Question

A settling tank in a water treatment plant is designed for a surface overflow rate of $30m^3/day.m^2$. Assume specific gravity of sediment particles $=2.65$, density of water ( p ) = $1000kg/m^3$, dynamic viscosity of water $\left(\mu \right)=0.001N.s/m^2$, and stoke's law is valid. The approximate minimum size of particles that would be completely removed is:

• A. $0.02mm$
• B. $0.04mm$
• C. $0.03mm$
• D. $0.01mm$

Answer 1

The correct option is A $0.02mm$

Surface over flow rate

$V=30m^3/day–m^2$

$=\frac{30m}{86400sec}$

$=3.47\times {10}^{-4}{m}^3/m^{2}sec$

Particles having settling velocity greater than or equal to over flow rate will be completely removed.

Assuming , stoke's law to be valid

$\Rightarrow V\le \frac{g{d}^2({\rho }_s-{\rho }_w)}{18\mu }$

$\Rightarrow d\ge \sqrt{\frac{18\mu V}{g({\rho }_s-{\rho }_w)}}$

$\ge \sqrt{\frac{18\times 0.001\times 3.47\times {10}^{-4}}{9.81\times (2.65-1)\times 1000}}$

$\ge 1.964\times {10}^{-5}m$
$\ge 1.964\times {10}^{-2}mm$
$d\ge 0.01964=0.02m$