Horizontal Pipeline Gas Flow Rate: 24-Pet-B2 Oil and Gas Evaluation and Economics, Undated paper, Q5

Question 5: Horizontal Pipeline Gas Flow Rate (10 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.

Find. Gas flow rate $q_{sc}$ through the pipeline (MMSCFD).

Horizontal gas transmission pipeline: 50 km of 0.3049 m ID line dropping from 7000 kPa to 5600 kPa.

Approach. Use the exam's own general (fully turbulent) flow-capacity equation, which needs a Darcy friction factor $f$; since $f$ depends on the Reynolds number, which itself depends on the flow rate being solved for, iterate: assume $f$, solve for $q_{sc}$ and $N_{Re}$, update $f$ from Colebrook's equation, repeat to convergence. Apply the stated pipeline efficiency $E$ as a final derating factor.

Convert to field units (as required by the formula sheet's constants). $$P_1=1015.3\ \text{psia}, \quad P_2=812.2\ \text{psia}, \quad P_{sc}=14.696\ \text{psia}, \quad T_{sc}=518.67\ ^\circ\text{R}$$ $$d=12.00\ \text{in}, \quad L=164{,}042\ \text{ft}, \quad T_m=540.27\ ^\circ\text{R}, \quad \varepsilon/d=1.312\times10^{-4}$$
Iterate friction factor via Colebrook. Starting from an assumed $f$, compute $q_{sc}$ from the general flow equation, then $N_{Re}$, then update $f$ from $1/\sqrt f=-2\log_{10}\!\big(\varepsilon/(3.7d)+2.51/(N_{Re}\sqrt f)\big)$, repeating to convergence: $$N_{Re}=7.86\times10^{6}\ \text{(fully turbulent)}, \qquad \boxed{f=0.01281}$$
Solve for pipeline capacity (before efficiency). $$q_{sc}=5.634\left(\frac{518.67}{14.696}\right)\sqrt{\frac{(1015.3^2-812.2^2)(12.00)^5}{0.01281(0.70)(0.79)(540.27)(164{,}042)}}$$ $$q_{sc}=76{,}307\ \text{MSCFD}=76.31\ \text{MMSCFD}$$
Apply the pipeline efficiency factor. $$q_{sc,actual}=E\cdot q_{sc}=0.92(76.31)=\boxed{70.20\ \text{MMSCFD}}\ \ (\approx1.99\times10^{6}\ \text{Sm}^3/\text{d})$$

Quantity	Result
Friction factor, $f$	0.01281 (Colebrook, converged)
Reynolds number, $N_{Re}$	$7.86\times10^{6}$
Capacity before efficiency	76.31 MMSCFD
Gas flow rate, $q_{sc}$ (with $E=0.92$)	70.20 MMSCFD ($\approx1.99\times10^6$ Sm$^3$/d)

Topic: Steady-state compressible gas flow in a horizontal pipeline — the general flow equation and friction-factor closure.

Key relations: $q_{sc}=5.634(T_{sc}/p_{sc})\sqrt{(p_1^2-p_2^2)d^5/(f\gamma_gZTL)}$; $N_{Re}=710.39(p_{sc}/T_{sc})(\gamma_gq_{sc})/(\mu d)$; Colebrook friction factor.

Why this works: The question explicitly permits any capacity equation (Basic/Weymouth/Panhandle/AGA); the exam's own formula sheet supplies exactly one, the general flow equation with an explicit Darcy friction factor — using it (rather than substituting an unlisted transmission-factor correlation) keeps every input traceable to the given data, and $f$ converges in 3–4 iterations because $q_{sc}\propto1/\sqrt f$ is a weak dependence.

Common pitfall: Skipping the pipeline efficiency factor $E$ — it is listed among the given data specifically to be applied as a derating multiplier on the theoretical (smooth-flow) capacity, not folded into the friction factor.

Source: Mohitpour et al., Pipeline Design and Construction, Ch. 3–4; Katz et al., Handbook of Natural Gas Engineering, Ch. 7; this exam's own formula sheet.

Question 5 of 7: Horizontal Pipeline Gas Flow Rate

Question 5: Horizontal Pipeline Gas Flow Rate (10 marks)