Settling tank 20 m long, 3.5 m deep; 0.02 and 0.01 mm

A continuous-flow ideal settling tank removes a particle completely if it settles through the full depth $H$ during the detention time $t$, i.e. $v_s \ge H/t$ (equivalently $v_s\ge$ overflow rate $v_0$). So the detention time for 100 % removal is $t = H/v_s$.

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

Take $S_s = 2.65$, $\nu_{25} = 0.00893$ cm²/s, $d=0.002$ cm, $H = 3.5$ m $=350$ cm.

$$v_s=\frac{g(S_s-1)d^2}{18\nu}=\frac{981\times1.65\times0.002^2}{18\times0.00893}=0.0403\ \text{cm/s}$$

$Re = v_sd/\nu = 0.009<1$ (Stokes' law valid).

$$t=\frac{H}{v_s}=\frac{350}{0.0403}=8689\ \text{s} = 2.41\ \text{h}$$

Recommended detention time $\approx 2.4$ h. (Horizontal velocity $=L/t = 2000/8689 = 0.230$ cm/s, well below the scour velocity.)

(b) Removal of 0.01 mm particles at 15 °C in the same tank

The tank has overflow rate $v_0 = H/t=0.0403$ cm/s. At 15 °C, $\nu=0.01139$ cm²/s, $d=0.001$ cm:

$$v_s=\frac{981\times1.65\times0.001^2}{18\times0.01139}=0.00790\ \text{cm/s}\quad (Re=0.0007<1)$$ $$\text{Removal}=\frac{v_s}{v_0}=\frac{0.00790}{0.0403}=0.196\Rightarrow 19.6\ \%$$

Answer: detention time = 2.41 h; removal of 0.01 mm particles at 15 °C = 19.6 %.