Kennedy design: Q 10, n 0.0245

Given: $Q = 10\ \text{m}^3/\text{s}$, $n = 0.0245$, $S = 0.0002$ (1 in 5000), $m = 1$, side slope 0.5H:1V.

Method (Kennedy's silt theory, trial and error)

1. Assume depth $D$.
2. Kennedy's critical velocity: $V_0 = 0.55\,m\,D^{0.64}$.
3. Area $A = Q/V_0$, bed width $B = (A - zD^2)/D$, perimeter $P = B + 2D\sqrt{1+z^2}$, $R = A/P$.
4. Find $V$ from Manning's formula, $V = \frac{1}{n}R^{2/3}S^{1/2}$.
5. Repeat until $V = V_0$.

Trials

The velocities $V$ and $V_0$ become equal at $D \approx 0.89$ m, so this is the required depth.

Final section

Adopt $D = 0.89$ m, $B = 21.6$ m.

$$\begin{aligned} A &= (21.6 + 0.5\times 0.89)\times 0.89 = 19.62\ \text{m}^2 \\ P &= 21.6 + 2\times 0.89\sqrt{1+0.5^2} = 23.59\ \text{m},\quad R = 0.832\ \text{m} \\ V &= 0.511\ \text{m/s},\quad V_0 = 0.55\times 1\times 0.89^{0.64} = 0.510\ \text{m/s} \\ Q &= AV = 19.62\times 0.511 = 10.02\ \text{m}^3/\text{s} \end{aligned}$$

Answer: Depth $D \approx 0.89$ m, bed width $B \approx 21.6$ m, side slope 0.5H:1V, bed slope 1 in 5000, velocity $\approx 0.51$ m/s (discharge check 10.0 m³/s, i.e. within about 0.2% of the design value).