Lecture 8

Report
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.
Lecture 8 – Tuesday 2/5/2013




Block 1:
Block 2:
Block 3:
Block 4:
Mole Balances
Rate Laws
Stoichiometry
Combine
 Pressure Drop
 Liquid Phase Reactions
 Gas Phase Reactions
 Engineering Analysis of Pressure Drop
2
Pressure Drop in PBRs
Concentration Flow System:
Gas Phase Flow System:
CA 
3
FA


CA 
FA

  0 1  X 
FA0 1  X 
C 1  X  T0 P
 A0
T P0
1  X  T P0
0 1  X 
T0 P
b 
b 


FA0   B  X  C A0   B  X 
FB
a 
a  T0 P


CB 


P
T
1  X  T P0
  1  X 
0
0
T0 P
T P0
T0 P
Pressure Drop in PBRs
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 Balances
Must use the differential form of the mole balance to
separate variables:
dX
FA0
  rA
dW
Rate Laws
Second order in A and irreversible:  rA  kC A2
4
Pressure Drop in PBRs
1  X  P T0
CA 
 CA 0

1 X  P0 T
FA
Stoichiometry
Isothermal, T=T0

Combine:

1  X  P
CA  CA 0
1 X  P0
dX kC 1  X   P 
 

2 
dW
FA0 1  X   P0 
2
A0
2
2
Need to find (P/P0) as a function of W (or V if you have a PFR)
5
Pressure Drop in PBRs
Ergun Equation:



dP
 G  1    1501   
 3  

 1
.75
G 


dz g c D p    
Dp
 TURBULENT 
 LAMINAR

Constant mass flow:
m  m 0
    0 0
0
  0

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FT P0 T
  0
FT 0 P T0
P0 T
  0 (1  X )
P T0
Pressure Drop in PBRs
Variable Density
dP
G

dz  0 g c D p
Let
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P T0 FT 0
  0
P0 T FT
 P0 T FT
 1    1501   
 3  
 1.75G 
Dp
   
 P T0 FT 0
G
0 
0 gc Dp

 1    1501   
 3  
 1.75G 
Dp
   

Pressure Drop in PBRs
Catalyst Weight
Where



Let
8
W  zAc b  zAc 1   c
b  bulk density
c  solid catalyst density
  porosity (a.k.a., void fraction )
(1   )  solid fraction
 0
P0 T FT
dP

dW Ac 1   c P T0 FT 0
2 0
1

Ac 1   c P0
Pressure Drop in PBRs
dy
 T FT

dW
2 y T0 FT 0
P
y
P0
We will use this form for single reactions:
d P P0 
 1 T
1  X 

dW
2 P P0  T0
dy
 T
1  X 

dW
2 y T0
9
dy

1  X 

dW
2y
Isothermal case
Pressure Drop in PBRs
kC 1  X  2
dX

y
2
dW
FA0 1  X 
2
A0
2
dX
dP
 f  X , P  and
 f  X , P  or
dW
dW
dy
 f  y, X 
dW
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.
10
Polymath will combine the Mole Balances, Rate Laws
and Stoichiometry.
Packed Bed Reactors
For   0
dy  
(1  X )

dW 2 y
When W  0 y  1
dy   dW
2
y  (1  W )
2
y  (1  W )1/ 2
11
1 Pressure Drop in a PBR
P
12
W
2 Concentration Profile in a PBR
CA
P
CA  CA 0 1  X 
P0

No P
P
W
13

3 Reaction Rate in a PBR
-rA
P
rA  kC  k (1  X )  
 P0 
2
A
2
No P

W
14
P
2
4 Conversion in a PBR
X
No P
P


W
15
5 Flow Rate in a PBR
P
For   0 :

f 
0
 P0 
  0  
P

1
No P

W
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P0 T
  0 1  X 
P T0
T  T0
P0
y
P
0
1
f  
 (1   X ) y
17
Example 1:
Gas Phase Reaction in PBR for =0
Gas Phase reaction in PBR with δ = 0 (Analytical
Solution)
A + B  2C
Repeat the previous one with equimolar feed of A
and B and:
kA = 1.5dm6/mol/kg/min CA0  CB0
α = 0.0099 kg-1
Find X at 100 kg
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C A0
CB0
X ?
Example 1:
Gas Phase Reaction in PBR for =0
1) Mole Balance
dX  r ' A

dW
FA0
2) Rate Law
 r ' A  kCACB
3) Stoichiometry
CA  CA0 1  X y
CB  CA0 1  X y
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Example 1:
Gas Phase Reaction in PBR for =0
dy


dW
2y
W 0
2, ydy  dW
y 2 1  W
, y 1
y  1  W 
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4) Combine
 rA  kC 1 X  y  kC 1  X  1  W 
2
A0
2
2
2
A0
dX kCA2 0 1  X  1  W 

dW
FA0
2
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2
Example 1:
Gas Phase Reaction in PBR for =0
kCA2 0
dX
1  W dW

2
1  X  FA0
kCA2 0 
X
W 2 
W 


1 X
FA0 
2 
W  0, X  0, W  W , X  X
X  0.6 with pressure drop
X  0.75 without pressure drop, i.e.   0
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Example 2:
Gas Phase Reaction in PBR for ≠0
The reaction
A + 2B  C
is carried out in a packed bed reactor in which there is
pressure drop. The feed is stoichiometric in A and B.
Plot the conversion and pressure ratio y = P/P0 as a
function of catalyst weight up to 100 kg.
Additional Information
kA = 6 dm9/mol2/kg/min
α = 0.02 kg-1
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Example 2:
Gas Phase Reaction in PBR for ≠0
A + 2B  C
1) Mole Balance
dX  rA

dW FA0
2) Rate Law
 rA  kCACB2
3) Stoichiometry: Gas, Isothermal
P0
  0 1  X 
P
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
1 X 
C A  C A0
y
1  X 
Example 2:
Gas Phase Reaction in PBR for ≠0

B  2 X 
4) CB  C A0 1  X  y
dy

1  X 

5)
dW
2y
 1  X 

6) f 
0
y
1
2
7)   y A0 [1  1  2]  [2]  
3
3
C A0  2, FA0  2, k  6,   0.02
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Initial values: W=0, X=0, y=1
Final values: W=100
Combine with Polymath.
If δ≠0, polymath must be used to solve.
Example 2:
Gas Phase Reaction in PBR for ≠0
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Example 2:
Gas Phase Reaction in PBR for ≠0
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T = T0
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Robert the Worrier wonders: What if we increase
the catalyst size by a factor of 2?
Pressure Drop
Engineering Analysis
0 =  ∗ 0 =
 ∗ 0
0
20
150 1 −  
=

+ 1.75

   02   3 
1
≈
0
29
2
Pressure Drop
Engineering Analysis
30
Pressure Drop
Engineering Analysis
31
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
32
End of Lecture 8
33
Pressure Drop - Summary
 Pressure Drop
 Liquid Phase Reactions
 Pressure Drop does not affect concentrations in lquid phase
reactions.
 Gas Phase Reactions
 Epsilon does not equal to zero
d(P)/d(W)=…
Polymath will combine with d(X)/d(W) =… for you
 Epsilon = 0 and isothermal
P=f(W)
Combine then separate variables (X,W) and integrate
 Engineering Analysis of Pressure Drop
34
Pressure Change – Molar Flow Rate
FT P0 T
0
FT 0 P T0
dP

dW
A c 1  c
FT T
0
FT 0 T0
dy

dW
yP0 A c 1  c
dy
 FT T

dW
2 y FT 0 T0
35
20

P0 A C 1  C
Use for heat effects,
multiple rxns
FT
 1  X  Isothermal: T = T0
FT 0
dX

  1  X 
dW
2y
Example 1:
Gas Phase Reaction in PBR for =0
A + B  2C
dm6
1
k  1.5
,   0.0099kg
m ol kg  min
Case 1:
W  100kg , X  ? , P  ?
Case 2: DP  2 DP1
C A0
CB0
36
, C B 0  C A0
1
, P02  P01 ,
2
X ? , P?
X ?
P?
PBR
dX
FA0
 r ' A
dW
rA  kC ACB
FA
CA 
y
FT
C A  CB
  0 and T  T0  y  (1  W )1/2
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