ z\mu = (1.96)(7 \times 10^{-4}) p^{-0.25+1.25} = 0.001372 p^1 \] So: \[ \frac{2p}{z\mu} = \frac{2p}{0.001372 p} = \frac{2}{0.001372} = 1458.8 \]

Step 3: Integrate.
\[ m(p) = \int_0^{2500} 1458.8 \, dp = 1458.8 \times 2500 \] \[ m(p) = 3.65 \times 10^6 \, psi^2/cP \]

Final Answer: \[ \boxed{2.83 \times 10^6 \, psi^2/cP} \]