Question

A water treatment plant treats $6000m^3$ of water per day. As a part of the treatment process, discrete particles are required to be settled in a clarifier. A column test indicates that an overflow rate of $1.5m$ per hour would produce the desired removal of particles through settling in the clarifier having a depth of $3.0m$. The volume of the required clarifier, (in $m^3$, round off to 1 decimal place) would be

• A. 500

Answer 1

The correct option is A 500
Given data :
Flow rate $=6000m^3/day$
Over flow rate = 1.5 m/hour $=1.5\times 24m/day=36m/day$
Flow area = $\frac{\text{Flow rate}}{\text{over flow rate}}=\frac{6000}{36}$
$=166.67m^2$
Volume required for clarifier
= flow area $\times$ depth $=166.67\times 3=500m^3$