Continuous flow sedimentation tank particle size

In an ideal continuous-flow (horizontal) sedimentation tank, a particle is removed if it settles the full depth $H$ before the water travels the length $L$. So the particle removed 100% has a settling velocity $v_s = v_0 = H/t_d$ with $t_d = L/v_h$.

Step 1: Detention time and critical settling velocity

$$\begin{aligned} t_d &= \frac{L}{v_h} = \frac{60}{0.012} = 5000\ \text{s}\\ v_s &= \frac{H}{t_d} = \frac{4}{5000} = 0.00080\ \text{m/s} = 0.080\ \text{cm/s} \end{aligned}$$

Step 2: Size of the particle (Stokes' law)

$$v_s = \frac{g(G-1)d^2}{18\nu}\ \Rightarrow\ d=\sqrt{\frac{18\,\nu\,v_s}{g(G-1)}}=\sqrt{\frac{18\times 0.01\times 0.080}{981\times(2.65-1)}} = 0.00298\ \text{cm} = 0.030\ \text{mm}$$

Check: $Re = v_s d/\nu = 0.024 < 1$, so Stokes' law is valid.

Step 3: Effect of halving the particle size

Since $v_s\propto d^2$, a particle of half the size has $v_s' = v_s/4 = 0.0200$ cm/s. In an ideal tank the fraction of particles of a given size that is removed is $v_s'/v_0$:

$$\text{Removal} = \frac{v_s'}{v_0} = \frac{0.000200}{0.00080} = 25\%$$

The particles of the original size are removed 100%, so the removal of the half-size particles falls from 100% to 25%, i.e. a decrease of 75%.

Answer: the particle size effectively (100%) removed is about 0.030 mm; if the size is halved, only 25% are removed (a 75% reduction).