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Chemical Engineering
Department
CDB 2043 KINETICS AND REACTOR DESIGN
CHAPTER 8. STEADY STATE NONISOTHERMAL REACTOR
1
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At the end of this chapter, student should be able to:
1. Explain the importance of energy balance in designing
reactors
2. Apply the concept of energy balance to reactor design
for both adiabatic reaction and reaction involving heat
exchange
3. Evaluate the optimum conversion in reactors where
chemical equilibrium is the limitations.
2
LEARNING OUTCOME OF CHAPTER 8
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LECTURE 1WHY WE NEED THE ENERGY BALANCE EQUATION?
HOW TO DEVELOP THE ENERGY BALANCE EQUATION
3
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Why we need energy balance equation?
What do we have so far?
Mole balance
(design) equation
0dV
dX
PFor
:
A
A
F
r
FRExample
Rate law &
Arrhenius
equation
21
1
11exp
TTR
Ekk
kCrAA
Stoichiometry
)1(0
00
0
0
XCC
vCF
vCFvv
AA
AA
AA
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Why energy balance?
Mole balance(design) equation
0dV
dX
PFor
:
A
A
F
r
FR
Example
Rate law &
Arrhenius
equation
TTR
Ekk
kCrAA
11exp1
1
Stoichiometry
)1(0
00
0
0
XCCvCF
vCF
vv
AA
AA
AA
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Why energy balance?
1st order, exothermic, adiabatic reaction
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Combining these three equations gives:
Why energy balance?
XCF
TTR
Ek
dV
dXA
A
1
11exp
0
0
1
1
For non-isothermal operation: k depends on T (affects final
conversion, X)
T varies with V (reactor length)
Need either
X=f(T) or
T=f(V)
0dV
dX
A
A
F
r
AA C
TTR
Ekr
11exp
1
1
)1(0 XCC AA
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Set valuefor X
CalculateT
Calculatek
CalculaterA
Calculate(FA0/-rA)
8
General approach in solving non-isothermal
reactor problem
Use energy
balance to
relate X
and T
Use Levenspiel
plot techniqueto solve
problem
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The energy balance
General energy balance for open system
outou tinin
..
^
sysEFEFWQ
dt
Ed
Zero for closed system
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The energy balance
For n species, the equation becomes:
outi
n
1i
iini
n
1i
i
..
^
sysEFEFWQ
dt
Ed
Qs is how can we
make this equationUSEABLE @ USER
FRIENDLY?
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Steps to a more user-friendly equation
Evaluate the work term to get the equation in terms of enthalpies
ou ti
n
1i
iini
n
1i
i
..
^
sysEFEFWQ
dt
Ed
isys
Hfdt
Ed
^
Dissect the steady state molar flow rate terms to get the equation in terms of heats of
reaction
Rxsys
Hfdt
Ed
^
Dissect the enthalpies terms to get the equation in terms of heat capacity.
pisys
Cfdt
Ed
^
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Resulting equation based on energy balance* for steady stateflow system:
where:
The energy balance
*Full derivation: pg 474 - 486, Fogler
0)( 01
0
..
THXFTTCFWQ RXAA
n
i
iopiiAs
RpRRXRX TTCTHTH 0
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How do we apply the equation in designing
reactors?
0)( 01
0
..
THXFTTCFWQ RXAA
n
i
iopiiAs
dtr
dXN
V
r
XFV
F
r
A
A
A
A
A
A
0
0
0dV
dX
Solve using
Numerical technique
ODE solver
Graphical method
e.g: Levenspiel plot
method
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8.3 Adiabatic energy balance
General Energy Balance Equation
Consider an adiabatic system with no work done.
)1(
)(
RpR
o
Rx
iopii
TTCTH
TTCX
0W,0Q s..
0)( 01
0
..
THXFTTCFWQ RXAA
n
i
iopiiAs
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Adiabatic tubular reactor
Rearrangement of equation 1 gives:
Couple with mole balance equation:
we can develop the temperature, conversion and
concentration profiles along the reactor.
)2(
0
PPii
RPPiiRRX
CXC
TCXTCTHX
T
),(0 TXrdV
dXF
AA
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Since the second term in the denominator is usually zero, aplot of XEB vs T gives:
X
T
Use together with mole balanceequation in solving reactionengineering problem
XEB XMB
Adiabatic energy balance
)( RoRxiopii
EBTH
TTCX
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CHECKPOINT!
For non-isothermal system, we need 2
equations to describe the conversion, X
XMB, which is based on Material
Balance
XEB, which is based on Energy
Balance
The final conversion is the value that satisfy
both these equations
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Example 8-2: Heat of reaction
Calculate the heat of reaction for the synthesis of ammonia from hydrogen
and nitrogen at 150oC in kcal/mol of N2 reacted and also in H2 reacted. The
reaction is given below:
N2 + 3H2 2NH3
Info:Heat of formation of NH3 at 25
oC = -11,020 cal/mol.N2 reacted
Heat of formation of elements at 25oC = 0
Heat capacities at 25oC
CP,NH3 = 8.92 cal/mol.K
CP,N2 = 6.984 cal/mol.K
CP,H2 = 6.992 cal/mol.K
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Heat of formation at reference temperature (25o
C):
Heat of reaction at any temperature:
19
Solution to Example 8.2
)()()()()( 00000RARBRcRDRRx
THTHa
bTH
a
cTH
a
dTH
RPRRxRx
TTCTHTH )()( 0
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Using these equations, we will get:
20
Solution to Example 8.2
reactedNmol
kcalKH
Rx
2
004.22298
KreactedNmol
calCp
2
12.10
2
3.23Nmol
kcalH
Rx
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Example 8-3: Liquid phase isomerisation of normal
butane
Additional Information:
HRx = -6900 J/mol butane
Activation Energy = 65.7 kJ/molKc = 3.03 at 60
oC
CA0 = 9.3 kmol/dm3
CPn-B = 141 J/mol.KCPi-B = 141 J/mol.K
CPi-P = 161 J/mol.K
Handout
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END OF LECTURE
22
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LECTURE 2STEADY STATE TUBULAR REACTOR WITH HEAT EXCHANGE
CASE 1: CONSTANT TEMPERATURE HEAT EXCHANGE
CASE 2: VARIABLE TEMPERATURE HEAT EXCHANGE
23
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Steady state tubular reactor with heat
exchange (Case 1)
d b l h h
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From:
Deriving the energy balance equation General energy balance equation, with no work done:
Steady state tubular reactor with heat
exchange (Case 1)
TTVUaQ
where
HFHFQ
a
vviivii
.
.
:
0
0
..
vviiviiS HFHFWQ
S d b l h h
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Deriving the energy balance equation After further substitution of the terms, the energy balance
equation for tubular reactor with heat exchange is:
Steady state tubular reactor with heat
exchange (Case 1)
Pii
aRxA
CF
TTUaHr
dV
dT
Heat generated Heat removed
Full derivation: Fogler, pg 496
VALID FOR CONSTANT
TEMPERATURE HEAT EXCHANGE
MEDIUM ONLY!!
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From Example 8.3, the vapour pressure at the exit to theadiabatic reactor at 360K is 1.5MPa for isobutene, which is
greater than the rupture pressure of the glass vessel used. Given
the information below, will the reactor temperature rise above
325K?
Reactor volume (10 banks): 6 m3 (volume of each reactor)
Ua: 5000 kJ/h.m3.K
Ta = 37oC
All other information : refer to Example 8.3
27
Example 8.4: Butane Isomerisation
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WHAT IF THETEMPERATURE OF THE
COOLING / HEATINGMEDIUM IS NOT A
CONSTANT?
28
S d b l i h h
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Energy balance on coolant fluid
Steady state tubular reactor with heat
exchange (Case 2)
St d t t t b l t ith h t
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Energy balance on coolant fluid For co-current case:
For counter-current case:
Steady state tubular reactor with heat
exchange (Case 2)
Pcc
aa
Cm
TTUa
dV
dT
.
Pcc
aa
Cm
TTUa
dV
dT
.
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One of the steps in the production of acetic anhydride is the vapourphase cracking of acetone to ketene and methane. The reaction is first
order with respect to acetone and the reaction rate is given below (T in
K).
The feed to the reactor consist og 7850 kg/hr acetone. The reactor
consist of a bank of 1000 tubes of 1 inch diameter. The inlet
temperature and pressure are the same of both cases at 1035 K and 162
kPa. Plot the conversion and temperature profile along the length of thereactor for the following cases:
Example 8-5: Production of acetic anhydride
Tk
CHCOCHCOCHCH
222,3434.34ln
4233
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1. The reactor is operated adiabatically
2. The reactor is surrounded by a heat exchanger where the heat
transfer coefficient is 110 J/m2.s.K and the temperature of the
heating medium is constant at 1150 K.
3. Same as part (2) but reactor is subjected to heating medium with
changing temperature
Example 8-5: Production of acetic anhydride
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END OF LECTURE 2
33
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LECTURE 3EQUILIBRIUM CONVERSION
34
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Equilibrium conversion
Will consider the following:A. How adiabatic temperature is related to equilibrium
conversion for the following cases:
1. Exothermic reaction
2. Endothermic reaction
B. How to find optimum feed temperature
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A. Adiabatic temperature vs Xe
Exothermic reaction
Equilibrium
TO TO1
TO1>TO
T
X
Xe
Adiabatic
temperature
RX
opiiEB
H
TTCX
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How do we increase Xe for exothermic reaction?
A. Adiabatic temperature vs Xe
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Endothermic reaction
A. Adiabatic temperature vs Xe
X
T
Xe
Adiabatic
temperature
TO
RX
opiiEB
H
TTCX
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How do we increase Xe for endothermic reaction?
A. Adiabatic temperature vs Xe
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Consider reversible & exothermic reaction under adiabaticoperation
B. Optimum feed temperature
Analysis on Xe vs. T(from energybalance) gives thefollowing profile:
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Analysis on conversion vs. reactor length gives the following profile:
B. Optimum feed temperature
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B. Optimum feed temperature
Therefore, if we analyse the achievable conversion in a reactor for a
given entering temperature, we will find:
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END OF LECTURE
43
STEP 1 Evaluate the work term to get the equation in terms
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44
STEP 1: Evaluate the work term to get the equation in terms
of enthalpies
STEP 2: Dissect the steady state molar flow rate terms to
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45
STEP 2: Dissect the steady state molar flow rate terms to
get the equation in terms of heats of reaction
STEP 2: Dissect the steady state molar flow rate terms to
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STEP 2: Dissect the steady state molar flow rate terms to
get the equation in terms of heats of reaction
STEP 3: Dissect the enthalpies terms to get the equation in
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STEP 3: Dissect the enthalpies terms to get the equation in
terms of heat capacity.
STEP 3: Dissect the enthalpies terms to get the equation in
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STEP 3: Dissect the enthalpies terms to get the equation in
terms of heat capacity.
STEP 3: Dissect the enthalpies terms to get the equation in
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STEP 3: Dissect the enthalpies terms to get the equation in
terms of heat capacity.