The liquid-phase conversion of a reactant A to product B occurs with irreversibl
ID: 475849 • Letter: T
Question
The liquid-phase conversion of a reactant A to product B occurs with irreversible, 1st-order kinetics and a rate constant of 1.00 min1 . Available are a feed stream at a volumetric flow rate of 1.00 L/min and a pair of CSTRs, one with a volume of 2.00 L and a second with a volume of 1.00 L.
Consider what happens when the reactors are placed in parallel and a fraction y of the feed is sent to the larger reactor, with the remainder of the feed sent to the smaller reactor. Mixing the outlet streams produces a final stream with conversion Xf
Plot Xf (on the vertical axis) as a function of y (on the horizontal axis). What behavior do you observe?
What is the maximum value for Xf ? What is the value of y required to achieve this maximum value? Does this result surprise you? What are the Damköhler numbers of the two reactors?
Compare the maximum conversion possible for the CSTRs in parallel (from above) with the conversion for the following three networks:
c) The CSTRs are placed in series, with the larger CSTR placed in front of the smaller CSTR.
d) The CSTRs are placed in series, with the larger CSTR placed behind the smaller CSTR.
e) A single 3.00-L CSTR.
Explanation / Answer
For a CSTR in 1st order system
Space time, T= CAO*XA/-rA
-rA = Rate of reaction, T =space time, XA= conversion
T= CAOXA/ KCAO*(1-XA)
T= XA/(1-XA) (1), k = 1
When connected in parallel,
Flow rate of y is sent to larger reactor while 1-f is sent to smaller reactor stream while the other reactor flow rate is =1-y
Since they are connected in parallel,
Conversion has to be same. From Eq.1, T has to be same
T= V/VO = 2/y =X1/(1-X1) , X1= conversion from 2 Liters reactor
Hence X1Y= 2-2X1 and X1= 2/(2+y)
Similarly for the second reactor, conversion X2 is related as
T= V/VO =1(/1-y)= X2/(1-X2)
X2= 1/(2-y)
Combined conversion, Xf = 2y/(y+2) + (1-y)/ 2-y
The data point along with the plot are shown below.
from the plot maximum conversion is achieved at y=0.6 and conversion =0.74
Damkoler number , KT
for the 1st reactor KT= K*V/VO= 1*2/0.6= 3.33 and for the second reactor damkohler number = 1*1/0.4= 2.5
2 when the larger reactor comes first in series arrangement
C1= outlet concentration from 1st reactor = CO/(1+KT)
T= V/VO= 2 =2
C1= CO/ 3 =0.333CO
C2= outlet concentration from 2nd reactor = C1/(1+KT),
T =1/1 = 1
C2= 0.333CO/(1+1*1/1)= 0.333CO/ 2 =0.1665CO
Overall conversion =1-C2/CO= 1-0.16655CO/CO= 0.8335
Less than the maximum conversion possible
3, for the second case where the smaller reactor comes first
T= V/vO= 1/1 =1
C1= CO/(1+1)= 0.5CO
For the second reactor, T=2/1=2
C2= C1/(1+2)= 0.5CO/3CO= 0.1667CO
Conversion overall = 1-0.1667 = 0.8333
More than the maximum conversion possible in parallel arrangement
3. When there is single reactor of volume 3 L, T= 3/1 =3
C1= CO/(1+3)= 0.25CO
X1= 1-C1/CO= 0.75
y 2y/(y+2) (1-y)/(2-y) y Xf 0 0 0.5 0 0.5 0.1 0.095238 0.473684211 0.1 0.568922 0.2 0.181818 0.444444444 0.2 0.626263 0.3 0.26087 0.411764706 0.3 0.672634 0.4 0.333333 0.375 0.4 0.708333 0.5 0.4 0.333333333 0.5 0.733333 0.6 0.461538 0.285714286 0.6 0.747253 0.7 0.518519 0.230769231 0.7 0.749288 0.8 0.571429 0.166666667 0.8 0.738095 0.9 0.62069 0.090909091 0.9 0.711599 1 0.666667 0 1 0.666667Related Questions
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