Lacey design: Q 15, f 1.1

Lacey's regime values

Given: $Q = 15\ \text{m}^3/\text{s}$, $f = 1.100$, side slope 0.5H:1V.

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{15\times 1.100^2}{140}\right)^{1/6} = 0.711\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{15}{0.711} = 21.08\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.711^2}{2\times 1.100} = 1.150\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{15} = 18.40\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.100^{5/3}}{3340\times 15^{1/6}} = 0.000223 = \frac{1}{4475} \end{aligned}$$

Trapezoidal section

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

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 18.40\,D + 21.08 &= 0 \\ D &= \frac{18.40 - \sqrt{18.40^2 - 4\times 1.7361\times 21.08}}{2\times 1.7361} = 1.307\ \text{m} \\ B &= P - 2.2361D = 18.40 - 2.2361\times 1.307 = 15.47\ \text{m} \end{aligned}$$

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

Check: $A = (15.47 + 0.5\times 1.307)\times 1.307 = 21.08\ \text{m}^2$ against required 21.08 $\text{m}^2$.

Answer: Bed width $B \approx 15.5$ m, depth of water $D \approx 1.31$ m, side slope 0.5H:1V, bed slope $S = 1$ in 4475 (0.000223), velocity $V = 0.71$ m/s, hydraulic mean radius $R = 1.15$ m.