Question

9. A CMFR with a detention time of 5 hours is being used to remove a pollutant with a k 6 mgl-hr. Prior to time zero, the influent concentration is 20 mg/L. Beginning at time zero, Cin is: Cin 60 10u(t 10) + 40u(t 20) with time in units of hours What in the effluent concentration after a long period of time in units of mg/? a. 10 mg/L b. 20 mg/L 40 mg/L ① 60 mg/L 2

Answer 1

Ans) Given,

Detention time = 5hr

K = 6 mg/L-hr

C_in = 20mg/L

We know,

t_d = (1/K C_out )( C_in/C_out) -1 )

5 = (1/ 6 C_out )( 20/C_out -1)

On solving, C_out = 60 mg/L

Now, concentration at long period of time, C

C = C_out e^-Kt

C = 60 e^{-6x 0}

C = 60 mg/L