Steady State Non Iso Thermal Reactor Design

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Steady State NonisothermalSteady State Nonisothermal

Reactor DesignReactor Design

Dicky DermawanDicky Dermawanwww.dickydermawan.net78.netwww.dickydermawan.net78.net

[email protected]@gmail.com

ITK-330 Chemical Reaction EngineeringITK-330 Chemical Reaction Engineering

http://www.dickydermawan.net78.net/


mailto:[email protected]






RationaleRationale

All reactions always accompanied by heat effect:exothermic reactions vs. endothermic reactions

Unless heat transfer system is carefully

designed, reaction mass temperature tend tochange Design of heat transfer system itself requires the

understanding of this heat effect

Energy balance is also needed, together withperformance equations derived from massbalance



ObjectivesObjectives

Describe the algorithm for CSTRs, PFRs, and PBRs thatare not operated isothermally.

Size adiabatic and nonadiabatic CSTRs, PFRs, andPBRs.

Use reactor staging to obtain high conversions for highlyexothermic reversible reactions. Carry out an analysis to determine the Multiple Steady

States (MSS) in a CSTR along with the ignition andextinction 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 afunction of position (PFR/PBR) and operating variables



Why Energy Balance?Why Energy Balance?

Imagine that we are designing a nonisothermal PFR for aImagine that we are designing a nonisothermal PFR for afirst order liquid phase exothermic reaction:first order liquid phase exothermic reaction:

PerformancePerformance

equation:equation: 0A

A

F

r

dV

dX −=

Kinetics:Kinetics: =− Ar k AC⋅

The temperaturewill increase withconversion down

the length of reactor

−⋅⋅=

T

1

T

1

R

Eexpk k

1

a1

Stoichiometry:Stoichiometry: 0υ=υ A0A CF ⋅υ=

0A00A CF ⋅υ=)X1(CC 0AA −⋅=

Combine:Combine:

0

X1

υ−

−⋅⋅=

1a1

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

1iIIDDCCBBAAii HFHFHFHFHFHFOut

At steady state:

dt

EdHFHFWQ

sysn

1i

ii

n

1i

0i0is =⋅−⋅+− ∑∑==

∑=+−

n

1i

sWQ 0iF 0iH ∑=

−n

1iiF iH 0=

Consider generalizedreaction:

DCBAad

ac

a b +→+

I0AI

ad

D0AD

a

c

C0AC

a b

B0AB

0AA

FF

)X(FF

)X(FF

)X(FF

)X1(FF

Θ⋅=

+Θ⋅=

+Θ⋅=

−Θ⋅=

−⋅=

Upon substitution:

( )ABa b

Cac

Dad

A0 HHHHXF- −−+⋅⋅

∑∑==

⋅−⋅n

1iii

n

1i0i0i HFHF

( ) ( ) ( )

( ) ( )

−⋅Θ+−⋅Θ+

−⋅Θ+−⋅Θ+−⋅=

CI0IID0DD

C0CCB0BBA0A0A

HHHH

HHHHHHF

∑∑ ==⋅−⋅

n

1iii

n

1i0i0i HFHF ( )∑= −⋅Θ⋅=

n

1ii0ii0A 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

piR oii

R

dTC)T(HH

From thermodynamics, we know that:

∫ ⋅+=0i

R

T

T

piR oi0i dTC)T(HH Thus: )TT(C

~dTCHH 0i pi

T

T

pi0ii

0i

−⋅=⋅=− ∫

0i

T

T

pi

piTT

dTC

C~ 0i

−

⋅

=∫

( )R pR oRxRx TTC)T(H)T(H −⋅∆+∆=∆

R

T

T

pi

piTT

dTC

C R

−

⋅∆

=∆∫

)T(H)T(H)T(H)T(H)T(H R oDR

oDa

bR

oDa

cR

oDa

dR

oRx −⋅−⋅+⋅=∆

pA pBa b

pCac

pDad

p CCCCC −⋅−⋅+⋅=∆



∑=

−⋅⋅Θ⋅−=n

1i0 pii0A )TT(C

~F

Energy Balance (cont’)Energy Balance (cont’)

∑∑==

⋅−⋅ n

1i

iin

1i

0i0i HFHF ( )∑=

−⋅Θ⋅−= n

1i0iii0A HHF )T(HXF Rx0A ∆⋅⋅−

Upon substitution:

∑∑==

⋅−⋅n

1iii

n

1i0i0i HFHF ( )]TTC)T(H[XF R pR

oRx0A −⋅∆+∆⋅⋅−

Finally….

0HFHFWQn

1i

ii

n

1i

0i0is =⋅−⋅+− ∑∑==

( ) 0TTC)T(HXF)TT(C~FWQ R pR oRx0A

n

1i

0i pii0As =−⋅∆+∆⋅⋅−−⋅⋅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 R pR oRx0A

n

1i

0i pii0A =−⋅∆+∆⋅⋅−−⋅⋅Θ⋅− ∑=

( )[ ]R pR oRx

n

1i0i pii

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

Eexpk k

1

a1

)X1(Ck

XFV

0A

0A

−⋅⋅⋅

=

Solve the energy balance for T

( )[ ]R pR oRx

n

1i

0i pii

TTC)T(H

)TT(C~

X−⋅∆+∆−

−⋅⋅Θ=

∑=

Calculate k

Calculate V using combining equation




)X1(CC 0AA −⋅=

Case B (Rating): V specified, calculate X (and T)


Kinetics:

Stoichiometry:

Mole balance:

A

0A

r

XFV

−⋅=

=− Ar k AC⋅

−⋅⋅=

T

1

T

1

R

Eexpk k

1

a1

)X1(Ck

XFV

mb0A

mb0A

−⋅⋅⋅

=

Energy balance:( )[ ]R pR

oRx

n

1i

0i pii

ebTTC)T(H

)TT(C~

X−⋅∆+∆−

−⋅⋅Θ= ∑=

Find X & T that satisfy BOTH the material balanceand energy balance,

viz. plot Xmb vs T and Xeb vs T in the same graph: theintersection is the solution




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 Aand 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/kcal41)K 273(H

mol/kcal15)K 273(H

mol/kcal20)K 273(H

oC

oB

oA

−=

−=

−=

cal/mol.K 30C

cal/mol.K 15C

cal/mol.K 15C

pC

pB

pA

=

=

=

cal/mol10000E

300at01.0k

a

smolL

=

= ⋅

mol/L1.0C 0A =



Application to Adiabatic CSTR DesignApplication to Adiabatic CSTR DesignCase A: Sizing: X specified, calculate V (and T)


Kinetics:

Stoichiometry:

Combine:

A

0Ar XFV − ⋅=

=− Ar k BA CC ⋅⋅

−⋅⋅=

T

1

T

1

R

Eexpk k

1

a1

20A

0

220A

0A

)X1(Ck

X

)X1(Ck

XFV

−⋅⋅⋅υ

=−⋅⋅

⋅=

Energy balance:

( )[ ]R pR oRx

n

1i

0i pii

TTC)T(H

)TT(C~

X−⋅∆+∆−

−⋅⋅Θ=

∑=

Calculate k

Calculate V using combining equation

)X1(CC 0AA −⋅=)X1(C)X(CC 0ABB0AB −⋅=⋅ ν−Θ⋅=

K cal/mol301515CCC~

pBB pA

n

1i

pii ⋅=+=⋅Θ+=⋅Θ∑=cal/mol6000-kcal/mol6152041HHH)273(H

o

B

0

A

o

C

o

Rx=−=++−=−−=∆

0151530CCCC pB pA pC p =−−=−−=∆

K 47020085.0300T200

300T

)6000(

)300T(3085.0 =⋅+=⇒

−=

−−−⋅

=

smol

L 317.4

470

1

300

1

987.1

10000exp01.0k

⋅=

−⋅⋅=

L175)85.01(1.0317.4

85.02V2=

−⋅⋅⋅=




)X1(CCC 0ABA −⋅==

( )[ ]R pR oRx

n

1i

0i pii

ebTTC)T(H

)TT(C~

X−⋅∆+∆−

−⋅⋅Θ

=

∑=

Case B (Rating): V specified, calculate X (and T)


Kinetics:

Stoichiometry:

Mole balance:

A

0A

r

XFV

−⋅=

−⋅⋅=

T

1

T

1

R

Eexpk k

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 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 0.933

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~

XR pR

o

Rx

n

1i

0i pii ⇒

−⋅∆+∆−

−⋅⋅Θ=

∑=

Example for First Order Reaction


Kinetics:

Stoichiometry:

Pressure drop:

−⋅⋅=

T1

T1

R Eexpk k

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(CC 0AA −⋅=Gas liquid

0A

A

F

r

dW

dX −=

[ ]

[ ]

[ ]

p

n

1i

pii

n

1i

0 piiR p

o

Rx

n

1i

0 piiR p

o

Rx p

n

1i

pii

n

1i

n

1i

0 pii piiR p p

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(k k

==

)P,X(CC )X(TT

)P,T,X(CCAA

AA =

==

)P,X(r r

])P,X[C],X[k (r r

AA

AAA

−=−−=−

)P,X(g)P,T,X(gdWdP

)P,X(f )r (f dW

dXA

==

=−= Thus

The combination results in 2 simultaneousdifferential equations



SampleSampleProblemProblem

for for

AdiabaticAdiabatic

PFRPFRDesignDesign

P8-6A



Sample Problem for Adiabatic PBRSample Problem for Adiabatic PBR

DesignDesign

NINA Di b ti R t D iNINA Di b ti R t D i



NINA = Diabatic Reactor DesignNINA = Diabatic Reactor DesignHeat Transfer Rate to the Reactor Heat Transfer Rate to the Reactor

∑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

TT

TTln

TTAUQ

⇒Combining:

⇓⇐⇐

⇓

⇓⇒ ⇒

0HXF)TT(C~

FW Rx0A

n

1i

0i pii0As =∆⋅⋅−−⋅⋅Θ⋅−−

∑=Q



NINA = Diabatic Reactor DesignNINA = Diabatic Reactor DesignHeat Transfer Rate to the Reactor (cont’)Heat Transfer Rate to the Reactor (cont’)

∑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

1i0i pii0As =∆⋅⋅−−⋅⋅Θ⋅−− ∑=

( )TTAU 1a −⋅⋅

The energy balance becomes:



SampleSample

Problem for Problem for DiabaticDiabatic

CSTRCSTR

DesignDesign

P8-4BP8-4B



Sample Problem for Diabatic CSTR DesignSample Problem for Diabatic CSTR Design



Application of Energy Balance to DiabaticApplication of Energy Balance to Diabatic

Tubular Reactor DesignTubular Reactor Design

Heat transfer in CSTR: ( )TTAUQ 1a −⋅⋅=

In PFR, T varies along the

reactor:

( ) ( ) dVTT

V

AUdATTUQ

V

a

A

a ⋅−⋅⋅=⋅−⋅= ∫ ∫

( )TTaUdV

Qda −⋅⋅=

Thus:

D

4

L

LD a

reaktor tabungvolume

reaktor tabungselimutluas

V

A

4D2

=⋅

⋅⋅π=

==

⋅π

For PBR: dW1

dVW

VV

W

bbb ⋅

ρ=⇒

ρ=⇔=ρ

Thus: ( )TTaU

dW

Qda

b

−⋅

ρ

⋅=

A li ti f E B l t Di b tiA li ti f E B l t Di b ti



Application of Energy Balance to DiabaticApplication of Energy Balance to Diabatic

Tubular Reactor DesignTubular Reactor DesignThe 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 =−⋅+⋅⋅−−⋅⋅⋅− ∑

=ΔΔΘ

0dTC)T(HXFdTCFQT

TpR

oRx0A

T

Tpii0A

Ro

=

⋅+⋅⋅−⋅⋅⋅− ∫ ∫ ∑ ΔΔΘ

Inserting

0dV

dXdTC)T(HF

dV

dTCXF

dV

dTCF

dV

Qd T

TpR

oRx0Ap0Apii0A

R

=⋅

⋅+⋅−⋅⋅⋅−⋅⋅⋅− ∫ ∑ ΔΔΔΘ

dV

dXFr 0AA ⋅=−

( )TTaU a −⋅⋅ ( ) dV

dT

CXCF ppii0A ⋅⋅+⋅⋅− ∑ ΔΘ ( ) )]T(H[r RxA Δ−⋅−+ 0=

( ) ( )

( )ppii0A

RxAa

CXCF

)]T(H[r TTaU

dV

dT

ΔΘ

Δ

⋅+⋅⋅

−⋅−+−⋅⋅=

∑

Coupled with0A

AFr

dVdX −=

)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 naiknaik reaksikan pada Tmax yang diperkenankan

KlnTRG ⋅⋅−=Δ

2

Rx

TR

H

dT

K)(lnd

⋅=Δ

Eksotermik:Eksotermik: K turun dengan kenaikan T XXeqeq turunturun 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-ExothermicDesign for Reversible Highly-Exothermic

ReactionsReactions

-r -r AA

= -r = -r AA

(X,T)(X,T)

Generally:Generally: Higher XHigher X slower reaction rateslower reaction rate

Higher THigher T faster ratefaster rate

At X = XAt X = Xeqeq :: -r -r

AA = 0= 0

f



Design for Equilibrium Highly-ExothermicDesign for Equilibrium Highly-Exothermic

ReactionsReactions

#1#1 Starting with R-free solution, between 0 dan 100Starting with R-free solution, between 0 dan 100ooC determine theC determine the

equilibrium conversion of A for the elementary aqueous reaction:equilibrium conversion of A for the elementary aqueous reaction:

AA RR

cal/mol18000H

cal/mol3375G0

298

0298

−=

−=Δ

Δ

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 equalIn addition, assume specific heats of all solutions are equal

to that of water to that of water Ccal/g.1C 0p =




Reactions:Reactions:

Reaction Rate in X – T DiagramReaction Rate in X – T 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 – T DiagramReaction Rate in The X – T Diagram

at Cat CA0A0 = 1 mol/L= 1 mol/L

0 10 20 30 40 50 60 70 80 90 100

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Suhu, C

K o n v e r s i

r A

− 0 01,

r A

− 0 025,

r A

− 0 05,

r A

− 0 1,

r A

− 0 25,

r A

− 0 5,

r A

− 1

r A

− 2

r A

− 4



Design for Equilibrium Highly-Exothermic ReactionsDesign for Equilibrium Highly-Exothermic Reactions::

Optimum Temperature ProgressionOptimum Temperature Progression

in Tubular Reactor in Tubular Reactor

#3#3

a.a. Calculate the space time needed for 80% conversion of a feed starting with initialCalculate 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/L

b.b. Plot the temperature and conversion profile along the length of the reactor Plot 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



EffectEffect

( )[ ]RpRoRx

n

1i0ipiia

0A

TTC)T(H

)TT(C~

)TT(F

AU

X

−⋅+−

−⋅⋅+−⋅⋅

=∑=

ΔΔ

Θ

D i f E ilib i Hi hl E th iD i f E ilib i Hi hl E th i




ReactionsReactions:: CSTR PerformanceCSTR Performance

D i f E ilib i Hi hl E th iDesign for Equilibrium Highly Exothermic




ReactionsReactions:: CSTR PerformanceCSTR Performance

#4#4 A concentrated aqueous A-solution of the previousA concentrated aqueous A-solution of the previousexamples, Cexamples, C

A0A0= 4 mol/L, F= 4 mol/L, F

A0A0= 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 25 ooC, what size of reactor is needed?C, what size of reactor is needed?

b.b. What is the optimum operating temperature for thisWhat is the optimum operating temperature for thispurpose?purpose?

c.c. What size of reactor is needed if feed enters at optimumWhat 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 theC to keep thereactor operation at its the optimum temperature?reactor operation at its the optimum temperature?



Interstage CoolingInterstage Cooling

R i E B l i CSTR O tiR i E B l i CSTR O ti



Review on Energy Balance in CSTR OperationReview on Energy Balance in CSTR Operation

Bila term kerja diabaikan dan ∆ HRx konstan:

∑=

−⋅⋅Θ⋅−−n

1i

0 pii0As )TT(C~

FW( )TTAU a −⋅⋅ 0HXF0Rx0A =∆⋅⋅−

XF 0A ⋅ ( ) ( )

−⋅

⋅+−⋅⋅Θ⋅=∆−⋅ ∑

=a

0A

n

1i

0 pii0A0Rx TT

F

AU)TT(C

~FH

Untuk CSTR: A

0A

r

XF

V −

⋅

=

( )Vr A ⋅− ( ) ( )

−⋅

⋅+−⋅⋅=∆−⋅ a

0A00 p0A

0Rx TT

F

AU)TT(CFH

Pembagian kedua ruas dengan FA0 :

( ) ⋅+−⋅=∆−⋅

⋅−0 p00 p

0Rx

0A

A C)TT(CHF

Vr

0 p0A CF

AU

⋅⋅

( )aTT −⋅

0 p0A CF

AU

⋅

⋅

=κ 1

TT

1

TT

CFAU

TAUTCF

Ta0

CFAU

aCFAU

0

0 p0A

a00 p0A

c00A

0 p0A

+κ

⋅κ +

=+

⋅+

=⋅+⋅

⋅⋅+⋅⋅

= ⋅⋅

⋅⋅

R i E B l i CSTR O tiR i E B l i CSTR O ti



Multiple SteadyMultiple Steady

State & Stability of State & Stability of

CSTR OperationCSTR Operation

1

TTT a0c +κ

⋅κ +=

κ )TT( a0 ⋅κ +( ) +−⋅=∆−⋅

⋅−)TT[(CHF

Vr

00 p

0

Rx0A

A

( )]TT a−⋅

)1(Tc +κ ⋅−+κ ⋅⋅= )1(T[C 0 p

)TT()1(C c0 p −⋅+κ ⋅=

−⋅κ +⋅= TT[C 0 p ]

]

( )0RxHX ∆−⋅ )TT()1(C c0 p −⋅+κ ⋅=

)T(G )T(R =

A

0A

r

XFV

−⋅

=

( )[ ]

TTC)T(H

)TT(C~

X

denganBandingkan

R pR oRx

n

1i

0 pii

−⋅∆+∆−

−⋅⋅Θ=

∑=

Review on Energy Balance in CSTR OperationReview on Energy Balance in CSTR Operation

Multiple SteadyMultiple Steady St t St bilit f CSTRState: Stability of CSTR



Multiple SteadyMultiple Steady State: Stability of CSTRState: 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 inSample Problem on Multiple Steady State in



Sample Problem on Multiple Steady State inSample Problem on Multiple Steady State in

CSTR OperationCSTR OperationP8-17BP8-17B

Steady State Non Iso Thermal Reactor Design

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