1. If 0.75 MGD was treated with a 15% chlorine solution and a dose of 10 ppm was desired, how many gallons of solution would be required per day?

2. How many pounds of BOD were applied to the surface area of a trickling filter if flow = 1.5 MGD and the BOD concentration were 200 mg/L if the filter measures 50 ft in diameter?
Lbs BOD/day/sqft = (lbs BOD/day) / (Sq ft)
or Lbs BOD/day/1000cuft = (lbs BOD/day) / (Cuft / 1000)

3. Determine organic loading on a trickling filter if flow were 1.5 MGD and the BOD concentration were 200 mg/L if the filter measures 50 ft in diameter and 10 feet deep?
Lbs BOD/day/sqft = (lbs BOD/day) / (Sq Ft)
or Lbs BOD/day/1000 cuft / = (lbs BOD/day) / (Cuft / 1000)

4. If the influent to an aeration basin was 3 MGD and had a BOD concentration of 120 mg/L and the aeration basin held 700,000 gallons with a MLVSS concentration of 1500 mg/L, what is the F/M ratio?
(lbs of food) / (lbs of microorganisms)
or (raw flow, MGD x BOD, mg/L x 8.34 lb/gal) / (Aer. tank vol. MGD x MLVSS, mg/L x 8.34 lb/gal)

5. If an aeration basin held 3 MG with a MLSS concentration of 850 mg/L and the primary clarifier effluent rate of 1.5 MGD had a suspended solids concentration of 205 mg/L, what is the sludge age?
(lbs of BOD entering aeration) / (lbs of MLSS in aeration)

Answers:

1. 0.72 MGD X 10 ppm X 8.34 lbs/gal = 62.55 lbs
(62.55 lbs) / (8.34 lbs/gal) = (7.5 gal) / (0.15) = 50 gal

2. 1.5MGD x 200mg/L x 8.34lb/gal = (2502lb/day)
then 0.785 x 50’ x 50’ = 1962.5 sqft
(2502 lb/day) / (1962.5 sqft) = 1.3 lbs/BOD/day/sqft

3. Use info from above, then 1962.5 sqft x 10 ft = (19625 cuft) / (1,000 cuft) = 19.6 cuft then (2502 lb/day) / (19.6 cuft) = 127.7 lbs BOD/day/1000 cuft

4. 3 mgd X 120 mg/L X 8.34 lb/gal = 360 lbs
0.7 mg X 1500 mg/L X 8.34 lb/gal = 1050 lbs
(360 lbs) / (1050 lbs) = 0.34

5. (1.5 mgd) x (205 mg/L) x (8.34lb /gal) = 307.5 lbs
(3 mg) x (850 mg/L) x (8.34 lb/gal) = 2550 lbs
(307.5 lbs) / (2550 lbs) = 0.12 days