Question

A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process followed by an activated sludge process composed of aeration tanks and a secondary clarifiers.

This problem focuses on the activated sludge process.

The activated sludge process receives a flow of 15 million gallons per day (MGD).

The primary sedimentation process effluent has a BOD₅ concentration of 180 mgBOD₅/L (this concentration is often referred as S). Before flowing into the activated sludge process, the primary effluent is mixed with the returned activated sludge (RAS) – concentrated biomass - that is continuously pumped out of the bottom of the secondary clarifiers. The returned activated sludge (RAS) ratio is 25%, in other word, an equivalent of 25% of plant flow (15 MGD) is pumped out of the bottom of the secondary clarifiers. The mixture of primary effluent and RAS forms make up the activated sludge, which is also called mixed liquor. The resulting mixed liquor has a total suspended solids (TSS) concentration of 2,500 mgTSS/L and a total volatile suspended solids (TVSS) concentration of 2,125 mgTVSS/L (this biomass concentration is often referred as X).

It flows into aeration thanks that have a total volume of 2.5 million gallons (Mgal) (this volume is often referred as V).

The mixed liquor then flow to secondary clarifiers that have a total volume of 5 million gallons.

Activated that settles at the bottom of the secondary clarifiers is pumped out of the tank. The majority is continuously sent back to the aeration tank as RAS but some is removed from the activated sludge process as waste activated sludge (WAS). The WAS has an average TSS concentration of 10,000 mg/L and a total mass of 10,000 lb of WAS TSS is removed every day.

The final effluent has an average BOD₅ concentration of 12 mgBOD₅/L and an average TSS concentration of 10 mg/L.

What is the activated sludge food to microorganism (F/M) ratio? (answer will be given in the unit of lb BODlb MLVSS . day)

The final effluent TSS is assumed to be composed of biomass that was not captured in the secondary clarification process. Therefore, the mass of final effluent TSS has to be included in the total amount of biomass removed (biomass leaving the system). With that said, what is the daily mass of TSS or biomass leaving the system with the final effluent? (answer will be given in the unit of lb FE TSS)

What is the activated system sludge retention time (also called sludge age or mean cell retention time) when only considering the mass of activated sludge present in the aeration tank? (use the biomass TSS values for this question) (answer will be given in the unit of day)

Answer 1