Lacey design: Q 30, f 1.1

Lacey's regime values

Given: $Q = 30\ \text{m}^3/\text{s}$, $f = 1.100$, side slope 0.5H:1V.

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{30\times 1.100^2}{140}\right)^{1/6} = 0.799\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{30}{0.799} = 37.57\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.799^2}{2\times 1.100} = 1.449\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{30} = 26.02\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.100^{5/3}}{3340\times 30^{1/6}} = 0.000199 = \frac{1}{5023} \end{aligned}$$

Trapezoidal section

For side slope $z = 0.5$: $A = (B + zD)D$ and $P = B + 2D\sqrt{1+z^2} = B + 2.2361D$.

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 26.02\,D + 37.57 &= 0 \\ D &= \frac{26.02 - \sqrt{26.02^2 - 4\times 1.7361\times 37.57}}{2\times 1.7361} = 1.619\ \text{m} \\ B &= P - 2.2361D = 26.02 - 2.2361\times 1.619 = 22.40\ \text{m} \end{aligned}$$

(The other root of the quadratic gives a very deep, narrow section, so the smaller root is taken.)

Check: $A = (22.40 + 0.5\times 1.619)\times 1.619 = 37.57\ \text{m}^2$ against required 37.57 $\text{m}^2$.

Answer: Bed width $B \approx 22.4$ m, depth of water $D \approx 1.62$ m, side slope 0.5H:1V, bed slope $S = 1$ in 5023 (0.000199), velocity $V = 0.80$ m/s, hydraulic mean radius $R = 1.45$ m.