Lacey design: Q 20, f 1.5

Lacey's regime values

Given: $Q = 20\ \text{m}^3/\text{s}$, $f = 1.500$, side slope 0.5H:1V.

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{20\times 1.500^2}{140}\right)^{1/6} = 0.828\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{20}{0.828} = 24.16\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.828^2}{2\times 1.500} = 1.142\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{20} = 21.24\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.500^{5/3}}{3340\times 20^{1/6}} = 0.000357 = \frac{1}{2800} \end{aligned}$$

Trapezoidal section

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

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 21.24\,D + 24.16 &= 0 \\ D &= \frac{21.24 - \sqrt{21.24^2 - 4\times 1.7361\times 24.16}}{2\times 1.7361} = 1.269\ \text{m} \\ B &= P - 2.2361D = 21.24 - 2.2361\times 1.269 = 18.40\ \text{m} \end{aligned}$$

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

Check: $A = (18.40 + 0.5\times 1.269)\times 1.269 = 24.16\ \text{m}^2$ against required 24.16 $\text{m}^2$.

Answer: Bed width $B \approx 18.4$ m, depth of water $D \approx 1.27$ m, side slope 0.5H:1V, bed slope $S = 1$ in 2800 (0.000357), velocity $V = 0.83$ m/s, hydraulic mean radius $R = 1.14$ m.