Question

A wastewater treatment plant treats 20 MGD of wastewater containing 950 mg/L of suspended solids in a primary clarifier that has a 20% suspended solids removal efficiency. The rate of sludge collection, the flow rate out the bottom of the clarifier, is 0.08 MGD.

a) What is the solids concentration (in mg/L) in the sludge leaving the clarifier?
b) What mass of solids (in kg/y) is removed annually by the primary clarifier?

Answer 1

a) 47500 mg/L

b) 5250366.444 kg/year

Step-by-step explanation:

Given data:

suspended solids removal efficiency = 20%

Flowrate in the primary clarifier ( Q ) = 20 MGD ( change to Liters/day

Q = 20* 10^6 * 3.785412 Liters /day

settled concentration ( St ) = 950mg/L * 0.2 = 190 mg/L

amount of settled solid = Q * St

= ( 20* 10^6 * 3.785412 ) * 190 = 14384.5656 kg/day

∴ Amount going into sludge with a flowrate of 0.08 MGD = 14384.5656 kg/day

a) concentration of solid in sludge ( leaving the clarifier )

= amount of settled solid / flow rate out of the clarifier in liters/day

= 14384.5656 / ( 0.08 * 10^6 * 3.785412 )

= 0.0475 kg/L

= 47500 mg/L

b) Determine mass of solids that is removed annually

= 14384.5656 kg/day * 365 days

= 5250366.444 kg/year