Kennedy design: Q 7 cumec

Given: $Q = 7\ \text{m}^3/\text{s}$, $n = 0.025$, $S = 0.0003$ (1 in 3333), $m = 1.1$, side slope 0.5H:1V. Bed slope $= 0.3$ m/km $= 0.0003$. 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 $D \approx 1.18$ m, so this is the required depth.

Final section

Adopt $D = 1.18$ m, $B = 8.2$ m.

$$\begin{aligned} A &= (8.2 + 0.5\times 1.18)\times 1.18 = 10.37\ \text{m}^2 \\ P &= 8.2 + 2\times 1.18\sqrt{1+0.5^2} = 10.84\ \text{m},\quad R = 0.957\ \text{m} \\ V &= 0.673\ \text{m/s},\quad V_0 = 0.55\times 1.1\times 1.18^{0.64} = 0.673\ \text{m/s} \\ Q &= AV = 10.37\times 0.673 = 6.98\ \text{m}^3/\text{s} \end{aligned}$$

Answer: Depth $D \approx 1.18$ m, bed width $B \approx 8.2$ m, side slope 0.5H:1V, bed slope 1 in 3333, velocity $\approx 0.67$ m/s (discharge check 7.0 m³/s, i.e. within about 0.3% of the design value).