Bass Lake has a volume of 20.0 x 10 6 m 3 and is fed by a polluted stream having

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

Flow rate of stream-1 = 7.5 m3/s,   =7.5*1000 l/s=7500 L/s

Concentration = 23 mg/L. Mass of pollutant= 7500*23 mg/s=172500 mg/s

Similarly : stream-2 = 2*1000*25= 50000 mg/s

Stream-3 = 4*1000*44= 176000 mg/s

Total mass = 172500+50000+176000=398500 mg/s

Volume of lake = 20*106 m3 =20*109 L

Concentration = 398500 mg/s/(7500+2000+4000)= 29.51 mg/L

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