A PFR and CMFR are arranged in parallel as shown below. A reaction degrades a ch

Question

A PFR and CMFR are arranged in parallel as shown below. A reaction degrades a chemical in a first order reaction with a reaction constant of k=0.05 min^-1. The influent is 20 mg/L and the flow is split evenly between the reactors. What is the steady state effluent exiting the system? Calculate the steady state flow rate (m^3/day) and concentration (mg/L) of lake effluent for the following system from Problem 1-3. Now consider a decay reaction that occurs in the well-mixed lake with a mass rate equal to the expression -kC_EV. The rate constant k = 0.1 day ^-1 and the volume of the lake v = 10^5 m ^3.

Explanation / Answer

Let the flow to PFR be X L/min

Then the flow to the CMFR will be 30-X L/min

For a PFR and first order reaction

-ln(CA/CAO)= KT where K= rate constant and T= space time= 30/X

-ln(CA/CAO)= (30/X )*K     (1)

-ln(1-XA)= K*(30/X)     (1)

For a CSTR

KT 1= XA/ (1-XA) where XA= conversion = 1-CA/CAO where CA= concentration at any time T and CAO= initial concentration

XA/(1-XA)= K*200/(30-X) (2)

XA wil have to be assumed and this sholud match the Eq.1 as well as Eq.2

Since XA has to be the same at the outlet of two reactors

Assume some X and match Eq.1 and Eq.2 with those value of X

Let X =10 L/min Eq.1 becomes -ln(1-XA)= 0.05*30/10= 0.15 , 1-XA=0.8607, XA=0.1393

Eq.2 becomes XA/(1-XA)= 0.05*200/(30-10 , XA=0.3934

Both XAs are not matching. Assume some other X and repeat until XA is same.

Trial and error gives X =1.41l/min, Volumetric flow rate to PFR and 30-X =30-1.41=28.59 L/min to CSTR.

XA =0.3441.

1-CA/CAO=0.3441, CA/CAO= 1-0.3441= 0.6559, CA=20*0.6559=13.12 mg/L

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