Activated Sludge Design. A Wastewater Treatment Plant Uses Activated Sludge For Secondary Treatment To Process 0.3 M3/S Of Primary Effluent With A BOD5 So = 220 Mg/L. The Mixed Liquor Has A Concentration Of X = 2,100 Mg VSS/L, And The Recycled Activated Sludge, RAS = Xr = 10,000 Mg VSS/L. The F/M Ratio = 0.52 Mg BOD5 Mg VSS-4 D-?, And The Mean Cell

 

5. Activated Sludge Design. A wastewater treatment plant uses activated sludge for secondary treatment to process 0.3 m/s of primary effluent with a BOD5 = So = 220 mg/L. The mixed liquor has a concentration of X = 2,100 mg VSS/L, and the recycled activated sludge, RAS = X = 10,000 mg VSS/L. The F/M ratio = 0.52 mg BOD5 mg VSS-1 d1, and the mean cell residence time, Oc = 9 days. Draw a labeled schematic of the activated sludge tank and secondary clarifier, and then answer the following: a. What is the required volume of the activated sludge tank? V = 5222 m b. What is the waste activated sludge (WAS) flow rate? (Hint: assume Xe = 0.) Qw = 121.8 m/d 1 c. What is the flow rate of the secondary treated effluent? Qe = 0.3 m/s d. What is the aeration period (i.e., hydraulic retention time) of the activated sludge tank? (Hint: you will need to perform a mass balance around the secondary clarifier, assuming Xe = 0.) e = 3.8 hr