lecture 19
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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 19
Today’s lectureGas Phase ReactionsTrends and Optimums
2
User Friendly Equations Relate T and X or Fi
1. Adiabatic CSTR, PFR, Batch, PBR achieve this:
0CW PS
Rx
0PiEB H
TTCX i
Rx
0P
HTTC~
X A
iPi
Rx0 C
XHTT
Last Lecture
3
2. CSTR with heat exchanger, UA(Ta-T) and a large coolant flow rate
Rx
0Pia0A
EB H
TTC~TTFUA
Xi
TTa
Cm
User Friendly Equations Relate T and X or Fi
4
3. PFR/PBR with heat exchange
FA0T0
CoolantTa
User Friendly Equations Relate T and X or Fi
3A. In terms of conversion, X
XCC~F
THrTTUa
dWdT
pPi0A
RxAaB
i5
User Friendly Equations Relate T and X or Fi
3B. In terms of molar flow rates, Fi
i
ij
Pi
RxAaB
CF
THrTTUa
dWdT
4. For multiple reactions
i
ij
Pi
RxijaB
CF
HrTTUa
dVdT
5. Coolant Balance
cPc
aA
CmTTUa
dVdT
6
Reversible Reactions
endothermic reaction
exothermic reaction
KP
T
endothermic reaction
exothermic reaction
Xe
T
7
Heat ExchangeExample: Elementary liquid phase reaction carried out in a PFR
FA0FI
Tacm Heat
Exchange Fluid
A B
The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1.8
Heat Exchangea) Adiabatic. Plot X, Xe, T and the rate of
disappearance as a function of V up to V = 40 dm3.
b) Constant Ta. Plot X, Xe, T, Ta and Rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature is constant at 300 K for V = 40 dm3. How do these curves differ from the adiabatic case.
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Heat Exchangec) Variable Ta Co-Current. Plot X, Xe, T,
Ta and Rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature varies along the length of the reactor for V = 40 dm3. The coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)?
d) Variable Ta Counter Current. Plot X, Xe, T, Ta and Rate of disappearance of A when there is a heat loss to the coolant and the coolant temperature varies along the length of the reactor for V = 20 dm3. The coolant enters at 300 K. How do these curves differ from those in the adiabatic case and part (a) and (b)?
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Example: PBR A ↔ B
5) Parameters….
Gas Phase Heat Effects
• For adiabatic:• Constant Ta:
• Co-current: Equations as is • Counter-current:
0dWdTa
T)-T toT-T flip(or )1(dWdT
aa
0UA
11
Reversible Reactions
)1( FrdWdX
0AAMole
Balance:
0A
A
0A
BA
b
Fr
Fr
dVdX
VW
12
Reversible Reactions
)3( T1
T1
REexpkk
11
)4( T1
T1
RHexpKK
2
Rx2CC
)2( KCCkr
C
BAA
Rate Law:
13
Reversible Reactions
Stoichiometry:
€
5( ) CA =CA 0 1 − X( )y T0 T( )
€
6( ) CB =CA 0Xy T0 T( )
€
dydW
=αyFTFT0
TT0
⎛ ⎝ ⎜
⎞ ⎠ ⎟= −
α2y
TT0
⎛ ⎝ ⎜
⎞ ⎠ ⎟
W = ρV
dydV
= −αρ b2y
TT0
⎛ ⎝ ⎜
⎞ ⎠ ⎟
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Reversible Reactions
b00A2Rx
2C110A
, ,T ,C ,T ,H
)15()7( ,K ,T ,R ,E ,k ,F
Parameters:
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Reversible Reactions
Example: PBR A ↔ B
3) Stoich (gas):
€
v = v0 1+εX( )P0
PTT0
5( ) CA =FA 0 1 − X( )v0 1+εX( )
PP0
T0
T=CA 0 1− X( )
1+εX( )yT0
T
6( ) CB =CA 0X1+εX( )
yT0
T
7( ) dydW
=−α2y
FTFT 0
TT0
=−α2y
1+εX( )TT0
Gas Phase Heat Effects
16
Reversible Reactions
€
KC =CBeCAe
=CA 0X eyT0 T
CA 0 1− X e( )yT0 T
8( ) Xe =KC
1+KC
Example: PBR A ↔ B
17
Gas Phase Heat EffectsReversible Reactions
Example: PBR A ↔ B
Gas Phase Heat Effects
18
Exothermic Case: Xe
T
KC
T
KC
T T
Xe~1
Endothermic Case:
Reversible Reactions
Adibatic EquilibriumConversion on Temperature Exothermic ΔH is negativeAdiabatic Equilibrium temperature (Tadia) and conversion (Xeadia)
PA
Rx0 C
XHTT
C
Ce K1
KX
X
Xeadi
a
Tadia T19
X2
FA0 FA1 FA2 FA3
T0X1 X3T0 T0
Q1 Q2
20
X
T
X3
X2
X1
T0
Xe XEB
RX
0Pii
HTTC
X
21
Gas Phase Heat Effects
22
€
dTdV
=−rA( ) −ΔHRx( ) −Ua T −Ta( )
∑FiCPi
€
∑FiCPi = FA 0 ∑ΘiCPi + ΔCPX[ ]
Case 1: Adiabtic and ΔCP=0
€
T = T0 +−ΔHRx( )X∑ΘiCPi
(16A)
Additional Parameters (17A) & (17B)
€
T0, ∑ΘiCPi =CPA + ΘICPI
Example A ↔ B
Heat effects:
€
dTdW
=−ra( ) −ΔHR( ) −
Ua
ρ bT −Ta( )
FA 0 θ iCPi∑ 9( )
23
Case 2: Heat Exchange – Constant Ta
Gas Phase Heat EffectsExample A ↔ B
)C17( TT 0V , Cm
TTUadVdT
aoaP
aa
cool
Case 3. Variable Ta Co-Current
Case 4. Variable Ta Counter Current Guess ?T 0V
CmTTUa
dVdT
aP
aa
cool
Guess Ta at V = 0 to match Ta0 = Ta0 at exit, i.e., V = Vf
24
Example A ↔ B
25
26
27
28
29
Endothermic
PFR
A B
dXdV
k 1 1 1KC
X
0 , Xe
KC1KC
XEB iCPi
T T0 HRx
CPA
ICPI T T0 HRx
T0
XXEB
Xe
TT0 HRx X
CPAICPI
30
31
Effects of Inerts In The Feed
What happens when we vary
k1I
32
Endothermic
As inert flow increases the conversion will increase. However as inerts increase, reactant concentration decreases, slowing down the reaction. Therefore there is an optimal inert flow rate to maximize X.
First Order Irreversible
33
T
X
Adiabatic T and Xe
T0
exothermic
T
X
T0
endothermic
PIIPA
R0 CC
XHTT
Gas Phase Heat EffectsTrends:-Adiabatic:
34
Adiabatic:• As T0 decreases the conversion X will increase,
however the reaction will progress slower to equilibrium conversion and may not make it in the volume of reactor that you have.
• Therefore, for exothermic reactions there is an optimum inlet temperature, where X reaches Xeq right at the end of V. However, for endothermic reactions there is no temperature maximum and the X will continue to increase as T increases.
Gas Phase Heat Effects
35
X
T
Xe
T0
X
T
X
T
Adiabatic:
Gas Phase Heat Effects
36
Effect of adding Inerts
X
T
V1 V2X
TT0
I
0I
Xe
X
€
X =T −T0( ) CpA +θ ICpI[ ]
−ΔHRx
37
Exothermic Adiabatic
As θI increase, T decrease and
€
dXdV
=k
υ 0 Hθ I( )
k
θI
38
39
Adiabatic
KC
V
XeX
V
T
V
Xe
V
k
V
Xe
XFrozen
V
OR
Endothermic
T
V
KC
V
Xe
V
XeX
V
k
V
Exothermic
40
Heat Exchange
T
V
KC
V
Xe
V
XeX
V
Endothermic
T
V
KC
V
Xe
V
XeX
V
Exothermic
41
Derive the Steady State Energy Balance (w/o Work)
0HFHFdWTTUaii0i0ia
W
0 B
0dWdHFH
dWdF0TTUa i
iii
aB
Differentiating with respect to W:
42
Aii
i rrdWdF
Mole Balance on species i:
T
TPiRii
R
dTCTHH
Enthalpy for species i:
Derive the Steady State Energy Balance (w/o Work)
43
Differentiating with respect to W:
dWdTC0
dWdH
Pii
0dWdTCFHrTTUa
PiiiiAaB
Derive the Steady State Energy Balance (w/o Work)
44
0dWdTCFHrTTUa
PiiiiAaB
THH Rii
XFF ii0Ai
Final Form of the Differential Equations in Terms of Conversion:A:
45
Final Form of terms of Molar Flow Rate:
Pii
AB
CF
HrUa
dWdT
B: T,Xg
Fr
dWdX
0A
A
46
(d) Gas Phase Counter Current Heat Exchange Vf = 20 dm3
Matches
47
End of Lecture 19
48
Example: PBR A ↔ B VW b
2) Rates:
4 T1
T1
RHexpkk
3 T1
T1
REexpkk
2 kCCkr
2
R2CC
11
CBA'A
1 F
rdWdX
0A
'A1) Mole balance:
Gas Phase Heat Effects
49
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