Respuesta :
Explanation:
The given data is as follows.
Volume of lake = [tex]15 \times 10^{6} m^{3}[/tex] = [tex]15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}[/tex]
Concentration of lake = 5.6 mg/l
Total amount of pollutant present in lake = [tex]5.6 \times 15 \times 10^{9} mg[/tex]
= [tex]84 \times 10^{9}[/tex] mg
= [tex]84 \times 10^{3}[/tex] kg
Flow rate of river is 50 [tex]m^{3} sec^{-1}[/tex]
Volume of water in 1 day = [tex]50 \times 10^{3} \times 86400 liter[/tex]
= [tex]432 \times 10^{7}[/tex] liter
Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are [tex]2.9792 \times 10^{10} mg[/tex] or [tex]2.9792 \times 10^{4} kg[/tex]
Flow rate of sewage = [tex]0.7 m^{3} sec^{-1}[/tex]
Volume of sewage water in 1 day = [tex]6048 \times 10^{4}[/tex] liter
Concentration of sewage = 300 mg/L
Total amount of pollutants = [tex]1.8144 \times 10^{10} mg[/tex] or [tex]1.8144 \times 10^{4}kg[/tex]
Therefore, total concentration of lake after 1 day = [tex]\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l[/tex]
= 6.8078 mg/l
[tex]k_{D}[/tex] = 0.2 per day
[tex]L_{o}[/tex] = 6.8078
Hence, [tex]L_{liquid}[/tex] = [tex]L_{o}(1 - e^{-k_{D}t}[/tex]
[tex]L_{liquid}[/tex] = [tex]6.8078 (1 - e^{-0.2 \times 1})[/tex]
= 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.
