Question

A plain sedimentation tank with a length of $20m$, width of $10m$, and a depth of $3m$ is used in a water treatment plant to treat 4 million litre of water per day (4 MLD). The average temperature of water is ${20}^{\circ }C$ .
The dynamic viscosity of water is $1.002\times {10}^{-3}Ns/m^2$ at ${20}^{\circ }C$.
Density of water is 998.2 $kg/m^3$.
Average specific gravity of particles is $2.65$.
What is the minimum diameter of the particles which can be removed with 100% efficiency in the above sedimentation tank?

• A. $16.0\times {10}^{-3}mm$
• B. $11.8\times {10}^{-3}mm$
• C. $160\times {10}^{-3}mm$
• D. $50\times {10}^{-3}mm$

Answer 1

The correct option is A $16.0\times {10}^{-3}mm$

The particles whose settling velocity $\left(v_1\right)\ge$ SOR will be removed with 100% efficiency.

For minimum diameter of particle which can be removed,

Settling velocity = surface overflow rate

$V_t=\frac{g{d}^2}{18\mu }({\rho }_s-{\rho }_w)$

$\frac{20}{86400}\frac{m}{s}=\frac{9.81d^2(2.65\times 998.2-998.2)}{18\times 1.002\times {10}^{-3}}$

$d=10.75\times {10}^{-6}m$

$\simeq 16\times {10}^{-3}mm$