The conventional activated sludge process requires an aeration tank detention time of __ to __ hours, and a F/M ratio of __ to __.

• A. 2 to 4 hours; 0.05 to 0.15
• B. 6 to 8 hours; 0.8 to 1.2
• C. 4 to 6 hours; 0.2 to 0.5
• D. 8 to 12 hours; 0.3 to 0.6

Answer: (C)

The operating ranges reflect how long the microbes need to treat the waste and how much food is available to them. For a conventional activated sludge aeration basin, about 4 to 6 hours of detention time gives enough contact between the wastewater and the active biomass to achieve substantial biodegradation without making the reactor impractically large. If the time is shorter, treatment efficiency drops because the microbes don’t get enough time to break down the organics; if it’s much longer, you’re often increasing reactor size and may not gain additional treatment performance for typical municipal wastewaters.

The food-to-microorganism (F/M) ratio describes how much substrate is available per unit mass of active biomass. A moderate F/M of roughly 0.2 to 0.5 per day indicates a balanced supply of organics that sustains microbial activity and stable biomass levels, while avoiding excessive sludge production. Values higher than this lead to faster growth and more sludge; values lower can starve the biomass and reduce treatment performance.

So the conventional ranges are about 4 to 6 hours for aeration-tank detention time and 0.2 to 0.5 per day for F/M, which is why that option is the best fit.