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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
Lecture 6 – Tuesday 1/29/2013
2
Block 1: Mole Balances
Block 2: Rate Laws
Block 3: Stoichiometry
Block 4: Combine
Review of Blocks 1, 2 and 3 Examples : 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 0
dX
dV rA
X
A
Ar
dXFV
0
0PFR
Vrdt
dXN AA 0
0
0
X
A
AVr
dXNtBatch
X
t
FA 0
dX
dW r A
X
A
Ar
dXFW
0
0PBR
X
W 3
Building Block 1: Mole Balances in terms of conversion, X
Review Lecture 2
Building Block 2: Rate Laws
4
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 order
BAAA CCkr 2
Review Lecture 3
5
Review Lecture 4
Building Block 3:
Stoichiometry
6
Review Lecture 5
Building Block 4: Combine
7
Review Lecture 5
Building Block 4:
Combine
Today’s lecture Example for Liquid Phase Undergraduate Laboratory
Experiment
(CH2CO)2O + H2O 2CH3COOH
A + B 2C Entering Volumetric flow rate v0 = 0.0033 dm3/s Acetic 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
8
Today’s lecture Example 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/s Find space time, 𝜏 with X = 0.9 Find reactor volume, V for X = 0.9
Case II Batch constant volume
Find the time, t, necessary to achieve 90% conversion. Compare 𝜏 and t.
9
Part 1: Mole Balances in terms of
Conversion
10
Algorithm for Isothermal Reactor Design
1. Mole Balances and Design Equation
2. Rate Laws
3. Stoichiometry
4. Combine
5. 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 Experiment Example: CH3CO2 + H20 2CH3OOH
1) Mole Balance: CSTR: A
A
r
XFV
0
MCB 2.510
MCA 10
?X
V 1 dm3
0 3.3 103 dm3
s
11
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
12
CSTR Laboratory Experiment
XC
XFFC A
AAA
1
10
0
0
XC
XFC BA
BAB
0
0
0
2.511
2.51B
000 2.512.51 BAAB CCXCC
13
CSTR Laboratory Experiment
XkCXC
k
Ckr AABA 11' 000
XkXV
X
kXV
XC
XkCV
A
A
1
1
1 000
00
k
kX
1
75.003.4
03.3X
14
CSTR Laboratory Experiment
?X
0.311 dm3
1) Mole Balance:
0A
A
F
r
dV
dX
2) Rate Law: BAA CkCr
3) Stoichiometry: XCC AA 10
0BB CC 15
A + B 2C
s
dm3
00324.0
PFR Laboratory Experiment
4) Combine: XkCXCCkr AABA 11' 000
kddVk
X
dX
C
XkC
dV
dX
A
A
0
00
0
1
1
kX
1
1ln
keX 1
sec 0.96sec00324.0
311.03
3
0
dm
dmV
61.0X16
1 01.0 sk
PFR Laboratory Experiment
2 NOCl 2 NO + Cl2
1) Mole Balance:
0A
A
F
r
dV
dX
2) Rate Law: 2
AA kCr
17
2A 2B + C
Gas Flow PFR Example
s
dm3
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
22
0
1
1
X
XkCr AA
X
XCC AA
1
10
A B + ½C
Da
kCVkC
dVkC
dXX
X
XC
XkC
dV
dX
AA
V
A
X
A
A
0
0
0
0 0
0
0
2
2
2
00
22
0
1
1
1
1
18`1
Gas Flow PFR Example
X
XXXkCA
1
11ln12
2
2
0
2
1
2
110
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
C
r
VN
r
N
Vr
dt
dX
2) Rate Law: 2
AA kCr
XC
V
XNC A
AA
1
10
0
0
22
0 1 XkCr AA
3) Stoichiometry:
(Gas Flow) 0VV
20
Gas Phase 2A 2B + C t=?
Constant Volume Batch Example
20
0
22
0 11
XkCC
XkC
dt
dXA
A
A
4) Combine:
dtkC
X
dXA02
1
tkCX
A01
1
sec 155t
20 1 XkCdt
dXA
21
Constant Volume Batch Example
Mole Balance
Rate Laws
Stoichiometry
Isothermal Design
Heat Effects
22
End of Lecture 6
23
