Given the following, what is the F/M ratio of this activated sludge process? Tank dimensions: 80ft long x 20ft wide x 12ft deep, Average flow rate is 300 gpm, Plant influent BOD is 180 mg/L, Primary effluent BOD is 150 mg/L, MLSS is 1350 mg/L.

Given the following, what is the F/M ratio of this activated sludge process? Tank dimensions: 80ft long x 20ft wide x 12ft deep, Average flow rate is 300 gpm, Plant influent BOD is 180 mg/L, Primary effluent BOD is 150 mg/L, MLSS is 1350 mg/L.

• A. 0.25
• B. 0.33
• C. 0.50
• D. 0.40

Answer: (D)

The F/M ratio is the rate of food (BOD) available to the microorganisms per unit mass of mixed liquor suspended solids (MLSS) in the aeration tank. It’s expressed as pounds of BOD per day per pound of MLSS. First, find the aeration tank volume: 80 ft × 20 ft × 12 ft = 19,200 ft³. Converting to gallons: 19,200 × 7.4805 ≈ 143,700 gallons. Convert to liters: 143,700 × 3.785 ≈ 543,000 L. Mass of MLSS in the tank: 543,000 L × 1350 mg/L = 733,050,000 mg ≈ 733,050 g ≈ 1,616 lb of MLSS. Food rate to the aeration basin (using plant influent BOD and flow): flow is 300 gpm → 432,000 gal/day. In liters: 432,000 × 3.785 ≈ 1,635,000 L/day. BOD concentration is 180 mg/L, so BOD inflow ≈ 1,635,000 L/d × 180 mg/L ≈ 294,300,000 mg/d ≈ 294,300 g/d ≈ 649 lb/d. Now the F/M ratio = food rate (lb BOD/day) / MLSS mass (lb) ≈ 649 / 1,616 ≈ 0.40. Therefore, the F/M ratio is about 0.40.

The F/M ratio is the rate of food (BOD) available to the microorganisms per unit mass of mixed liquor suspended solids (MLSS) in the aeration tank. It’s expressed as pounds of BOD per day per pound of MLSS.

First, find the aeration tank volume: 80 ft × 20 ft × 12 ft = 19,200 ft³. Converting to gallons: 19,200 × 7.4805 ≈ 143,700 gallons. Convert to liters: 143,700 × 3.785 ≈ 543,000 L.

Mass of MLSS in the tank: 543,000 L × 1350 mg/L = 733,050,000 mg ≈ 733,050 g ≈ 1,616 lb of MLSS.

Food rate to the aeration basin (using plant influent BOD and flow): flow is 300 gpm → 432,000 gal/day. In liters: 432,000 × 3.785 ≈ 1,635,000 L/day. BOD concentration is 180 mg/L, so BOD inflow ≈ 1,635,000 L/d × 180 mg/L ≈ 294,300,000 mg/d ≈ 294,300 g/d ≈ 649 lb/d.

Now the F/M ratio = food rate (lb BOD/day) / MLSS mass (lb) ≈ 649 / 1,616 ≈ 0.40.

Therefore, the F/M ratio is about 0.40.