Steady State Steady State NonisothermalNonisothermal
Reactor DesignReactor Design
DickyDicky DermawanDermawanwww.dickydermawan.net78.netwww.dickydermawan.net78.net
[email protected]@gmail.com
ITKITK--330 Chemical Reaction Engineering330 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
A
F
r
dV
dX −=
Kinetics:Kinetics: =− Ar k AC⋅
The temperature will increase with conversion down
the length of reactor
−⋅⋅=T
1
T
1
R
Eexpkk
1
a1
Stoichiometry:Stoichiometry: 0υ=υ A0A CF ⋅υ=
0A00A CF ⋅υ=)X1(CC 0AA −⋅=
Combine:Combine:
0
X1
υ
−
−⋅⋅=
1
a1
T
1
R
Eexpk
dV
dX
T
1)V,T(XX =
)V(TT =
)X(TT =)V(XX =
Energy BalanceEnergy Balance
∑=
⋅+⋅+⋅+⋅+⋅=⋅n
1i
0I0I0D0D0C0C0B0B0A0A0i0i HFHFHFHFHFHF:In
∑=
⋅+⋅+⋅+⋅+⋅=⋅n
1i
IIDDCCBBAAii HFHFHFHFHFHFOut
At steady state:
dt
EdHFHFWQ
sysn
1i
ii
n
1i
0i0is =⋅−⋅+− ∑∑==
&&
∑=
+−n
1i
sWQ &&0iF 0iH ∑
=
−n
1iiF iH 0=
Consider generalized reaction:
DCBAad
ac
ab +→+
I0AI
ad
D0AD
ac
C0AC
ab
B0AB
0AA
FF
)X(FF
)X(FF
)X(FF
)X1(FF
Θ⋅=
+Θ⋅=
+Θ⋅=
−Θ⋅=
−⋅=
Upon substitution:
( )ABab
Cac
Dad
A0 HHHHXF- −−+⋅⋅
∑∑==
⋅−⋅n
1i
ii
n
1i
0i0i HFHF( ) ( ) ( )
( ) ( )
−⋅Θ+−⋅Θ+
−⋅Θ+−⋅Θ+−⋅=
CI0IID0DD
C0CCB0BBA0A0A HHHH
HHHHHHF
∑∑==
⋅−⋅n
1i
ii
n
1i
0i0i HFHF ( )∑=
−⋅Θ⋅=n
1i
i0ii0A HHF )T(HXF Rx0A ∆⋅⋅−
Energy Balance (cont’)Energy Balance (cont’)
∑∑==
⋅−⋅n
1i
ii
n
1i
0i0i HFHF ( )∑=
−⋅Θ⋅−=n
1i
0iii0A HHF )T(HXF Rx0A ∆⋅⋅−
∫ ⋅+=T
T
piRoii
R
dTC)T(HH
From thermodynamics, we know that:
∫ ⋅+=0i
R
T
T
piRoi0i dTC)T(HH Thus: )TT(C
~dTCHH 0ipi
T
T
pi0ii
0i
−⋅=⋅=− ∫
0i
T
T
pi
piTT
dTC
C~ 0i
−
⋅
=
∫
( )RpRoRxRx TTC)T(H)T(H −⋅∆+∆=∆
R
T
T
pi
piTT
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 −⋅−⋅+⋅=∆
∑=
−⋅⋅Θ⋅−=n
1i
0pii0A )TT(C~
F
Energy Balance (cont’)Energy Balance (cont’)
∑∑==
⋅−⋅n
1i
ii
n
1i
0i0i HFHF ( )∑=
−⋅Θ⋅−=n
1i
0iii0A HHF )T(HXF Rx0A ∆⋅⋅−Upon substitution:
∑∑==
⋅−⋅n
1i
ii
n
1i
0i0i HFHF ( )]TTC)T(H[XF RpRoRx0A −⋅∆+∆⋅⋅−
Finally;.
0HFHFWQn
1iii
n
1i0i0is =⋅−⋅+− ∑∑
==
&&
( ) 0TTC)T(HXF)TT(C~
FWQ RpRoRx0A
n
1i
0ipii0As =−⋅∆+∆⋅⋅−−⋅⋅Θ⋅−− ∑=
&&
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
1i
0ipii0A =−⋅∆+∆⋅⋅−−⋅⋅Θ⋅− ∑=
( )[ ]RpRoRx
n
1i
0ipii
TTC)T(H
)TT(C~
X−⋅∆+∆−
−⋅⋅Θ
=∑=
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
r
XFV
−
⋅=
=− Ar k AC⋅
−⋅⋅=T
1
T
1
R
Eexpkk
1
a1
)X1(Ck
XFV
0A
0A
−⋅⋅
⋅=
Solve the energy balance for T
( )[ ]RpRoRx
n
1i
0ipii
TTC)T(H
)TT(C~
X−⋅∆+∆−
−⋅⋅Θ
=∑=
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
r
XFV
−
⋅=
=− Ar k AC⋅
−⋅⋅=T
1
T
1
R
Eexpkk
1
a1
)X1(Ck
XFV
mb0A
mb0A
−⋅⋅
⋅=
Energy balance:( )[ ]RpR
oRx
n
1i
0ipii
ebTTC)T(H
)TT(C~
X−⋅∆+∆−
−⋅⋅Θ
=∑=
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 Design
Example: 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
r
XFV
−
⋅=
=− Ar k BA CC ⋅⋅
−⋅⋅=T
1
T
1
R
Eexpkk
1
a1
20A
022
0A
0A
)X1(Ck
X
)X1(Ck
XFV
−⋅⋅
⋅υ=
−⋅⋅
⋅=
Energy balance:
( )[ ]RpRoRx
n
1i
0ipii
TTC)T(H
)TT(C~
X−⋅∆+∆−
−⋅⋅Θ
=∑=
Calculate k
Calculate V using combining equation
)X1(CC 0AA −⋅=
)X1(C)X(CC 0ABB0AB −⋅=⋅ν−Θ⋅=
Kcal/mol 301515CCC~
pBBpA
n
1i
pii ⋅=+=⋅Θ+=⋅Θ∑=
cal/mol 6000- kcal/mol 6152041HHH)273(H oB
0A
oC
oRx =−=++−=−−=∆
0151530CCCC pBpApCp =−−=−−=∆
K47020085.0300T200
300T
)6000(
)300T(3085.0 =⋅+=⇒
−=
−−−⋅
=
smol
L 317.4
470
1
300
1
987.1
10000exp01.0k
⋅=
−⋅⋅=
L 175)85.01(1.0317.4
85.02V
2=
−⋅⋅
⋅=
Application to Adiabatic CSTR DesignApplication to Adiabatic CSTR Design
)X1(CCC 0ABA −⋅==
( )[ ]RpRoRx
n
1i
0ipii
ebTTC)T(H
)TT(C~
X−⋅∆+∆−
−⋅⋅Θ
=∑=
Case B (Rating): V specified, calculate X (and T)
Performance equation:
Kinetics:
Stoichiometry:
Mole balance:
A
0A
r
XFV
−
⋅=
−⋅⋅=T
1
T
1
R
Eexpkk
1
a1
2mb0A
mb0
)X1(Ck
XV
−⋅⋅
⋅υ=
Energy balance:
=− Ar k BA CC ⋅⋅
2mb
mb
)X1(1.0T
1
300
1
987.1
10000exp01.0
X2500
−⋅⋅
−⋅⋅
⋅=
200
300T
)6000(
)300T(30Xeb
−=
−−
−⋅=
0
0.2
0.4
0.6
0.8
1
300 350 400 450 500
Xmb
Xeb
T 300 310 320 330 340 350 360 370 380 390 400 410 420 430 440 450 460 470 480 482 484 485 490 500
Xmb 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 0.933
Xeb 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
T
T
P
P
X1
X1CC 0
0
0AA ⋅⋅⋅ε+
−⋅=
( )[ ]
TTC)T(H
)TT(C~
XRpR
o
Rx
n
1i
0ipii ⇒
−⋅∆+∆−
−⋅⋅Θ=
∑=
Example for First Order Reaction
Performance equation:
Kinetics:
Stoichiometry:
Pressure drop:
−⋅⋅=T
1
T
1
R
Eexpkk
1
a1
Energy balance:
=− Ar k AC⋅
for PFR/small ∆P: P/P0 = 1
)X1(P/P
P
T
T
2dW
dP
0
0
0
⋅ε+⋅⋅⋅α
−=
)X1(CC0AA
−⋅=Gas liquid
0A
A
F
r
dW
dX −=
[ ]
[ ]
[ ]
p
n
1i
pii
n
1i
0piiRp
o
Rx
n
1i
0piiRp
o
Rxp
n
1i
pii
n
1i
n
1i
0piipiiRpp
o
Rx
CXC~
TC~
TCXHX
T
TC~
TCXHXTCXTC~
TC~
TC~
TCXTCXHX
∆⋅+⋅Θ
⋅⋅Θ+⋅∆⋅+∆−⋅=
⋅⋅Θ+⋅∆⋅+∆−⋅=⋅∆⋅+⋅⋅Θ
⋅⋅Θ−⋅⋅Θ=⋅∆⋅+⋅∆⋅−∆−⋅
∑
∑
∑∑
∑ ∑
=
=
==
= =
)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(gdW
dP
)P,X(f )r(fdW
dXA
==
=−= Thus
The 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
)TT(CFW1i0ipii0As−−⋅⋅Θ⋅−−∑=&
⇒
Rate of energy transferred between the reactor and the coolant:
The rate of heat transfer from the exchanger to the reactor:
−
−
−⋅⋅=
2a
1a
2a1a
TT
TTln
TTAUQ&
⇒Combining:
⇓⇐⇐
⇓
⇓⇒ ⇒
0HXF)TT(C~
FW Rx0A
n
1i
0ipii0As =∆⋅⋅−−⋅⋅Θ⋅−− ∑=
&Q&
NINA = Diabatic Reactor DesignNINA = Diabatic Reactor DesignHeat Transfer Rate to the Reactor (cont’)Heat Transfer Rate to the Reactor (cont’)
)TT(CFW1i0ipii0As−−⋅⋅Θ⋅−−∑=&
⇒⇓
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
1i
0ipii0As =∆⋅⋅−−⋅⋅Θ⋅−− ∑=
&( )TTAU 1a −⋅⋅
The energy balance becomes:
SampleSample
Problem forProblem for
DiabaticDiabatic
CSTRCSTR
DesignDesign
P8P8--4B4B
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:
( ) ( ) dVTTV
AUdATTUQ
V
a
A
a ⋅−⋅⋅=⋅−⋅= ∫∫&
( )TTaUdV
Qda −⋅⋅=
&
Thus:
D
4
L
LD a
areaktor tabung volume
reaktor tabung selimut luas
V
A
4D
2=
⋅
⋅⋅π=
==
⋅π
For PBR: dW1
dVW
VV
W
bbb ⋅
ρ=⇒
ρ=⇔=ρ
Thus: ( )TTaU
dW
Qda
b
−⋅ρ⋅
=&
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:
( )TTaUdV
Qda −⋅⋅=
&
and recalling that
Or:
( ) 0TTC)T(HXF)TT(C~
FQ RpRoRx0A
n
1i0ipii0A =−⋅+⋅⋅−−⋅⋅⋅− ∑
=∆∆Θ&
0dTC)T(HXFdTCFQT
TpR
oRx0A
T
Tpii0A
Ro
=
⋅+⋅⋅−⋅⋅⋅− ∫∫ ∑ ∆∆Θ&
Inserting
0dV
dXdTC)T(HF
dV
dTCXF
dV
dTCF
dV
Qd T
T
pRoRx0Ap0Apii0A
R
=⋅
⋅+⋅−⋅⋅⋅−⋅⋅⋅− ∫∑ ∆∆∆Θ
&
dV
dXFr 0AA ⋅=−
( )TTaU a −⋅⋅ ( )dV
dTCXCF ppii0A ⋅⋅+⋅⋅− ∑ ∆Θ ( ) )]T(H[r
RxA ∆−⋅−+ 0=
( ) ( )( )ppii0A
RxAa
CXCF
)]T(H[rTTaU
dV
dT
∆Θ
∆
⋅+⋅⋅
−⋅−+−⋅⋅=
∑
Coupled with 0A
A
F
r
dV
dX −=
)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 XXeqeq naik naik reaksikan
pada Tmax yang diperkenankan
K lnTRG ⋅⋅−=∆
2
Rx
TR
H
dT
K) (lnd
⋅=
∆
Eksotermik: Eksotermik: K turun dengan kenaikan T XXeqeq 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 HighlyDesign for Reversible Highly--Exothermic Exothermic
ReactionsReactions
--rrAA = = --rrAA (X,T)(X,T)
Generally:Generally: Higher X Higher X �� slower reaction rateslower reaction rate
Higher T Higher T �� faster ratefaster rate
At X = At X = XXeqeq : : --rrAA = 0= 0
Design for Equilibrium HighlyDesign for Equilibrium Highly--Exothermic Exothermic
ReactionsReactions
#1#1 Starting with RStarting with R--free solution, between 0 free solution, between 0 dandan 100100ooC C
determine the equilibrium conversion of A for the elementary determine the equilibrium conversion of A for the elementary
aqueous reaction:aqueous reaction:
A A �� RR cal/mol 18000H
cal/mol 3375G0298
0298
−=
−=
∆
∆
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 0A
0R
==
Assume ideal solution, in which case:Assume ideal solution, in which case: CA
R
0AA
0RR
KC
C
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 waterCcal/g. 1C
0p =
Design for Equilibrium HighlyDesign for Equilibrium Highly--Exothermic Exothermic
Reactions:Reactions:
Reaction Rate in X Reaction Rate in X –– T DiagramT Diagram
k T( ) 0.0918exp 5859−1
T
1
298−
⋅
⋅:=
rA X T,( ) k T( )− CA0⋅ 1 X−X
K T( )−
⋅:=
Reaction Rate in The X Reaction Rate in The X –– T DiagramT Diagram
at 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
Konversi
rA
− 0 01,
rA
− 0 025,
rA
− 0 05,
rA
− 0 1,
rA
− 0 25,
rA
− 0 5,
rA
− 1
rA
− 2
rA
− 4
Design for Equilibrium HighlyDesign for Equilibrium Highly--Exothermic ReactionsExothermic Reactions::
Optimum Temperature ProgressionOptimum Temperature Progression
in Tubular Reactorin Tubular Reactor
#3#3
a.a. Calculate the space time needed for 80% conversion of a feed Calculate the space time needed for 80% conversion of a feed
starting with initial concentration of A of 1 mol/Lstarting with initial concentration of A of 1 mol/L
b.b. Plot the temperature and conversion profile along the length of Plot the temperature and conversion profile along the length of
the reactorthe 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−⋅+−
−⋅⋅+−⋅⋅
=∑=
∆∆
Θ
Design for Equilibrium HighlyDesign for Equilibrium Highly--Exothermic Exothermic
ReactionsReactions:: CSTR PerformanceCSTR Performance
Design for Equilibrium HighlyDesign for Equilibrium Highly--Exothermic Exothermic
ReactionsReactions:: CSTR PerformanceCSTR Performance
#4#4 A concentrated aqueous AA concentrated aqueous A--solution of the previous 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:
∑=
−⋅⋅Θ⋅−−n
1i
0pii0As )TT(C~
FW&( )TTAU a −⋅⋅ 0HXF 0Rx0A =∆⋅⋅−
XF 0A ⋅ ( ) ( )
−⋅
⋅+−⋅⋅Θ⋅=∆−⋅ ∑
=a
0A
n
1i
0pii0A0Rx TT
F
AU)TT(C
~FH
Untuk CSTR:A
0A
r
XFV
−⋅
=
( )VrA ⋅− ( ) ( )
−⋅
⋅+−⋅⋅=∆−⋅ a
0A00p0A
0Rx TT
F
AU)TT(CFH
Pembagian kedua ruas dengan FA0:
( ) ⋅+−⋅=∆−⋅
⋅−0p00p
0Rx
0A
A C)TT(CHF
Vr
0p0A CF
AU
⋅⋅
( )aTT −⋅
0p0A CF
AU
⋅⋅
=κ1
TT
1
TT
CFAU
TAUTCFT a0
CFAU
aCFAU
0
0p0A
a00p0Ac
0p0A
0p0A
+κ⋅κ+
=+
⋅+=
⋅+⋅
⋅⋅+⋅⋅=
⋅⋅
⋅⋅
Multiple Steady Multiple Steady
State & Stability of State & Stability of
CSTR OperationCSTR Operation
1
TTT a0c +κ
⋅κ+=
κ
)TT( a0 ⋅κ+
( ) +−⋅=∆−⋅
⋅−)TT[(CH
F
Vr00p
0Rx
0A
A ( )]TT a−⋅
)1(Tc +κ⋅−+κ⋅⋅= )1(T[C 0p
)TT()1(C c0p −⋅+κ⋅=
−⋅κ+⋅= TT[C 0p ]
]
( )0RxHX ∆−⋅ )TT()1(C c0p −⋅+κ⋅=
)T(G )T(R=
A
0A
r
XFV
−
⋅=
( )[ ] TTC)T(H
)TT(C~
X
dengan Bandingkan
RpRoRx
n
1i0pii
−⋅∆+∆−
−⋅⋅Θ
=∑=
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 state
Lower steady state
Ignition temperature
Extinction temperature
Runaway Reaction
Sample Problem on Multiple Steady State in Sample Problem on Multiple Steady State in
CSTR OperationCSTR OperationP8P8--17B17B