1. An aeration basin is 90 ft long, 25 ft wide and 13 feet deep. If it receives a flow of 0.24 MGD at a BOD concentration of 225 mg/L, what is the basin’s organic loading in pounds of BOD per day per 1,000 cubic feet?

2. If an activated sludge aeration basin receives a flow of 0.69 MGD at a BOD concentration of 175 mg/L, how many pounds of BOD enter the aeration basin each day?

3. If the aeration basin from question #2 is 130 feet long, 35 feet wide and 15 feet deep, what is its organic loading in pounds of BOD/day/1,000 cuft?

Answers:

1. 15.4 lbs/day/1000 cuft
Organic Loading = (Lbs of BOD to aeration tank) / (Volume of aeration tank)
Find pounds of BOD to aeration tank: Flow, MGD X Conc, mg/L X 8.34 lbs/gal (Davidson Pie Chart)
0.24 MGD X 225 mg/L X 8.34 lbs/gal = 450.36 lbs/day
Find Volume of aeration tank: L, ft X W, ft X H, ft or 90 ft X 25 ft X 13 ft = 29,250 cuft
(Problem asks for units per 1000 cuft or divide by 1000 or shuffle to the left 3 places) (29,250 cuft) / (1,000 cuft/day) = 29.25 1000 cuft
Work problem (450.36 lbs/day) / (29.25/1000 cuft) =

2. 1007.06 pounds/day
Flow, MGD X Conc, mg/L X 8.34 lbs/gal = 0.69 MGD X 175 mg/L X 8.34 lbs/gal =

3. 14.75 lbs BOD/day/1,000 cuft
0.69 MGD X 175 mg/L X 8.34 lbs/gal = 1007.06 lbs/day
130 ft X 35 ft X 15 ft = (68,250 cuft) / (1,000 cuft) = 68.3/1000 cuft
(1007.06 lbs/day) / (68.3/1000 cuft) =