Gas Transmission Pipeline Diameter Sizing: 24-Pet-B2 Oil and Gas Evaluation and Economics, May 2015, Q3

Question 3: Gas Transmission Pipeline Diameter Sizing (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. $\gamma_g=0.68$; $L=20\text{ km}=65{,}616.8$ ft; $p_1=1500$ psia, $p_2=1000$ psia; $\bar T=77^{\circ}\text{F}=536.67^{\circ}\text{R}$; $q_{sc}=220{,}000$ MSCFD; $\epsilon=0.0012$ ft; $\mu=0.015$ cP; $\bar Z=0.85$; $T_{sc}=520^{\circ}\text{R}$, $p_{sc}=14.7$ psia (formula-sheet general flow equation, field units).

Find. Required internal diameter $d$ (in).

Approach. The general flow equation $q_{sc}=5.634(T_{sc}/p_{sc})\sqrt{(p_1^2-p_2^2)d^5/(f\gamma_gZ_{av}\bar TL)}$ has $d$ on both sides implicitly (through the friction factor $f$, which depends on the Reynolds number $N_{Re}\propto1/d$ and on relative roughness $\epsilon/d$) — solve $d$ and $f$ simultaneously by iteration (Colebrook equation).

Step 1 of the iteration (shown per the question’s instruction). Assume a starting friction factor $f^{(0)}=0.015$. Solving the general flow equation for $d$: $d^5=\left[\dfrac{q_{sc}}{5.634(T_{sc}/p_{sc})}\right]^2\dfrac{f\gamma_gZ_{av}\bar TL}{p_1^2-p_2^2}=\left[\dfrac{220{,}000}{5.634(520/14.7)}\right]^2\dfrac{0.015(0.68)(0.85)(536.67)(65{,}616.8)}{1500^2-1000^2}$, giving $d^{(0)}=12.72$ in. The Reynolds number at this trial diameter, $N_{Re}=710.39(p_{sc}/T_{sc})(\gamma_gq_{sc})/(\mu d)$, comes out at $N_{Re}^{(0)}\approx1.57\times10^7$ — deep into fully-rough turbulent flow ($N_{Re}\gg3500/r$) — so the Colebrook equation $1/\sqrt f=-2\log_{10}(r/3.7+2.51/(N_{Re}\sqrt f))$ updates $f$ to $f^{(1)}=0.0201$, and $d$ is re-solved from that new $f$.
Converge the iteration. Repeating this $d\to N_{Re}\to f\to d$ loop to convergence: $\boxed{f=0.02008}$, $N_{Re}=1.52\times10^7$, relative roughness $\epsilon/d=0.0012/(13.19/12)=1.09\times10^{-3}$, and $\boxed{d=13.19\ \text{in}}$.

Quantity	Value
Converged friction factor, $f$	0.02008
Reynolds number, $N_{Re}$	$1.52\times10^7$ (turbulent)
Relative roughness, $\epsilon/d$	$1.09\times10^{-3}$
Required internal diameter, $d$	13.19 in

Check: the required hydraulic diameter (13.19 in) is a calculated minimum; in practice this would be rounded up to the nearest standard pipe schedule (e.g., NPS 14 or 16, whose actual internal diameter depends on the selected wall thickness/schedule) to provide margin and match commercially available pipe.

Topic: Steady-state compressible gas flow in a transmission pipeline (general flow equation), sized here for diameter rather than capacity.

Key relations: General flow equation $q_{sc}\propto d^{2.5}\sqrt{(p_1^2-p_2^2)/(f\gamma_gZ\bar TL)}$; Colebrook $1/\sqrt f=-2\log_{10}(r/3.7+2.51/(N_{Re}\sqrt f))$; $N_{Re}=710.39(p_{sc}/T_{sc})(\gamma_gq_{sc})/(\mu d)$.

Why this works: The driving force for compressible gas flow is $p_1^2-p_2^2$ (not $p_1-p_2$); because friction factor and Reynolds number both depend on the unknown diameter, sizing for diameter (as opposed to solving for capacity at a known diameter) always requires iteration, not a direct algebraic solution.

Common pitfall: Assuming a fixed friction factor and solving $d$ algebraically in one step — $f$ typically shifts by 30–40% between the first guess and convergence in this problem, which would size the line meaningfully wrong.

Source: Mohitpour et al., Pipeline Design and Construction, 3rd ed., Ch. 2; CSA Z662, Oil and Gas Pipeline Systems.

Question 3 of 7: Gas Transmission Pipeline Diameter Sizing

Question 3: Gas Transmission Pipeline Diameter Sizing (20 marks)