Civil Engineering > GATE 2026 SET-1 > Sewage Treatment
An Activated Sludge Process (ASP) has an inlet wastewater flowrate of 20000 m3/day with a Biochemical Oxygen Demand (BOD) concentration of 250 mg/l. It produces treated wastewater containing 20 mg/l BOD. The aeration tank has a working volume of 6000 m3 and a biomass concentration of 3000 mg/l. Biological Sludge Residence Time (BSRT) of the system is 6 days. The influent wastewater and the treated effluent from the system have negligible concentrations of biomass. The sludge recycle line from the bottom of the Secondary Sedimentation Tank (SST) to the inlet of the aeration tank has a flowrate of 6000 m3/day.
To maintain equilibrium, the flowrate (in m3/day) of sludge that is to be wasted from the system is ______ (in integer).
To maintain equilibrium, the flowrate (in m3/day) of sludge that is to be wasted from the system is ______ (in integer).
Correct : 1000
The Biological Sludge Residence Time (BSRT) is defined as the total mass of biomass in the system divided by the mass of biomass wasted per day.
BSRT = (V × X) / (Qw × Xw)
When waste sludge is drawn directly from the aeration tank, the waste sludge concentration Xw equals the aeration tank biomass concentration X = 3000 mg/L. The BSRT formula simplifies to:
BSRT = V / Qw
Substituting the known values:
6 = 6000 / Qw
Qw = 6000 / 6 = 1000 m³/day
This is the flowrate of sludge that must be wasted daily to maintain the system at equilibrium with BSRT = 6 days, aeration tank volume = 6000 m³, and biomass concentration = 3000 mg/L.
Correct answer: 1000 m³/day ✓
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