A radioactively contaminated aqueous stream (100 L/hr) with a half-life of 25 ho
ID: 505329 • Letter: A
Question
A radioactively contaminated aqueous stream (100 L/hr) with a half-life of 25 hours needs to be processed. Three options are available:
Option A: Two CSTRs (V = 45,000 L each) connected in series
Option B: Three CSTRs (V = 45,000 L each) connected in parallel
Option C: One plug flow reactor (V = 5,000 L)
Calculate the percentage of the original radioactivity that will be removed using each option. (You may assume radioactive decay is a first-order process.) Which option would you suggest if maximum decontamination is the goal? (XA = 0.005, 0.974, and 0.75, respectively)
Explanation / Answer
1. For 1st order reaction in series
C =CO/(1+KT)n, n = no of reactors in series
K= Rate constant for 1st order reaction =0.693/half life= 0.693/25= 0.02772/hr
T= space time = V/Vo =45000/100 = 450 hr
C= CO/(1+0.02772*450)2= 0.0055CO
XA=1-C/CO= 1-0.0055= 0.9945
2. When corrected parallelly, the volumetric flow rate gets reduced to 1/3
Hence T= 45000/(100/3)= 1350 hr
For 1 parallel reactor , C/CO= 1/(1+KT)= 1/(1+0.02772*1350)= 0.026
C/CO= 0.074, XA= conversion =1-C/CO= 1-0.026=0.974
3. When single plug flow reactor is used
KT= -ln(1-XA)
T= 5000/100= 50hr
-ln(1-XA)=KT = 50*0.02772, XA=0.75
Hence two CSTRS in series gives higher conversion.