Steady State Steady State Nonisothermal Nonisothermal Reactor Design Reactor Design Dicky Dicky Dermawan Dermawan www.dickydermawan.net78.net www.dickydermawan.net78.net [email protected][email protected]ITK ITK-330 Chemical Reaction Engineering 330 Chemical Reaction Engineering
37
Embed
Steady State Steady State NonisothermalNonisothermal ...libvolume2.xyz/biotechnology/semester7/bioreactor... · Steady State Steady State NonisothermalNonisothermal Reactor Design
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Steady State Steady State NonisothermalNonisothermal
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
#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