Lacey: max discharge at slope 1/4000

Given: $S = 1/4000$, $f = 0.9$, side slope 0.5H:1V. The maximum discharge is the one for which the channel is in regime at this slope.

Discharge from Lacey's slope equation

$$S = \frac{f^{5/3}}{3340\,Q^{1/6}} \Rightarrow Q = \left(\frac{f^{5/3}}{3340\,S}\right)^6 = \left(\frac{0.9^{5/3}}{3340/4000}\right)^6 = 1.029\ \text{m}^3/\text{s}$$

Maximum discharge $Q \approx 1.03$ m³/s (used for the section below).

Section for this discharge:

Lacey's regime values

Given: $Q = 1.03\ \text{m}^3/\text{s}$, $f = 0.900$, side slope 0.5H:1V.

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{1.03\times 0.900^2}{140}\right)^{1/6} = 0.426\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{1.03}{0.426} = 2.42\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.426^2}{2\times 0.900} = 0.504\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{1.03} = 4.82\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{0.900^{5/3}}{3340\times 1.03^{1/6}} = 0.000250 = \frac{1}{4001} \end{aligned}$$

Trapezoidal section

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

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 4.82\,D + 2.42 &= 0 \\ D &= \frac{4.82 - \sqrt{4.82^2 - 4\times 1.7361\times 2.42}}{2\times 1.7361} = 0.657\ \text{m} \\ B &= P - 2.2361D = 4.82 - 2.2361\times 0.657 = 3.35\ \text{m} \end{aligned}$$

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

Check: $A = (3.35 + 0.5\times 0.657)\times 0.657 = 2.42\ \text{m}^2$ against the required 2.42 $\text{m}^2$.

Answer: Bed width $B \approx 3.4$ m, depth of water $D \approx 0.66$ m, side slope 0.5H:1V, bed slope $S = 1$ in 4000 (0.000250), velocity $V = 0.43$ m/s, hydraulic mean radius $R = 0.50$ m; maximum discharge $\approx 1.03$ m³/s.