Overflow rate 4000 L/m2/h at 10°C

Specific gravity of the particles is not given, so the usual value for silt/sand, $S_s = 2.65$, is assumed. Particles whose settling velocity $v_s$ is less than the overflow rate $v_0$ are removed in the proportion $v_s/v_0$; those with $v_s \ge v_0$ are removed completely.

Data

• Overflow rate $v_0 = 4000$ l/m²/h $= 4$ m³/m²/h
• $v_0 = \dfrac{4\times100}{3600} = 0.1111$ cm/s
• Temperature 10 °C: $\nu = 0.01306$ cm²/s (kinematic viscosity of water at 10 °C), $g = 981$ cm/s²

Settling velocity (Stokes' law)

$$v_s = \frac{g\,(S_s-1)\,d^2}{18\,\nu}$$

(a) d = 0.05 mm = 0.005 cm

$$v_s = \frac{981\times1.65\times0.005^2}{18\times0.01306} = 0.1721\ \text{cm/s}$$

$Re = v_s d/\nu = 0.066 < 1$, so Stokes' law applies. Since $v_s = 0.1721$ > $v_0 = 0.1111$ cm/s, the removal is 100 %.

(b) d = 0.02 mm = 0.002 cm

$$v_s = \frac{981\times1.65\times0.002^2}{18\times0.01306} = 0.0275\ \text{cm/s}$$

$Re = 0.0042 < 1$ (valid). Here $v_s < v_0$:

$$\text{Removal} = \frac{v_s}{v_0} = \frac{0.0275}{0.1111} = 0.248 = 24.8\ \%$$

Answer: (a) 0.05 mm particles: 100 % removed; (b) 0.02 mm particles: 24.8 % removed.