A secondary clarifier is 45 ft in diameter and 12 ft deep. If the influent flow is 1.5 MGD and the MLSS is 2,415 mg/L, what is the loading rate in lbs/day/ft²?

• A. 9.5
• B. 19.0
• C. 53.7
• D. 106.0

Answer: (B)

Loading rate to a secondary clarifier is the mass of solids entering the clarifier per day per unit area of the clarifier surface. To find it, calculate how many pounds of solids come in each day and divide by the clarifier’s surface area.

First convert the flow and concentration to a daily solids load:

• Flow: 1.5 MGD = 1.5 million gallons per day ≈ 5.678 million liters per day.

• Solids: MLSS = 2415 mg/L, so daily solids = 5.678e6 L/d × 2415 mg/L ≈ 1.371e10 mg/d.

• Convert to pounds: 1.371e10 mg/d ÷ 453,592 mg/lb ≈ 30,200 lb/d.

Next compute the clarifier surface area. For a circular tank with diameter 45 ft:

• Radius = 22.5 ft, area = πr^2 ≈ 3.1416 × (22.5)^2 ≈ 1590 ft^2.

Finally, loading rate = 30,200 lb/d ÷ 1590 ft^2 ≈ 19.0 lb/day/ft^2.

So the loading rate is about 19.0 lb/day/ft^2.