Transcript Lec8_non
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.
2
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.
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
3 Concentration Flow System:
C A
F
A
Gas Phase Flow System: 0 1
X
T T
0
P
0
P C A
F A
F A
0 1 0 1
X X
T T
0
P
0
P
C A
1 0 1
X X
T
0
T P P
0
C B
F B
F A
0
B
b
0 1
X a
T T
0
X P
0
P
C A
0 1
B
b
X a
X T
0
T P P
0
4 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:
F A
0
dX dW
r A
Second order in A and irreversible:
r A
kC A
2
Stoichiometry: Isothermal, T=T 0 Combine:
C A
F A
C A
0 1 1
X
X
P P
0
T
0
T C A
C A
0 1 1
X
X
P P
0
dX dW
kC A
2 0
F A
0 1 1
X X
2 2
P P
0 2 5 Need to find (P/P 0 ) as a function of W (or V if you have a PFR)
6 Ergun Equation: Constant mass flow:
dP dz
G
g c D p
0 0 0 0 0 1 3 150 1
D
p
LAMINAR
G TURBULENT
0
F T P
0
T F T
0
P
0 ( 1
X
)
T
0
P
0
P T T
0
7 Variable Density 0
P P
0
T
0
T F T
0
F T dP
dz
G
0
g c D p
3 150 1
D
p
1 .
75
G
P
0
P T T
0
F T F T
0 Let 0
G
0
g c D p
1 3 150 1
D
p
1 .
75
G
CatalystWeight
W
zA c
b
zA c
1
c
8 Where
c
b
bulk
solid density catalyst density porosity
(
a
.
k
.
a
.,
void fraction
)
dP dW
A c
1 0
c P
0
T P T
0
F T F T
0 Let
A c
1 2 0
c
1
P
0
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dy dW
2
y T T
0
F T F T
0
y
P P
0 We will use this form for single reactions:
d
P dW P
0 2
P
1
P
0
T
0 1
X
dy dW
2
y T T
0 1
X
dy dW
2
y
1
X
Isothermal case
10
dX dW
2
kC A
0
F A
0 1 1
X X
2 2
y
2
dX dW
f
X
,
P
and
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.
Polymath will combine the .
, and
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For dy dW When
0 ( 1
X
) 2
y W
0
y
1
dy
2
dW y
2 ( 1
W
)
y
( 1
W
) 1 / 2
1
P 12 W
13
2
C A
C A
C A
0 1
X
P P
0 W
No
P
P
14
3
-r A
r A
kC A
2 W
No
P
P
15
4
X W
No
P
P
16
5
For
0 0 :
P
0
P
1.0
P
No
P
W
17 0 1
X
P
0
P T T
0
f T
T
0
y
P
0
P
0 1 ( 1
X
)
y
18 Gas Phase Reaction in PBR with δ A + B 2C = 0 (Analytical Solution) Repeat the previous one with equimolar feed of A and B and
k
A
α
= 1.5dm
6 /mol/kg/min = 0.0099 kg -1 Find X at 100 kg
C A
0
C B
0
C A
0
C B
0
X
?
19 1) Mole Balance: 2) Rate Law: 3) Stoichiometry:
dX dW
r
'
A F A
0
r
'
A
kC A C B C A
C A
0 1
X
y C B
C A
0 1
X
y
20
dy dW
2
y W
0 ,
y
1 , 2
ydy
dW y
2 1
W y
1
W
1 2 4) Combine:
r A
kC
2
A
0 1
X
2
y
2
kC
2
A
0 1
X
1
W
dX dW
2
kC A
0 1
X
1
W
F A
0
21 1
dX X
2
kC A
2 0 1
W
dW F A
0 1
X X
kC A
2 0
F A
0
W
W
2 2
W
0 ,
X
0 ,
W
W
,
X
X X X
0 .
6
with
0 .
75
pressure
without drop pressure
drop
,
i
.
e
.
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.
22 Plot the conversion and pressure ratio y = P/P catalyst weight upto 100 kg.
0 as a function of Additional Information k A = 6dm 9 /mol 2 /kg/min α = 0.02 kg -1
23 A + 2B C 1) Mole Balance:
dX dW
r A
F A
0 2) Rate Law:
r A
kC A C B
2 3) Stoichiometry: Gas, Isothermal 0 1
C A
C A
0
X
P
0
P
1 1
X
X
y
24 4) 5) 6)
f C B dy dW
C A
0 1
B
2
X X
y
0 2
y
1 1
y
X
X
7) 2 3 ,
C A
0 2 ,
F A
0 2 ,
k
6 , 0 .
02 Initial values: W=0, X=0, y=1 W=100 Combine with Polymath.
If δ ≠0, polymath must be used to solve .
25
26
27
T = T 0
29
30
31
32 Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance
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End of Lecture 8
34 dP dW 0 F T A c F T 1 0 P 0 T P T 0 c dy dW 0 F T T yP 0 A c F T 1 0 T 0 c dy dW 2 y F T F T 0 T T 0 F T F T 0 1 X Isothermal: T = T 0 P 0 A C 2 1 0 C Use for heat effects, multiple rxns dX dW 2 y 1 X
35 A + B 2C
k
1 .
5
dm
6
mol
kg
min Case 1: ,
W
100
kg
0 .
0099
kg
1 ,
X
?
, ,
P
?
C B
0
C A
0 Case 2:
D P
2
D P
1 ,
P
02 1 2
P
01 ,
X
?
,
P
?
C A
0
C B
0
X
?
36
r F
A
0
A
dX
r
'
A
dW
kC
A
C
B
C
A
C
A
y
F F
T
C
B A
y
( 1
W
) 1 / 2