24-Pet-B2 Oil and Gas Evaluation and Economics · May 2016
Question 2 of 7: Gas Properties and Pipeline Velocity
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2016, 98-Pet-B2, Natural Gas Engineering — 3 hours, closed book (non-communicating calculator permitted), 7 questions of 20 marks each. NOTES item 5 states only the first five questions in the answer book are marked; all 7 are solved.
Reference texts: Katz et al., Handbook of Natural Gas Engineering; Lee & Wattenbarger, Gas Reservoir Engineering (SPE Textbook Series Vol. 5); Ahmed, Reservoir Engineering Handbook, 5th ed.; Mohitpour et al., Pipeline Design and Construction, 3rd ed. (ASME Press); McCain, The Properties of Petroleum Fluids, 3rd ed.
Question 2: Gas Properties and Pipeline Velocity (20 marks)
Find. (a) $T_{pc},p_{pc}$; (b) gas density $\rho$; (c) $B_g$; (d) average gas velocity $v$.
Approach. Build the apparent molecular weight and gas gravity from composition, get pseudocriticals from the exam’s own Standing-type correlation with the N$_2$ correction (only MW is tabulated per component, not individual critical properties, so Kay’s rule cannot be applied directly here), solve for $Z$ via Dranchuk–Abu-Kassem, then chain into density, $B_g$, and velocity.
Apparent molecular weight and gas gravity.
$$M_a=\sum y_iM_i = 0.90(16.04)+0.05(30.07)+0.05(28.01)$$
$$\boxed{M_a = 17.34\ \text{lb}_m/\text{lb-mole}}, \qquad \gamma_g=\frac{M_a}{28.97}=\boxed{0.5986}$$
(a) Pseudocritical properties (Standing correlation + N$_2$ correction). The formula sheet gives $T_{pc}=169.2+349.5\gamma_g-74.0\gamma_g^2$ and $p_{pc}=756.8-131.0\gamma_g-3.6\gamma_g^2$, with the N$_2$/CO$_2$/H$_2$S correction $T_{pc}'=T_{pc}-250y_{N_2}$, $p_{pc}'=p_{pc}-170y_{N_2}$ (no CO$_2$/H$_2$S present):
$$T_{pc}=169.2+349.5(0.5986)-74.0(0.5986)^2=351.9\ ^\circ\text{R}, \quad p_{pc}=756.8-131.0(0.5986)-3.6(0.5986)^2=677.1\ \text{psia}$$
$$\boxed{T_{pc}'=351.9-250(0.05)=339.4\ ^\circ\text{R}}, \qquad \boxed{p_{pc}'=677.1-170(0.05)=668.6\ \text{psia}}$$
(b) Z-factor and density at 600 psia, 100°F. $T=100+459.67=559.67\ ^\circ\text{R}$, so
$$T_r=\frac{559.67}{339.4}=1.649, \qquad p_r=\frac{600}{668.6}=0.897$$
Solving the Dranchuk–Abu-Kassem equation of state for these reduced coordinates gives $Z=0.938$. Density follows from $\rho=pM_a/(ZRT)$, $R=10.732$ psi·ft$^3$/(lb-mole·$^\circ\text{R}$):
$$\rho=\frac{600(17.34)}{0.938(10.732)(559.67)}=\boxed{1.847\ \text{lb}_m/\text{ft}^3}$$
(c) Gas formation volume factor.
$$B_g=0.02827\,\frac{ZT}{p}=0.02827\,\frac{0.938(559.67)}{600}=\boxed{0.02474\ \text{ft}^3/\text{SCF}}$$
(d) Average gas velocity. Convert the 10 MMSCFD standard-condition rate to flowing-condition (actual) volume via the real-gas law, then divide by the 16-in pipe area:
$$Q_{actual}=q_{sc}\frac{p_{sc}ZT}{T_{sc}p}=10\times10^6\times\frac{14.7(0.938)(559.67)}{519.67(600)}=2.481\times10^5\ \text{ft}^3/\text{day}$$
$$A=\frac{\pi}{4}\left(\frac{16}{12}\right)^2=1.396\ \text{ft}^2$$
$$v=\frac{Q_{actual}}{A\times86400\ \text{s/day}}=\boxed{2.05\ \text{ft/s}}$$
