Question

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?

Answer 1

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.