What is the Food to Microorganism ratio for an aeration tank with a volume of 0.37 MGD, MLVSS of 2214 mg/L, influent flow of 0.85 MGD, and BOD of 176 mg/L?

Study for the Virginia Wastewater Class 4 Test. Use flashcards and multiple choice questions, each question with hints and explanations. Prepare for success in your exam!

To determine the Food to Microorganism (F:M) ratio for the aeration tank, you need to calculate both the food input (represented as BOD) and the biomass (represented as MLVSS) present in the tank.

First, calculate the influent BOD loading. This is done by multiplying the influent flow rate by the influent BOD concentration:

  • Influent Flow (MGD): 0.85 MGD

  • BOD (mg/L): 176 mg/L

Convert the flow rate from MGD to gallons per day, and then convert that to liters per day to maintain consistent units:

[

0.85 \text{ MGD} = 0.85 \times 8.34 \text{ (gallons per pound)} \times 1000 \text{ (liters)} \approx 3544.58 \text{ L/day}

]

Calculate the daily BOD load:

[

\text{BOD Load} = \text{Flow} \times \text{BOD} = 0.85 \text{ (MGD)} \times 176 \text{ (mg/L)} = 149.6 \

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