Efficient trapezoid with unequal side slopes

The bed slope is not given in the question. It is assumed to be $S_0=1$ in 1000 $=0.001$, and the working is shown so that another slope can be substituted (the depth varies as $S_0^{-3/16}$).

Given data

$z_1=0.5$, $z_2=1.5$ (H:V), $Q=12$ m³/s, $n=0.014$.

Condition for most efficient section

The most efficient section has the least wetted perimeter for a given area. As for a symmetric trapezoid, this gives the condition that the top width equals the sum of the two sloping sides, and $R=y/2$. Let $s=\sqrt{1+z_1^2}+\sqrt{1+z_2^2}$.

$$A=by+\frac{(z_1+z_2)}{2}y^2=by+y^2$$ $$P=b+y\left(\sqrt{1+z_1^2}+\sqrt{1+z_2^2}\right)=b+y(1.1180+1.8028)=b+2.9208\,y$$

Writing $P=\dfrac{A}{y}-\dfrac{z_1+z_2}{2}y+s\,y$ and setting $dP/dy=0$ (with $A$ constant) gives

$$b=y\left(\sqrt{1+z_1^2}+\sqrt{1+z_2^2}-(z_1+z_2)\right)=(2.9208-2)\,y=0.9208\,y$$

Then

$$A=1.9208\,y^2,\qquad P=3.8416\,y,\qquad R=\frac{A}{P}=\frac y2$$

Normal depth from Manning's equation

$$Q=\frac1nAR^{2/3}S_0^{1/2}\;\Rightarrow\;12=\frac{1}{0.014}(1.9208\,y^2)\left(\frac y2\right)^{2/3}(0.001)^{1/2}$$ $$y^{8/3}=4.3905\;\Rightarrow\;y=1.742\ \text{m}$$ $$b=0.9208\times1.742=1.604\ \text{m},\qquad A=5.826\ \text{m}^2,\qquad P=6.690\ \text{m}$$

Top width $T=b+(z_1+z_2)y=5.087$ m.

Critical depth

Critical flow condition: $\dfrac{Q^2}{g}=\dfrac{A_c^3}{T_c}$ with $A_c=by_c+y_c^2$ and $T_c=b+2y_c$:

$$\frac{12^2}{9.81}=14.679=\frac{(1.604\,y_c+y_c^2)^3}{1.604+2y_c}$$

Solving by trial: $y_c=1.349$ m. Check: $A_c=3.982$ m², $T_c=4.301$ m, $V_c=3.014$ m/s, $Fr=1$.

Answer (for $S_0=0.001$): the most efficient section has $b=1.60$ m and $y=1.74$ m, with side slopes 0.5H:1V and 1.5H:1V, and the critical depth is $y_c=1.35$ m. Since $y_n>y_c$, the flow is subcritical.