Question 9 of 13: Pipe Sizing from the Moody Chart

Question 9 — Pipe Sizing from the Moody Chart (5 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.

Find. A pipe diameter (0.2–0.4 m range) that delivers 0.3 m³/s under the available 70 m head.

Approach. Write f and Re as functions of trial D, evaluate two or three trial diameters against the Moody chart (equivalent here to the Colebrook equation the chart plots), and interpolate for the D whose (Re, e/D) point gives an f that satisfies the Darcy–Weisbach head-loss equation with the required 70 m.

1. Express f and Re in terms of D. With V=4Q/(πD²): $$Re = \frac{{\rho VD}}{{\mu}} = \frac{{4\rho Q}}{{\pi\mu D}} = \frac{{4(800)(0.3)}}{{\pi(0.048)D}} = \frac{{6366}}{{D}}$$ $$h_f = f\frac{{L}}{{D}}\frac{{V^2}}{{2g}} \;\Rightarrow\; f = \frac{{h_f\, D\, 2g}}{{L\,V^2}} = \frac{{h_f\, 2g\, \pi^2D^5}}{{16\,L\,Q^2}}$$
2. Trial diameters (bracketing the required point, as the chart method asks). At D=0.30 m: Re≈21,200, e/D=0.0002, chart f≈0.026, giving h_f≈39.8 m (too low — pipe too large). At D=0.25 m: Re≈25,500, e/D=0.00024, chart f≈0.025, giving h_f above 70 m (pipe too small). The required point lies between these trials.
3. Solve the Colebrook/Darcy–Weisbach system directly (equivalent to reading the intersection of the plotted line through the trial points with the D-implied f–Re curve on the Moody chart): $$\boxed{{D \approx 0.267\text{{ m (267 mm)}}}},\quad V=5.38\text{{ m/s}},\quad Re=2.39\times10^4,\quad f=0.0253$$