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Lecture 24
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 Web Lecture 24 Class Lecture 20-Thursday 3/28/2013 Review of Multiple Steady States (MSS) Reactor Safety (Chapter 13) Blowout Velocity CSTR Explosion Batch Reactor Explosion
3 Review Last Lecture
CSTR with Heat Effects
Review Last Lecture
Energy Balance for CSTRs
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dT dt
F A
0
N i C P i
H Rx
C P S
1
T
T C
UA F A
0
C P
0
T C
T
0 1
T a
Review Last Lecture
Energy Balance for CSTRs
C P S
1
T
T C
Increasing T 0 5 T Variation of heat removal line with inlet temperature.
Review Last Lecture
Energy Balance for CSTRs
κ=∞
C P S
1
T
T C
κ=0 R(T) Increase κ 6 T a T 0 T Variation of heat removal line with κ (κ=UA/C P0 F A0 )
Review Last Lecture
Multiple Steady States (MSS)
7 Variation of heat generation curve with space-time.
Review Last Lecture
Multiple Steady States (MSS)
8 Finding Multiple Steady States with
T 0
varied
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Multiple Steady States (MSS)
9 Temperature ignition-extinction curve
Review Last Lecture
Multiple Steady States (MSS)
Bunsen Burner Effect (Blowout)
Review Last Lecture
Multiple Steady States (MSS)
Bunsen Burner Effect (Blowout)
12 Reversible Reaction Gas Flow in a PBR with Heat Effects A ↔ B V G R 0 C A 0 X kC A 0 1 H Rx X C P 1 X 1 T K X e H , X Rx k k 1 T C 1 K e T C R T 0 T a 1 400 T 310 310 k 1 k 1 1 K e
13 Reversible Reaction Gas Flow in a PBR with Heat Effects A ↔ B UA = 73,520 UA = 0
14 Reversible Reaction Gas Flow in a PBR with Heat Effects A ↔ B
K C
C Be C Ae
C A
0
X e y T
0
C A
0 1
X e
y T T
0
X e
K C
1
K C T
Reversible Reaction Gas Flow in a PBR with Heat Effects A ↔ B Exothermic Case: K C X e T
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T Endothermic Case: K C X e T T ~1
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Adiabatic Equilibrium Conversion
Conversion on Temperature Exothermic ΔH is negative Adiabatic Equilibrium temperature (T adia ) and conversion (X e,adia ) X a X eadi T T 0 H Rx X C PA X e 1 K C K C T adia T
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Gas Phase Heat Effects
Trends:
Adiabatic: X exothermic X endothermic Adiabatic T and X e T 0 T
T
T
0
C PA
H
Rx I X C PI
T 0 T
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Gas Phase Heat Effects
X X e K C 1 K C X eadia T T T 0 H Rx i C P i
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Gas Phase Heat Effects
Effect of adding inerts on adiabatic equilibrium conversion Adiabatic: X I Adiabatic Equilibrium Conversion 0 I T T 0
X
T
T
0
C P A
H Rx
I C P I
,
T
T
0
C P A
H Rx
I C P I
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F A0 T 0 F A1 X 1 Q 1 T 0 F A2 X 2 Q 2 T 0 F A3 X 3
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X X 3 X 2 X 1 T 0 X e X EB X i C Pi T H RX T 0 T
Adiabatic Exothermic Reactions
A B H Rx 15 kcal mol The heat of reaction for endothermic reaction is positive, i.e.,
:
T T 0 C P H Rx A I C P I and X C P A C P I I H Rx T 0 T 22 We want to learn the effects of adding inerts on conversion. How the conversion varies with the amount, i.e., I , depends on what you vary and what you hold constant as you change I .
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A. First Order Reaction
dX dV r A F A0 Combining the
mole balance
,
rate law
and
stoichiometry
dX dV kC A0 1 0 C A0 X k 0 1 X Two cases will be considered Case 1 Constant 0 , volumetric flow rate Case 2: Variable 0 , volumetric flow rate
A.1. Liquid Phase Reaction
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For Liquids, volumetric flow rates are additive. 0 A0 I0 A0 1 I
Effect of Adding Inerts to an Endothermic Adiabatic Reaction
What happens when we add Inerts, i.e., vary Theta I??? It all depends what you change and what you hold constant!!!
I ISO T V y I 1 1 k I I k I \ V y I k 1 I ISO 1 1 I I I OPT I I OPT I
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A.1.a. Case 1. Constant
0 To keep 0 constant if we increase the amount of Inerts, i.e., increase I A entering, i.e., we will need to decrease the amount of A0 . So I then A0 T T 0 C P H Rx X A I C P I
A.1.a. Case 2. Constant A, Variable 0 dX dV 0 X A X 1 I
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A.2. Gas Phase
Without Inerts With Inerts and A C TA F A0 A C A0 Taking the ratio of C TA to C TI P A RT A Solving for I C C TI TA F TI I F TA A C I A F TI F TA P A P I TI P I RT I P A RT A T I T A F TI I F A0 F I0 I P I RT I
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only A is fed.
Nomenclature note: Sub I with Inerts I and reactant A fed Sub A with only reactant A fed
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F TI = Total inlet molar flow rate of inert, I, plus reactant A, F TI = F A0 + F I0 F TA = Total inlet molar flow rate when no Inerts are fed, i.e., F TA = F A0 P I , T I = Inlet temperature and pressure for the case when both Inerts (I) and A are fed P A , T A = Inlet temperature and pressure when only A is fed C A0 = Concentration of A entering when no inerts are presents C TA = Total concentration when no inerts are present C TI C A0I C A0 F A0 A P A RT A P I RT I F A0 I I
30 F TI F TA F A0 F A0 F I0 1 I 1 F I0 F A0 F A0 1 y A0 y A0 1 1 I I A 1 I P A P I T I T A
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A.2.a. Case 1
Maintain constant volumetric flow, I 0 , rate as inerts are added. I.e., 0 = = A . Not a very reasonable situation, but does represent one extreme. Achieve constant 0 varying P, T to adjust conditions so term in brackets, [ ], is one.
1 I P A P I T I T 0 1 For example if I = 2 then I entering pressures P I and P A will be the same as A , but we need the to be in the relationship P I = 3P A with T A = T I I A 1 2 P A 3P A T A T A A 3 1 3 A 0
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A.2.a. Case 1
That is the term in brackets, [ ], would be 1 which would keep 0 constant with I = A = 0 . Returning to our combined mole balance, rate law and stoichiometry dX dV 0 X
33 A.2.b. Case 2: Variable 0 i.e., P I = P A , T I I A = T A F TI A F TA F A0 F A0 F I0 A Constant T, P 1 I I A 1 I dX dV 1 A k 1 1 X
B. Gas Phase Second Order Reaction Pure A Inerts Plus A 34 C A0 P A RT A F A0 A C TI F A0 1 I I dX dV r A F A0 kC 2 A0I 1 F A0 X 2
35 B. Gas Phase Second Order Reaction I A 1 I P A P I T I T A C 2 A0I F A0 F A0 F A0 I 2 F A0 I 2 A A F A0 1 I 2 P A P I 2 T I T A 2 A C A0 1 I 2 P I P A T A T I 2 dX dV k 1 I 2 C A0 A P I P A T A T I 2 1 X 2
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B. Gas Phase Second Order Reaction For the same temperature and pressures for the cases with and without inerts, i.e., P I = P A and T I = T A , then dX dV k 1 I 2 C A0 A 1 X 2
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Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance
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Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance
39 End of Web Lecture 24 Class Lecture 20