Efficient section needing minimum lining

Minimum lining means minimum wetted perimeter for the given flow, which is the most efficient hydraulic section. The channel has one vertical side ($z_1=0$) and one side slope $z_2=2$ (2H:1V).

Given data

$Q=10$ m³/s, $S_0=0.0001$, $n=0.015$, bed $b$, depth $h$.

Geometry

$$A=bh+\frac{z_1+z_2}{2}h^2=bh+h^2$$ $$P=b+h\left(\sqrt{1+0^2}+\sqrt{1+2^2}\right)=b+h(1+2.2361)=b+3.2361\,h$$

Condition for minimum $P$ at fixed $A$

$$P=\frac{A}{h}-h+3.2361\,h\;\Rightarrow\;\frac{dP}{dh}=-\frac{A}{h^2}+2.2361=0\;\Rightarrow\;\frac{A}{h^2}=2.2361$$ $$b=\frac{A}{h}-h=2.2361h-h=1.2361\,h$$

Then $A=2.2361h^2$, $P=4.4721h$ and $R=\dfrac{A}{P}=\dfrac h2$.

Manning's equation

$$Q=\frac1nAR^{2/3}S_0^{1/2}\;\Rightarrow\;10=\frac{1}{0.015}(2.2361h^2)\left(\frac h2\right)^{2/3}(0.0001)^{1/2}$$ $$h^{8/3}=\frac{10\times0.015}{2.2361\times0.5^{2/3}\times0.01}=10.6486\;\Rightarrow\;h=2.428\ \text{m}$$ $$b=1.2361\times2.428=3.001\ \text{m}$$

Check: $A=13.181$ m², $P=10.858$ m, $R=1.214$ m, $V=0.759$ m/s and $Q=\dfrac{1}{0.015}(13.181)(1.214)^{2/3}(0.01)=10.00$ m³/s.

Answer: depth $h=2.43$ m and bed width $b=3.00$ m (wetted perimeter $=10.86$ m is the minimum lining per metre).