L13-1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Nonisothermal Reactor DesignSteady–state total energy balance (TEB):
n nsys
s A0 i0 i i RX A0 Ai 1 i 1
ˆdE0 Q W F H H H T F X
dt
For a SS nonisotherm flow reactor:
Tn
s A0 i p,i RX A0 Ai 1Ti0
0 Q W F C dT H T F X
“Simplified” TEB:
n
s A0 i p,i i0 RX A0 Ai 1
0 Q W F C T T (T)FH X Constant (average) heat capacities :
Can rearrange this equation to solve for T
RX Rn
s A0 i p,i i P R0 A0 Ai 1
H (T )0 Q W F ˆ TC T T XT FC
T = reaction temp Ti0 = initial (feed) temperature TR= reference temp
L13-2
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Solve TEB for Conversion
ns A0 i p,i i0 RX R P R A0 A
i 1ˆ0 Q W F C T T H (T ) C T T F X
Solve for XA:
nA0 i p,i i0 s
i 1A
RX R P R A0
F C T T W QX
H (T ) C T T F
Plug in Q for the specific type of reactor
ni p,i i0
i 1A
RX R P R
C T TX
H (T ) C T T
T = reaction temp Ti0 = initial (feed) temperature TR= reference temp
For an adiabatic reaction (Q=0) and shaft work can be neglected (ẆS=0)
L13-3
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Application to CSTRa) Solve TEB for T at the exit (Texit = Tinside reactor)b) Calculate k = Ae-E/RT where T was calculated in step ac) Plug the k calculated in step b into the design equation to calculate VCSTR
Case 1: Given FA0, CA0, A, E, Cpi, H°I, and XA, calculate T & V
a) Solve TEB for T as a function of XA (make a table of T vs XA using EB)b) Solve CSTR design equation for XA as a function of T (plug in k = Ae-E/RT )
(use design eq to make a table of XA vs T)c) Plot XA,EB vs T & XA,MB vs T on the same graph. The intersection of these 2
lines is the conditions (T and XA) that satisfies the energy & mass balance
Case 2: Given FA0, CA0, A, E, Cpi, H°I, and V, calculate T & XA
XA,EB = conversion determined from the TEB equationXA,MB = conversion determined using the design equation
XA
T
XA,EB
XA,MB
XA,exit
Texit
Intersection is T and XA that satisfies both equations
L13-4
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review: Application to a SS PFR
PFRFA0 FA
distance
TXA
Negligible shaft work (ẆS=0) and adiabatic (Q=0)
a) Use TEB to construct a table of T as a function of XA
b) Use k = Ae-E/RT to obtain k as a function of XA
c) Use stoichiometry to obtain –rA as a function of XA
d) Calculate:
XA AA0
A AXA0
dXV Fr X ,T
May use numerical methods
L13-5
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L13: Equilibrium Conversion in Nonisothermal Reactor Design
• The highest conversion that can be achieved in reversible reactions is the equilibrium conversion– For reversible reactions, the equilibrium conversion is
usually calculated first• The equilibrium conversion increases with increasing
temperature for endothermic reactions• The equilibrium conversion decreases with increasing
temperature for exothermic reactions
L13-6
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Review of Equilibrium Kinetics
KC: equilibrium constant (capital K):
c dC D
C a bA B
C CK
C C
kAk A
aA b B c C d D
Gas-phase reaction:
products
reactants
C raised to stoichiometric coefficientsC raised to stoichiometric coefficients
If KC is given at a single temperature T2, & CP can
be neglected then:
RX RC C 2
2
H T 1 1K T K T expR T T
KP: equilibrium constant in terms of partial pressures Pi:
c dC D
P a bA B
P PK
P P i iP CRT
For ideal gases, KP = KC(RT)Δn where n c d b a
Temp dependence of KP is given by van’t Hoff’s equation:
Rx Rx R P RP2 2
H T H T C T TdlnKdT RT RT
RX RP P 2
2
H T 1 1K T K T expR T T
If CP can be
neglected then:
L13-7
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Equilibrium Conversion XAe
T
XA,e
0
1
A B heatexothermic
endothermicA heat B
Example) AB CA0=1 CB0=0
A0 AeBeC
Ae A0 Ae
C 0 XCK
C C 1 X
AeC
Ae
XK
1 X
C Ae AeK 1 X X
Rearrange to solve in terms of XAe
C Ae C AeK X K X C Ae CK X 1 K
C
AeC
KX
1 K
This equation enables us to express Xae as a function of T
L13-8
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
XAe and Temperature
CAe
C
KX
1 K
RX RC C 2
2
H T 1 1K T K T expR T T
Substitute for KC:
RX RC 2
2Ae
RX RC 2
2
H T 1 1K T expR T T
XH T 1 11 K T exp
R T T
Divide numerator & denominator by KC
Ae
RX R
C 2 2
1XH T1 1 1exp 1
K T R T T
XX1 e
e
Changed sign
R
RX RAe
2X
HExot
T 1 1 exp and XR T
hermic: HT
0, when T
R
RX RAe
2X
H TEndo 1thermic: H 0, when T 1 exp and X
R T T
L13-9
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
XAe and Temperature
Ae
RX R
C 2 2
1XH T1 1 1exp 1
K T R T T
RX RAe
2P RX R
H T 1 1 when T eExothermic & C =0: H T 0 xp &, XR T T
p R
RX R
2R AeX
H T 1 1, when T Endothermic & C 0: H T 0 exp & XR T T
A B heatMakes sense from Le Chatelier’s principle
Exothermic rxn produces heat→ increasing temp adds heat (product) & pushes rxn to left (lower conversion)
Makes sense from Le Chatelier’s principle
Heat is a reactant in an endothermic rxn→ increasing temp adds reactant (heat) & pushes rxn to right (higher conversion)
A heat B
L13-10
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
For the elementary solid-catalyzed liquid-phase reaction
A B1. Make a plot of equilibrium conversion as a function of temperature.2. Determine the adiabatic equilibrium temperature and conversion when
pure A is fed to the reactor at a temperature of 298 K.
A
B
PA
PB
e
H (298K) 40000 cal / mol
H (298K) 60000 cal / molC 50 cal / mol K
C 50 cal / mol K
K 100,000 at 298K
Adiabatic Equilibrium T Example
BA A
C
Cr k CK
L13-11
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Rate law:CBr k CA A Ke
equilibrium
-rA = 0Be
Aee
CC
K
A0 e
eA0 e
C XK (T)
C 1 X
ee
e
K TX
1 K T
e e 1R
1
oX 1 1K K (T )exp -
R T TH
R
TpTX RR
oRXH T C dTH T
B Ap p pC C - C 0
o o oRX B AH H - H -20000
The Van’t Hoff equation: e
2d lnK
dT TH
R
1T =298K
Xe = f (T) Xe only depends on thermodynamics!Nothing to do with the energy balance!
A B A B
P P eA B
H (298K) 40000 cal / mol H (298K) 60000 cal / molC 50 cal / mol K C 50 cal / mol K K 100000 at 298K
L13-12
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
nA0 i p,i i0 s
i 1A
ARX R P R 0
F C T T W Q
ˆH (T ) C T FTX
Reaction is carried out adiabatically with an inlet temp of 298 K, CPA = 50 cal/mol∙K, & the heat of reaction = 20,000 cal/mol. The energy balance is:
ni pi 0
i 1EB
RX
C T TX
H T
A
ni pi p
i 1C 1 C
AP 0EB
RX
C T TX
H T
EB
50 T 298X
20000
From thermodynamicsX
T
From energy balance
TRX RX R PTR
H T H (T ) C dT
0 0
How to increase the conversion?
C
AeC
KX
1 K
AP 0EB
RX
C T TX
H T
A B
ns A i pi i A RX R p Ri
ˆQ W F C ( T T ) F X H ( T ) C ( T T ) Rearrang e
0 0 01
0
L13-13
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Does increasing the entering temperature increase XA?
XAe
XA
T (K)XA,EB for adiabatic operation
(slants up for exothermic rxn)
Equilibrium curve
for exothermic rxn
p,A A0A,EB
RX R
C T TX
H (T )
Ae
RX R
C 2 2
1XH T1 1 1exp 1
K T R T T
Higher T moves XA,EB curve to the right
Adiabatic op, XA,EB at T i0,2
XA,e (Tadiabatic) decreases for an exothermic reaction
XA,e at Ti0,1
XA,e at Ti0,2
L13-14
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Optimum Feed TemperatureFor a reversible and exothermic reaction, the feed temperature
should be optimized to maximize the conversion
From thermodynamicsXEB
T600500350
0.15
0.33
0.75
XA
W
T0 = 600T0 = 500
T0 = 350
There is an optimum inlet temperature!
High T0: reaction reaches equilibrium fast, but XA is low
Low T0 would give high XA,e but the specific reaction rate k is so small that most of the reactant passes through the reactor without reacting (never reach XA,e)
L13-15
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
How does one increase XA for adiabatic operation of an exothermic reaction?
with interstage cooling!
XEB
T
final conversion
XA,EB1
cooling process
XA,EB2
XA,EB3
XA,EB4
T0 Reactor 4Reactor 1
Cooling, C1 C2 C3
Reactor 3Reactor 2
Each reactor operates adiabatically
Endothermic reactions are similar, but with heating instead of cooling
L13-16
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
The equilibrium conversion increases with increasing temperature, so use interstage heating to increase the conversion
Endothermic Reactions
XEB
T
heating process
final conversion
Red lines are from the energy balance, slant backwards because H°RX >0 for endothermic reaction
L13-17
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Suppose pure A enters a reactor at 298K . What is the maximum XA achievable in an adiabatic reactor? Assume ẆS=0, and CP = 0, CP.A=60
J/molK, H°RX(TR)= 20,000 J/mol, KC=10 exp[2405T-7.2]
Solve TEB for XA:
ns A0 i p,i i0 RX R P R A0 A
i 10 Q W F C T T H (T ) C T T F X
n
A0 i p,i i0 RX R P R A0 Ai 1
F C T T H (T ) C T T F X
0 0
AA 1 and A is only sp solve ecies for, X :
p,A A0A
RX R P R
C T TX
H (T ) C T T
p,A A0P A
RX R
C T TC 0 so X
H (T )
Plot XEB vs T and XA,e vs T to compute the maximum XA graphically
Ae
RX R
C 2 2
1XH T1 1 1exp 1
K T R T T
L13-18
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Suppose pure A enters a reactor at 298K . What is the maximum XA achievable in an adiabatic reactor? Assume ẆS=0, and CP = 0, CP.A=60
J/molK, H°RX(TR)= 20,000 J/mol, KC=10 exp[2405T-7.2]
p,A A0A,EB
RX R
C T TX
H (T )
Ae
RX R
C 2 2
1XH T1 1 1exp 1
K T R T T
T XA,EB
200 0.294250 0.144298 0325 -0.081350 -0.156375 -0.231400 -0.306425 -0.381450 -0.456500 -0.606550 -0.756600 -0.906
T XAe
200 0.000149250 0.001634298 0.007612325 0.014746350 0.024722375 0.038481400 0.056318425 0.078262450 0.10407500 0.165203550 0.234315600 0.305584
L13-19
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
200 250 300 350 400 450 500 550 600
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
Energy balance conversion
Equilibrium conversion
T (K)
XASuppose pure A enters a reactor at 298K . What is the maximum XA
achievable in an adiabatic reactor? Assume ẆS=0, and CP = 0, CP.A=60 J/molK, H°RX(TR)= 20,000 J/mol (endothermic), KC=10 exp[2504T-7.2]
p,A A0A,EB
RX R
C T TX
H (T )
Ae
RX R
C 2 2
1XH T1 1 1exp 1
K T R T T
XAe
EB for adiabatic operation,
(slants down for endothermic rxn)
Tadiabatic
Nearly 0 conversion, not good!
Tadiabatic: Outlet T if reactor had an infinite volumeXA,e at Tadiabatic is max achievable XA in adiabatic reactor
L13-20
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
200 250 300 350 400 450 500 550 600
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
Energy balance conversionEquilibrium conversion
T (K)
XADoes increasing the inlet temperature to 600K improve the conversion of this
reaction? ẆS=0, and CP = 0, CP.A=60 J/molK, H°RX(TR)= 20,000 J/mol (endothermic), & KC=10 exp[2504T-7.2]
p,A A0A,EB
RX R
C T TX
H (T )
Ae
RX R
C 2 2
1XH T1 1 1exp 1
K T R T T
XAe
Tadiabatic: Outlet T if reactor had an infinite volume
EB for adiabatic operation (endothermic) T0=298K
Tadiabatic
XA,e at Tadiabatic is max achievable XA in adiabatic reactor
Nearly 0 conversion
L13-21
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
200 250 300 350 400 450 500 550 600
-1
-0.75
-0.5
-0.25
0
0.25
0.5
0.75
1
Equilibrium conversion
T (K)
XADoes increasing the inlet temperature to 600K improve the conversion of this
reaction? ẆS=0, and CP = 0, CP.A=60 J/molK, H°RX(TR)= 20,000 J/mol (endothermic), & KC=10 exp[2504T-7.2]
p,A A0A,EB
RX R
C T TX
H (T )
Ae
RX R
C 2 2
1XH T1 1 1exp 1
K T R T T
XAe ≈ 0 at T0 = 298K
EB for adiabatic operation (endothermic) T0=298K
Tadiabatic
EB for adiabatic operation T0=600KXAe ≈ 0.2 when T0 = 600K
Tadiabatic
(T0=600K)
Yes, higher conversion is achieved