Bed slope for critical normal depth

Given: rectangular channel, $b = 4\ \text{m}$, normal depth $y_n = 1.25\ \text{m}$ equals the critical depth, $n = 0.015$.

Step 1: Critical flow velocity

$$V = \sqrt{g\,y} = \sqrt{9.81 \times 1.25} = 3.502\ \text{m/s}$$

Step 2: Geometry

$$A = 4 \times 1.25 = 5\ \text{m}^2,\quad P = 4 + 2(1.25) = 6.5\ \text{m},\quad R = \frac{5}{6.5} = 0.769\ \text{m}$$

The discharge is $Q = AV = 17.51\ \text{m}^3/\text{s}$.

Step 3: Manning's equation, $V = \dfrac{1}{n}R^{2/3}S_0^{1/2}$:

$$S_0 = \left(\frac{nV}{R^{2/3}}\right)^2 = \left(\frac{0.015 \times 3.502}{0.769^{2/3}}\right)^2 = 0.003915$$

Answer: Bed slope $S_0 \approx 0.0039$ (about 1 in 255). This is the critical slope for this channel and discharge.