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A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant concentration of 5.6 mg/L. At time zero the concentration in the lake is 5.6 mg/L before an industry begins discharging waste with a flow of 0.7 m3/s also into the lake with a pollutant concentration of 300 mg/L. The decay coefficient for the pollutant in the lake is 0.2 per day. What is concentration leaving the lake one day after the pollutant is added?

User Nakeema
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1 Answer

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Step-by-step explanation:

The given data is as follows.

Volume of lake =
15 * 10^(6) m^(3) =
15 * 10^(6) m^(3) * (10^(3) liter)/(1 m^(3))

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake =
5.6 * 15 * 10^(9) mg

=
84 * 10^(9) mg

=
84 * 10^(3) kg

Flow rate of river is 50
m^(3) sec^(-1)

Volume of water in 1 day =
50 * 10^(3) * 86400 liter

=
432 * 10^(7) liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are
2.9792 * 10^(10) mg or
2.9792 * 10^(4) kg

Flow rate of sewage =
0.7 m^(3) sec^(-1)

Volume of sewage water in 1 day =
6048 * 10^(4) liter

Concentration of sewage = 300 mg/L

Total amount of pollutants =
1.8144 * 10^(10) mg or
1.8144 * 10^(4)kg

Therefore, total concentration of lake after 1 day =
(131936 * 10^(6))/(1.938 * 10^(10))mg/ l

= 6.8078 mg/l


k_(D) = 0.2 per day


L_(o) = 6.8078

Hence,
L_(liquid) =
L_(o)(1 - e^{-k_(D)t}


L_(liquid) =
6.8078 (1 - e^(-0.2 * 1))

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

User Bendecoste
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