Question 8 of 13: Elevation Drop for Uniform Flow in a Trapezoidal Irrigation Canal

Question 8: Elevation Drop for Uniform Flow in a Trapezoidal Irrigation Canal (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 elevation drop needed to maintain Q = 3 m³/s over 8 km, using general pipe-flow relations with an equivalent (hydraulic) diameter.

Approach. Compute the flow area, wetted perimeter and hydraulic radius at y=1.5 m, convert to an equivalent (hydraulic) diameter D_e=4R_h as instructed by the Nomenclature page, then treat the canal as a "pipe" of that diameter and apply the Darcy–Weisbach/Moody relation to find the friction head loss over 8 km — which, for steady uniform flow, equals the required elevation drop.

1. Flow area, wetted perimeter, hydraulic radius and equivalent diameter at y=1.5 m. At 45° side slopes the horizontal offset equals the depth, so the water-surface width is 0.5+2(1.5)=3.5 m: $$A = \tfrac12(0.5+3.5)(1.5) = \boxed{3.00\ \text{m}^2}, \qquad P = 0.5+2(1.5\sqrt2) = 4.743\ \text{m}$$ $$R_h = A/P = 0.6326\ \text{m}, \qquad D_e = 4R_h = \boxed{2.530\ \text{m}}$$
2. Velocity, Reynolds number and relative roughness. $$V = Q/A = 3/3.00 = \boxed{1.00\ \text{m/s}}, \qquad Re = \frac{\rho V D_e}{\mu} = \frac{1000(1.00)(2.530)}{1.0\times10^{-3}} = 2.53\times10^{6}$$ $$\epsilon/D_e = 0.001/2.530 = 3.95\times10^{-4}$$
3. Friction factor (Colebrook, equivalent to reading the Moody chart at this Re and ε/D). $$\frac{1}{\sqrt f} = -2\log_{10}\left(\frac{\epsilon/D_e}{3.7}+\frac{2.51}{Re\sqrt f}\right) \quad\Rightarrow\quad f = \boxed{0.0161}$$
4. Friction head loss over 8 km — equals the required elevation drop for uniform flow. $$\Delta z = h_f = f\frac{L}{D_e}\frac{V^2}{2g} = 0.0161\left(\frac{8000}{2.530}\right)\frac{1.00^2}{2(9.81)} = \boxed{2.59\ \text{m}}$$