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Steady State Nonisothermal Steady State Nonisothermal Reactor DesignReactor Design
Dicky DermawanDicky Dermawanwww.dickydermawan.net78.net
ITK-330 Chemical Reaction EngineeringITK-330 Chemical Reaction Engineering
RationaleRationale All reactions always accompanied by heat effect:
exothermic reactions vs. endothermic reactions Unless heat transfer system is carefully
designed, reaction mass temperature tend to change
Design of heat transfer system itself requires the understanding of this heat effect
Energy balance is also needed, together with performance equations derived from mass balance
ObjectivesObjectives Describe the algorithm for CSTRs, PFRs, and PBRs that
are not operated isothermally. Size adiabatic and nonadiabatic CSTRs, PFRs, and
PBRs. Use reactor staging to obtain high conversions for highly
exothermic reversible reactions. Carry out an analysis to determine the Multiple Steady
States (MSS) in a CSTR along with the ignition and extinction temperatures.
Analyze multiple reactions carried out in CSTRs, PFRs, and PBRs which are not operated isothermally in order to determine the concentrations and temperature as a function of position (PFR/PBR) and operating variables
Why Energy Balance?Why Energy Balance?Imagine that we are designing a nonisothermal PFR for a Imagine that we are designing a nonisothermal PFR for a
first order liquid phase exothermic reaction:first order liquid phase exothermic reaction:
Performance Performance equation:equation: 0A
AF
rdVdX
Kinetics:Kinetics: Ar k AC
The temperature will increase with conversion down
the length of reactor
T1
T1
RE
expkk1
a1
Stoichiometry:Stoichiometry: 0 A0A CF
0A00A CF )X1(CC 0AA
Combine:Combine:
0
X1
1
a1 T
1RE
expkdVdX
T1
)V,T(XX
)V(TT
)X(TT )V(XX
Energy BalanceEnergy Balance
n
1i0I0I0D0D0C0C0B0B0A0A0i0i HFHFHFHFHFHF:In
n
1iIIDDCCBBAAii HFHFHFHFHFHFOut
At steady state:
dtEd
HFHFWQ sysn
1iii
n
1i0i0is
n
1isWQ
0iF 0iH
n
1iiF iH 0
Consider generalized reaction:
DCBA ad
ac
ab
I0AI
ad
D0AD
ac
C0AC
ab
B0AB
0AA
FF
)X(FF
)X(FF
)X(FF
)X1(FF
Upon substitution:
ABab
Cac
Dad
A0 HHHHXF-
n
1iii
n
1i0i0i HFHF
CI0IID0DD
C0CCB0BBA0A0A HHHH
HHHHHHF
n
1iii
n
1i0i0i HFHF
n
1ii0ii0A HHF )T(HXF Rx0A
Energy Balance (cont’)Energy Balance (cont’)
n
1iii
n
1i0i0i HFHF
n
1i0iii0A HHF )T(HXF Rx0A
T
TpiR
oii
R
dTC)T(HH
From thermodynamics, we know that:
0i
R
T
TpiR
oi0i dTC)T(HH Thus: )TT(C~dTCHH 0ipi
T
Tpi0ii
0i
0i
T
Tpi
pi TT
dTC
C~ 0i
RpRoRxRx TTC)T(H)T(H
R
T
Tpi
pi TT
dTC
C R
)T(H)T(H)T(H)T(H)T(H RoDR
oDa
bR
oDa
cR
oDa
dR
oRx
pApBab
pCac
pDad
p CCCCC
n
1i0pii0A )TT(C~F
Energy Balance (cont’)Energy Balance (cont’)
n
1iii
n
1i0i0i HFHF
n
1i0iii0A HHF )T(HXF Rx0A
Upon substitution:
n
1iii
n
1i0i0i HFHF ]TTC)T(H[XF RpR
oRx0A
Finally….
0HFHFWQn
1iii
n
1i0i0is
0TTC)T(HXF)TT(C~FWQ RpRoRx0A
n
1i0ipii0As
So what?
Energy Balance (cont’)Energy Balance (cont’)For adiabatic reactions:
The energy balance at steady state becomes:
After rearrangement:
0Q
When work is negligible: 0Ws
0TTC)T(HXF)TT(C~F RpRoRx0A
n
1i0ipii0A
RpRoRx
n
1i0ipii
TTC)T(H
)TT(C~
X
This is the X=X(T) we’ve been looking for!
Application to Adiabatic CSTR DesignApplication to Adiabatic CSTR Design
)X1(CC 0AA
Case A: Sizing: X specified, calculate V (and T)
Performance equation:
Kinetics:
Stoichiometry:
Combine:
A
0A
rXF
V
Ar k AC
T1
T1
RE
expkk1
a1
)X1(CkXF
V0A
0A
Solve the energy balance for T
RpRoRx
n
1i0ipii
TTC)T(H
)TT(C~
X
Calculate k
Calculate V using combining equation
Application to Adiabatic CSTR DesignApplication to Adiabatic CSTR Design
)X1(CC 0AA
Case B (Rating): V specified, calculate X (and T)
Performance equation:
Kinetics:
Stoichiometry:
Mole balance:
A
0A
rXF
V
Ar k AC
T1
T1
RE
expkk1
a1
)X1(CkXFV
mb0A
mb0A
Energy balance: RpR
oRx
n
1i0ipii
ebTTC)T(H
)TT(C~
X
Find X & T that satisfy BOTH the material balance and energy balance,
viz. plot Xmb vs T and Xeb vs T in the same graph: the intersection is the solution
Application to Adiabatic CSTR DesignApplication to Adiabatic CSTR DesignExample: P8-5A
The elementary irreversible organic liquid-phase reaction:A + B C
is carried out adiabatically in a CSTR. An equal molar feed in A and B enters at 27oC, and the volumetric flow rate is 2 L/s.
(a) Calculate the CSTR volume necessary to achieve 85% conversion
(b) Calculate the conversion that can be achieved in one 500 L CSTR and in two 250 L CSTRs in series
mol/kcal 41)K273(H
mol/kcal 15)K273(H
mol/kcal 20)K273(H
oC
oB
oA
cal/mol.K 30C
cal/mol.K 15C
cal/mol.K 15C
pC
pB
pA
cal/mol 10000E
K 300at 01.0k
a
smolL
mol/L 1.0C 0A
Application to Adiabatic CSTR DesignApplication to Adiabatic CSTR DesignCase A: Sizing: X specified, calculate V (and T)
Performance equation:
Kinetics:
Stoichiometry:
Combine:
A
0A
rXF
V
Ar k BA CC
T1
T1
RE
expkk1
a1
20A
022
0A
0A
)X1(CkX
)X1(CkXFV
Energy balance: RpR
oRx
n
1i0ipii
TTC)T(H
)TT(C~
X
Calculate k
Calculate V using combining equation
)X1(CC 0AA )X1(C)X(CC 0ABB0AB
Kcal/mol 301515CCC~ pBBpA
n
1ipii
cal/mol 6000- kcal/mol 6152041HHH)273(H oB
0A
oC
oRx
0151530CCCC pBpApCp
K47020085.0300T200
300T)6000(
)300T(3085.0
smolL 317.4
4701
3001
987.110000exp01.0k
L 175)85.01(1.0317.4
85.02V 2
Application to Adiabatic CSTR DesignApplication to Adiabatic CSTR Design
)X1(CCC 0ABA
RpRoRx
n
1i0ipii
ebTTC)T(H
)TT(C~
X
Case B (Rating): V specified, calculate X (and T)
Performance equation:
Kinetics:Stoichiometry:Mole balance:
A
0A
rXF
V
T1
T1
RE
expkk1
a1
2mb0A
mb0
)X1(CkXV
Energy balance:
Ar k BA CC
2mb
mb
)X1(1.0T1
3001
987.110000exp01.0
X2500
200300T
)6000()300T(30Xeb
0
0.2
0.4
0.6
0.8
1
300 350 400 450 500
Xmb
Xeb
T 300 310 320 330 340 350 360 370 380 390 400 410 420 430 440 450 460 470 480 482 484 485 490 500Xmb 0.172 0.245 0.325 0.406 0.482 0.552 0.613 0.666 0.711 0.750 0.783 0.810 0.834 0.854 0.871 0.885 0.898 0.908 0.918 0.919 0.921 0.922 0.926 0.933Xeb 0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500 0.550 0.600 0.650 0.700 0.750 0.800 0.850 0.900 0.910 0.920 0.925 0.950 1.000
Application to Adiabatic PFR/PBR DesignApplication to Adiabatic PFR/PBR Design
TT
PP
X1X1CC 0
00AA
TTC)T(H
)TT(C~
XRpR
oRx
n
1i0ipii
Example for First Order Reaction
Performance equation:
Kinetics:Stoichiometry:
Pressure drop:
T1
T1
RE
expkk1
a1
Energy balance:
Ar k AC
for PFR/small P: P/P0 = 1
)X1(P/P
PTT
2dWdP
0
0
0
)X1(CC 0AA Gas liquid
0A
A
Fr
dWdX
p
n
1ipii
n
1i0piiRp
oRx
n
1i0piiRp
oRxp
n
1ipii
n
1i
n
1i0piipiiRpp
oRx
CXC~
TC~TCXHXT
TC~TCXHXTCXTC~
TC~TC~TCXTCXHX
)X(TT
Combine:)X(k
)X(TT)T(kk
)P,X(CC )X(TT
)P,T,X(CCAA
AA
)P,X(rr])P,X[C],X[k(rr
AA
AAA
)P,X(g)P,T,X(gdWdP
)P,X(f )r(fdWdX
A
ThusThe combination results in 2 simultaneous
differential equations
Sample Sample Problem Problem for for Adiabatic Adiabatic PFR PFR DesignDesign
P8-6A
Sample Problem for Adiabatic PBR Sample Problem for Adiabatic PBR DesignDesign
NINA = Diabatic Reactor DesignNINA = Diabatic Reactor DesignHeat Transfer Rate to the ReactorHeat Transfer Rate to the Reactor
0TTC)T(HXF)TT(C~FW RpRoRx0A
n
1i0ipii0As
Q
Rate of energy transferred between the reactor and the coolant:
The rate of heat transfer from the exchanger to the reactor:
2a
1a
2a1a
TTTT
ln
TTAUQ
Combining:
0HXF)TT(C~FW Rx0A
n
1i0ipii0As
Q
NINA = Diabatic Reactor DesignNINA = Diabatic Reactor DesignHeat Transfer Rate to the Reactor (cont’)Heat Transfer Rate to the Reactor (cont’)
0TTC)T(HXF)TT(C~FW RpRoRx0A
n
1i0ipii0As
Q
At high coolant flow rates the exponential term will be small, so we can expand the exponential term as a Taylor Series, where the terms of second order or greater are neglected:Then:
0HXF)TT(C~FW Rx0A
n
1i0ipii0As
TTAU 1a
The energy balance becomes:
SampleSampleProblem forProblem forDiabaticDiabaticCSTRCSTRDesignDesign
P8-4BP8-4B
Sample Problem for Diabatic CSTR DesignSample Problem for Diabatic CSTR Design
Application of Energy Balance to Diabatic Application of Energy Balance to Diabatic Tubular Reactor DesignTubular Reactor Design
Heat transfer in CSTR: TTAUQ 1a
In PFR, T varies along the reactor:
dVTTVAUdATTUQ
V
a
A
a
TTaUdVQd
a
Thus:
D4
L
LD a
areaktor tabung volume
reaktor tabung selimut luasVA
4D2
For PBR: dW1dVWVVW
bbb
Thus: TTaUdW
Qda
b
Application of Energy Balance to Diabatic Tubular Application of Energy Balance to Diabatic Tubular Reactor DesignReactor Design
The steady state energy balance, neglecting work term:
Differentiation with respect to the volume V:
TTaUdVQd
a
and recalling that
Or:
0TTC)T(HXF)TT(C~FQ RpRoRx0A
n
1i0ipii0A
ΔΔΘ
0dTC)T(HXFdTCFQT
TpR
oRx0A
T
Tpii0A
Ro
ΔΔΘ
Inserting
0dVdX
dTC)T(HFdVdT
CXFdVdT
CFdVQd T
TpR
oRx0Ap0Apii0A
R
ΔΔΔΘ
dVdXFr 0AA
TTaU a dVdT
CXCF ppii0A ΔΘ )]T(H[r RxA Δ 0
ppii0A
RxAa
CXCF
)]T(H[rTTaU
dVdT
ΔΘ
Δ
Coupled with 0A
AF
rdVdX
)T,X(g
)T,X(f
Form 2 differential with 2 dependent variables X & T
Sample Problem for Diabatic Tubular Reactor Sample Problem for Diabatic Tubular Reactor DesignDesign
Design for Reversible ReactionsDesign for Reversible Reactions
Endotermik: Endotermik: K naik dengan kenaikan T X Xeqeq naik naik reaksikan pada Tmax yang diperkenankan
K lnTRG Δ
2Rx
TR
HdT
K) (lnd
Δ
Eksotermik: Eksotermik: K turun dengan kenaikan T X Xeqeq turun turun reaksikan pada T rendah
Laju reaksi lambat pada T rendah!
Ada trade off antara aspek termodinamika dan kinetika
Xeq = Xeq (K)
= Xeq (T)
Design for Reversible Highly-Exothermic Design for Reversible Highly-Exothermic ReactionsReactions
-r-rAA = -r = -rAA (X,T) (X,T)
Generally:Generally: Higher X Higher X slower reaction rate slower reaction rateHigher T Higher T faster rate faster rate
At X = XAt X = Xeqeq : : -r-rAA = 0 = 0
Design for Equilibrium Highly-Exothermic Design for Equilibrium Highly-Exothermic ReactionsReactions
#1#1 Starting with R-free solution, between 0 dan 100 Starting with R-free solution, between 0 dan 100ooC determine the C determine the equilibrium conversion of A for the elementary aqueous reaction:equilibrium conversion of A for the elementary aqueous reaction:
A A R R
cal/mol 18000Hcal/mol 3375G
0298
0298
ΔΔ
The reported data is based on the following standard states of The reported data is based on the following standard states of reactants and products:reactants and products: 1mol/LCC 0
A0R
Assume ideal solution, in which case:Assume ideal solution, in which case: CA
R0AA
0RR K
CC
C/C
C/CK
In addition, assume specific heats of all solutions are equal In addition, assume specific heats of all solutions are equal to that of waterto that of water
Ccal/g. 1C 0p
Design for Equilibrium Highly-Exothermic Design for Equilibrium Highly-Exothermic Reactions:Reactions:
Reaction Rate in X – T DiagramReaction Rate in X – T Diagram
k T( ) 0.0918exp 58591T
1298
rA X T( ) k T( ) CA0 1 XX
K T( )
Reaction Rate in The X – T DiagramReaction Rate in The X – T Diagramat Cat CA0A0 = 1 mol/L = 1 mol/L
0 10 20 30 40 50 60 70 80 90 1000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Suhu, C
Kon
vers
i
rA 0 01
rA 0 025
rA 0 05
rA 0 1
rA 0 25
rA 0 5
rA 1
rA 2rA 4
Design for Equilibrium Highly-Exothermic ReactionsDesign for Equilibrium Highly-Exothermic Reactions:: Optimum Temperature ProgressionOptimum Temperature Progression
in Tubular Reactorin Tubular Reactor#3#3a.a. Calculate the space time needed for 80% conversion of a feed starting with initial Calculate the space time needed for 80% conversion of a feed starting with initial
concentration of A of 1 mol/Lconcentration of A of 1 mol/Lb.b. Plot the temperature and conversion profile along the length of the reactorPlot the temperature and conversion profile along the length of the reactor
Let the maximum operating allowable temperature be 95Let the maximum operating allowable temperature be 95ooCC
Design for Reversible Reactions: Heat EffectDesign for Reversible Reactions: Heat Effect
RpRoRx
n
1i0ipiia
0A
TTC)T(H
)TT(C~)TT(F
AU
X
ΔΔ
Θ
Design for Equilibrium Highly-Exothermic Design for Equilibrium Highly-Exothermic ReactionsReactions:: CSTR Performance CSTR Performance
Design for Equilibrium Highly-Exothermic Design for Equilibrium Highly-Exothermic ReactionsReactions:: CSTR Performance CSTR Performance
#4#4 A concentrated aqueous A-solution of the previous A concentrated aqueous A-solution of the previous examples, Cexamples, CA0A0 = 4 mol/L, F = 4 mol/L, FA0A0 = 1000 mol/min, is to be 80% = 1000 mol/min, is to be 80% converted in a mixed reactor. converted in a mixed reactor.
a.a. If feed enters at 25If feed enters at 25ooC, what size of reactor is needed?C, what size of reactor is needed?
b.b. What is the optimum operating temperature for this What is the optimum operating temperature for this purpose?purpose?
c.c. What size of reactor is needed if feed enters at optimum What size of reactor is needed if feed enters at optimum temperature?temperature?
d.d. What is the heat duty if feed enters at 25What is the heat duty if feed enters at 25ooC to keep the C to keep the reactor operation at its the optimum temperature?reactor operation at its the optimum temperature?
Interstage CoolingInterstage Cooling
Review on Energy Balance in CSTR OperationReview on Energy Balance in CSTR Operation
Bila term kerja diabaikan dan HRx konstan:
n
1i0pii0As )TT(C~FW TTAU a 0HXF 0
Rx0A
XF 0A
a
0A
n
1i0pii0A
0Rx TT
FAU)TT(C~FH
Untuk CSTR:A
0Ar
XFV
VrA
a
0A00p0A
0Rx TT
FAU)TT(CFH
Pembagian kedua ruas dengan FA0:
0p00p
0Rx
0A
A C)TT(CHF
Vr0p0A CF
AU aTT
0p0A CFAU
1TT
1
TT
CFAUTAUTCF
T a0
CFAU
aCFAU
0
0p0A
a00p0Ac
0p0A
0p0A
Multiple Steady Multiple Steady State & Stability of State & Stability of CSTR OperationCSTR Operation
1TTT a0
c
)TT( a0
)TT[(CHF
Vr00p
0Rx
0A
A ]TT a
)1(Tc )1(T[C 0p
)TT()1(C c0p
TT[C 0p ]
]
0RxHX )TT()1(C c0p
)T(G )T(R
A
0Ar
XFV
TTC)T(H
)TT(C~
X
dengan Bandingkan
RpRoRx
n
1i0pii
Review on Energy Balance in CSTR OperationReview on Energy Balance in CSTR Operation
Multiple Steady Multiple Steady State: Stability of CSTR State: Stability of CSTR OperationOperation
Temperature Ignition – Extinction CurveFinding Multiple Steady State: Varying To
Upper steady stateLower steady stateIgnition temperatureExtinction temperatureRunaway Reaction
Sample Problem on Multiple Steady State in Sample Problem on Multiple Steady State in CSTR OperationCSTR Operation
P8-17BP8-17B
