A sedimentation basin is used to remove suspended solids from a wastewater. The

Question

A sedimentation basin is used to remove suspended solids from a wastewater. The rate of flow of wastewater into the tank is 10 L/s, and the influent concentration of suspended solids is 200 mg/L. The removal efficiency of the settling tank for suspended solids is 60%. Calculate the amount of suspended solids (sludge) accumulating in the sludge zone each day. You may assume that the amount of water withdrawn when pumping out the sludge from the sludge zone is very small compared to the inflow of wastewater to the tank and can therefore be ignored.

Explanation / Answer

sedimentation basin is used to remove suspended solids from a wastewater.

the rate of flow of wastewater into the tank = 10 L/s or (10*10-3 m-3/s = 10-2m3/s)

the influent concentration of suspended solids = 200 mg/L or 200 g/m3

the removal efficiency of the settling tank for suspended solids = 60%

means, 60% suspended solid can settledown in a sedimenation tank= 60*200/100 = 120 g/m3

120 g of suspended solids can be removed in 1 m3 of volume.

flow of wastewater = 0.01 m3/s

mass of solids per second removed from the sedimentation tank = 120*0.01 = 1.2 g

means, 1.2 g of suspended solids accumulated in 1 second.

then, mass of suspended solids in one day = 1.2*(60*60*24) = 103680 g or 103.68 Kg.

hence, 103.68 Kg of suspended solids (sludge) accumulating in the sludge zone each day.

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