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Activated Sludge Design. (+8 total) A wastewater treatment plant uses activated sludge forsecondary treatment to process 0.3 m³/s of primary effluent with a BOD5 = S0 = 230 mg/L. The mixedliquor has a concentration of X = 2,150 mg VSS/L, and the recycled activated sludge, RAS = X₁ = 10,000mg VSS/L. The F/M ratio = 0.52 mg BOD5 mg VSS-1 d-¹, and the mean cell residence time, c = 9 days.Draw a labeled schematic of the activated sludge tank and secondary clarifier, and then answer thefollowing: a. What is the required volume of the activated sludge tank? V = 5332 m³ What is the waste activated sludge (WAS) flow rate? (Hint: assume Xe = 0.) . What is the flow rate of the secondary treated effluent? . What is the aeration period (i.e., hydraulic retention time) of the activated sludge tank? (H Qw = 127.4 m³/d? Qe = 0.298 m³/s e = 3.9 hr

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