Concrete-lined canal: Q 70, slope 1 in 4000

Given: $Q = 70\ \text{m}^3/\text{s}$, $S = 0.000250$ (1 in 4000), $n = 0.015$, side slope 0.5H:1V. Manning's $n$ is not given; $n = 0.015$ for concrete lining is assumed.

Most economical (best hydraulic) section

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

$$B = 2D(\sqrt{1+0.5^2} - 0.5) = 1.2361D,\qquad A = D^2\left(2\sqrt{1+z^2} - z\right) = 1.7361D^2$$

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

$$Q = \frac{1}{n}\,1.7361D^2\left(\frac{D}{2}\right)^{2/3}S^{1/2} \Rightarrow D^{8/3} = \frac{Q n\,2^{2/3}}{1.7361\sqrt{S}}$$ $$D = 4.664\ \text{m},\quad B = 5.765\ \text{m},\quad A = 37.76\ \text{m}^2,\quad R = 2.332\ \text{m},\quad V = 1.854\ \text{m/s}$$

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

Answer: Full supply depth $D \approx 4.66$ m, bed width $B \approx 5.76$ m, side slope 0.5H:1V, $V \approx 1.85$ m/s. Add freeboard (0.5-0.75 m) and lining thickness as required.