Question 7 of 13: Flow Rate in a Gravity-Fed Concrete Pipeline

Question 7: Flow Rate in a Gravity-Fed Concrete Pipeline (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. The steady flow rate Q.

Approach. With no pump and no minor losses, all of the elevation drop is consumed by pipe friction. Guess a velocity, read (or here, compute from the Colebrook equation that the Moody chart itself plots) the matching friction factor, and iterate until the friction-loss equation balances the given 40 m — exactly the trial procedure the question describes, done by root-finding instead of by eye on the chart.

1. Head-loss balance (Darcy–Weisbach, no minor losses): $$\Delta z = h_f = f\frac{L}{D}\frac{V^2}{2g}$$
2. Relative roughness: $$\frac{\epsilon}{D}=\frac{0.001}{1.0}=0.001$$
3. Iterate V against f (the Moody chart's own governing curve, the Colebrook equation, solved simultaneously with the head-loss balance): $$\frac{1}{\sqrt{f}}=-2\log_{10}\!\left(\frac{\epsilon/D}{3.7}+\frac{2.51}{Re\sqrt f}\right),\qquad Re=\frac{VD}{\nu}$$ Converging (equivalent to reading the ε/D = 0.001 curve at increasing Re until f·L/D·V²/2g = 40 m is satisfied) gives f ≈ 0.0198 at Re ≈ 1.99×10⁶ (fully rough turbulent flow, consistent with the flat right-hand end of the ε/D=0.001 curve on the chart).
4. Velocity and flow rate: $$V=\boxed{1.99\ \text{m/s}}, \qquad Q=VA=1.99\left(\frac{\pi}{4}(1.0)^2\right)=\boxed{1.56\ \text{m}^3/\text{s}}$$