Kennedy design: Q 5 cumec

Given: $Q = 5\ \text{m}^3/\text{s}$, $n = 0.025$, $S = 0.0005$ (1 in 2000), $m = 1.1$, side slope 0.5H:1V. Bed slope $= 0.5$ m/km $= 0.0005$. 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 $D \approx 1.69$ m, so this is the required depth.

Final section

Adopt $D = 1.69$ m, $B = 2.7$ m.

$$\begin{aligned} A &= (2.7 + 0.5\times 1.69)\times 1.69 = 5.99\ \text{m}^2 \\ P &= 2.7 + 2\times 1.69\sqrt{1+0.5^2} = 6.48\ \text{m},\quad R = 0.925\ \text{m} \\ V &= 0.849\ \text{m/s},\quad V_0 = 0.55\times 1.1\times 1.69^{0.64} = 0.846\ \text{m/s} \\ Q &= AV = 5.99\times 0.849 = 5.09\ \text{m}^3/\text{s} \end{aligned}$$

Answer: Depth $D \approx 1.69$ m, bed width $B \approx 2.7$ m, side slope 0.5H:1V, bed slope 1 in 2000, velocity $\approx 0.85$ m/s (discharge check 5.1 m³/s, i.e. within about 1.7% of the design value).