Regime channel design by Lacey's method

Given: $Q = 75\ \text{m}^3/\text{s}$, $d = 0.65$ mm. Assume side slope 0.5H : 1V (Lacey's usual value).

Step 1: Silt factor

$$f = 1.76\sqrt{d_{mm}} = 1.76\sqrt{0.65} = 1.419$$

Step 2: Velocity

$$V = \left(\frac{Qf^2}{140}\right)^{1/6} = \left(\frac{75 \times 2.014}{140}\right)^{1/6} = 1.013\ \text{m/s}$$

Step 3: Area, perimeter, hydraulic radius, slope

$$A = \frac{Q}{V} = 74.06\ \text{m}^2,\qquad P = 4.75\sqrt{Q} = 41.14\ \text{m}$$ $$R = \frac{5V^2}{2f} = \frac{5(1.0256)}{2(1.419)} = 1.81\ \text{m}\quad (A/P = 1.80\ \text{m, consistent})$$ $$S = \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.419^{5/3}}{3340(75)^{1/6}} = 2.61 \times 10^{-4}\ (\approx 1\text{ in }3830)$$

Step 4: Bed width and depth. For side slope 0.5H : 1V:

$$A = (b + 0.5y)y = 74.06,\qquad P = b + 2y\sqrt{1 + 0.25} = b + 2.236y = 41.14$$

Substituting $b = 41.14 - 2.236y$:

$$(41.14 - 1.736y)y = 74.06 \;\Rightarrow\; 1.736y^2 - 41.14y + 74.06 = 0 \;\Rightarrow\; y = 1.96\ \text{m}$$ $$b = 41.14 - 2.236(1.963) = 36.75\ \text{m}$$

Answer: $V = 1.01$ m/s, $b \approx 36.7$ m, $y \approx 1.96$ m, side slope 0.5H : 1V, bed slope $\approx 1$ in 3830 (add freeboard).