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24-Pet-B2 Oil and Gas Evaluation and Economics · December 2014

Question 3 of 7: Gas Pipeline Flow Capacity and Velocity

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams December 2014, 98-Pet-B2, Natural Gas Engineering — 3 hours, closed book (Casio/Sharp approved calculators only), 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 3: Gas Pipeline Flow Capacity and Velocity (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. $D=48\text{ in}=1.2192$ m; $L=20\text{ km}=20{,}000$ m; $\gamma_g=0.7$; $p_1=2500$ kPa, $p_2=1500$ kPa; $\epsilon=5.85\times10^{-4}$ m; $T_{sc}=288$ K, $p_{sc}=101.325$ kPa; $\bar T=290$ K; $\mu=0.015$ cp; $\bar Z=0.85$.

Find. Pipeline flow capacity $q$ (std m$^3$/day) and gas velocity $V$ inside the pipe.

Approach. The general flow equation needs the Darcy friction factor $f$, which itself depends on the Reynolds number computed from $q$ — solve $q$ and $f$ simultaneously by iteration (Colebrook equation), then convert the standard-condition rate to an actual volumetric rate at mean line conditions to get velocity.

  1. Relative roughness. $r=\epsilon/D=5.85\times10^{-4}/1.2192=\boxed{r=4.80\times10^{-4}}$.
  2. Iterate capacity & friction factor. Starting from an assumed $f$, compute $q=\dfrac{1.149\times10^6\,T_{sc}}{p_{sc}}\left[\dfrac{p_1^2-p_2^2}{\gamma_g fT Z L}\right]^{0.5}d^{2.5}$, then $N_{Re}=17.96\gamma_g q/(\mu d)$, then update $f$ from the Colebrook equation $1/\sqrt f=-2\log_{10}(r/3.7+2.51/(N_{Re}\sqrt f))$, repeating to convergence. At convergence $N_{Re}\gg 3500/r\approx7.3\times10^6$ (comfortably past the “complete turbulence, rough pipe” threshold on the Moody chart regardless of the exact $N_{Re}$ value), so the Colebrook equation collapses to its fully-rough limit $1/\sqrt f=-2\log_{10}(r/3.7)$, giving $\boxed{f=0.01655}$ (the Moody chart’s printed form $1/\sqrt f=1.14-2\log_{10}r$ gives 0.01653, changing $q$ by under 0.1%).
  3. Flow capacity. Substituting the converged $f$: $\boxed{q=44.86\times10^6\ \text{std m}^3/\text{day}}$ ($=44.86$ MMm$^3$/d).
  4. Mean line pressure. $\bar p=\dfrac{2}{3}\left(p_1+p_2-\dfrac{p_1p_2}{p_1+p_2}\right)=\dfrac{2}{3}\left(2500+1500-\dfrac{2500(1500)}{4000}\right)=\boxed{\bar p=2041.7\ \text{kPa}}$.
  5. Actual volumetric rate and velocity. Converting the standard-condition rate to actual conditions at $\bar p,\bar T$: $Q_{actual}=q\left(\dfrac{p_{sc}}{\bar p}\right)\left(\dfrac{\bar T}{T_{sc}}\right)\bar Z$, then $V=Q_{actual}/(A\times86{,}400)$ with $A=\pi D^2/4$: $\boxed{V=18.9\ \text{m/s}}$.
QuantityValue
Relative roughness, $r$$4.80\times10^{-4}$
Friction factor, $f$0.01655
Pipeline flow capacity, $q$44.86 × 10$^6$ std m$^3$/day
Mean line pressure, $\bar p$2041.7 kPa
Gas velocity, $V$18.9 m/s
Check: the flow is fully turbulent, rough-pipe regime at these conditions regardless of exactly how the Reynolds-number formula’s stated units ($q$ in std m$^3$/day, $\mu$ in Pa·s) are read, since $N_{Re}$ lands orders of magnitude past the $3500/r$ complete-turbulence threshold either way — $f$ is therefore governed by the roughness-only asymptote and is insensitive to that ambiguity. Velocity is evaluated at the mean line pressure/temperature (standard pipeline-hydraulics practice for a single representative velocity along a pipe with varying local pressure), not at either end point.