### Lecture 5

```Lecture 5
Kjemisk reaksjonsteknikk
Chemical Reaction Engineering
 Isothermal reaction design algorithm
 Algorithm to estimate the time or conversion for
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different reactor
 Examples
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Reaction Engineering
Stoichiometry
Rate Laws
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2
Mole Balance
Isothermal reactor design
These topics build upon one another
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For Gas Phase Flow Systems
A
b
B
a
Cj 
Fj


c
C
a
d
D
a
F A 0  j   j X
 0 1   X


T P0

C A 0  j   j X  T 0 P
1   X 
T P0
T0 P
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Cj=f(Fj, T, P) =f(x, T,P)
νj stochiomentric number, b/a, c/a, d/a
Isothermal T=T0
Liquid phase: ε=0
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Batch reactor
1) Mole balance
2) Rate law
N A0
Second-order
 rA  kC A
 rA  k 2 C A
t
2
C A  C A 0 1  X 
dx

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dt
5. Evaluation
dt
  r AV
First-order
3) Stoichiometry:
4) Combine:
dX
X
t 
1
k
k 1 
ln
X
1
1 x
x  1  exp(  k 1 t )
dx
dt
t 
x 

k 2 C A 0 1 
2

X
x
k 2 C A 0 (1  x )
tk 2 C A 0
1  tk 2 C A 0
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CSTR reactor
V 
1) Mole balance
2) Rate law
3) Stoichiometry:
4) Combine:
FA 0 X
 rA
First-order
Second-order
 rA  kC A
 rA  k 2 C A
2
F A  F A 0 1  X
 
V
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


C A0 x
kC A 0 (1  x )
CA 
FA
 
V




FA 0 1  X 
0
 C A 0 1  X 
x
kC A 0 (1  x )
2
5. Evaluation
x 
k
1  k
x 
1  2 kC A 0 
1  4 kC A 0
2 kC A 0
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PFR reactor
FA 0
1) Mole balance
2) Rate law
3) Stoichiometry:
4) Combine:
dX
dV
  rA
First-order
Second-order
 rA  kC A
 rA  k 2 C A
2
F A  F A 0 1  X
dx
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d (V /  )


k 1  X

1   X 
CA 
C A 0 1  X 
1   X 
dx
d (V /  )

k 2 C A 0 1  X

2
1   X 2
5. Evaluation
k   ln
1
1 x
 x
kC A 0   2  1    ln 1  X    X 
2
1   2 X
1 X
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 Examples:
Liquid Phase Laboratory Experiment
(CH2CO)2O + H2O  2CH3COOH
A
+ B

2C
Entering
Volumetric flow rate
v0 = 0.0033 dm3/s
Acetic Anhydride
7.8% (1M)
Water
92.2% (51.2M)
Elementary with k’
1.95x10-4 dm3/(mol.s)
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Case I
Case II
CSTR
PFR
V = 1dm3
V = 0.311 dm3
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Part 1: Mole Balances in Terms of Conversion
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Algorithm for Isothermal Reactor Design
1. Mole Balance and Design Equation
2. Rate Law
3. Stoichiometry
4. Combine
5. Evaluate
A. Graphically (Chapter 2 plots)
B. Numerical (Quadrature Formulas Chapter 2 and
appendices)
C. Analytical (Integral Tables in Appendix)
D. Software Packages (Appendix- Polymath)
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CSTR Laboratory Experiment
Example: CH3CO2 + H20  2CH3OOH
C A 0  1M
C B 0  51 . 2 M
X  ?
V  1 dm
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3
 0  3.3 10
3
dm
3
s
A + B 2C
1) MoleBalance:
CSTR:
V 
FA 0 X
 rA
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CSTR Laboratory Experiment
 rA  k A C A C B
1) Rate Law:
1) Stoichiometry:
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A
FA0
-FA0X
FA=FA0(1-X)
B
FA0ΘB
-FA0X
FB=FA0(ΘB-X)
C
0
2FA0X
FC=2FA0X
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CSTR Laboratory Experiment
CA 
CB 
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B 
FA


FA 0 1  X 
0
FA 0  B  X 
0
51 . 2
 C A 0 1  X 
 C A 0  B  X 
 51 . 2
1
C B  C A 0 51 . 2  X   C A 0 51 . 2   C B 0
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CSTR Laboratory Experiment
 rA  k ' C B 0 C A 0 1  X   kC A 0 1  X 

k
V 
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X 
0 C A 0 X


kCA 0 1  X 
V
X
V
kX

  

0 1  X  k
 0 1  X 
k
1  k
X 
3 . 03
 0 . 75
4 . 03
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PFR Laboratory Experiment
A + B  2C
0.311 dm
1) Mole Balance:

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2) Rate Law:
3) Stoichiometry:
dX
X
3

dV
 rA
FA 0

 rA  kC A C B
C A  C A 0 1  X 
C B  C B0
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PFR Laboratory Experiment
4) Combine:  rA  k ' C B 0 C A 0 1  X   kC A 0 1  X 
dX

kC A 0 1  X 
C A00
dV
dX
1  X 
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ln
k

1
1 X
X  1 e
0
dV  kd 
 k
 k
  94 sec
k  0 . 01 s
1
X  0 . 61
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 Examples:
Gas Phase : PFR and Batch Calculation
2NOCl  2NO + Cl2
2A
 2B + C
Pure NOCl fed with CNOCl,0 = 0.2 mol/dm3 follows an
elementary rate law with k = 0.29 dm3/(mol.s)
Case I
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PFR with
v0 = 10 dm3/s
Find space time, with X = 0.9
Find reactor volume, V for X = 0.9
Case II
Batch constant volume
Find the time, t, necessary toachieve 90% conversion.
Compare
and t.

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Example (Gas Flow, PFR)
2 NOCl  2 NO + Cl2
2A  2B + C
 0  10
dm
3
k  0 . 29
s
T  T0
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2) Rate Law:
3
C A 0  0 .2
mol  s
P  P0
1) Mole Balance:

dm
mol
L
X  0.9
dX
dV


 rA
FA 0
 rA  kC A
2
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Example (Gas Flow, PFR)
   0 1   X 
3) Stoich: Gas
CA 
C A 0 1  X 
1   X 
A  B + C/2
kC A 0 1  X 
2
4) Combine:
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 rA 
dX
dV
1   X 
 1  X 2
0
X

2
V
dX 

0

1   X 
2
2
1  X 2
2
C A 0  0 1   X 
kC
2
A0
Damköhler number
kC
A0
0
Da

kC A 0 V
dV 
 kC A 0 
0
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Example (Gas Flow, PFR)
kC A 0   2  1    ln 1  X    X 
2
1   
2
X
1 X
1 1
  y A 0   1   
2
2
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kC A 0   17 . 02

17 . 02
 294 sec
kC A 0
V  V0   2940 L
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Gas Phase 2A  2B + C
Example Constant Volume (Batch)
1) Mole Balance: dX   rA V 0 
dt
2) Rate Law:
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3) Stoich:
N A0
 rA
N A 0 V0

 rA
C A0
 rA  kC A
2
  0
Gas V  V0
CA 
N A 0 1  X 
 rA  kC
V0
2
A0

V

 C A 0 1  X 
1  X 2
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Example Constant Volume (Batch)
4) Combine:
dX

kC
dt
dX
d
1  X 
2
CA0
 kC A 0 1  X 
dX
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2
A0
1  X 
1
1 X
2
 kC A 0 1  X 
2
2
 kC A 0 dt
 kC A 0 t
t  155 sec
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