Settling tank 15 m x 7 m x 3 m, 95% removal

In an ideal tank the fraction of particles removed is $v_s/v_0$ for $v_s<v_0$, and 100 % for $v_s\ge v_0$, where $v_0=Q/A$ is the overflow rate.

Step 1: Settling velocity of 0.035 mm particles at 20 °C

$d=0.0035$ cm, $S_s=2.65$, $\nu=0.01007$ cm²/s.

$$v_s=\frac{981\times1.65\times0.0035^2}{18\times0.01007}=0.1094\ \text{cm/s}\quad(Re=0.038<1)$$

Step 2: Overflow rate for 95 % removal

$$\frac{v_s}{v_0}=0.95\Rightarrow v_0=\frac{0.1094}{0.95}=0.1151\ \text{cm/s}=99.5\ \text{m}^3/\text{m}^2/\text{day}$$

Step 3: Flow and flow velocity

Surface area $A=15\times7=105$ m²; cross-section $=7\times3=21$ m².

$$Q=v_0A=0.001151\times105=0.12091\ \text{m}^3/\text{s}\ (=10446\ \text{m}^3/\text{day})$$ $$\text{Flow velocity}=\frac{Q}{B\,H}=\frac{0.12091}{21}=0.00576\ \text{m/s}=0.5757\ \text{cm/s} =0.345\ \text{m/min}$$

This is slightly above the usual guideline of about 0.3 m/min for avoiding scour of settled solids. The question only asks for the value, but in practice the tank would be made wider or deeper to reduce the velocity.

Step 4: Check for 100 % removal of 0.040 mm particles

$d=0.004$ cm:

$$v_s=\frac{981\times1.65\times0.004^2}{18\times0.01007}=0.1429\ \text{cm/s}\quad(Re=0.057<1)$$

$v_s=0.1429$ cm/s $>v_0=0.1151$ cm/s ($v_s/v_0=1.24>1$), so these particles are removed completely.

Answer: overflow rate = 0.1151 cm/s (99.5 m³/m²/day); flow velocity = 0.5757 cm/s (0.345 m/min); yes, the tank is sufficient to remove 100 % of 0.040 mm particles.