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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
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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
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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
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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 =
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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 ∆⋅⋅−
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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 −⋅−⋅+⋅=∆
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∑=
−⋅⋅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?
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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!
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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
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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
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
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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 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 =
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Application to Adiabatic CSTR DesignApplication to Adiabatic CSTR DesignCase A: Sizing: X specified, calculate V (and T)
Performance equation:
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=
−⋅⋅⋅=
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Application to Adiabatic CSTR DesignApplication to Adiabatic CSTR Design
)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)
Performance equation:
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
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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
Performance equation:
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
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SampleSampleProblemProblem
for for
AdiabaticAdiabatic
PFRPFRDesignDesign
P8-6A
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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
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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
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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:
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SampleSample
Problem for Problem for DiabaticDiabatic
CSTRCSTR
DesignDesign
P8-4BP8-4B
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Sample Problem for Diabatic CSTR DesignSample Problem for Diabatic CSTR Design
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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
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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
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Sample Problem for Diabatic Tubular Reactor Sample Problem for Diabatic Tubular Reactor
DesignDesign
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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)
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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
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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 =
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Design for Equilibrium Highly-ExothermicDesign for Equilibrium Highly-Exothermic
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( )−
⋅:=
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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
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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
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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
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Design for Equilibrium Highly-ExothermicDesign for Equilibrium Highly-Exothermic
ReactionsReactions:: CSTR PerformanceCSTR Performance
D i f E ilib i Hi hl E th iDesign for Equilibrium Highly Exothermic
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Design for Equilibrium Highly-ExothermicDesign 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?
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Interstage CoolingInterstage Cooling
R i E B l i CSTR O tiR i E B l i CSTR O ti
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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
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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
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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