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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 6
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 6
Lecture 6 – Tuesday 1/29/2013
2
Block 1: Mole BalancesBlock 2: Rate LawsBlock 3: StoichiometryBlock 4: Combine
Review of Blocks 1, 2 and 3Examples : Undergraduate Reactor Experiments CSTR PFR BR
Gas Phase Reaction with Change in the Total Number of Moles
Reactor Differential Algebraic Integral
V FA 0X rA
CSTR
FA 0dXdV
rA
X
AA r
dXFV0
0PFR
VrdtdXN AA 0
0
0
X
AA Vr
dXNtBatch
X
t
FA 0dXdW
r A
X
AA r
dXFW0
0PBR
X
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Building Block 1: Mole Balancesin terms of conversion, X
Review Lecture 2
Building Block 2: Rate Laws
4
Power Law Model:BAA CkCr
βαβα
OrderRection OverallBin 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
5
Review Lecture 4
Building Block 3: Stoichiometry
6
Review Lecture 5
Building Block 4: Combine
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Review Lecture 5
Building Block 4: Combine
Today’s lectureExample for Liquid Phase Undergraduate Laboratory
Experiment(CH2CO)2O + H2O 2CH3COOH
A + B 2CEnteringVolumetric flow rate v0 = 0.0033 dm3/sAcetic Anhydride 7.8% (1M)Water 92.2% (51.2M)Elementary with k’ 1.95x10-4 dm3/(mol.s)
Case I CSTR V = 1dm3
Case II PFR V = 0.311 dm3
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Today’s lectureExample for Gas Phase : PFR and Batch Calculation 2NOCl 2NO + Cl2
2A 2B + C Pure NOCl fed with CNOCl,0 = 0.2 mol/dm3 follows an elementary rate law with k = 0.29 dm3/(mol.s)
Case I PFR with v0 = 10 dm3/sFind space time, with X = 0.9 Find reactor volume, V for X = 0.9
Case II Batch constant volumeFind the time, t, necessary to achieve
90% conversion. Compare and t.
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Part 1: Mole Balances in terms of Conversion
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Algorithm for Isothermal Reactor Design1. Mole Balances and Design Equation2. Rate Laws3. Stoichiometry4. Combine5. Evaluate
A. Graphically (Chapter 2 plots)B. Numerical (Quadrature Formulas Chapter 2 and appendices)C. Analytical (Integral Tables in Appendix)D. Software Packages (Appendix- Polymath)
CSTR Laboratory ExperimentExample: CH3CO2 + H20 2CH3OOH
1) Mole Balance: CSTR:A
A
rXFV
0
MCB 2.510
MCA 10
?X
V 1 dm3
0 3.310 3 dm3
s
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A + B 2C
2) Rate Law: BAAA CCkr
3) Stoichiometry:
B FA0ΘB -FA0X FB=FA0(ΘB-X)
A FA0 -FA0X FA=FA0(1-X)
C 0 2FA0X FC=2FA0X
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CSTR Laboratory Experiment
XCXFFC AAA
A
110
0
0
XCXFC BABA
B
00
0
2.511
2.51B
000 2.512.51 BAAB CCXCC
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CSTR Laboratory Experiment
XkCXC
k
Ckr AABA 11' 000
XkXV
XkXV
XCXkCV
A
A
1
1
1 000
00
kkX
1
75.003.403.3
X
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CSTR Laboratory Experiment
?X
0.311 dm3
1) Mole Balance:0A
A
Fr
dVdX
2) Rate Law:BAA CkCr
3) Stoichiometry: XCC AA 10
0BB CC 15
A + B 2C
sdm3
00324.0
PFR Laboratory Experiment
4) Combine: XkCXCCkr AABA 11' 000
kddVkX
dXC
XkCdVdX
A
A
0
00
0
1
1
kX
11ln
keX 1
sec 0.96sec00324.0
311.03
3
0
dmdmV
61.0X16
1 01.0 sk
PFR Laboratory Experiment
2 NOCl 2 NO + Cl2
1) Mole Balance:0A
A
Fr
dVdX
2) Rate Law: 2AA kCr
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2A 2B + C
Gas Flow PFR Example
sdm3
0 10smol
dmk
3
29.0L
molCA 2.00
0TT
P P0
X 0.9 ?V
3) Stoichiometry: (Gas Flow)
X 10
4) Combine: 2
220
11
XXkCr A
A
X
XCC AA
110
A B + ½C
DakCVkCdVkCdX
XX
XCXkC
dVdX
AA
VA
X
A
A
00
0
0 0
0
02
2
200
220
11
11
18`1
Gas Flow PFR Example
X
XXXkCA
1
11ln122
20
21
2110
Ay
02.170 AkC
sec 29402.17
0
AkC
LVV 29400 19
Gas Flow PFR Example
1) Mole Balance:0000
0
A
A
A
A
A
A
Cr
VNr
NVr
dtdX
2) Rate Law: 2AA kCr
XCV
XNC AA
A
110
0
0
220 1 XkCr AA
3) Stoichiometry: (Gas Flow)
0VV
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Gas Phase 2A 2B + C t=?
Constant Volume Batch Example
20
0
220 11 XkCC
XkCdtdX
AA
A
4) Combine:
dtkC
XdX
A021
tkCX A01
1
sec 155t
20 1 XkC
dtdX
A
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Constant Volume Batch Example
Mole Balance
Rate Laws
Stoichiometry
Isothermal Design
Heat Effects
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End of Lecture 6
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