Question

Find the diameter of particle in "mm" that will be removed 80% from water from a sedimentation tank of size $20m\times 10m\times 3m$ designed to treat 5 MLD of water. Take specific gravity of particle 2.65 and kinematic viscosity of water is ${10}^{-2}c{m}^2/sec$.

• A. 0.16
• B. 0.2
• C. 0.02
• D. 0.016

Answer 1

The correct option is D 0.016

$V_0=\frac{Q}{A_s}=\frac{5\times {10}^6/{10}^3\times 24\times 60\times 60}{20\times 10}$

$V_s=\eta \times V_0=\frac{80}{100}\times 1000$

$V_s=\frac{g[S-1]d^2}{18\nu }$

$\Rightarrow \frac{80}{100}\times \frac{5\times {10}^6/{10}^3\times 24\times 60\times 60}{20\times 10}=\frac{9.81\times [265-1]}{18\times {10}^{-2}\times \frac{1}{(100{)}^2}}d=0.016mm$