Bass Lake has a volume of 20.0 x 106 m3 and is fed by a polluted stream having a

Question

Bass Lake has a volume of 20.0 x 106 m3 and is fed by a polluted stream having a flow rate of 7.5m3/s and a pollution concentration equal to 23.0 mg/l. There are also two sewage outfalls that enter the lake: the first has a flow of 2.0m3/s and a pollutant concentration of 25 mg/l; the second outfall has a flow of 4.0m3/s and a pollutant concentration of 44.0 mg/l. Assuming the pollution is completely mixed in the lake and assuming no evaporation or other water losses or gains, find the steady state concentration in the lake?

Explanation / Answer

Steady state occurs in the lake implies net incoming water = net outgoing water.

Let steady state conecentration = x

net inflow = 7.5 + 2 + 4 = 13.5 m3/s = net outflow

Steady state implies:
Net outgoing pollutants = Net incoming pollutants

13.5*x = (7.5*23 + 2*25 + 4*44)

So x = 29.52 mg/l

The steady state pollutant concentration of the lake is 29.52 mg/l.

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