What is the F/M ratio on an aeration tank if 1,500 pounds of BOD are added per day and 5,000 pounds of solids are under aeration?

The F/M ratio, or Food to Microorganism ratio, is calculated by dividing the mass of biochemical oxygen demand (BOD) entering the aeration tank by the mass of mixed liquor suspended solids (MLSS) in the tank. In this case, you have 1,500 pounds of BOD added per day and 5,000 pounds of solids under aeration.

To find the F/M ratio, you can use the formula:

F/M ratio = BOD added per day (in pounds) / MLSS (in pounds).

Substituting the values into the equation:

F/M ratio = 1,500 pounds BOD / 5,000 pounds MLSS = 0.3.

This indicates that there are 0.3 pounds of BOD provided per pound of solids. A value of 0.3 is within an acceptable range for many activated sludge systems, as it suggests a balance between the substrate available for the microorganisms and the biomass present, which is critical for effective treatment in aerobic processes.

The other options represent values that would imply different ratios of food to microorganisms, which would affect the efficiency and stability of the treatment process differently. Therefore, the calculated F/M ratio of 0.3 is