Regime canal for 90 cumecs

Average particle size $d = 0.2$ cm $= 2$ mm. No side slope is given, so 0.5H:1V is assumed.

Silt factor

$$f = 1.76\sqrt{d_{mm}} = 1.76\sqrt{2.0} = 2.489$$

Lacey's regime values

Given: $Q = 90\ \text{m}^3/\text{s}$, $f = 2.489$, side slope 0.5H:1V.

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{90\times 2.489^2}{140}\right)^{1/6} = 1.259\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{90}{1.259} = 71.48\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 1.259^2}{2\times 2.489} = 1.592\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{90} = 45.06\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{2.489^{5/3}}{3340\times 90^{1/6}} = 0.000647 = \frac{1}{1547} \end{aligned}$$

Trapezoidal section

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

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 45.06\,D + 71.48 &= 0 \\ D &= \frac{45.06 - \sqrt{45.06^2 - 4\times 1.7361\times 71.48}}{2\times 1.7361} = 1.697\ \text{m} \\ B &= P - 2.2361D = 45.06 - 2.2361\times 1.697 = 41.27\ \text{m} \end{aligned}$$

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

Check: $A = (41.27 + 0.5\times 1.697)\times 1.697 = 71.48\ \text{m}^2$ against required 71.48 $\text{m}^2$.

Answer: Bed width $B \approx 41.3$ m, depth of water $D \approx 1.70$ m, side slope 0.5H:1V, bed slope $S = 1$ in 1547 (0.000647), velocity $V = 1.26$ m/s, hydraulic mean radius $R = 1.59$ m.