Activated Sludge Process a. Derive the formula for the following. - F/M ratio - Hydraulic Retention Time - Sludge Retention Time 4 Sludge Volume Index - MLVSS/MLSS ratio b. If the MLSS concentration is 1,500mg/L and the aeration tank volume is 2,400m3, how many pounds of suspended solids are in the aeration tank? c.' The acration tank contains 4,000mg/ - of MLSS. Lab tests indicate the MLSS is 80% volatile matter. What is the MLVSS concentration in the aeration tank? d. A wastewater treatment plant has 126g/L of MLVSS in an aeration tank with a volume of 633m3. If the primary effluent has a BOD of 180mg/L and a flowrate of 5MGD, what is the F/M ratio of the tank? e. A plant has a flow of 850,000 gallons/d. The BOD of the wastewater is 200mg/L. You want to achieve a F/M ratio of 0.6 for optimum performance. Determine the quantity of microbes needed per day. f. A sludge sample is tested and it is determined that the settled sludge volume is 420mL/L. and the MLSS is 2850mg/L. What is the SVI? What does this indicate about the treatment plant