In: Civil Engineering
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.
1. Draw a schematic of the activated sludge process and insert the different values given in the question.
What is the average RAS flowrate? (answer will be given in the unit of !!"#$" ?? ???) %&'
What is the average aeration process hydraulic detention time? Remember to include the RAS flowrate in your calculation (answer will be given in the unit of (hr))
What is the aeration process average BOD5 loading rate? Remember that the RAS data is
not included in this calculation (answer will be given in the unit of ! $(*+, ")
-... /0!1/ 23&4012 . %&'
What is the total mass of biomass present in the aeration tanks? (answer will be given in the unit of (lb TVSS))
6. What is the activated sludge food to microorganism (F/M) ratio? (answer will be given in the unit of ! $( *+, ") )
67 89:;; . %&'
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))
Based on the ”general loading and operational parameters for activated sludge process” table below, which activated process is the plant operating? (answer will be one of the 5 processes listed in the table)
10. Knowing the mass of WAS TSS removed and the WAS TSS concentration, what is the daily volume of WAS removed from the system? (answer will be given in the unit of (gallon/day))
Will the sludge age increase if the volume and mass of WAS removed from the system increase? (answer will be given as yes or no)
Will the type of bacteria present in the activated sludge and its performance likely to change as the sludge age is increased or decreased? (answer will be given as yes or no)