Kennedy design: Q 40, slope 1 in 5000

Given: $Q = 40\ \text{m}^3/\text{s}$, $n = 0.0225$, $S = 0.0002$ (1 in 5000), $m = 1.0$. The side slope is not given, so 0.5H:1V is assumed. Manning's formula is used with the given $n$.

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 $D \approx 2.50$ m, so this is the required depth.

Final section

Adopt $D = 2.50$ m, $B = 15.0$ m.

$$\begin{aligned} A &= (15.0 + 0.5\times 2.50)\times 2.50 = 40.62\ \text{m}^2 \\ P &= 15.0 + 2\times 2.50\sqrt{1+0.5^2} = 20.59\ \text{m},\quad R = 1.973\ \text{m} \\ V &= 0.989\ \text{m/s},\quad V_0 = 0.55\times 1.0\times 2.50^{0.64} = 0.989\ \text{m/s} \\ Q &= AV = 40.62\times 0.989 = 40.17\ \text{m}^3/\text{s} \end{aligned}$$

Answer: Depth $D \approx 2.50$ m, bed width $B \approx 15.0$ m, side slope 0.5H:1V, bed slope 1 in 5000, velocity $\approx 0.99$ m/s (discharge check 40.2 m³/s, i.e. within about 0.4% of the design value).