Kennedy design: Q 50, m 0.9

Given: $Q = 50\ \text{m}^3/\text{s}$, $n = 0.0225$, $S = 0.0002$ (1 in 5000), $m = 0.9$, side slope 0.5H:1V. Kutter's formula is used because Kutter's $n$ is specified.

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 Kutter's formula, $V = C\sqrt{RS}$ with $C = \dfrac{23 + \frac{1}{n} + \frac{0.00155}{S}}{1 + \left(23 + \frac{0.00155}{S}\right)\frac{n}{\sqrt R}}$.
5. Repeat until $V = V_0$.

Trials

The velocities $V$ and $V_0$ become equal at $D \approx 3.67$ m, so this is the required depth.

Final section

Adopt $D = 3.67$ m, $B = 10.1$ m.

$$\begin{aligned} A &= (10.1 + 0.5\times 3.67)\times 3.67 = 43.80\ \text{m}^2 \\ P &= 10.1 + 2\times 3.67\sqrt{1+0.5^2} = 18.31\ \text{m},\quad R = 2.393\ \text{m} \\ V &= 1.137\ \text{m/s},\quad V_0 = 0.55\times 0.9\times 3.67^{0.64} = 1.138\ \text{m/s} \\ Q &= AV = 43.80\times 1.137 = 49.78\ \text{m}^3/\text{s} \end{aligned}$$

Answer: Depth $D \approx 3.67$ m, bed width $B \approx 10.1$ m, side slope 0.5H:1V, bed slope 1 in 5000, velocity $\approx 1.14$ m/s (discharge check 49.8 m³/s, i.e. within about 0.4% of the design value).