A water treatment plant treats 6000 m3 of water per day. As a part of the treatment process, discrete particles are required to be settled

Q. A water treatment plant treats 6000 m³ 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 1.5 m per hour would produce the desired removal of particles through settling in the clarifier having a depth of 3.0 m. The volume of the required clarifier, (in m³, round off to 1 decimal place) would be    

Solution:

Flow rate = 6000 m3 / d a y

Over flow rate = 1.5 m/hour = 1.5 × 24 m / d a y = 36 m / d a y

Flow area = Flow rate/over flow rate

= 6000/36

= 166.67 m 2 Volume required for clarifier

= flow area × depth

= 166.67 × 3

= 500 m 3