9. The following data are given for an activated sludge process: BOD5 (influent) = 250 mg/L...

9. The following data are given for an activated sludge process: BOD5 (influent) = 250 mg/L (soluble) BOD5 (effluent) = 20 mg/L (soluble) MCRT = 10 days MLSS in the aeration tank = 2500 mg/L Concentration of SS in the recycle line = 10,000 mg/L Concentration of SS in the effluent = 30 mg/L (70% biodegradable) BOD5/BODu = 0.75, No nutrient limitations. Yield co-efficient, Y = 0.65 Endogenouse decay coefficient, kd = 0.06/day , Q=5MGD

Determine:
(a) Find soluble BOD5 (in mg/L) in the effluent, BOD5 treatment efficiency (in %) and overall efficiency (in %)
(b) Find hydraulic detention time (in hours) and the volume of tank (in MG).
(c) Yobs and biomass production (in lb VSS/day) (d) Qw (in MGD) when the wasting from the AT and from recycle line. (e) Recirculation ratio (f) Find Specific Utilization rate (U) and food to microorganism (F/M) ratio (g) Calculate volumetric loading rate (lb BOD5 /1000 ft3)

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Ally Wells answered 8 months ago

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