Lacey design: Q 15, d 0.33 mm

Silt factor

$$f = 1.76\sqrt{d_{mm}} = 1.76\sqrt{0.33} = 1.011$$

Lacey's regime values

Given: $Q = 15\ \text{m}^3/\text{s}$, $f = 1.011$, side slope 0.5H:1V.

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{15\times 1.011^2}{140}\right)^{1/6} = 0.692\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{15}{0.692} = 21.69\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.692^2}{2\times 1.011} = 1.183\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{15} = 18.40\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.011^{5/3}}{3340\times 15^{1/6}} = 0.000194 = \frac{1}{5150} \end{aligned}$$

Trapezoidal section

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

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 18.40\,D + 21.69 &= 0 \\ D &= \frac{18.40 - \sqrt{18.40^2 - 4\times 1.7361\times 21.69}}{2\times 1.7361} = 1.351\ \text{m} \\ B &= P - 2.2361D = 18.40 - 2.2361\times 1.351 = 15.38\ \text{m} \end{aligned}$$

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

Check: $A = (15.38 + 0.5\times 1.351)\times 1.351 = 21.69\ \text{m}^2$ against required 21.69 $\text{m}^2$.

Answer: Bed width $B \approx 15.4$ m, depth of water $D \approx 1.35$ m, side slope 0.5H:1V, bed slope $S = 1$ in 5150 (0.000194), velocity $V = 0.69$ m/s, hydraulic mean radius $R = 1.18$ m.