Question

For a total design flow of 67 MGD and an overflow rate of 1300 gal ft2...

For a total design flow of 67 MGD and an overflow rate of 1300 gal ft2 · day, what is the minimum area of each of 11 sedimentation basins (n - 1 = 10) in parallel? Answer in units of ft2

For a depth of 13 ft, what is the volume of each basin? Answer in units of ft3

What is the design detention time of each basin? Answer in units of hr.

For a L/D ratio of 25, what is the length of each sedimentation basin?

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Homework Answers

Answer #1

1 US gallon = 0.1336 ft3

Total Design flow = Q = 67 MGD

= 67 x 106 x 0.1336 ft3/d

= 8.95 x 106 ft3/d

Surface overflow rate = vs = 1300 gal/ft2/day

= 1300 x .1336 ft3/ft2/day

= 173.7 ft/day

Surface area = At = Q/vs

= (8.95 x 106)/173.7

= 51525.6 ft2

Area of one basin = A = 51525.6/11

= 4684.14 ft2

= 4684 ft2

Volume of basin = V = Axd = 4684x13

= 60892 ft3

Total Volume of all basins = Vt = 11 x 60892 = 669812 ft3

Detention time = t = Vt/Q

= 669812/(8.95x106)

= 0.0748 day

= 0.0748 x24 hr

= 1.8 hrs

L/D = 25

L = 25D = 25x13 = 325 ft

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