Tractive force design for 10 cumecs, 3 cm particles

Given: $Q = 10\ \text{m}^3/\text{s}$, $S_0 = 0.0016$, 50% of particles $\ge 3$ cm ($d_{50} = 0.03$ m), trapezoidal, $\phi = 34^\circ$, $K_2 = 0.75$ (ratio of permissible bed shear to critical shear).

Assumptions: $\tau_c = 0.06(\gamma_s - \gamma)d$ with $\gamma_s = 2.65\gamma$; maximum shear on bed $\gamma y S_0$, on sides $0.75\gamma y S_0$; Strickler's $n = 0.047(0.03)^{1/6} = 0.0262$.

Step 1: Critical and permissible shear on the bed

$$\tau_c = 0.06(1.65)(9810)(0.03) = 29.14\ \text{N/m}^2,\qquad K_2\tau_c = 0.75(29.14) = 21.85\ \text{N/m}^2$$

Step 2: Side slope. Choose $z = 2$ ($\theta = 26.57^\circ < \phi = 34^\circ$):

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

Step 3: Depth

• Bed: $y \le \dfrac{21.85}{9810 \times 0.0016} = 1.39$ m
• Sides: $y \le \dfrac{0.600 \times 21.85}{0.75 \times 9810 \times 0.0016} = 1.11$ m

The sides govern: $y = 1.11$ m.

Step 4: Base width from Manning's equation ($n = 0.0262$, $S_0 = 0.0016$):

$$A = (b + 2.23)(1.114),\quad P = b + 2(1.114)\sqrt5 = b + 4.98$$

Trial $b = 4.62$ m: $A = 7.63\ \text{m}^2$, $P = 9.61$ m, $R = 0.795$ m.

$$Q = \frac{1}{0.0262}(7.63)(0.795)^{2/3}(0.04) = 10.0\ \text{m}^3/\text{s} \checkmark$$

Step 5: Check

$$V = 1.31\ \text{m/s},\quad T = 4.62 + 4(1.114) = 9.08\ \text{m},\quad Fr = \frac{1.31}{\sqrt{9.81 \times 7.63/9.08}} = 0.46 < 1$$

Answer: Base width $b \approx 4.6$ m, side slope 2H : 1V, depth $y \approx 1.11$ m (subcritical flow, $V = 1.31$ m/s).