Old tank 11 m x 5 m x 3 m as settling tank

In an ideal settling tank, particles with $v_s\ge v_0$ are fully removed, and those with $v_s<v_0$ are removed in the ratio $v_s/v_0$, where $v_0=Q/A$ is the overflow rate.

Step 1: Settling velocity of the 0.025 mm particles

$d=0.0025$ cm, $S_s=2.65$, $\nu_{20}=0.01007$ cm²/s.

$$v_s=\frac{981\times1.65\times0.0025^2}{18\times0.01007}=0.0558\ \text{cm/s}\quad(Re=0.014<1)$$

Step 2: Overflow rate

For 93 % removal:

$$v_0=\frac{v_s}{0.93}=\frac{0.0558}{0.93}=0.0600\ \text{cm/s}=51.9\ \text{m}^3/\text{m}^2/\text{day}\ (=2.16\ \text{m}^3/\text{m}^2/\text{h})$$

Surface area $A=11\times5=55$ m², so the tank can treat

$$Q=v_0A=0.000600\times55=0.03301\ \text{m}^3/\text{s}=2852\ \text{m}^3/\text{day}$$

Horizontal velocity $=Q/(B\,H)=0.03301/(5\times3)=0.2200$ cm/s $=0.132$ m/min.

Step 3: Check for 99 % removal of 0.05 mm particles

$d=0.005$ cm:

$$v_s=\frac{981\times1.65\times0.005^2}{18\times0.01007}=0.2232\ \text{cm/s}\quad(Re=0.111<1)$$

$v_s=0.2232$ cm/s is greater than $v_0=0.0600$ cm/s ($v_s/v_0=3.72$), so these particles are removed 100 %, which exceeds 99 %.

Answer: overflow rate = 0.0600 cm/s (51.9 m³/m²/day); yes, the tank is sufficient to remove 99 % of 0.05 mm particles (it removes 100 %).