Lacey design: Q 18, f 1.5

Lacey's regime values

Given: $Q = 18\ \text{m}^3/\text{s}$, $f = 1.500$, side slope 0.5H:1V.

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{18\times 1.500^2}{140}\right)^{1/6} = 0.813\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{18}{0.813} = 22.13\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.813^2}{2\times 1.500} = 1.102\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{18} = 20.15\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.500^{5/3}}{3340\times 18^{1/6}} = 0.000364 = \frac{1}{2751} \end{aligned}$$

Trapezoidal section

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

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 20.15\,D + 22.13 &= 0 \\ D &= \frac{20.15 - \sqrt{20.15^2 - 4\times 1.7361\times 22.13}}{2\times 1.7361} = 1.228\ \text{m} \\ B &= P - 2.2361D = 20.15 - 2.2361\times 1.228 = 17.41\ \text{m} \end{aligned}$$

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

Check: $A = (17.41 + 0.5\times 1.228)\times 1.228 = 22.13\ \text{m}^2$ against required 22.13 $\text{m}^2$.

Answer: Bed width $B \approx 17.4$ m, depth of water $D \approx 1.23$ m, side slope 0.5H:1V, bed slope $S = 1$ in 2751 (0.000364), velocity $V = 0.81$ m/s, hydraulic mean radius $R = 1.10$ m.