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Common Data for Questions 60 and 61:An activated sludge system (sketched below) is operating at equilibrium with the following information. Wastewater related data: flow rate = 500 m3/hour, influent BOD = 150 mg/L, effluent BOD = 10 mg/L. Aeration tank related data: hydraulic retention time = 8 hours, mean-cell-residence time = 240 hours, volume = 4000 m3, mixed liquor suspended solids = 2000 mg/L.The food-to-biomass (F/M) ratio (in kg BOD per kg biomass per day) for the aeration tank isa)0.015b)0.210c)0.225d)0.240Correct answer is option 'A'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Common Data for Questions 60 and 61:An activated sludge system (sketched below) is operating at equilibrium with the following information. Wastewater related data: flow rate = 500 m3/hour, influent BOD = 150 mg/L, effluent BOD = 10 mg/L. Aeration tank related data: hydraulic retention time = 8 hours, mean-cell-residence time = 240 hours, volume = 4000 m3, mixed liquor suspended solids = 2000 mg/L.The food-to-biomass (F/M) ratio (in kg BOD per kg biomass per day) for the aeration tank isa)0.015b)0.210c)0.225d)0.240Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Common Data for Questions 60 and 61:An activated sludge system (sketched below) is operating at equilibrium with the following information. Wastewater related data: flow rate = 500 m3/hour, influent BOD = 150 mg/L, effluent BOD = 10 mg/L. Aeration tank related data: hydraulic retention time = 8 hours, mean-cell-residence time = 240 hours, volume = 4000 m3, mixed liquor suspended solids = 2000 mg/L.The food-to-biomass (F/M) ratio (in kg BOD per kg biomass per day) for the aeration tank isa)0.015b)0.210c)0.225d)0.240Correct answer is option 'A'. Can you explain this answer?.

Solutions for Common Data for Questions 60 and 61:An activated sludge system (sketched below) is operating at equilibrium with the following information. Wastewater related data: flow rate = 500 m3/hour, influent BOD = 150 mg/L, effluent BOD = 10 mg/L. Aeration tank related data: hydraulic retention time = 8 hours, mean-cell-residence time = 240 hours, volume = 4000 m3, mixed liquor suspended solids = 2000 mg/L.The food-to-biomass (F/M) ratio (in kg BOD per kg biomass per day) for the aeration tank isa)0.015b)0.210c)0.225d)0.240Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Civil Engineering (CE). Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free.

Here you can find the meaning of Common Data for Questions 60 and 61:An activated sludge system (sketched below) is operating at equilibrium with the following information. Wastewater related data: flow rate = 500 m3/hour, influent BOD = 150 mg/L, effluent BOD = 10 mg/L. Aeration tank related data: hydraulic retention time = 8 hours, mean-cell-residence time = 240 hours, volume = 4000 m3, mixed liquor suspended solids = 2000 mg/L.The food-to-biomass (F/M) ratio (in kg BOD per kg biomass per day) for the aeration tank isa)0.015b)0.210c)0.225d)0.240Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Common Data for Questions 60 and 61:An activated sludge system (sketched below) is operating at equilibrium with the following information. Wastewater related data: flow rate = 500 m3/hour, influent BOD = 150 mg/L, effluent BOD = 10 mg/L. Aeration tank related data: hydraulic retention time = 8 hours, mean-cell-residence time = 240 hours, volume = 4000 m3, mixed liquor suspended solids = 2000 mg/L.The food-to-biomass (F/M) ratio (in kg BOD per kg biomass per day) for the aeration tank isa)0.015b)0.210c)0.225d)0.240Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Common Data for Questions 60 and 61:An activated sludge system (sketched below) is operating at equilibrium with the following information. Wastewater related data: flow rate = 500 m3/hour, influent BOD = 150 mg/L, effluent BOD = 10 mg/L. Aeration tank related data: hydraulic retention time = 8 hours, mean-cell-residence time = 240 hours, volume = 4000 m3, mixed liquor suspended solids = 2000 mg/L.The food-to-biomass (F/M) ratio (in kg BOD per kg biomass per day) for the aeration tank isa)0.015b)0.210c)0.225d)0.240Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Common Data for Questions 60 and 61:An activated sludge system (sketched below) is operating at equilibrium with the following information. Wastewater related data: flow rate = 500 m3/hour, influent BOD = 150 mg/L, effluent BOD = 10 mg/L. Aeration tank related data: hydraulic retention time = 8 hours, mean-cell-residence time = 240 hours, volume = 4000 m3, mixed liquor suspended solids = 2000 mg/L.The food-to-biomass (F/M) ratio (in kg BOD per kg biomass per day) for the aeration tank isa)0.015b)0.210c)0.225d)0.240Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Common Data for Questions 60 and 61:An activated sludge system (sketched below) is operating at equilibrium with the following information. Wastewater related data: flow rate = 500 m3/hour, influent BOD = 150 mg/L, effluent BOD = 10 mg/L. Aeration tank related data: hydraulic retention time = 8 hours, mean-cell-residence time = 240 hours, volume = 4000 m3, mixed liquor suspended solids = 2000 mg/L.The food-to-biomass (F/M) ratio (in kg BOD per kg biomass per day) for the aeration tank isa)0.015b)0.210c)0.225d)0.240Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Civil Engineering (CE) tests.