Concrete-lined canal: Q 45, 45° slopes

Given: $Q = 45\ \text{m}^3/\text{s}$, $S = 0.000143$ (1 in 7000), $n = 0.015$, side slope 1H:1V. A side slope of 45° means 1H:1V.

Most economical (best hydraulic) section

For a trapezoidal section of least perimeter: $R = D/2$ and $B = 2D(\sqrt{1+z^2} - z)$.

$$B = 2D(\sqrt{1+1^2} - 1) = 0.8284D,\qquad A = D^2\left(2\sqrt{1+z^2} - z\right) = 1.8284D^2$$

Using Manning's equation with $R = D/2$:

$$Q = \frac{1}{n}\,1.8284D^2\left(\frac{D}{2}\right)^{2/3}S^{1/2} \Rightarrow D^{8/3} = \frac{Q n\,2^{2/3}}{1.8284\sqrt{S}}$$ $$D = 4.305\ \text{m},\quad B = 3.566\ \text{m},\quad A = 33.88\ \text{m}^2,\quad R = 2.152\ \text{m},\quad V = 1.328\ \text{m/s}$$

The velocity is within the safe limit for concrete lining (about 2.5 m/s).

Answer: Full supply depth $D \approx 4.30$ m, bed width $B \approx 3.57$ m, side slope 1H:1V, $V \approx 1.33$ m/s. Add freeboard (0.5-0.75 m) and lining thickness as required.