Stable alluvial canal, Q 50, d 0.5 mm

Alluvium with $d_{mean} = 0.5$ mm carrying silt is designed on the regime principle (Lacey). Side slope 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: $Q = 50\ \text{m}^3/\text{s}$, $f = 1.245$, side slope 0.5H:1V.

$$\begin{aligned} V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{50\times 1.245^2}{140}\right)^{1/6} = 0.906\ \text{m/s} \\ A &= \frac{Q}{V} = \frac{50}{0.906} = 55.19\ \text{m}^2 \\ R &= \frac{5V^2}{2f} = \frac{5\times 0.906^2}{2\times 1.245} = 1.649\ \text{m} \\ P &= 4.75\sqrt{Q} = 4.75\sqrt{50} = 33.59\ \text{m} \\ S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.245^{5/3}}{3340\times 50^{1/6}} = 0.000225 = \frac{1}{4452} \end{aligned}$$

Trapezoidal section

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

Eliminating $B$:

$$\begin{aligned} 1.7361\,D^2 - 33.59\,D + 55.19 &= 0 \\ D &= \frac{33.59 - \sqrt{33.59^2 - 4\times 1.7361\times 55.19}}{2\times 1.7361} = 1.813\ \text{m} \\ B &= P - 2.2361D = 33.59 - 2.2361\times 1.813 = 29.53\ \text{m} \end{aligned}$$

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

Check: $A = (29.53 + 0.5\times 1.813)\times 1.813 = 55.19\ \text{m}^2$ against required 55.19 $\text{m}^2$.

Answer: Bed width $B \approx 29.5$ m, depth of water $D \approx 1.81$ m, side slope 0.5H:1V, bed slope $S = 1$ in 4452 (0.000225), velocity $V = 0.91$ m/s, hydraulic mean radius $R = 1.65$ m.

Sketch of designed section

      FSL
  ____|~~~~~~~~~~~~~~~~~~~~~~~|____
     /<--------- T --------->\
    /  D = 1.81 m            \  1V : 0.5H
   /___________________________\
        B = 29.5 m

Top width $T = B + 2zD = 31.35$ m (freeboard to be added above FSL).