Efficient trapezoidal section and triangular channel depth

Condition for the most efficient trapezoidal section

For bottom width $b$, depth $y$, side slope $n$H : 1V, minimising $P = \dfrac{A}{y} - ny + 2y\sqrt{1+n^2}$ at constant $A$ gives:

• $\dfrac{b + 2ny}{2} = y\sqrt{1+n^2}$: half the top width equals one sloping side.
• Hydraulic radius $R = \dfrac{y}{2}$.
• The sides are tangent to a semicircle drawn with centre on the water surface. If the side slope is also free, the best is $60^\circ$ with the horizontal (half a regular hexagon).

Normal depth of triangular channel

Data: side slope $m = 2$ (2H : 1V), $S = 0.0005$, $Q = 0.5$ m³/s, $n = 0.04$.

$$\begin{aligned} A &= my^2 = 2y^2 \\ P &= 2y\sqrt{1+m^2} = 2y\sqrt{5} = 4.472y \\ R &= \frac{2y^2}{4.472y} = 0.4472y \end{aligned}$$

Manning's equation:

$$\begin{aligned} Q &= \frac{1}{n}AR^{2/3}S^{1/2} \\ 0.5 &= \frac{1}{0.04}(2y^2)(0.4472y)^{2/3}(0.0005)^{1/2} \\ 0.5 &= 25 \times 2 \times 0.5848 \times 0.02236\, y^{8/3} = 0.6538\,y^{8/3} \\ y^{8/3} &= 0.7648 \;\Rightarrow\; y = 0.904\ \text{m} \end{aligned}$$

Check: $A = 2(0.904)^2 = 1.636$ m², $R = 0.404$ m, $Q = 25 \times 1.636 \times 0.404^{2/3} \times 0.02236 = 0.50$ m³/s.

Answer: Normal depth $y_n \approx 0.90$ m.