Lacey and Kennedy bed width, Q 40

Side slope is not stated, so 0.5H:1V is assumed.

Lacey's design

Given: $Q = 40\ \text{m}^3/\text{s}$, $f = 1.000$, side slope 0.5H:1V.

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{40\times 1.000^2}{140}\right)^{1/6} = 0.812\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{40}{0.812} = 49.29\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.812^2}{2\times 1.000} = 1.647\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{40} = 30.04\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.000^{5/3}}{3340\times 40^{1/6}} = 0.000162 = \frac{1}{6177} \end{aligned}$$

Trapezoidal section

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

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 30.04\,D + 49.29 &= 0 \\ D &= \frac{30.04 - \sqrt{30.04^2 - 4\times 1.7361\times 49.29}}{2\times 1.7361} = 1.835\ \text{m} \\ B &= P - 2.2361D = 30.04 - 2.2361\times 1.835 = 25.94\ \text{m} \end{aligned}$$

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

Check: $A = (25.94 + 0.5\times 1.835)\times 1.835 = 49.29\ \text{m}^2$ against required 49.29 $\text{m}^2$.

Answer: Bed width $B \approx 25.9$ m, depth of water $D \approx 1.84$ m, side slope 0.5H:1V, bed slope $S = 1$ in 6177 (0.000162), velocity $V = 0.81$ m/s, hydraulic mean radius $R = 1.65$ m.

Kennedy's method with $m = 1$ and the same $B/D$

Lacey's $B/D = 25.94/1.835 = 14.13$. Let $B = 14.13D$, so $A = (B + 0.5D)D = 14.63D^2$.

Kennedy's critical velocity: $V_0 = 0.55\,m\,D^{0.64}$ with $m = 1$, and $Q = V_0A$:

$$\begin{aligned} 40 &= 0.55D^{0.64}\times 14.63D^2 = 8.048\,D^{2.64} \\ D^{2.64} &= 4.970 \\ D &= 1.836\ \text{m} \\ B &= 14.13\times 1.836 = 25.94\ \text{m} \\ V_0 &= 0.55\times 1.836^{0.64} = 0.811\ \text{m/s} \end{aligned}$$

Answer: Lacey: $B \approx 25.9$ m, $D \approx 1.84$ m, $V = 0.81$ m/s, $S$ = 1 in 6177. Kennedy (with $m = 1$, same $B/D$): bed width $\approx 25.9$ m, $D \approx 1.84$ m, essentially the same as Lacey's because $V_0 = 0.811$ m/s is almost equal to Lacey's $V = 0.812$ m/s.