Question

A water treatment plant is designed to remove 12.0 μm diameter particles with a density of 1.2 g/mL. The water is at 10°C. The plant takes in water at a rate of 0.10 m3 /s. The rectangular sedimentation tank has a depth of 4 m and a length: width ratio of 5:1. Calculate the basin’s minimum width (in m) required to remove the particles.

Answer 1

Volume of tank$= 4 × 3.33 × 0.66 = 8.79 m^3$

Minimum width = $\frac{1 }{6}×4=\frac{2}{3}m$