lecture 4
DESCRIPTION
Lecture 4. 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 4 – Tuesday 1/22/2013. Block 1 Mole Balances Size CSTRs and PFRs given – r A =f(X) Block 2 Rate Laws - PowerPoint PPT PresentationTRANSCRIPT
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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 4
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Lecture 4 – Tuesday 1/22/2013
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Block 1Mole BalancesSize CSTRs and PFRs given –rA=f(X)
Block 2Rate LawsReaction OrdersArrhenius Equation
Block 3StoichiometryStoichiometric TableDefinitions of ConcentrationCalculate the Equilibrium Conversion, Xe
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Reactor Differential Algebraic Integral
V FA 0X
rA
CSTR
FA 0
dX
dV rA
X
AA r
dXFV
0
0PFR
Vrdt
dXN AA 0
0
0
X
AA Vr
dXNtBatch
X
t
FA 0
dX
dW r A
X
AA r
dXFW
0
0PBR
X
W3
Reactor Mole Balances Summaryin terms of conversion, X
Review Lecture 2
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Levenspiel Plots
4
FA 0
rA
X
Review Lecture 2
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PFRReview Lecture 2
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Reactors in Series
reactorfirst tofedA of moles
ipoint toup reactedA of molesX i
Only valid if there are no side streams
Review Lecture 2
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Reactors in SeriesReview Lecture 2
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iA Cgr Step 1: Rate Law
XhCi Step 2: Stoichiometry
XfrA Step 3: Combine to get
Two steps to get
rA f X
Review Lecture 2
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Building Block 2: Rate Laws
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Power Law Model:BAA CkCr
βα
β
α
OrderRection Overall
Bin order
Ain order
CBA 32 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 orderBAAA CCkr 2
Review Lecture 3
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Arrhenius Equation
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E = Activation energy (cal/mol)
R = Gas constant (cal/mol*K)
T = Temperature (K)
A = Frequency factor (same units as rate constant k)
(units of A, and k, depend on overall reaction order)
RTEAek
T k A
T 0 k 0
A 1013
T
k
Review Lecture 3
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Reaction Engineering
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These topics build upon one another
Mole Balance Rate Laws Stoichiometry
Review Lecture 3
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Algorithm
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iA Cgr Step 1: Rate Law
XhCi Step 2: Stoichiometry
XfrA Step 3: Combine to get
How to find XfrA
Review Lecture 3
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Building Block 3: Stoichiometry
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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 stoichiometric 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).
A is the limiting reactant.
Da
dC
a
cB
a
bA
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Stoichiometry
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NA NA 0 NA 0X
NB NB 0 b
aNA 0 NA 0
NB 0
NA 0
b
aX
For every mole of A that reacts, b/a moles of B react. Therefore moles of B remaining:
Let ΘB = NB0/NA0
Then:
X
a
bNN BAB 0
NC NC 0 c
aNA 0X NA 0 C
c
aX
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Batch System - Stoichiometry Table
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Species Symbol Initial Change Remaining
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:0
0
0
0
00
00
0
0
A
i
A
i
A
i
A
ii y
y
C
C
C
C
N
N
1a
b
a
c
a
dand
C C NC0=NA0ΘC +c/aNA0X
NC=NA0(ΘC+c/aX)
D D ND0=NA0ΘD +d/aNA0X
ND=NA0(ΘD+d/aX)
δ = change in total number of mol per mol A reacted
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Stoichiometry Constant Volume Batch
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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
V
NA 0 1 X V0
CA 0 1 X
CB NB
V
NA 0
V0
B b
aX
CA 0 B
b
aX
etc.
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Stoichiometry Constant Volume Batch
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Suppose
rA kACA2CB
Batch: 0VV
X
a
bXCkr BAAA
220 1
Equimolar feed: 1B
Stoichiometric feed:a
bB
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Stoichiometry Constant Volume Batch
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rA f X and we have
If BAAA CCkr 2 , then
X
a
bXCr BAA
230 1 Constant Volume Batch
Ar1
X
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Batch Reactor - Example
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Consider the following elementary reaction with KC=20 dm3/mol and CA0=0.2 mol/dm3. Find Xe for both a batch reactor and a flow reactor.
Calculate the equilibrium conversion for gas phase reaction, Xe .
C
B2AAA K
CCkr
BA2
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Batch Reactor - Example
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Step 1:
dX
dt
rAV
NA 0
30 2.0 dmmolCA
moldmKC3 20
Step 2: rate law: BB2AAA CkCkr
Calculate Xe
B
AC k
kK
C
B2AAA K
CCkr
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Batch Reactor - Example
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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
BeAe K
CC 20
Ke CBe
CAe2
CAe NAe
VCA 0 1 X e
CBe CA 0
Xe
2
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Batch Reactor - Example
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Species Initial Change Remaining
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/BA Stoichiometry:
Constant Volume:
Batch
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Batch Reactor - Example
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2e0A
e2
e0A
e0A
eX1C2
X
X1C2
XC
K
82.0202
X1
XCK2 2
e
e0Ae
X eb 0.703
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Flow System – Stoichiometry Table
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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:
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Flow System – Stoichiometry Table
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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:
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 System
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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 – Stoichiometry Table
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Stoichiometry
A
A
FCConcentration Flow System:
0Liquid Phase Flow System:
CA FA
FA 0 1 X 0
CA 0 1 X
CB NB
NA 0
0
B b
aX
CA 0 B
b
aX
Flow Liquid Phase
etc.
We will consider CA and CB for gas phase reactions in the next lecture
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Mole Balance
Rate Laws
Stoichiometry
Isothermal Design
Heat Effects
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End of Lecture 4
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