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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 mass (in kg/day) of solids wasted from the system isa)24000b)1000c)800d)33Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared
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the Civil Engineering (CE) exam syllabus. Information about 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 mass (in kg/day) of solids wasted from the system isa)24000b)1000c)800d)33Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam.
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Here you can find the meaning of 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 mass (in kg/day) of solids wasted from the system isa)24000b)1000c)800d)33Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
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 mass (in kg/day) of solids wasted from the system isa)24000b)1000c)800d)33Correct answer is option 'C'. Can you explain this answer?, a detailed solution for 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 mass (in kg/day) of solids wasted from the system isa)24000b)1000c)800d)33Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of 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 mass (in kg/day) of solids wasted from the system isa)24000b)1000c)800d)33Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice 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 mass (in kg/day) of solids wasted from the system isa)24000b)1000c)800d)33Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Civil Engineering (CE) tests.