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

Question 8 — Elevation Drop for 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 Δz required over the 8 km reach.

Approach. Compute the trapezoid's geometric properties at the ACTUAL flowing depth (1.5 m, not the full 2 m design depth) to get the hydraulic diameter, then apply the same Colebrook–White / Darcy–Weisbach procedure as Q7, with the wetted perimeter excluding the free surface. For steady uniform flow the required elevation drop equals the friction head loss.

1. Geometry at the flowing depth. Top width at y = 1.5 m: T = b + 2y(cot45°) = 0.5+2(1.5)(1) = 3.5 m: $$A = \frac{b+T}{2}y = \frac{0.5+3.5}{2}(1.5) = \boxed{3.00\text{ m}^2}, \qquad P = b+2y\sqrt2 = 0.5+2(1.5)(1.4142) = 4.743\text{ m}$$
2. Hydraulic diameter and mean velocity. $$D_e = \frac{4A}{P} = \frac{4(3.00)}{4.743} = \boxed{2.530\text{ m}}, \qquad V = \frac{Q}{A} = \frac{3}{3.00} = 1.00\text{ m/s}$$
3. Reynolds number, relative roughness, friction factor. ν_water = 1.0×10⁻⁶ m²/s: $$Re = \frac{D_e V}{\nu} = 2.53\times10^{6}, \qquad \frac{e}{D} = \frac{0.001}{2.530} = 3.95\times10^{-4} \;\Rightarrow\; \boxed{f = 0.0161 \text{ (Colebrook--White)}}$$
4. Required elevation drop. $$\Delta z = h_L = f\left(\frac{L}{D_e}\right)\!\left(\frac{V^2}{2g}\right) = 0.0161\times\frac{8000}{2.530}\times\frac{1^2}{19.62} = \boxed{2.59\text{ m}}$$