1 - 17/04/2015 department of chemical engineering lecture 4 kjemisk reaksjonsteknikk chemical...
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1 - 04/18/23
Departm
ent of Chem
ical E
ngineering
Lecture 4
Kjemisk reaksjonsteknikk
Chemical Reaction Engineering
Review of previous lectures Stoichiometry Stoichiometric Table Definitions of Concentration Calculate the Equilibrium Conversion, Xe
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Departm
ent of Chem
ical E
ngineering
Reactor Mole Balances in terms of conversion
Reactor Differential Algebraic Integral
A
0A
r
XFV
CSTR
A0A rdV
dXF
X
0 A0A r
dXFVPFR
Vrdt
dXN A0A
Vr
dXNt
X
0 A0A Batch
X
t
A0A rdW
dXF
X
0 A0A r
dXFWPBR
X
W2
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Departm
ent of Chem
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dDcCbBaA
d
r
c
r
b
r
a
r DCBA
Da
dC
a
cB
a
bA
Last LectureRelative Rates of Reaction
3
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Departm
ent of Chem
ical E
ngineering
BAA CkCr
βα
β
α
OrderRection Overall
Bin order
Ain order
Last LectureRate Laws - Power Law Model
4
C3BA2 A reactor follows an elementary rate law if the reaction orders just happens to agree with the stoichiometric coefficients for the reaction as written.e.g. If the above reaction follows an elementary rate law
2nd order in A, 1st order in B, overall third order
B2AAA CCkr
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Departm
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ngineering
Last LectureArrhenius Equation
k Ae E RT
k is the specific reaction rate (constant) and is given by the Arrhenius Equation.
Where:
T k A
T 0 k 0
A1013
k
T5
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Departm
ent of Chem
ical E
ngineering
These topics build upon one another
Mole Balance Rate Laws Stoichiometry
Reaction Engineering
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Departm
ent of Chem
ical E
ngineering
How to find
rA f X
rA g Ci Step 1: Rate Law
Ci h X Step 2: Stoichiometry
rA f X Step 3: Combine to get
7
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Departm
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We shall set up Stoichiometry Tables using species A as our basis of calculation in the following reaction. We will use the stochiometric tables to express the concentration as a function of conversion. We will combine Ci = f(X) with the appropriate rate law to obtain -rA = f(X).
Da
dC
a
cB
a
bA
A is the limiting Reactant.
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Departm
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NA NA 0 NA 0X
Xa
b
N
NNXN
a
bNN
A
BAABB
0
0000
For every mole of A that react, b/a moles of B react. Therefore moles of B remaining:
Let ΘB = NB0/NA0
Then:
NB NA 0 B b
aX
NC NC 0 c
aNA 0X NA 0 C
c
aX
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Departm
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Species Symbol Initial Change Remaining
Batch System Stoichiometry Table
B B NB0=NA0ΘB -b/aNA0X NB=NA0(ΘB-b/aX)
A A NA0 -NA0X NA=NA0(1-X)
Inert I NI0=NA0ΘI ---------- NI=NA0ΘI
FT0 NT=NT0+δNA0X
Where: 0A
0i
0A
0i
00A
00i
0A
0ii y
y
C
C
C
C
N
N
1a
b
a
c
a
dan
d
C C NC0=NA0ΘC +c/aNA0X NC=NA0(ΘC+c/aX)
D D ND0=NA0ΘD +d/aNA0X ND=NA0(ΘD+d/aX)
10δ = change in total number of mol per mol A reacted
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Departm
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Constant Volume Batch
Note: If the reaction occurs in the liquid phaseor
if a gas phase reaction occurs in a rigid (e.g. steel) batch reactor
V V0Then
CA NA
VNA 0 1 X
V0
CA 0 1 X
CB NB
VNA 0
V0
B b
aX
CA 0 B
b
aX
etc.11
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Departm
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Suppose
rA kACA2CB
Batch: 0VV
Stoichiometry
12
rA kACA 02 1 X 2 B
b
aX
Equimolar feed:
B 1
Stoichiometric feed:
B b
a
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Departm
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A A FA0 -FA0X FA=FA0(1-X)
Species Symbol Reactor Feed Change Reactor Effluent
B B FB0=FA0ΘB -b/aFA0X FB=FA0(ΘB-b/aX)
i Fi0
FA 0
Ci00
CA 00
Ci0
CA 0
y i0
yA 0
Where:
Flow System Stochiometric Table
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Departm
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Species Symbol Reactor Feed Change Reactor Effluent
0A
0i
0A
0i
00A
00i
0A
0ii y
y
C
C
C
C
F
F
Where:
Flow System Stochiometric Table
Inert I FI0=A0ΘI ---------- FI=FA0ΘI
FT0 FT=FT0+δFA0X
C C FC0=FA0ΘC +c/aFA0X FC=FA0(ΘC+c/aX)
D D FD0=FA0ΘD +d/aFA0X FD=FA0(ΘD+d/aX)
1a
b
a
c
a
dand
A
A
FCConcentration – Flow System14
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Departm
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Species Symbol Reactor Feed Change Reactor Effluent
A A FA0 -FA0X FA=FA0(1-X)
B B FB0=FA0ΘB -b/aFA0X FB=FA0(ΘB-b/aX)
C C FC0=FA0ΘC +c/aFA0X FC=FA0(ΘC+c/aX)
D D FD0=FA0ΘD +d/aFA0X FD=FA0(ΘD+d/aX)
Inert I FI0=FA0ΘI ---------- FI=FA0ΘI
FT0 FT=FT0+δFA0X
0A
0i
0A
0i
00A
00i
0A
0ii y
y
C
C
C
C
F
F
1a
b
a
c
a
dWhere: and
A
A
FCConcentration – Flow System
Flow System Stochiometric Table
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Departm
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A
A
FCConcentration Flow System:
0Liquid Phase Flow System:
X1CX1FF
C 0A0
0AAA
X
a
bCX
a
b
V
N
V
NC B0AB
0
0ABB
Flow Liquid Phase
etc.16
Liquid Systems
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Departm
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BAA CkCr If the rate of reaction were
X
a
bX1Cr B
20AAthen we would have
XfrA This gives us
FA 0
rA
X17
Liquid Systems
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Departm
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For Gas Phase Flow Systems
We obtain:
0
0
0T
T0 T
T
P
P
F
F
Combining the compressibility factor equation of state with Z = Z0
CT P
ZRT
CT 0 P0
Z0R0T0
FT CTV
FT 0 CT 0V0
Stoichiometry:
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Departm
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T
T
P
P
F
FF
T
T
P
P
FF
FFC
T
jT
T
T
jjj
0
00
00
0
00
CB CT 0
FB
FT
P
P0
T0
T
CT 0 FT 0 0
For Gas Phase Flow Systems
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Departm
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XFFF 0A0TT The total molar flow rate is:
For Gas Phase Flow Systems
P
P
T
T
F
XFF 0
00T
0A0T0
P
P
T
TX
F
F1 0
00T
0A0
P
P
T
TXy1 0
00A0
P
P
T
TX1 0
00
Substituting FT gives:
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Departm
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ngineering
21
For Gas Phase Flow Systems
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Departm
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Gas Phase Flow System:
Concentration Flow System:
0
00A
0
00
0AAA P
P
T
T
X1
X1C
PP
TT
X1
X1FFC
A
A
FC
P
P
T
TX1 0
00
0
0B0A
0
00
B0AB
B P
P
T
T
X1
Xab
C
P
P
TT
X1
Xab
FF
C
22
For Gas Phase Flow Systems
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Departm
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For Gas Phase Flow Systems
0
00
0
00
0
11 P
P
T
T
X
Xab
C
PP
TT
X
Xab
FF
CjAjA
jj
Cj=f(Fj, T, P) =f(x, T,P)
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Departm
ent of Chem
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2
0
0
B2
0AAA T
T
P
P
X1
Xab
X1
X1Ckr
If –rA=kCACB
This gives us
FA0/-rA
X24
For Gas Phase Flow Systems
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Departm
ent of Chem
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Consider the following elementary reaction with KC=20 dm3/mol and CA0=0.2 mol/dm3. Calculate Equilibrium Conversion or both a batch reactor (Xeb) and a flow reactor (Xef).
Example: Calculating the equilibrium conversion for gas phase reaction in a flow reactor, Xef
C
B2AAA K
CCkr
BA2
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Departm
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Departm
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Consider the following elementary reaction with KC=20 m3/mol and CA0=0.2 mol/m3. Xe’ for both a batch reactor and a flow reactor.
Calculating the equilibrium conversion for gas phase reaction,Xe
C
B2AAA K
CCkr
BA2
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Batch Reactor Example
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Departm
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Step 1:
dX
dt rAVNA 0
Lmol 2.0C 0A
KC 20 L mol
Step 2: rate law, BB2AAA CkCkr
Calculate Xe
B
AC k
kK
C
B2AAA K
CCkr
28
Batch Reactor Example
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Departm
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Symbol Initial Change Remaining
B 0 ½ NA0X NA0 X/2
A NA0 -NA0X NA0(1-X)
Totals: NT0=NA0 NT=NA0 -NA0 X/2
@ equilibrium: -rA=0 C
Be2Ae K
CC0
Ke CBe
CAe2
CAe NAe
VCA 0 1 Xe
CBe CA 0
X e
229
Calculate Xe
Batch Reactor Example
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Departm
ent of Chem
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A NA0 -NA0X NA=NA0(1-X)
B 0 +NA0X/2 NB=NA0X/2
NT0=NA0 NT=NA0-NA0X/2
Solution:
2Ae
BeC C
CK
C
Be2AeAA K
CCk0rAt equilibrium
0VV 2/BAStoichiometry
Constant volumeBatch
Calculating the equilibrium conversion for gas phase reaction
30
Batch Reactor Example
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Departm
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2e0A
e2
e0A
e0A
eX1C2
X
X1C2
XC
K
82.0202
X1
XCK2 2
e
e0Ae
X eb 0.703
BR Example Xeb
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Departm
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Gas Phase Example Xef
2A B
B2
1A
X ef ?
Rate law:
C
B2AAA K
CCkr
X eb 0.703
32
Solution:
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Departm
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Species Fed Change Remaining
A FA0 -FA0X FA=FA0(1-X)
B 0 +FA0X/2 FB=FA0X/2
FT0=FA0 FT=FA0-FA0X/2
Gas Flow Example Xef
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Departm
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A FA0 -FA0X FA=FA0(1-X)
B 0 FA0X/2 FB=FA0X/2
Stoichiometry: Gas isothermal T=T0, isobaric P=P0
V V0 1X
CA FA 0 1 X V0 1X
CA 0 1 X 1X
CB FA 0 X 2
V0 1X CA 0 1 X 2 1X
Gas Flow Example Xef
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Departm
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Pure A yA0=1, CA0=yA0P0/RT0, CA0=P0/RT0
C
0A
2
0AAA KX12
XC
X1
X1Ckr
2e
ee0AC
X1
X1XCK2
yA 0 1 1
2 1
1
2
@ eq: -rA=0
Gas Flow Example Xef
35
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Departm
ent of Chem
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8dm
mol2.0
mol
dm202CK2
3
3
0AC
yA 0 11
2 1
1
2
8 Xe 0.5Xe
2
1 2Xe Xe2
8.5X e2 17X e 8 0
Gas Flow Example Xef
X ef 0.757Flow: Recall
X eb 0.70Batch:36