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An average data for conventional activated sludge treatment plant is as follows:Wastewater flow = 35000 m3/dVolume of aeration tank = 10500 m3Influent BOD = 230 mg/LEffluent BOD = 20 mg/LMixed Liquor suspended solids (MLSS) = 2200 mg/LEffluent suspended solids = 30 mg/LWaste sludge suspended solids = 9500 mg/LQuantity of waste sludge = 200 m3/dayAssume MLVSS = 75% MLSSBased on the above information, the oxygen requirement of the aeration tank is _________ tonnes/day.a)7.22b)8.11c)6.02d)4.63Correct answer is option 'B'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An average data for conventional activated sludge treatment plant is as follows:Wastewater flow = 35000 m3/dVolume of aeration tank = 10500 m3Influent BOD = 230 mg/LEffluent BOD = 20 mg/LMixed Liquor suspended solids (MLSS) = 2200 mg/LEffluent suspended solids = 30 mg/LWaste sludge suspended solids = 9500 mg/LQuantity of waste sludge = 200 m3/dayAssume MLVSS = 75% MLSSBased on the above information, the oxygen requirement of the aeration tank is _________ tonnes/day.a)7.22b)8.11c)6.02d)4.63Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An average data for conventional activated sludge treatment plant is as follows:Wastewater flow = 35000 m3/dVolume of aeration tank = 10500 m3Influent BOD = 230 mg/LEffluent BOD = 20 mg/LMixed Liquor suspended solids (MLSS) = 2200 mg/LEffluent suspended solids = 30 mg/LWaste sludge suspended solids = 9500 mg/LQuantity of waste sludge = 200 m3/dayAssume MLVSS = 75% MLSSBased on the above information, the oxygen requirement of the aeration tank is _________ tonnes/day.a)7.22b)8.11c)6.02d)4.63Correct answer is option 'B'. Can you explain this answer?.

Solutions for An average data for conventional activated sludge treatment plant is as follows:Wastewater flow = 35000 m3/dVolume of aeration tank = 10500 m3Influent BOD = 230 mg/LEffluent BOD = 20 mg/LMixed Liquor suspended solids (MLSS) = 2200 mg/LEffluent suspended solids = 30 mg/LWaste sludge suspended solids = 9500 mg/LQuantity of waste sludge = 200 m3/dayAssume MLVSS = 75% MLSSBased on the above information, the oxygen requirement of the aeration tank is _________ tonnes/day.a)7.22b)8.11c)6.02d)4.63Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE. Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free.

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