Tennessee Wastewater Operator Practice Exam

In a conventional activated sludge system that focuses solely on BOD removal, the F/M ratio typically falls within which range?

1. 0.10-0.20

2. 0.20-0.40

3. 0.40-0.60

4. 0.60-0.80

The correct answer is: 0.20-0.40

In a conventional activated sludge system aimed primarily at biochemical oxygen demand (BOD) removal, the food-to-microorganism (F/M) ratio is a critical parameter that impacts the efficiency of the biological treatment process. The F/M ratio represents the amount of organic matter (food) available for the microorganisms present in the system per unit of biomass. The range of 0.20 to 0.40 is considered optimal for effective BOD removal in such systems. When the F/M ratio falls within this range, it indicates a favorable balance where there is enough organic material to sustain microbial growth without overwhelming the system. This balance allows the microorganisms to efficiently metabolize the organic matter, leading to effective BOD reduction while minimizing the production of excess biomass. If the ratio were lower than 0.20, it might result in insufficient food for the microorganisms, jeopardizing their ability to remove BOD effectively. Conversely, if the F/M ratio were higher than 0.40, it could lead to rapid growth of microorganisms, resulting in a high volume of excess activated sludge and potentially poorer effluent quality due to washout of cells and other undesirable consequences. Thus, the range of 0.20 to 0.40 ensures both operational stability