Tractive force design for 15 cumecs, 5 cm particles

Given: $Q = 15\ \text{m}^3/\text{s}$, $S_0 = 0.0015$, 50% of particles $\ge 5$ cm ($d_{50} = 0.05$ m), trapezoidal, $\phi = 34^\circ$, ratio of bed shear to critical shear $K_2 = 0.7$.

Assumptions: $\tau_c = 0.06(\gamma_s - \gamma)d$ with $\gamma_s = 2.65\gamma$; the design bed shear is limited to $K_2\tau_c$; maximum shear on bed $\gamma y S_0$ and on sides $0.75\gamma y S_0$; Strickler's $n = 0.047\,d^{1/6} = 0.0285$.

Step 1: Critical and permissible shear on the bed

$$\tau_c = 0.06(1.65)(9810)(0.05) = 48.56\ \text{N/m}^2,\qquad K_2\tau_c = 0.7(48.56) = 33.99\ \text{N/m}^2$$

Step 2: Side slope. Choose $z = 2$ (the side slope must be flatter than $\phi = 34^\circ$, i.e. $z > 1.48$):

$$\theta = 26.57^\circ,\qquad K = \sqrt{1 - \frac{0.2}{\sin^2 34^\circ}} = \sqrt{1 - \frac{0.2}{0.3127}} = 0.600$$

Step 3: Depth

• Bed: $y \le \dfrac{33.99}{9810 \times 0.0015} = 2.31$ m
• Sides: $y \le \dfrac{0.600 \times 33.99}{0.75 \times 9810 \times 0.0015} = 1.85$ m

The sides govern: $y = 1.85$ m.

Step 4: Base width from Manning's equation ($n = 0.0285$):

$$A = (b + 3.70)(1.85),\quad P = b + 2(1.85)\sqrt5 = b + 8.27$$

Trial $b = 2.14$ m: $A = 10.79\ \text{m}^2$, $P = 10.40$ m, $R = 1.037$ m.

$$Q = \frac{1}{0.0285}(10.79)(1.037)^{2/3}(0.03873) = 15.0\ \text{m}^3/\text{s} \checkmark$$

Step 5: Check

$$V = 1.39\ \text{m/s},\quad T = 2.14 + 4(1.85) = 9.54\ \text{m},\quad Fr = \frac{1.39}{\sqrt{9.81 \times 10.79/9.54}} = 0.42 < 1$$

Answer: Base width $b \approx 2.1$ m, side slope 2H : 1V ($\theta = 26.6^\circ$), depth $y \approx 1.85$ m. The narrow base is a consequence of the high permissible depth; the section may be widened (with lower depth) if a shallower channel is preferred.