Question

Consider a system with two CSTRs in series (i.e., the output from the first reactor becomes the input to the second reactor). Assume a pollutant enters the system and undergoes first-order decay with a rate constant k.

(a) What mean residence time, θ = V/Q, is required in the first reactor to achieve 90 percent removal in that reactor (i.e., C1/Cin = 0.1, where C1 is the steady-state outlet concentration for a fixed inlet concentration Cin)? Express your answer as θ = f(k).

(b) If the two reactors are designed so that a total of 90% removal is attained from the inlet of reactor 1 to the outlet of reactor 2, what is the mean residence time θ required for each tank? (The total system residence time is 2θ.) Express your answer as θ = f(k).

Answer 1

We know that for first order reaction carried out in a CSTR

(C in -C1)/ C1= kT

Cin/C1-1 =KT (1)

Where K= rate constant , T= space time =V/Q

C1= concentration at the outlet of first reactor

Cin= concentration at the inlet to the first reactor

Given C1/Cin =0.1 or Cin/C1=10

10-1 =KT or T= 9/K

For two reactors in series

(Cin/C1- 1 =KT and C1/C2- 1 = KT

Cin/C1= 1+KT and C1/C2= 1+KT

(Cin/C1)* (C1/C2)= (1+KT)²

Cin/C2= (1+KT)²

Given Cin/C2= 10 or

10 = (1+KT)²   or (1+KT)= sqrt(10)=3.162

KT= 2.162 or T= 2.162/K