Consider two CSTRs in series. A liquid phase reaction A B is conducted in these reactors
under isothermal conditions. Pure A is given in the feed. For the following conditions, determine
the optimum volume of the reactors, i.e. minimize the total volume of the reactor.
For second order reaction (or in general for any order) we have to use the values and find the
solution. For example, Q = 25 lit-min-1, CA-in = 0.04 mol/lit, rA  kCA2 , k= 0.075 lit/mol/min,
xout = 0.85
Using the same procedure as we have done in the previous problem, we get
V1 
F
kC
x
Ain 1
2
2
Ain
1
1  x 
and V2 
FAin x1
Therefore, V 
FAin  x2  x1 
kC A2 in 1  x2 

2
FAin  x2  x1 

FAin
kC A2in
kC A2in 1  x1  kC A2in 1  x2 
Differentiate wrt x1 and set it to zero.
2
FAin  1  x1   2 1  x1  x1
V
1  FAin
 2 


4
2
x1 kC Ain 
1  x1 
1  x2   kCAin
 1 x

F
1
1
 Ain 

0
3
2
kC Ain  1  x1  1  x2  
I.e. 1  x2 
1  x2 
2
2
1  x1 

1  x12
1  x1 

2
 x
 x2  x1  
1


2
2 
 1  x1  1  x2  
 1  x2
1 
1


4
2
 1  x1  1  x2  
4
,
2
1  x12
 1  x 2 
1

x2  1 
2 

1

x
1 

 x

1
For the given values, FA-in = 1 mol/min, V  8333  

44.4

0.85

x


1 
2
 1  x1 

Taking the derivative and setting it to zero, we get
 1  x1 2  2 1  x1  x1


44.4

0
4
1  x1 


Solving it, using numerical tools, we get
X1 = 0.665
V1 = 49.4 kL and V2 = 67 kL. Total volume = 117 kL.
Instead, if we have used equal sized reactors, what would be the total volume to achieve the
same conversion?
V1 
FAin x1
kC A2 in 1  x1 
2
 V2 
FAin  x2  x1 
kC A2 in 1  x2 
2
Eliminating the common factors, we get
x  x 
x1
 2 12 .
2
1  x1  1  x2 
Rearranging this, we get 1  x2  x1   x2  x1 1  x1  .
2
2
Substitute x2 = 0.85, we can solve this equation using numerical tools and get x 1 = 0.69
For this value of x1, V1 = 60 kL = V2.
Total volume = 120 kL. Although it is not ‘optimal’, it is only slightly larger than the optimal
volume. Other factors such as availability and stock of spare parts will influence the choice of
reactors and frequently a plant will use equal sized reactor even if it is not optimal from the point
of view of total volume.