Canal in alluvium, Q 40, d 0.5 mm

No side slope is given, so 0.5H:1V is assumed.

Silt factor

$$f = 1.76\sqrt{d_{mm}} = 1.76\sqrt{0.5} = 1.245$$

Lacey's regime values

Given: $Q = 40\ \text{m}^3/\text{s}$, $f = 1.245$, side slope 0.5H:1V.

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{40\times 1.245^2}{140}\right)^{1/6} = 0.873\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{40}{0.873} = 45.82\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.873^2}{2\times 1.245} = 1.531\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{40} = 30.04\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.245^{5/3}}{3340\times 40^{1/6}} = 0.000233 = \frac{1}{4290} \end{aligned}$$

Trapezoidal section

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

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 30.04\,D + 45.82 &= 0 \\ D &= \frac{30.04 - \sqrt{30.04^2 - 4\times 1.7361\times 45.82}}{2\times 1.7361} = 1.690\ \text{m} \\ B &= P - 2.2361D = 30.04 - 2.2361\times 1.690 = 26.26\ \text{m} \end{aligned}$$

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

Check: $A = (26.26 + 0.5\times 1.690)\times 1.690 = 45.82\ \text{m}^2$ against required 45.82 $\text{m}^2$.

Answer: Bed width $B \approx 26.3$ m, depth of water $D \approx 1.69$ m, side slope 0.5H:1V, bed slope $S = 1$ in 4290 (0.000233), velocity $V = 0.87$ m/s, hydraulic mean radius $R = 1.53$ m.