Settling tank 30 m x 3 m, 0.02 mm at 25°C; 0.01 mm at 10°C

For 100 % removal a particle must settle the full depth in the detention time: $t=H/v_s$. For smaller particles, the removal is $v_s/v_0$ where $v_0=H/t$.

(a) Detention time for 0.02 mm at 25 °C

$d=0.002$ cm, $S_s=2.65$, $\nu_{25}=0.00893$ cm²/s, $H=300$ cm.

$$v_s=\frac{981\times1.65\times0.002^2}{18\times0.00893}=0.0403\ \text{cm/s}\quad(Re=0.009<1)$$ $$t=\frac{300}{0.0403}=7448\ \text{s}=2.07\ \text{h}$$

(b) Removal of 0.01 mm particles at 10 °C

The tank's overflow rate is $v_0=H/t=0.0403$ cm/s. At 10 °C, $\nu=0.01306$ cm²/s, $d=0.001$ cm:

$$v_s=\frac{981\times1.65\times0.001^2}{18\times0.01306}=0.00689\ \text{cm/s}\quad(Re=0.0005<1)$$ $$\text{Removal}=\frac{0.00689}{0.0403}=0.171\Rightarrow17.1\ \%$$

Answer: detention time = 2.07 h; removal of 0.01 mm particles at 10 °C = 17.1 %.