Lecture 5

Report
Lecture 5
Kjemisk reaksjonsteknikk
Chemical Reaction Engineering
 Isothermal reaction design algorithm
 Algorithm to estimate the time or conversion for
Department of Chemical Engineering
different reactor
 Examples
1 - 07/04/2015
Reaction Engineering
Stoichiometry
Rate Laws
Department of Chemical Engineering
2
Mole Balance
Isothermal reactor design
These topics build upon one another
2 - 07/04/2015
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
Department of Chemical Engineering
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
3 - 07/04/2015
Department of Chemical Engineering
4
4 - 07/04/2015
Department of Chemical Engineering
5
5 - 07/04/2015
Department of Chemical Engineering
6
6 - 07/04/2015
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

Department of Chemical Engineering
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
7 - 07/04/2015
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
Department of Chemical Engineering



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
8 - 07/04/2015
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
Department of Chemical Engineering
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
9 - 07/04/2015
 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)
Department of Chemical Engineering
Case I
Case II
CSTR
PFR
V = 1dm3
V = 0.311 dm3
10
10 - 07/04/2015
Part 1: Mole Balances in Terms of Conversion
Department of Chemical Engineering
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)
11
11 - 07/04/2015
CSTR Laboratory Experiment
Example: CH3CO2 + H20  2CH3OOH
C A 0  1M
C B 0  51 . 2 M
X  ?
V  1 dm
Department of Chemical Engineering
12
3
 0  3.3 10
3
dm
3
s
A + B 2C
1) MoleBalance:
CSTR:
V 
FA 0 X
 rA
12 - 07/04/2015
CSTR Laboratory Experiment
 rA  k A C A C B
1) Rate Law:
1) Stoichiometry:
Department of Chemical Engineering
A
FA0
-FA0X
FA=FA0(1-X)
B
FA0ΘB
-FA0X
FB=FA0(ΘB-X)
C
0
2FA0X
FC=2FA0X
13
13 - 07/04/2015
CSTR Laboratory Experiment
CA 
CB 
Department of Chemical Engineering
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
14
14 - 07/04/2015
CSTR Laboratory Experiment
 rA  k ' C B 0 C A 0 1  X   kC A 0 1  X 

k
V 
Department of Chemical Engineering
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
15
15 - 07/04/2015
PFR Laboratory Experiment
A + B  2C
0.311 dm
1) Mole Balance:

Department of Chemical Engineering
16
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
16 - 07/04/2015
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 
Department of Chemical Engineering
17
ln
k

1
1 X
X  1 e
0
dV  kd 
 k
 k
  94 sec
k  0 . 01 s
1
X  0 . 61
17 - 07/04/2015
 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
Department of Chemical Engineering
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.

18
18 - 07/04/2015
Example (Gas Flow, PFR)
2 NOCl  2 NO + Cl2
2A  2B + C
 0  10
dm
3
k  0 . 29
s
T  T0
Department of Chemical Engineering
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
19
19 - 07/04/2015
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:
Department of Chemical Engineering
20
 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
20 - 07/04/2015
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
Department of Chemical Engineering
kC A 0   17 . 02

17 . 02
 294 sec
kC A 0
V  V0   2940 L
21
21 - 07/04/2015
Gas Phase 2A  2B + C
Example Constant Volume (Batch)
1) Mole Balance: dX   rA V 0 
dt
2) Rate Law:
Department of Chemical Engineering
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
22
22 - 07/04/2015
Example Constant Volume (Batch)
4) Combine:
dX

kC
dt
dX
d
1  X 
2
CA0
 kC A 0 1  X 
dX
Department of Chemical Engineering
23
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
23 - 07/04/2015

similar documents