Question

A small town produces 0.15 MGD of wastewater. How many acres in size would an aerated lagoon need to be to treat this wastewater assuming a depth of 8 feet and a detention time of 6 days? (7.48 gal/ft3; 43,560 ft2/acre)

Answer #1

Theory: We need to determine the total voume of water for the complete detention time. Aerate lagoon must have this much of volume.

SOlution: Water flow rate = 0.15 MGD = milliongallon /day

1 million = 10^{6}

water mass flow rate = 0.15x10^{6} gallon/day.

Water volume flow rate = mass flow rate /density =
0.15x10^{6}/7.48 = 20053.48 ft3/day.

Total volume of loagoon needed = water volumetric flow rate* detention time

=200053.48*6 = 120320.86 ft3

Total surface area of loagoon needed = volume/depth = 120320.86/8 = 15040.11 ft2

as given 1 arcre = 43560 ft2 or 1 ft2 = 1/43560

Total surface area of loagoon needed = 15040.11/43560 = 0.345 acre.

I hope i was able to clear your doubt. Let me know if you still have any query. If you don't then kindly rate me.

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