Settling tank 15 x 4 x 2.5 m, 96% of 0.015 mm

In an ideal settling tank the fraction of particles removed is $v_s/v_0$ (for $v_s<v_0$), where $v_0=H/t$ is the overflow rate; larger particles are removed completely.

(a) Detention time for 96 % removal of 0.015 mm particles at 15 °C

$d=0.0015$ cm, $S_s=2.65$, $\nu_{15}=0.01139$ cm²/s, $H=250$ cm.

$$v_s=\frac{981\times1.65\times0.0015^2}{18\times0.01139}=0.01776\ \text{cm/s}\quad(Re=0.0023<1)$$ $$\frac{v_s}{v_0}=0.96\Rightarrow v_0=\frac{0.01776}{0.96}=0.01850\ \text{cm/s}$$ $$t=\frac{H}{v_0}=\frac{250}{0.01850}=13511\ \text{s}=3.75\ \text{h}$$

For this detention time the tank (15 m $\times$ 4 m $\times$ 2.5 m, volume 150 m³) can pass a flow $Q=V/t=40.0$ m³/h $=959$ m³/day (surface overflow rate $v_0=16.0$ m³/m²/day).

(b) Removal of 99 % of 0.020 mm particles

$d=0.002$ cm at the same temperature:

$$v_s=\frac{981\times1.65\times0.002^2}{18\times0.01139}=0.03158\ \text{cm/s}\quad(Re=0.0055<1)$$

Compare with the overflow rate $v_0=0.01850$ cm/s: $v_s/v_0=1.71>1$, so these particles are removed 100 %, which is more than 99 %.

Answer: detention time = 3.75 h (about 225 min); the tank is sufficient, since 0.020 mm particles are removed 100 % (more than the 99 % required).