Most efficient trapezoidal channel (3H:4V)

For the most efficient trapezoidal section, $R = y/2$ and half the top width equals one sloping side. Velocity is known, so Manning's equation gives $R$ directly.

Data

Side slope 3H : 4V, so $n = 3/4 = 0.75$ horizontal per vertical; $\sqrt{1+n^2} = 1.25$. Manning's $n_m = 0.012$, $S = 1/2000$, $V = 0.5$ m/s.

Hydraulic radius from Manning

$$\begin{aligned} V &= \frac{1}{n_m}R^{2/3}S^{1/2} \\ R^{2/3} &= \frac{0.5 \times 0.012}{(1/2000)^{1/2}} = \frac{0.006}{0.02236} = 0.2683 \\ R &= 0.2683^{1.5} = 0.1390\ \text{m} \end{aligned}$$

Depth and bottom width

$$\begin{aligned} y &= 2R = 0.278\ \text{m} \\ b &= 2y\left(\sqrt{1+n^2} - n\right) = 2 \times 0.278 \times (1.25 - 0.75) = 0.278\ \text{m} \end{aligned}$$

(With 3H : 4V slopes, $b = y$.)

Check

$A = (0.278 + 0.75 \times 0.278) \times 0.278 = 0.1352$ m², $P = 0.278 + 2 \times 0.278 \times 1.25 = 0.973$ m, $R = 0.139$ m. Discharge $Q = AV = 0.1352 \times 0.5 = 0.0676$ m³/s.

Answer: Depth ≈ 0.278 m, bottom width ≈ 0.278 m (top width ≈ 0.695 m); it carries about 0.068 m³/s at 0.5 m/s. The section is small because the velocity is low and the channel is smooth.