Question

An activated sludge plant receive 5.0 MGD of wastewater with a BOD of 220 mg/L. The primary clarifier removes 35% of the BOD. The sludge is aerated for 6 hr. The food-to-microorganism ratio is 0.30. The recirculation ratio is 0.2. The surface loading rate of the secondary clarifier is 800 gal/day-ft2. The final effluent has a BOD of 15 mg/L. What are the (a) BOD removal efficiency of the activated sludge treatment processes (secondary BOD removal), (b) aeration tank volume, (c) MLSS, and (d) secondary clarifier surface area? If 0.5 pound of oxygen is required for each pound of BOD entering the aeration tank and the density of air is approximately 0.075 lb/ft3, and the air is 20.9% oxygen by volume, calculate air requirements per day.

Answer 1

Air required = 172403.792 ft3

since, density of air = 0.075lb/ft3

air required = 0.075*172403.792 lb

= 12930.284 lb

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