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A conventional activated sludge treatment plant has to treat wastewater of 2250000 litres/day, having a five day BOD of 225 mg/l. The mixed liquor suspended solids concentration in the aeration tank is 2250 mg/l and concentration of biomass leaving the system is 22.5 g/m3. If the volume of the aeration tank is 225 m3, then the average time (in days) for which the biomass stays in the system will be
    Correct answer is '10'. Can you explain this answer?
    Verified Answer
    A conventional activated sludge treatment plant has to treat wastewate...
    The average time for which the biomass stays in the system is called sludge age

    where V = volume of the tank
    X = mixed liquor suspended solids concentration in the aeration tank
    Qw = flow of wastewater
    Xu = Biomass leaving the aeration tank per day
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    Most Upvoted Answer
    A conventional activated sludge treatment plant has to treat wastewate...
    Given information:
    - Wastewater flow rate = 2250000 litres/day
    - Five-day BOD (Biochemical Oxygen Demand) = 225 mg/l
    - Mixed liquor suspended solids (MLSS) concentration = 2250 mg/l
    - Biomass leaving the system concentration = 22.5 g/m3
    - Aeration tank volume = 225 m3

    Calculating the biomass:
    To calculate the biomass, we need to determine the mass of MLSS in the aeration tank. The volume of MLSS in the aeration tank can be calculated as follows:

    Volume of MLSS = MLSS concentration * Aeration tank volume
    = 2250 mg/l * 225 m3
    = 506,250 g

    Converting the mass to kg:
    Volume of MLSS = 506,250 g / 1000
    = 506.25 kg

    Calculating the biomass leaving the system:
    The concentration of biomass leaving the system is given as 22.5 g/m3. To calculate the mass of biomass leaving the system, we need to determine the volume of MLSS leaving the system.

    Volume of MLSS leaving the system = Biomass leaving the system concentration * Aeration tank volume
    = 22.5 g/m3 * 225 m3
    = 5062.5 g

    Converting the mass to kg:
    Volume of MLSS leaving the system = 5062.5 g / 1000
    = 5.0625 kg

    Calculating the average time for which the biomass stays in the system:
    The average time for which the biomass stays in the system can be calculated using the formula:

    Average time = Volume of MLSS / Volume of MLSS leaving the system
    = 506.25 kg / 5.0625 kg
    = 100 days

    Therefore, the average time for which the biomass stays in the system is 100 days, which is equivalent to 10 days (as per the answer options).
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    A conventional activated sludge treatment plant has to treat wastewater of 2250000 litres/day, having a five day BOD of 225 mg/l. The mixed liquor suspended solids concentration in the aeration tank is 2250 mg/l and concentration of biomass leaving the system is 22.5 g/m3. If the volume of the aeration tank is 225 m3, then the average time (in days) for which the biomass stays in the system will beCorrect answer is '10'. Can you explain this answer?
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    A conventional activated sludge treatment plant has to treat wastewater of 2250000 litres/day, having a five day BOD of 225 mg/l. The mixed liquor suspended solids concentration in the aeration tank is 2250 mg/l and concentration of biomass leaving the system is 22.5 g/m3. If the volume of the aeration tank is 225 m3, then the average time (in days) for which the biomass stays in the system will beCorrect answer is '10'. 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 A conventional activated sludge treatment plant has to treat wastewater of 2250000 litres/day, having a five day BOD of 225 mg/l. The mixed liquor suspended solids concentration in the aeration tank is 2250 mg/l and concentration of biomass leaving the system is 22.5 g/m3. If the volume of the aeration tank is 225 m3, then the average time (in days) for which the biomass stays in the system will beCorrect answer is '10'. 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 A conventional activated sludge treatment plant has to treat wastewater of 2250000 litres/day, having a five day BOD of 225 mg/l. The mixed liquor suspended solids concentration in the aeration tank is 2250 mg/l and concentration of biomass leaving the system is 22.5 g/m3. If the volume of the aeration tank is 225 m3, then the average time (in days) for which the biomass stays in the system will beCorrect answer is '10'. Can you explain this answer?.
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