Texas Wastewater Class B Practice Exam

How is the F/M Ratio calculated in the activated sludge process?

1. Multiplying the pounds of Incoming BOD by the pounds of MLVSS

2. Dividing the pounds of Incoming BOD by the pounds of MLVSS in the aeration tank

3. Adding the pounds of Incoming BOD and MLVSS

4. Subtracting the pounds of BOD from the pounds of MLVSS

The correct answer is: Dividing the pounds of Incoming BOD by the pounds of MLVSS in the aeration tank

The correct choice highlights that the F/M (Food to Microorganism) ratio is calculated by dividing the pounds of incoming Biochemical Oxygen Demand (BOD) by the pounds of Mixed Liquor Volatile Suspended Solids (MLVSS) present in the aeration tank. This ratio is crucial in the activated sludge process because it provides insight into the balance between the amount of organic material available for the microorganisms to consume (the food) and the amount of active biomass (the microorganisms) available to process that food. A high F/M ratio indicates that there is more food than microorganisms, which can lead to conditions that may promote washout of the biomass or insufficient treatment. Conversely, a low F/M ratio can signify that there are too many microorganisms for the amount of food available, potentially leading to poor growth and inefficient treatment. Therefore, correct calculation of the F/M ratio is essential for optimizing the operation of the activated sludge process. The other options do not represent the correct method for calculating the F/M ratio, as they either involve incorrect mathematical operations or misinterpret the relationship between BOD and MLVSS.