Bed motion check and non-moving gravel size

Given: $d_{50} = 0.35$ mm, $S_0 = 1.5 \times 10^{-4}$; trapezoidal channel $b = 3$ m, $z = 1.5$ (1.5H : 1V), $y = 0.25$ m.

Critical shear stress (empirical, $d$ in mm):

$$\tau_c = 0.155 + \frac{0.409\,d^2}{\sqrt{1 + 0.177\,d^2}}\ \ \text{N/m}^2$$

a) Will the bed particles move?

Hydraulic radius:

$$A = (3 + 1.5 \times 0.25)(0.25) = 0.844\ \text{m}^2$$ $$P = 3 + 2(0.25)\sqrt{1 + 1.5^2} = 3.901\ \text{m},\qquad R = \frac{0.844}{3.901} = 0.216\ \text{m}$$

Bed shear stress:

$$\tau_0 = \gamma R S_0 = 9810 \times 0.216 \times 1.5 \times 10^{-4} = 0.318\ \text{N/m}^2$$

Critical shear stress for $d = 0.35$ mm:

$$\tau_c = 0.155 + \frac{0.409(0.1225)}{\sqrt{1 + 0.177(0.1225)}} = 0.155 + \frac{0.0501}{1.0108} = 0.205\ \text{N/m}^2$$

Since $\tau_0 = 0.318 > \tau_c = 0.205\ \text{N/m}^2$, the bed particles will be in motion.

b) Minimum size that will not move

Set $\tau_c = \tau_0 = 0.318$:

$$0.155 + \frac{0.409\,d^2}{\sqrt{1 + 0.177\,d^2}} = 0.318 \;\Rightarrow\; \frac{0.409\,d^2}{\sqrt{1 + 0.177\,d^2}} = 0.163$$

Trial: $d = 0.64$ mm gives $\dfrac{0.409(0.4096)}{\sqrt{1.0725}} = 0.1617$; $d = 0.643$ mm gives $0.1630$. Hence

$$d_{min} \approx 0.64\ \text{mm}$$

Check: $\tau_c(0.643) = 0.318\ \text{N/m}^2 = \tau_0$.

Answer: (a) Particles move ($\tau_0 = 0.318 > \tau_c = 0.205\ \text{N/m}^2$). (b) Grains of about 0.64 mm and larger will not move.