Design stable channel in 8 mm gravel

Given: $Q = 15\ \text{m}^3/\text{s}$, gravel $d = 8$ mm (rounded), $S_0 = 0.0009$, side slope 2H : 1V ($z = 2$), $\phi = 32^\circ$.

Assumptions: the Shields value $\tau_* = 0.06$ (fully rough), $\gamma_s = 2.65\gamma$; maximum shear on the bed $\gamma y S_0$ and on the sides $0.75\gamma y S_0$; Manning's $n$ from Strickler's formula $n = 0.047\,d^{1/6}$ ($d$ in m).

Step 1: Critical shear stress on a level bed

$$\tau_c = 0.06(\gamma_s - \gamma)d = 0.06(1.65)(9810)(0.008) = 7.77\ \text{N/m}^2$$

Step 2: Side slope factor

$$\theta = \tan^{-1}\frac{1}{2} = 26.57^\circ,\qquad K = \sqrt{1 - \frac{\sin^2 26.57^\circ}{\sin^2 32^\circ}} = \sqrt{1 - \frac{0.2}{0.2808}} = 0.536$$

Step 3: Limiting depth

• Bed: $\gamma y S_0 \le \tau_c \Rightarrow y \le \dfrac{7.77}{9810 \times 0.0009} = 0.880$ m
• Sides: $0.75\gamma y S_0 \le K\tau_c \Rightarrow y \le \dfrac{0.536 \times 7.77}{0.75 \times 9810 \times 0.0009} = 0.629$ m

The side slope governs: $y = 0.63$ m.

Step 4: Base width. Strickler: $n = 0.047(0.008)^{1/6} = 0.021$. Manning's equation with $y = 0.629$ m, $z = 2$:

$$Q = \frac{1}{n}AR^{2/3}S_0^{1/2}, \quad A = (b + 2 \times 0.629)(0.629)$$

Trial gives $b = 22.5$ m: $A = 14.93\ \text{m}^2$, $P = 25.27$ m, $R = 0.591$ m, and

$$Q = \frac{1}{0.021}(14.93)(0.591)^{2/3}(0.03) = 15.0\ \text{m}^3/\text{s} \checkmark$$

Step 5: Checks

$$V = \frac{15}{14.93} = 1.00\ \text{m/s},\quad Fr = \frac{V}{\sqrt{gA/T}} = 0.41 < 1\ \text{(subcritical)}$$

Answer: Trapezoidal section with $b \approx 22.5$ m, $y \approx 0.63$ m, side slope 2H : 1V, $S_0 = 0.0009$. The channel is wide and shallow because the small gravel cannot withstand more shear stress. Add a freeboard (about 0.3 m).