Final answer:
To determine the solids concentration of primary clarifier sludge, the amount of solids removed is divided by the sludge flow rate, resulting in approximately 41,918.93 mg/L.
Step-by-step explanation:
The student is dealing with a problem related to solids processing in a wastewater treatment plant. To find the solids concentration of the primary clarifier sludge, we need to calculate the total amount of solids removed by the clarifier and divide that by the sludge flow rate.
The total daily amount of solids entering the clarifier is found by multiplying the influent flow rate by the influent solids concentration:
- (20 million gallons/day) x (800 mg/L) x (3.785 L/gallon) = 60,640,000 mg/day or 60,640 g/day
An 18 percent removal efficiency means:
- (60,640 g/day) x (0.18) = 10,915.2 g/day removed
To find the concentration of solids in the sludge, we divide the total amount of solids removed by the sludge flow rate:
- (10,915.2 g/day) / (0.070 million gallons/day x 3.785 L/gallon) = 41,918.93 mg/L
The solids concentration of the primary clarifier sludge is therefore approximately 41,918.93 mg/L.