Lec8_non

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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 2/1/2011
 Block 1:
on
s and
Must Use the Differential Form of Mole Balance
 Block 2:
 Block 3:
 Block 4:
 Pressure Drop:
Liquid Phase Reactions:
Pressure Drop does not affect the concentrations in liquid phase rxn.
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
Engineering Analysis of Pressure Drop
Concentration Flow System:
Gas Phase Flow System:
CA 
3
FA


CA 
FA

  0 1  X 
T P0
T0 P
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
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
Must use the differential form of the mole balance to separate
variables:
dX
FA0
4
dW
  rA
2


r

kC
Second order in A and irreversible:
A
A
1  X  P T0
CA 
 CA 0

1 X  P0 T
FA
Stoichiometry:
1  X  P
CA  CA 0
1 X  P0
Isothermal, T=T0

Combine:
5

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)
Ergun Equation:
Constant mass flow:



dP
 G  1    1501   
 3  

 1
.75
G 


dz g c D p    
Dp
 TURBULENT 
 LAMINAR

m  m 0
    0 0
0
  0

6
FT P0 T
  0
FT 0 P T0
P0 T
  0 (1  X )
P T0
Variable Density
dP
G

dz  0 g c D p
Let
7
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
   

Catalyst Weight W  zAc b  zAc 1   c
Where b  bulk density
c  solid catalyst density
  porosity (a.k.a., void fraction )



Let
8
 0
P0 T FT
dP

dW Ac 1   c P T0 FT 0
2 0
1

Ac 1   c P0
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
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
.
,
and
For   0
dy  
(1  X )

dW 2 y
When W  0 y  1
dy   dW
2
y 2  (1  W )
y  (1  W )1/ 2
11
1
P
12
W
2
CA
P
CA  CA 0 1  X 
P0

No P
P
W
13

3
-rA
rA  kC
2
A
No P


W
14
P
4
X
No P
P


W
15
5
P

For   0 :
 P0 
  0  
P

1.0
No P

W
16
P0 T
  0 1  X 
P T0
T  T0
P0
y
P
0
1
f 

 (1  X ) y
17
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
α = 0.0099 kg-1
Find X at 100 kg
C A0
CB0
18
CA0  CB0
X ?
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
19
dy


dW
2y
W 0
2 ydy  dW
,
y 2 1  W
, y 1
y  1  W 
12
4) Combine:
 rA  kC 1 X  y  kC 1  X  1  W 
2
A0
2
2
A0
dX kCA2 0 1  X  1  W 

dW
FA0
2
20
2
2
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
21
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
22
A + 2B  C
1) Mole Balance:
dX  rA

dW FA0
2) Rate Law:
 rA  kCACB2
3) Stoichiometry: Gas, Isothermal
P0
  0 1  X 
P
23

1 X 
C A  C A0
y
1  X 

B  2 X 
y
4) CB  C A0
1  X 
5)
dy

1  X 

dW
2y
6) f 
7)
 1  X 

0
y
2
   , C A0  2 , FA0  2 , k  6 ,   0.02
3
Initial values: W=0, X=0, y=1  W=100
Combine with Polymath.
24
If δ≠0, polymath must be used to solve.
25
26
T = T0
27
29
30
31
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
32
End of Lecture 8
33
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
34
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
A + B  2C
dm6
1
k  1.5
,   0.0099kg
m ol kg  min
Case 1:
Case 2:
C A0
CB0
35
, C B 0  C A0
W  100kg , X  ? , P  ?
DP  2 DP1
1
, P02  P01 ,
2
X ? , P?
X ?
dX
FA0
 r ' A
dW
rA  kCAC B
FA
CA 
y
FT
C A  CB
y  (1  W )
Param eters
1/ 2
36

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