Transcript Steady State Nonisothermal Reactor Design

ITK-330 Chemical Reaction Engineering
Steady State Nonisothermal
Reactor Design
Dicky Dermawan
www.dickydermawan.net78.net
[email protected]
Rationale




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
Objectives





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?
Imagine that we are designing a nonisothermal PFR for a
first order liquid phase exothermic reaction:
Performance
equation:
Kinetics:
dX  rA

dV FA 0
 rA 
k  CA
E
k  k1  exp  a
R
 1 1 
   
 T1 T 
X  X( V)
T  T(V)
T  T(X)
FA  0  C A
Stoichiometry:   0
Combine:
The temperature
will increase with
conversion down
the length of
reactor
C A  C A0  (1  X)
FA0  0  C A0
 Ea
dX
 k1  exp 
dV
R
1
T
 1
  
 T1
 1  X

 0
X  X ( T, V )
Energy Balance
At steady state:
n
 W
  F
Q
s
i0
i 1
n
n
dEˆ sy s
i 1
i 1
dt
 W
  F H  F H 
Q
s
i0
i0
i
i
n
H i0   Fi H i 
i 1
0
n
In :  Fi0  H i0  FA0  H A0  FB0  H B0  FC0  H C0  FD0  H D0  FI0  H I0
i 1
n
Out  Fi  H i  FA  H A  FB  H B  FC  H C  FD  H D  FI  H I
i 1
Consider
generalized
reaction:
A  ba B  ac C  da D
FA  FA 0  (1  X)
FB  FA 0  ( B  ba X)
FC  FA 0  ( C  ac X)
Upon substitution:
n
 Fi0  H i0
i 1
H  H A   B  H B0  H B   C  H C0  H C 
  Fi  H i  FA0   A0

 D  H D0  H D   I  H I0  H CI 

i 1
n
- FA0  X 
n
n
i 1
i 1
 Fi0  H i0   Fi  H i
n
da H D  ac H C  ba H B  H A 
 FA0    i  H i 0  H i 
i 1
FD  FA 0  ( D  da X)
FI  FA 0   I
 FA0  X  HRx (T)
Energy Balance (cont’)
n
n
i 1
i 1
 Fi0  H i0   Fi  H i
n
 FA 0   i  Hi  Hi 0 
i 1
 FA0  X  HRx (T)
From thermodynamics, we know that:
T
H i  H i (TR )   C pi  dT
o
TR
Ti 0
H i0  H i (TR )   C pi  dT
o
Thus:
TR
T
T
~
Hi  Hi0   Cpi  dT  Cpi  (T  Ti0 )
Ti 0
~
C pi 
ˆ  T  T 
H Rx (T )  H oRx (TR )  C
p
R
H oRx (TR )  da  H oD (TR )  ac  H oD (TR )  ba  H oD (TR )  H oD (TR )
T
ˆ 
C
pi
 C pi  dT
TR
T  TR
C p 
d
a
 C pD 
c
a
 C pC 
b
a
 C pB  C pA
 C pi  dT
Ti 0
T  Ti0
Energy Balance (cont’)
Upon substitution:
n
n
i 1
i 1
n
n
n
i 1
i 1
i 1
 Fi0  H i0   Fi  H i
n
 FA 0   i  Hi  Hi 0 
i 1
 FA0  X  HRx (T)
~
F

H

F

H


F



C
 i0 i0  i i
A0  i
pi  (T  T0 )
 FA 0  X  [H oRx (TR )  Cˆ p  T  TR ]
Finally….
n
n
i 1
i 1
 W
  F  H  F  H  0
Q
s
i0
i0
i
i
n
~

ˆ  T  T   0

Q  Ws  FA0    i  C pi  (T  Ti0 )  FA0  X  H oRx (TR )  C
p
R
i 1
So what?
Energy Balance (cont’)
For adiabatic reactions: Q  0
 0
When work is negligible: W
s
The energy balance at steady state becomes:

n

~
ˆ  T  T   0
 FA0   i  Cpi  (T  Ti0 )  FA0  X  H oRx (TR )  C
p
R
i 1
After rearrangement:
n
X

~


C
 i pi  (T  Ti0 )
i 1
ˆ
H oRx (TR )  C
p


 T  TR 
This is the X=X(T) we’ve been looking for!
Application to Adiabatic CSTR Design
Case A: Sizing: X specified, calculate V (and T)
FA 0  X
 rA
Performance equation:
V
Kinetics:
 rA 
Stoichiometry:
CA  CA0  (1  X)
Combine:
V
k
 CA
E
k  k1  exp a
R
 1 1 
   
 T1 T 
FA0  X
k  C A0  (1  X)
Solve the energy balance for T
Calculate k
n
X

~
 i  C pi  (T  Ti0 )
i 1
ˆ
H oRx (TR )  C
p

Calculate V using combining equation

 T  TR 
Application to Adiabatic CSTR Design
Case B (Rating): V specified, calculate X (and T)
FA 0  X
 rA
Performance equation:
V
Kinetics:
 rA 
Stoichiometry:
CA  CA0  (1  X)
Mole balance:
V
k
 CA
E
k  k1  exp a
R
 1 1 
   
 T1 T 
FA0  X mb
k  CA0  (1  X mb)
n
Energy balance:
X eb 

~


C
 i pi  (T  Ti0 )
i 1
ˆ 
H oRx (TR )  C
p

T  TR 
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 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)
(b)
Calculate the CSTR volume necessary to achieve 85%
conversion
Calculate the conversion that can be achieved in one 500 L
CSTR and in two 250 L CSTRs in series
H oA (273K)  20 kcal/ mol C pA  15 cal/mol.K C A0  0.1 mol/L
H oB (273K)  15 kcal/ mol C pB  15 cal/mol.K k  0.01 mLols at 300 K
H oC (273K)  41kcal/ mol C pC  30 cal/mol.K E a  10000cal/mol
Application to Adiabatic CSTR Design
Case A: Sizing: X specified, calculate V (and T)
FA 0  X
 rA
Performance equation:
V
Kinetics:
 rA 
Stoichiometry:
CA  CA0  (1  X)
k
 CA  CB
E
k  k1  exp a
R
 1 1 
   
 T1 T 
CB  CA0  (B   B  X)  CA0  (1  X)
V
Combine:
FA0  X
k  CA02  (1  X) 2

0  X
k  CA0  (1  X) 2
~
H oRx (273)  H oC  H 0A  H oB  41 20  15  6 kcal/mol - 6000cal/mol
 i  C pi  C pA   B  C pB  15  15  30 cal/mol K
n
ˆ  C  C  C  30  15  15  0
C
p
pC
pA
pB
i 1
Energy balance:
n
X
Calculate k

~
 i  C pi  (T  Ti0 )
i 1
ˆ
H oRx (TR )  C
p

0.85 

30  (T  300) T  300

 T  300  0.85  200  470K
 (6000)
200
 T  TR 
10000  1
1 
L
k  0.01 exp


  4.317
mol s
 1.987  300 470
Calculate V using combining equation
V
2  0.85
4.317  0.1  (1  0.85) 2
 175 L
Application to Adiabatic CSTR Design
Case B (Rating): V specified, calculate X (and T)
FA 0  X
 rA
Performance equation:
V
Kinetics:
Stoichiometry:
Mole balance:
 rA 
k
 CA  CB
E
k  k1  exp a
R
CA  CB  CA0  (1  X)
V
0  X mb
k  CA0  (1  X mb) 2
500 
Energy balance:
1
n
Xmb
0.8
Xeb
X eb 
0.6
0.4
 1 1 
   
 T1 T 

10000  1
1 
0.01 exp

   0.1  (1  X mb ) 2
 1.987  300 T 
~
 i  Cpi  (T  Ti0 )
i 1
ˆ
H oRx (TR )  C
p

2  X mb

 T  TR 
X eb 
30  (T  300) T  300

 (6000)
200
0.2
0
300
350
400
450
500
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 Design
Example for First Order Reaction
dX  rA

dW FA 0
Performance equation:
 rA 
Kinetics:
Stoichiometry:
Gas
Energy balance:
n
~
 i  Cpi  (T  Ti0 )
 H
Combine:
i 1
o
Rx


ˆ  T  T 
(TR )  C
p
R
 rA  rA (X, P)
1  X P T0
 
1    X P0 T
CA  CA0  (1  X)
liquid


n
n
~
~
ˆ  T  X  C
ˆ T   C
X   H oRx  X  C

T

 i pi  i  Cpi  T0
p
p
R
i 1
i 1

n

n
~
~
ˆ  T  X   H o  X  C
ˆ T   C


C

T

X


C
 i pi
 i pi  T0
p
Rx
p
R
i 1
k  k (T ) 
C  CA (X, T, P)
k (X) A
CA  CA (X, P)
T  T ( X) 
T  T ( X)

 rA  rA (k[X], CA [X, P])
CA  CA0 
 CA
 1 1 
   
 T1 T 
dP
 T P
    0  (1    X) for PFR/small P:
P/P0 = 1
dW
2 T0 P / P0
Pressure drop:
X
k
 Ea
k  k1  exp
R
i 1
T  T(X)

X   H
T
o
Rx

n
~
ˆ
 X  C p  TR   i  C pi  T0
i 1
n
~
 i  Cpi X  Cˆ p
i 1
dX
 f ( rA )  f ( X, P) Thus
dW
The combination results in 2 simultaneous
dP
 g ( X, T, P )  g ( X, P )
differential equations
dW
P8-6A
Sample
Problem
for
Adiabatic
PFR
Design
Sample Problem for Adiabatic PBR
Design
NINA = Diabatic Reactor Design
Heat Transfer Rate to the Reactor
n
~



W

F



C
Q
s
A0  i
pi  (T  Ti 0 )  FA 0  X  H Rx  0
i 1
The rate of heat transfer from the
exchanger to the reactor:
n
 FW   C~ (T T )  HXF oRx(T    TTCˆ) 0
s A0 i p i i0 A0 R p R
i1
  U  A  Ta1  Ta 2
Q
 T  Ta1
ln 
 T  Ta 2
Rate of energy transferred between the reactor and the coolant:
Combining:












NINA = Diabatic Reactor Design
Heat Transfer Rate to the Reactor (cont’)
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:
n
  FW C~ (T )  HXF (T Cˆ) T  0
s A0 i pi i0 A0 Rx Ro p R
i1
Then:


The energy balance becomes:
n
~
  F  C
U  A  Ta1  T   W
s
A0
i
pi  (T  Ti 0 )  FA 0  X  H Rx  0
i 1
P8-4B
Sample
Problem for
Diabatic
CSTR
Design
Sample Problem for Diabatic CSTR Design
Application of Energy Balance to Diabatic
Tubular Reactor Design
  U  A  T  T
Q
a1
Heat transfer in CSTR:
In PFR, T varies along the
reactor:
Thus:
A
V
A

Q   U  Ta  T   dA   U   Ta  T   dV
V

A luas selimut tabung reaktor
dQ

 U  a  Ta  T 
V
volume tabung reaktor
dV
a 
For PBR:
b 
W
W
1
V
 dV 
 dW
V
b
b
Thus:

dQ
Ua

 Ta  T 
dW
b
 D L
D 2
4
L

4
D
a
Application of Energy Balance to Diabatic Tubular
Reactor Design
The steady state energy balance, neglecting work term:
n
~
o

ˆ  T  T   0
Q  FA 0   Θ i  C pi  (T  Ti0 )  FA 0  X  ΔHRx
(TR )  ΔC
p
R
i1
T
T


o

Q  FA 0    Θ i  C pi  dT  FA 0  X   ΔHRx (TR )   ΔC p dT   0


To
TR


Differentiation with respect to the volume V:
T

 dX

dQ
dT
dT
o
 FA 0   Θ i  C pi 
 FA 0  X  ΔC p 
 FA 0   ΔHRx (TR )   ΔC p dT  
0
dV
dV
dV
dV


TR



dQ
dX
Inserting
 U  a  Ta  T  and recalling that  rA  FA 0 
dV
dV
dT
  rA   [ΔHRx (T)]  0
U  a  Ta  T   FA 0   Θ i  C pi  X  ΔC p 
dV

Or:

dT U  a  Ta  T    r A   [  ΔHRx (T )]
 g( X, T )

dV
FA 0   Θ i  C pi  X  ΔC p

Coupled with

dX  rA
 f ( X, T)

dV FA 0
Form 2
differential with 2
dependent
variables X & T
Sample Problem for Diabatic Tubular Reactor
Design
Design for Reversible Reactions
ΔG  R  T  ln K
d(ln K) ΔHRx

dT
R  T2
Xeq = Xeq (K)
= Xeq (T)
Endotermik: K naik dengan kenaikan T  Xeq naik  reaksikan
pada Tmax yang diperkenankan
Eksotermik: K turun dengan kenaikan T  Xeq turun  reaksikan
pada T rendah
Laju reaksi lambat pada T rendah!
Ada trade off antara aspek termodinamika dan kinetika
Design for Reversible Highly-Exothermic
Reactions
-rA = -rA (X,T)
Generally:
At X = Xeq :
Higher X  slower reaction rate
Higher T  faster rate
-rA = 0
Design for Equilibrium Highly-Exothermic
Reactions
#1 Starting with R-free solution, between 0 dan 100oC
determine the equilibrium conversion of A for the elementary
aqueous reaction:
0
A  R
ΔG 298  3375 cal/mol
ΔH0298  18000 cal/mol
The reported data is based on the following standard states of
0
CR
 C 0A  1mol/L
reactants and products:
Assume ideal solution, in which case:
0
CR / CR
C
K
 R  KC
CA
C A / C 0A
In addition, assume specific heats of all solutions are equal
to that of water
0
Cp  1 cal/g. C
Design for Equilibrium Highly-Exothermic
Reactions:
Reaction Rate in X – T Diagram
1
1 


k( T)  0.0918 exp5859  


 T 298 
X 

rA( X T)  k( T)  CA0  1  X 

K( T) 

Reaction Rate in The X – T Diagram
at CA0 = 1 mol/L
1
0.9
0.8
r
0  01
A
r
0.7
0  025
A
r
A
Konversi
0.6
0  05
r
A
0.5
0 1
r
A
0.4
0  25
r
A
0 5
0.3
r
A
1
r
0.2
A
2
r
A
0.1
0
0
10
20
30
40
50
Suhu, C
60
70
80
90
100
4
Design for Equilibrium Highly-Exothermic Reactions:
Optimum Temperature Progression
in Tubular Reactor
#3
a.
b.
Calculate the space time needed for 80% conversion of a feed
starting with initial concentration of A of 1 mol/L
Plot the temperature and conversion profile along the length of
the reactor
Let the maximum operating allowable temperature be 95oC
Design for Reversible Reactions: Heat Effect
n
U A
~
 (T  Ta )   Θ i  C pi  (T  Ti0 )
F
i1
X  A0
ˆ  T  T 
 ΔHo (T )  ΔC

Rx
R
p
R

Design for Equilibrium Highly-Exothermic
Reactions: CSTR Performance
Design for Equilibrium Highly-Exothermic
Reactions: CSTR Performance
#4 A concentrated aqueous A-solution of the previous
examples, CA0 = 4 mol/L, FA0 = 1000 mol/min, is to be 80%
converted in a mixed reactor.
a.
If feed enters at 25oC, what size of reactor is needed?
b.
What is the optimum operating temperature for this
purpose?
c.
What size of reactor is needed if feed enters at optimum
temperature?
d.
What is the heat duty if feed enters at 25oC to keep the
reactor operation at its the optimum temperature?
Interstage Cooling
Review on Energy Balance in CSTR Operation
U  A  Ta  T 
n
~
0
  F  C

F

X


H
W

(
T

T
)
s
A0
i
pi
0
A0
Rx  0
i 1
Bila term kerja diabaikan dan HRx konstan:
FA0  X
Untuk
CSTR: V 

 H 0Rx

 n

UA
~

 FA 0    i  C pi  (T  T0 ) 
 T  Ta 
FA 0
 i1

FA 0  X
 rA
 rA  V 
 H 0Rx



UA

 FA0   Cp0  (T  T0 ) 
 T  Ta 
FA0


Pembagian kedua ruas dengan FA0:


UA
  rA  V 
0
 T  Ta 

   H Rx  Cp0  (T  T0 )  Cp0 
F

C
A0
p0
 FA0 

UA
FA 0  C p0
Tc 
FA 0  C p0  T0  U  A  Ta
U  A  FA 0  C p0

T0  F UCA  Ta
A0
UA
F C
p0
1

T0    Ta
 1
Review on Energy Balance in CSTR Operation

Tc 

  rA  V 

   H 0Rx  Cp0  [(T  T0 )    T  Ta ]
 FA0 
V
T0    Ta
 1
 Cp0  [T    T  (T0    Ta )]
 Cp0  [T  (   1) Tc  (   1)]
FA 0  X
 rA
 Cp0  (   1)  (T  Tc )

X   H0Rx

 Cp0  (   1)  (T  Tc )
G ( T )  R (T )
Bandingkan dengan
n
X

~
 i  Cpi  (T  T0 )
i 1
ˆ 
H oRx (TR )  C
p

T  TR 
Multiple Steady
State & Stability of
CSTR Operation
Multiple Steady State: Stability of CSTR
Operation
Finding Multiple Steady State: Varying To
Temperature Ignition – Extinction Curve
Upper steady state
Lower steady state
Ignition temperature
Extinction temperature
Runaway Reaction
Sample Problem on Multiple Steady State in
CSTR Operation
P8-17B