Three parallel flow flocculation basins are to be used to treat a flow rate of 3 MGD. What is the design volume, in cubic feet, of one of these tanks if the detention time is 30 minutes

2800ft³

Explanation:

The formula used to calculate the volume of one tank is:

Vt = Q x t

We have Q as

Q = (3x10⁶gal/d)/3 tanks

Q = 1x10⁶gals/d per tank

To get volume of one tank

[(30min)x(1x10⁶)x(0.1337)]/1440 min

= 30x1000000x0.1337/1440

= 4011000/1440

= 2,785ft³ or 2800ft³

In conclusion, 2800ft³ is the design volume, in cubic feet, of one of these tanks with detention time as 30 minutes

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