Lacey design: Q 40, f 0.9

No side slope is given, so 0.5H:1V is assumed.

Lacey's regime values

Given: $Q = 40\ \text{m}^3/\text{s}$, $f = 0.900$, side slope 0.5H:1V (assumed).

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{40\times 0.900^2}{140}\right)^{1/6} = 0.784\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{40}{0.784} = 51.05\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.784^2}{2\times 0.900} = 1.705\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{40} = 30.04\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{0.900^{5/3}}{3340\times 40^{1/6}} = 0.000136 = \frac{1}{7362} \end{aligned}$$

Trapezoidal section

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

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 30.04\,D + 51.05 &= 0 \\ D &= \frac{30.04 - \sqrt{30.04^2 - 4\times 1.7361\times 51.05}}{2\times 1.7361} = 1.910\ \text{m} \\ B &= P - 2.2361D = 30.04 - 2.2361\times 1.910 = 25.77\ \text{m} \end{aligned}$$

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

Check: $A = (25.77 + 0.5\times 1.910)\times 1.910 = 51.05\ \text{m}^2$ against the required 51.05 $\text{m}^2$.

Answer: Bed width $B \approx 25.8$ m, depth of water $D \approx 1.91$ m, side slope 0.5H:1V, bed slope $S = 1$ in 7362 (0.000136), velocity $V = 0.78$ m/s, hydraulic mean radius $R = 1.71$ m.