Lecture 8
Chemical Reaction Engineering (CRE) is the
field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
Today’s lecture
 Block 1: Mole Balances on PFRs and PBR
 Must Use the Differential Form
 Block 2: Rate Laws
 Block 3: Stoichiometry
 Pressure Drop:
Liquid Phase Reactions:
Pressure Drop does not affect the concentrations in liquid phase rx.
2
Gas Phase Reactions:
Epsilon not Equal to Zero
d(P)/d(W)=..
Polymath will combine with d(X)/f(W)=..for you
Epsilon = 0 and Isothermal
P=f(W)
Combine then Separate Variables (X,W) and Integrate
Reactor Mole Balances in terms of
conversion
Reactor
Differential
Algebraic
Integral
X
Batch
N A0
dX
dt
  r AV
0
V 
CSTR
PFR
t  N A0 
FA 0
dX
dV
dX
 rA V
t
FA 0 X
 rA
X
  rA
X
V  FA 0 
0
dX
 rA
X
X
PBR
3
FA 0
dX
dW
  rA
W  FA 0 
0
dX
 rA
W
Concentration Flow System:
CA 
CB
4


FA 0 1  X 
 0 1   X 
T P0

   0 1   X 
Gas Phase Flow System:
FA
CA 
FA

T P0
T0 P
C A 0 1  X  T0 P
1   X 
T P0
T0 P
b 
b 


FA 0   B  X 
C A0  B  X 
FB
a 
a  T0 P





P
T
1   X 

T P0
0
 0 1   X 
T0 P
Pressure Drop in Packed Bed Reactors
Note: Pressure drop does NOT affect liquid phase reactions
Sample Question:
Analyze the following second order gas phase reaction that occurs
isothermally in a PBR:
AB
Mole Balance:
Must use the differential form of the mole balance to separate
variables:
dX
FA 0
5
dW
  rA 
Rate Law:
2


r

kC
Second order in A and irreversible:
A
A
Pressure Drop in Packed Bed Reactors
1  X  P T0
CA 
 CA0

1   X  P0 T
FA
Stoichiometry:
1  X  P
CA  CA0
1   X  P0
Isothermal, T=T0

Combine:
6

dX
dW

kC
2
A0
FA 0
1  X 
1   X 2
2
 P 


P 
 0
2
Need to find (P/P0) as a function of W (or V if you have a PFR)
Pressure Drop in Packed Bed Reactors
Ergun Equation:
Constant mass flow:

dP
 G  1     150 1   
 3  

 1
. 75
G

dz
g cD p   
Dp
   TURBULENT
 LAMINAR
 m
0
m
   0  0
  0
  0
7
0

FT P0 T
FT 0 P T 0
   0 (1   X )
P0 T
P T0





Pressure Drop in Packed Bed Reactors
Variable Density
  0
P T 0 FT 0
P0 T FT
P T F
 1     150 1   
T
 3  

 1 . 75 G  0
dz
 0 g c D p    
Dp
 P T 0 FT 0
dP
Let
8
G

 1     150 1   
 3  
0 
 1 . 75 G 
 0 g c D p    
Dp

G
Pressure Drop in Packed Bed Reactors
Catalyst Weight W  zA c  b  zA c 1    c
Where  b  bulk density
 c  solid catalyst
density
  porosity ( a.k.a., void



Let
9
dP
dW

 0
fraction )
P0 T FT
A c 1    c P T0 FT 0
 
2 0
1
A c 1    c P0
Pressure Drop in Packed Bed Reactors
dy

dW
 T FT
P
y
2 y T 0 FT 0
P0
We will use this form for single reactions:
d  P P0 

dW
dy
dW
dy
10
dW
 

1
2  P P0  T 0
 T
2 y T0

T

2y
1   X 
1   X 
1   X 
Isothermal case
Pressure Drop in Packed Bed Reactors
dX
dW
dX
dW
kC A 0 1  X 
2

FA 0 1   X 
2
2
 f  X , P  and
y
2
dP
dW
 f  X , P  or
dy
dW
 f y, X 
The two expressions are coupled ordinary differential
equations. We can only solve them simultaneously using an
ODE solver such as Polymath. For the special case of
isothermal operation and epsilon = 0, we can obtain an
analytical solution.
11
Polymath will combine the mole balance, rate law and
stoichiometry.
PBR
AB
1) Mole Balance:
dX
dW
C A  C A0

 rA
FA 0
1  X   P 
1  X 

  CA0
y


1   X   P0 
1   X 
 1  X   2

 y
 1   X  
2
2) Rate Law:
12
 rA  kC
2
A0
PBR
For
0
dy


dW
2y
When
dy
2
W 0
   dW
y  (1   W )
2
y  (1   W )
13
1/ 2
y 1
1
P
y  1   W 
1 2
14
W
2
CA
C A  C A 0 1  X 
P
P0

No  P
P
W
15

3
-rA
2
 rA  kC A
No  P

P

W
16

4
X
No  P
P


W
17
   0 (1   X )
T  T0 ,
f 
18
0


P0 T
y
P T0
P0
P
1
(1   X ) y
5
P


1.0
No P

W
19
Example 1: Gas Phase Reaction in PBR
for δ = 0
Gas Phase Reaction in PBR with δ = 0 (Polymath Solution)
A + B  2C
Repeat the previous one with equil molar feed of A and B and
kA = 1.5dm9/mol2/kg/min
α = 0.0099 kg-1
Find X at 100 kg
20
Example 1: Gas Phase Reaction in PBR
for δ = 0
A + B  2C
k  1 .5
dm
6
mol kg min
Case 1:
Case 2:
21
W  100 kg
D P  2 D P1
  0 . 0099 kg
P ?
X ?
P02 
1
1
2
P01
X ?
P ?
Example 1: Gas Phase Reaction in PBR
for δ = 0
1) Mole Balance:
dX

dW
FA 0
2) Rate Law:
 r ' A  kC A C B
3)
C A  C A 0 1  X y
4)
C B  C A 0 1  X y
W 0
22
 r 'A
y 1
Example 1: Gas Phase Reaction in PBR
for δ = 0
5)
dy

dW

2 ydy    dW
2y
y  1  W
2
y  1   W 
1 2
 rA  kC A 0 1  X  y  kC A 0 1  X  1   W 
2
dX
dW
23
2
2
kC A 0 1  X  1   W 
2

2
2
FA 0
2
Example 1: Gas Phase Reaction in PBR
for δ = 0
2
dX
1  X 
2

kC A 0
FA 0
1   W dW
2
2
kC A 0 
W
W 

1 X
FA 0 
2
X




W  0, X  0, W  W , X  X
X  0 . 6  with pressure
X  0 . 75  without
24
drop
pressure

drop , i.e .   0 
Example A + B → 2C
25
Example A + B → 2C
26
Example 2: Gas Phase Reaction in PBR
for δ ≠ 0
Polymath Solution
A + 2B  C
is carried out in a packed bed reactor in which there is pressure
drop.The fed is stoichiometric in A and B.
Plot the conversion and pressure ratio y = P/P0 as a function of
catalyst weight upto 100 kg.
Additional Information
kA = 6dm9/mol2/kg/min
α = 0.02 kg-1
27
Example 2: Gas Phase Reaction in PBR
for δ ≠ 0
A + 2B  C
1) Mole Balance:
dX
dW
2) Rate Law:

 r 'A
FA 0
 r ' A  kC A C B
2
3) Stoichiometry: Gas, Isothermal
V  V 0 1   X 
CA  C A0
28
P0
P
1  X 
y
1   X 
Example 2: Gas Phase Reaction in PBR
for δ ≠ 0
q B - 2X )
(
4)
CB = CA0
y
(1+ e X )
5)
dy

dW
6) f =
7)

2y
1   X 
u (1+ e X )
=
u0
y

2
3
C A 0  2 FA 0  2 k  6   0 . 02
Initial values: W=0, X=0, y=1  W=100
Combine with Polymath.
29
If δ≠0, polymath must be used to solve.
Example 2: Gas Phase Reaction in PBR
for δ ≠ 0
30
Example 2: Gas Phase Reaction in PBR
for δ ≠ 0
31
T = T0
32
Engineering Analysis
33
Engineering Analysis
34
Engineering Analysis
35
Pressure Change – Molar Flow Rate
0
dP
 
dW

dW
dy

dW
36
FT 0
FT 0 P T 0
 A c 1    c
0
dy
FT
FT P0 T
FT T
 
FT 0 T 0
yP 0 A c 1    c
 FT T
2 y FT 0 T 0
 1   X 
2 0
P0 A C 1    C
Use for heat effects, multiple rxns
Isothermal: T = T0
dX
dW


2y
1   X 
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
37
Gas Phase Reaction with Pressure Drop
Concentration Flow System:
CA 
CB
38


FA 0 1  X 
 0 1   X 
T P0

   0 1   X 
Gas Phase Flow System:
FA
CA 
FA

T P0
T0 P
C A 0 1  X  T0 P
1   X 
T P0
T0 P
b 
b 


FA 0   B  X 
C A0  B  X 
F
a 
a  T0 P


 B 

T P0
1   X 

T P0
 0 1   X 
T0 P
Example 2: Gas Phase Reaction in PBR
for δ ≠ 0
A + 2B  C
1) Mole Balance:
dX -rA¢
=
dW FA0
2) Rate Law:
-rA¢ = kCAC2B
3) Stoichiometry: Gas, Isothermal
39
(
)
P0
u = u 0 1+ eX
P
1- X)
(
CA = CA0
y
(1+ eX)
End of Lecture 8
40