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Isothermal Reactor Design: 1. Batch Operation

1. This lecture discusses isothermal reactor design for batch, continuous stirred tank reactors (CSTR), and CSTRs in series and parallel. 2. For a batch reactor, the time needed to reduce the reactant concentration from an initial to final value is given by equation 1. The total time for batch operation includes filling, emptying, and cleaning time as given by equation 2. 3. For a CSTR, the dimensionless residence time τ is given by equation 1. For CSTRs in series, the concentration leaving the last reactor is given by equation 2. The conversion in n CSTRs in series is given by equation 3.

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448 views3 pages

Isothermal Reactor Design: 1. Batch Operation

1. This lecture discusses isothermal reactor design for batch, continuous stirred tank reactors (CSTR), and CSTRs in series and parallel. 2. For a batch reactor, the time needed to reduce the reactant concentration from an initial to final value is given by equation 1. The total time for batch operation includes filling, emptying, and cleaning time as given by equation 2. 3. For a CSTR, the dimensionless residence time τ is given by equation 1. For CSTRs in series, the concentration leaving the last reactor is given by equation 2. The conversion in n CSTRs in series is given by equation 3.

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Lecture 6

Isothermal Reactor Design


1. Batch Operation
In modeling the batch reactor, we have assumed that there is no inflow or
outflow of material and that the reactor is well mixed. There is small or
no change in density.

1  dN A 
   rA
V  dt 
in terms of concentration
dC A
 rA
dt
For reactions
dC A
 rA
dt
For irreversible second order reaction in which A is being consumed, the
rate law is
 rA  kC A2
We combine the rate law and the mole balance to obtain
dC A
  kC A2
dt
dC
 A2  dt
kC A

Initially, CA=CA0 at t=0. for isothermal reaction


C t
1 A dC A
    dt
k C A 0 C A2 0

1 1 1 
   t .....1
k CA CA0 
This is time needed to reduce the reactant concentration from an initial
value CA0 to some specified value CA in a batch reactor. The total time in
any batch operations considerably longer, the total time needed for batch
operation is tt

tt  t f  te  tc  t ....2
Where tf= time for filling
te=time for empty
tc=time for cleaning

1
Lecture 6

1. Design of CSTR's
The design equation of CSTR is:
FA 0 X
V 
 rA  exit
int erms of 
V C C A
   A0
0 rA
For first order reaction  rA  kC A
C C A
  A0
kC A
Solving the effluent concentration of A, CA
CA0
CA 
1 k
For the case we are considering, there is no volume change during the
course of the reaction
C A 0 C A
X 
CA0
Combining
k
X  .....1
1 k
The product  k is sometimes referred to as the Damkohler number

CSTR's in series
For first order reaction with no change in volume is to be carried out in
two CSTR's placed in series. The effluent concentration of A from reactor
1 is
CA0
C A1 
1   1k 1
From a material balance on reactor 2,
FA 1  FA 2 0  C A 1  C A 2 
V2  
rA 2 k 2C A 2
Solving for CA2, the concentration exiting the second reactor, we get
C A1 CA0
CA2  
1   2 k 2  1   2 k 2   1   1k 1 
If instead of two CSTR's in series we had n equal size CSTR's connected
in series   1   2  .....   n    operated at the same temperature
 k 1  k 2  ....  k n  k  , the concentration leaving the last reactor is
CA0
C An  .......2
 1 k 
n

2
Lecture 6

1
X  1 ....3
 1 k 
n

A plot of the conversion as function of the number of reactor of reactors


in series for a first order reaction is shown in figure.
The rate of disappearance of A in the nth order

Conversion X
reactor is:
k 1
CA0 1.0  k  0.5
 rAn  kC An  k
 1 k 
n

 k  0.1

0
Number of tank, n 13
CSTR in parallel
The volume of 1st reactor is the same for all reactors in parallel if we have
the same flow rate and reaction conditions
V
Vi 
n
FA 0
FA 0i 
n
V F  X 
 A0  i 
n n  rAi 
 X 
V  FA 0  i 
 rAi 

For second order reaction


For second order liquid phase reaction, the combination of the rate law
and design equation yields:
FA 0 X
V 
kC 2 A
For constant density   0
V C A 0 C A X
  
0 kC A 0  1  X 
2 2
kC A
Solve the last equation for X
 1  2 kC A 0    1  2 kC A 0    2 kC A 0 
2 2

X 
2 kC A 0


 1  2 kC A 0   1  4 kC A 0
........4
2 kC A 0

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