ERT 316
ISOTHERMAL REACTOR DESIGN
START
1. The general
mole equation
Algorithm for
Isothermal Reactor
END
2. Design Equations:
• Batch
• CSTR
• PFR
YES
3. Is –
rA=f(X)
given?
NO
Evaluate the
algebraic (CSTR)
or integral (PFR)
equations
4. Determine the rate law in
terms of the concentration
of the reacting species
5. Use Stoichiometry to express
concentration as a function of conversion
• Liquid phase or Gas phase
• Constant Volume Batch
• Constant P and T
Copyright Cheng 05
6. Combine steps 4 and
5 to obtain –rA=f(X)
4.1 Design Algorithm for Isothermal Reactors
• To design an isothermal reactors, the following sequence is
highly recommended.
4.1 Design Algorithm for Isothermal Reactors
• To carry out the evaluation, the following
method can be used:
Graphically (Chapter 2 plot)
Numerical (Quadrature Formulas Chapter
2 and Appendix A4)
Analytical (Integral Tables)
Software (Polymath)
Algorithm for isothermal
Reactor (PFR reactor volume
For 1st order gas-phase rxn
Copyright Cheng 05
Scale up of Batch Reactor to
the design of CSTR
-scale up
-pilot plant-costly
-instead-build microplant(laboratory
bench scale)
-example: analyze data from a
laboratory batch reactor-determine
specific reaction rate, k-use it to design
full-scale flow reactor
Batch Operation
- Calculation of time taken to achieve a given conversion X
AB
• Step 1: Write the mole balance
dX
N A0
  rAV
dt
• Step 2: Write the rate law
 rA  kCA2
Example for second order
reaction
• Step 3: Derive concentration term from stoichiometry
C A  C A0 1  X 
Batch Operation
- Calculation of time taken to achieve a given conversion X
AB
• Step 4: Combine equation from step 1,2,3
dX
2
 kCA0 1  X 
dt
• Step 5: Evaluate
t
1
0 dt  kCA0
X
dX
0 1  X 2
1  X 
t


kCA0  1  X 
This is the reaction time
or tR
Algorithm to estimate reaction time
Mole Balance
Rate Law
N A0
dX
  rAV
dt R
First order
Second order
 rA  kCA
Stoichiometry
Combine
Evaluate
(integrate)
CA 
 rA  kCA2
NA
 C A0 1  X 
V0
dX
 k 1  X 
dt R
tR 
1  1 
ln 

k 1 X 
dX
2
 kCA0 1  X 
dt R
1  X 
tR 


kCA0  1  X 
Batch Operation
- Calculation of time taken to achieve a given conversion X
AB
• To reach 90% conversion in a constant-volume batch
reactor scales:
tR 

1  1 
ln 

k 1 X 
1  1 
ln 

k  1  0.9 
if k = 10-4 s-1
tR 
2.3
 23000s  6.4h
 4 1
10 s
For first order
4.2 Design of CSTR
• Step 1: Write the mole balance of CSTR
Design equation for CSTR is V  FAO X  v0C AO X
(rA ) exit
(rA ) exit
If volumetric flow rate does not change with the
reaction, (i.e. v = v0), then
V v0C A0 X C A0 X
t 

v0
 rAv0
 rA
where t is the space time
4.2 Design of CSTR
Step 2: Write the rate law
• For 1st order irreversible reaction,  rA  kCA
Step 3: Derive concentration in terms of conversion
(from stoichiometry)
C A  C A0 1  X 
Step 4: Combine eq from step 1,2, 3
1 X 
t 

k 1 X 
Rearranging;
tk
X 
1  tk
tk is often referred to as
Damköhler number (for
1st order)
4.2 Design of CSTR
Step 2: Write the rate law
• For 1st order irreversible reaction,  rA  kCA
Step 3: Derive concentration in terms of conversion
(from stoichiometry)
C A0
C A  C A0 1  X 
CA 
1  tk
Step 4: Combine eq from step 1,2, 3
1 X 
t 

k 1 X 
Rearranging;
tk
X 
1  tk
Damköhler number
• Is the ratio of the rate of reaction of A to the rate of
convective transport of A at the entrance to the
reactor.
 rA0V
Da 
 rate of reaction at entrance
entering flow rate of A
FA0
• For first order irreversible reaction;
 r V kC V
Da  A0  A0  tk
FA0
v0C A0
• For second order irreversible reaction;
 rA0V kCA2 0V
Da 

 tkCA0
FA0
v0C A0
How to estimate degree of conversion for a
CSTR?
• By using Damkohler number,
 rA0V
Da 
FA0
If Da  0.1, X < 0.1
If Da  10, X > 0.9
If first degree order, Da = tk
If second degree order, Da =tkCA0
Rule of thumb
4.2 Design of CSTR (for first order)
• For CSTRs in series, conversion as a function of
the number of tanks in series:
1
X  1
n
(1  tk )
• For CSTRs in parallel, conversion is:
tk
X 
1  tk
Just like a single
CSTR
Example: Producing 200 Million
Pounds per Year in a CSTR
It is desired to produce 200 million pounds per year
of ethylene glycol (EG). The reactor is to be operated
isothermally. A 1lb mol/ft3 solution of ethylene oxide
(EO) in water is fed to the reactor shown in figure
together with an equal volumetric solution of water
containing 0.9 wt% of the catalyst H2SO4. The specific
reaction rate constant is 0.311 min-1 .
(a) If 80% conversion is to be achieved, determine
the necessary CSTR volume.
(b) If 800-gal reactors were arranged in parallel,
what is the corresponding conversion?
(c) If 800-gal reactors were arranged in series, what
is the corresponding conversion?
A  B catalyst
 C
Example: Producing 200 Million
Pounds per Year in a CSTR
Extract the given information:
FC = 2 x 108 lbm/yr x 1 yr/365 days x 1day/24 h x 1hr/60 min x 1lbmol/62lbm
= 6.137 lbmol/min
From reaction stoichiometry,
FC = FA0X
FA0 
FC 6.137
lbmol

 7.67
X
0.8
min
Example: Producing 200 Million
Pounds per Year in a CSTR
STEP 1: Design equation of CSTR
V
FAO X
( rA ) exit
STEP 2: Rate Law
 rA  kCA
STEP 3: Stoichiometry (Liquid phase, v = v0 )
C A  C A0 1  X 
STEP 4: Combining;
V
FA0 X
v0 X

( rA ) exit k (1  X )
Example: Producing 200 Million
Pounds per Year in a CSTR
v0VV
STEP 5: Evaluate
The entering volumetric flowrate of stream A, with CA01 = lb mol/ft3 before
mixing is;
v A0
FA0 7.67lbmol / min
ft 3


 7.67
3
C A01
1lbmol / ft
min
From the problem statement,
vB 0  v A0
Thus, the total entering volumetric flow rate of liquid is
v0  v A0  vB 0
ft 3
 7.67  7.67  15.34
min
Substituting all the values to calculate volume of reactor;
FA0 X
v0 X
ft 3
0.8
3
V

 15.34

197
.
3
ft
(rA ) exit k (1  X )
min 0.311 min 1 1  0.8


Example: Producing 200 Million
Pounds per Year in a CSTR
b) CSTR in parallel.
Rearranging the equation of volume in part a)
v0 X
k (1  X )
V
X

v0 k (1  X )
V
t
X
k (1  X )
X
tk
Da

(1  tk )
(1  Da )
tk 
X
(1  X )
Example: Producing 200 Million
Pounds per Year in a CSTR
b) CSTR in parallel.
X
tk
Da

(1  tk ) (1  Da )
V
1 ft 3
1
t   800 gal 

 13.94 min
3
v0
7.48 gal 7.67 ft / min
Da  tk  13.94 min  0.311 min 1  4.34
X
4.34
 0.81
(1  4.34)
Example: Producing 200 Million
Pounds per Year in a CSTR
c) CSTR in series
1
X  1
(1  tk ) n
V
1 ft 3
1
t   800 gal 

 6.97 min
3
v0
7.48 gal 15.34 ft / min
Da  tk  6.97 min  0.311 min 1  2.167
1
X  1
 0.90
2
1  2.167 
4.3 PFR
• Assume no dispersion and no radial gradients in
either temperature, velocity, or concentration and in
the absence of pressure drop or heat exchange.
STEP 1: Write the mole balance of PFR:
X
dX
V  FA0 
 rA
0
STEP 2: Write the rate law
Eg: For second order,
 rA  kC
2
A
4.3 PFR
STEP 3: Write concentration in terms of conversion
(from stoichiometry)
C A  C A0 1  X 
For liquid phase
1 X 
C A  C A0 

 1  x 
For gas phase
4.3 PFR
STEP 4: Combine all the equations
F
V  A20
kCA0
v0  X 
dX

0 1  X 2 kCA0  1  X 
X
Rearranging,
X
FA0
V 2
kCA0
v
 0
kCA0
For liquid phase
tkCA0
Da 2

1  tkCA0 1  Da 2
1  X 2 dX
0 1  X 2
X

(1   ) 2 X 
2
2 (1   ) ln( 1  X )   X 

1 X 

For gas phase
Design a PFR: summary
In case of 2nd order rxn,
liquid phase, isothermal
dX
FA0
 rA
dV
• mole balance
No pressure drop
No heat exchange
V  FA0 
X
0
dX
 rA
 rA  kCA2
• rate laws
C A  C A0 (1  X )
• Stoichiometry
FA0
V 2
kCA0
• combination
or

X
0
v0C A0  X 
dX


2
2 
1  X  kCA0  1  X 
tkCA0
Da 2
X

1  tkCA0 1  Da 2
Damköhler number for 2nd-order reaction
Design a PFR: summary
In case of 2nd order rxn, gas phase,
isothermal
• mole balance
• rate laws
• Stoichiometry
• combination
dX
FA0
 rA
dV
No pressure drop
No heat exchange
V  FA0 
X
0
dX
 rA
 rA  kCA2
FA0 (1  X )
FA
FA
(1  X )
CA 


 C A0
v v0 (1  X ) v0 (1  X )
(1  X )
2

1  X 
V  FA0 
dX
2
2
0
kCA0 1  X 
X
v0
V
kCA0
2



1


X
2
2 (1   ) ln( 1  X )   X 

1 X 
