1. An influent to an aeration basin averages 1,150 gpm at a BOD concentration of 125 mg/L. If the aeration basin is 125 ft long, 40 ft wide and 15 feet deep, what is its organic loading in pounds of BOD/day/1000 cuft?

2. Calculate the aeration loading in pounds of BOD/day/1,000 cuft for the following plant. Influent flow is 1.46 MGD, Influent BOD is 235 mg/L, primary clarifier BOD removal efficiency is 30 % and there are two aeration basins at 140 feet long, 40 feet wide and 15 feet side water depth each.

3. An activated sludge aeration basin is 80 feet long, 20 feet wide and 12 feet deep. If the flow rate to it is 0.43 MGD at a CBOD concentration of 150 mg/L and the MLVSS concentration is 1,350 mg/L, what is the F/M ratio?

Answers:

1. 23 lbs BOD/day/1,000 cuft
Convert flow to MGD = 1,150 gpm X 1440 gal/day = 1,656,000 gpd (shuffle to get MGD) = 1.66 MGD
Find pounds of BOD to Aeration tank = 1.66 MGD X 125 mg/L X 8.34 lbs/gal = 1720.125 lbs/day
Find Vol of Tank = 125 ft X 40 ft X 15 ft = 75,000 cuft, now shuffle to get 1,000 cuft of tank = 75/1000 cuft
(1720.125 lbs/day) / (75/1,000 cuft) =

2. 11.96 lbs BOD/day/1,000 cuft
Find how much BOD reaches the aertion basin: 30% = 0.3,235 X 0.3=70.5, 235-70 = 165 mg/L
1.46 MGD X 165 mg/L X 8.34 lb/gal = 2009.106 lbs/day
140 ft X 40 ft X 15 ft X 2 = (168,000 cuft) / (1,000 cuft) = 168/1,000 cuft
(2009.106 lbs/cuft) / (168/1,000 cuft) =

3. 0.33 / day
(Lbs of food to aeration tank) / (lbs of solids under aeration) – 2 Davidson Pie Charts
Food to aeration: 0.43 MGD X 150 mg/L X 8.34 lbs/gal =
Aeration tank volume in MGD = 80 ft X 20 ft X 12 ft = 19200 cuft X 7.48 gal/cuft = 143,616 gal (shuffle for MGD) = 0.14 MGD
Find lbs of solids under aeration: 0.14 MGD X 1350 mg/L X 8.34 lbs/gal = 1616.97 lbs
( 537.93 lbs/day) / (1616.97 lbs) =