Answer :
Answer:
[tex]2.54\ \text{mg/L}[/tex]
Explanation:
C = Allowable concentration = 1.1 mg/L
[tex]Q_1[/tex] = Flow rate of river = [tex]130\ \text{m}^/\text{s}[/tex]
[tex]Q_2[/tex] = Discharge from plant = [tex]37\ \text{m}^3/\text{s}[/tex]
[tex]C_1[/tex] = Background concentration = 0.69 mg/L
[tex]C_2[/tex] = Maximum concentration that of the pollutant
The concentration of the mixture will be
[tex]C=\dfrac{Q_1C_1+Q_2C_2}{Q_1+Q_2}\\\Rightarrow C_2=\dfrac{C(Q_1+Q_2)-Q_1C_1}{Q_2}\\\Rightarrow C_2=\dfrac{1.1(130+37)-130\times 0.69}{37}\\\Rightarrow C_2=2.54\ \text{mg/L}[/tex]
The maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant is [tex]2.54\ \text{mg/L}[/tex].