Lacey: max discharge at slope 1/6000

Given: $S = 1/6000$, $f = 1.0$, side slope 0.5H:1V (assumed where not stated). The maximum discharge is the one for which the channel is in regime at this slope.

Discharge from Lacey's slope equation

$$S = \frac{f^{5/3}}{3340\,Q^{1/6}} \Rightarrow Q = \left(\frac{f^{5/3}}{3340\,S}\right)^6 = \left(\frac{1.0^{5/3}}{3340/6000}\right)^6 = 33.607\ \text{m}^3/\text{s}$$

Maximum discharge $Q \approx 33.61$ m³/s (used for the section below).

Section for this discharge:

Lacey's regime values

Given: $Q = 33.61\ \text{m}^3/\text{s}$, $f = 1.000$, side slope 0.5H:1V.

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{33.61\times 1.000^2}{140}\right)^{1/6} = 0.788\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{33.61}{0.788} = 42.63\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.788^2}{2\times 1.000} = 1.554\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{33.61} = 27.54\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.000^{5/3}}{3340\times 33.61^{1/6}} = 0.000167 = \frac{1}{6000} \end{aligned}$$

Trapezoidal section

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

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 27.54\,D + 42.63 &= 0 \\ D &= \frac{27.54 - \sqrt{27.54^2 - 4\times 1.7361\times 42.63}}{2\times 1.7361} = 1.739\ \text{m} \\ B &= P - 2.2361D = 27.54 - 2.2361\times 1.739 = 23.65\ \text{m} \end{aligned}$$

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

Check: $A = (23.65 + 0.5\times 1.739)\times 1.739 = 42.63\ \text{m}^2$ against required 42.63 $\text{m}^2$.

Answer: Bed width $B \approx 23.6$ m, depth of water $D \approx 1.74$ m, side slope 0.5H:1V, bed slope $S = 1$ in 6000 (0.000167), velocity $V = 0.79$ m/s, hydraulic mean radius $R = 1.55$ m.