Question

Surface overflow rate (SOR) of a primary settling tank (discrete settling) is $20000litre/m^2$ per day. Kinematic viscosity of water in the tank is $1.01\times {10}^{-2}c{m}^2/s$. Specific gravity of the settling particles is 2.64. Acceleration due to gravity is $9.81m/s^2$. The minimum diameter (in $\mu m$, round off to one decimal place) of the particles that will be removed with 80% efficiency in the tank is

• A. 14.4

Answer 1

The correct option is A 14.4
Given,

For primary settling tank (discrete settling)

Surface overflow rate (SOR)

$=20000lt/m^{2}perday$

$=\left(\frac{20000\times {10}^{-3}{m}^3}{86400sec{m}^2}\right)$

$=2.315\times {10}^{-4}m/se{c}^{-4}$

Kinematic viscosity $\left(v\right)=1.01\times {10}^{-2}c{m}^2/sec$

$=1.01\times {10}^{-6}{m}^2/sec$

Specific gravity of settling particle $\left(G_s\right)=2.64$

Gravitational acceleration (g) = $9.81m/se{c}^2$

Efficiency $\left(\eta \right)=\left[\frac{V}{SOR}\right]$

$\Rightarrow V=(\eta \times SOR)$

$=(0.8\times 2.315\times {10}^{-4})m/sec$

$=1.852\times {10}^{-4}m/sec$

As per stoke's law
$V=\frac{(G_s-1)g{d}^2}{18v}$

d= diameter of particle in metre

$\Rightarrow 1.852\times {10}^{-4}=\frac{(2.64-1)\times 9.81\times d^2}{18\times 1\times {10}^{-6}}$

$\Rightarrow d=1.4395\times {10}^{-5}m=14.395\mu m$