Economical lined channel: Q 20

Given: $Q = 20\ \text{m}^3/\text{s}$, $S = 0.000300$ (1 in 3333), $n = 0.017$, side slope 1.5H:1V, limiting velocity 1.5 m/s. Slope 30 cm/km $= 0.0003$.

Most economical (best hydraulic) section

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

$$B = 2D(\sqrt{1+1.5^2} - 1.5) = 0.6056D,\qquad A = D^2\left(2\sqrt{1+z^2} - z\right) = 2.1056D^2$$

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

$$Q = \frac{1}{n}\,2.1056D^2\left(\frac{D}{2}\right)^{2/3}S^{1/2} \Rightarrow D^{8/3} = \frac{Q n\,2^{2/3}}{2.1056\sqrt{S}}$$ $$D = 2.747\ \text{m},\quad B = 1.663\ \text{m},\quad A = 15.89\ \text{m}^2,\quad R = 1.373\ \text{m},\quad V = 1.259\ \text{m/s}$$

Velocity 1.26 m/s is within the limiting velocity of 1.5 m/s.

Answer: Full supply depth $D \approx 2.75$ m, bed width $B \approx 1.66$ m, side slope 1.5H:1V, $V \approx 1.26$ m/s. Add freeboard (0.5-0.75 m) and lining thickness as required.