Transcript Document

제2장
Conversion
and Reactor Sizing
Chemical Reaction Engineering 1
반응공학 1
Objectives
a. To define conversion (X) and space time (t)
b. To rewrite the mole balances in terms of
conversion for a batch reactor, CSTR, PFR, and
PBR.
c. In expressing -rA as a function of conversion (X),
a number of reactors and reaction system can be
sized or a conversion be calculated from a given
reactor size.
- To relate the relative rates of reaction of reactants
and products.
Definition of Conversion
Consider the general equation
aA  bB  cC  dD
Choose A as our basis of calculation
(The basis of calculation is most always the limiting reactant )
b
A B 
a
c
d
C D
a
a
Question
- How can we quantify how far a reaction has progressed ?
- How many moles of C are formed for every mole A consumed ?
The convenient way to answer these question is to define conversion.
XA 
moleof A reacted
moleof A fed
Design Equations
Batch system
The longer a reactant is in the reactor, the more reactant is converted to product
until either equilibrium is reached or the reactant is exhausted. consequently, the
conversion X is a function of reaction time
m oleof A m oleof A  m olesof A reacted 
consum ed   fed    m oleof A fed 

 
 

m oleof A
N A0  
X 
consum ed 


The number of moles of A that remain in the reactor after a time t
m olesof A  m olesof A   m olesof A that 
 in reactor    initially fed to    havebeen consum ed

 
 

 at tim e t  reactor at t  0 by chem icalreaction
N A  
N A0 
N A0 X 

The number of moles of A in the reactor after a conversion X
N A  N A0  N A0 X  N A0 (1  X )
The mole balance on species A for a batch system
dN A
 rAV
dt
In term of conversion by differentiating equation
dN A
dX
 0  N A0
dt
dt
The design equation for a batch reactor in differential form is
N A0
dX
  rAV
dt
The differential form
for a batch reactor
For constant-volume batch reactor

1 dN A
d N A / V 
dC

  A  rA
V dt
dt
dt
For a variable volume batch reactor
Vdt  N A0
dX
 rA
or dt  N A0
dX
 rAV
When the volume is varied by

t
0
V dt  N A0 
X
0
dX
 rA
For the most common batch reactors where volume is not predetermined,
the time necessary to achieve a conversion X is
t  N A0 
X t 
0
dX
 rAV
The integral form
for a batch reactor
Flow systems
If FA0 is the molar flow rate of species A fed to a system at steady state,
the molar rate at which species A is reacting within the entire system will be FA0X.
m oles of A fed m olesof A reacted
FA0  X  

tim e
m olesof A fed
FA0  X 

m olesof A reacted
tim e
The molar flow rate
 m olar flow rate  m olarrate at which  m olar flow rate 
 at which A is   
   at which A 
consum
ed

 
 

 fed to the system  within the system  leaves the system
FA0 
FA0 X 
FA 


Rearranging gives
FA  FA0 1  X 
The entering molar flow rate, FA0 (mol/s)
FA0  CA0v0
CA0 : the entering concentration
v0 : the entering volumetric flow rate
• For liquid systems : CA0 is commonly given in term of molarity
• For gas systems : CA0 can be calculated from the entering T and P
using the ideal gas law or some other gas law
CA0  f T , P
• For an ideal gas (see Appendix B) :
C A0
PA0 y A0 P0


RT0
RT0
FA0
CSTR or Back-mixing Reactor
-
FA
The design equation for a CSTR
FA0  FA
V
 rA
-
conversion of flow system
FA0  FA  FA0 X
-
(2-11)
Combining (2-12) with (2-11)
(2-12)
1
-rA
FA0 X
V
 rA exit
(2-13)
Area
design equation
for a CSTR
X
Equation to determine the CSTR volume necessary to achieve a specified
conversion X. Since the exit composition from the reactor is identical to the
composition inside the reactor, the rate of reaction is evaluated at the exit condition.
Tubular Flow Reactor (PFR)
-
FA0
FA
General mole balance equation
dFA

  rA
dV
-
(2-14)
conversion of flow system
FA  FA0  FA0 X
-
The differential form of the design equation
FA0
-
dX
  rA
dV
(2-15)
1
-rA
Volume to achieve a specified conversion X
Area
V  FA0 
X
0
dX
 rA
(2-16)
X
Packed-Bed Reactor (PBR)
-
FA0
FA
General mole balance equation
dF A
 rA'
dW
-
conversion of flow system
FA  FA0  FA0 X
-
The differential form of the design equation
FA 0
dX
  rA'
dW
(2-17)
-The catalyst weight W to achieve a specified conversion X
W  FA0 
X
0
dX
 rA'
(2-18)
Summary of Design Equation
t  N A0 
X t 
dX
 rAV
0
FA0
FA0 X
V
 rA exit
FA
FA0
FA0
FA
FA
V  FA0 
X
W  FA0 
X
0
0
Design equation
for a batch reactor
Design equation
for a CSTR
dX
 rA
Design equation
for a PFR
dX
 rA'
Design equation
for a PBR
Summary of Design Equation
t  N A0 
X t 
dX
 rAV
0
반응시간은
- NA0에 비례
- X에 비례
- 반응속도 (rA)에 반비례
FA0
FA0 X
V
 rA exit
FA
- 반응기 부피에 반비례
반응기 부피(촉매의 무게)는
FA0
FA
V  FA0 
X
0
dX
 rA
- FA0에 비례
- X에 비례
- 반응속도 (rA)에 반비례
FA0
FA
W  FA0 
X
0
dX
 rA'
Applications of the design equation
for continuous-flow reactor
The rate of disappear of A, -rA, is almost always a function of the
concentrations of the various species present. When a single reaction is
occurring, each of the concentrations can be expressed as a function of the
conversion x; consequently, -rA, can be expressed as a function of X.
FA0
For a first-order reaction :
V  FA0 
FA
X
0
dX
 rA
 rA  kCA  kCA0 1  X 
V=
FA0
kCA
0

X
0
dX
1-X
= -
FA0
kCA
0
ln (1-X)
How to use the raw data of chemical reaction rate?
Consider the isothermal gas-phase isomerization
A
B
The laboratory measurements give
the chemical reaction rate as a function of conversion.
(at T=500K, 830kPa(=8.2atm), Reactant=Pure A)
X
0
0.1
0.2
0.4
0.6
0.7
0.8
-rA (mol/m3-sec)
0.45
0.37
0.30
0.195
0.113
0.079
0.050
1/-rA (m3-sec/mol)
FAo/-rA (m3)
2.22
2.70
3.33
5.13
8.85
12.7
20.0
0.89
1.08
1.33
2.05
3.54
5.06
8.0
Levenspiel Plot

Small rate
Greatest rate
How to use the raw data of chemical reaction rate?
-
For irreversible reactions,
the maximum value of X is that for complete conversion, i.e. X=1.0.
A BC
1


rA
-
as
X 1
For reversible reactions,
the maximum value of X is the equilibrium conversion, i.e. X=Xe.
A  BC
1
   as X  X e
rA
Reactor Size
• Given –rA as a function of conversion.
• Constructing a Levenspiel plot.
• Here we plot either
For
FA0
 rA
FA0
 rA
or
1
 rA
as a function of X.
vs. X, the volume of a CSTR and the volume of a PFR
can be represented as the shaded areas in the Levenspiel plots.
Example 2-2. Sizing a CSTR
The reaction, AB, is carried out in a CSTR. Molar flow rate of A is 0.4 mol/sec.
(1) Using data in the previous Table, calculate the reactor volume necessary to achieve 80%
conversion in a CSTR
(2) Shade a area in Figure 2-2 that would give the CSTR volume necessary to achieve 80% conversion
(1)
FA0 X
V
 rA exit
(
1
rA
)
= 20 m3-sec/mol
X=0.8
V=(0.4 mol/sec)(0.8)(20 m3-sec/mol)
=6.4 m3=6400 liter
(2)
Example 2-3. Sizing a PFR
The reaction, AB, is carried out in a PFR. Molar flow rate of A is 0.4 mol/sec.
(1) Using data in the previous Table, calculate the reactor volume necessary to achieve 80%
conversion in a PFR
(2) Shade a area in Figure 2-2 that would give the PFR volume necessary to achieve 80% conversion
(3) Make qualitative sketches of conversion (X) and rate of reaction (-rA) with respect to reactor volume
(1)
V  FA0 
X
0
V=
FA0

dX
 rA
0.8 dX
-rA
=

0.8
FA0
dX
0 -rA
By applying Appendix A-23 (Five Point Quadrature Formula): X=0.8/4=0.2
0
V=
0.2
( 3 ) [0.89+4(1.33)+2(2.05)+4(3.54)+8] =2.165m =2165 liter
3
Example 2-3. Sizing a PFR
(b)
Example 2-3. Sizing a PFR
(c) By applying Simpson’s rule in Appendix A.4, we can calculate V for X=0.2, 0.4, 0.6, 0.8
(See the text, page 52). The results are as follows.
X
-rA (mol/m3-sec)
V (dm3)
0
0.45
0
0.2
0.30
218
0.4
0.195
551
0.6
0.113
1093
0.8
0.05
2165
전환율을 조금
더 높이기 위해
서는 반응기 부
피가 많이 늘어
나야 한다.
Example 2-4. Comparing CSTR and PFR Sizes
For isothermal reaction of greater than zero order, the PFR will always require a
smaller volume than the CSTR to achieve. What if zero order reaction?
Reactors in series
Define conversion
The conversion X defined as the “total number of moles” of A that
have reacted up to that point per mole of A fed to the “first” reactor.
(assumption : no side stream withdrawn and the feed stream enters
only the first reactor in the series)
total m olesof A reacted up to pointi
Xi 
m olesof A fed to first reactor
PFR-CSTR-PFR in series
X=0
FA0
V1
X1
FA1
V2
X2
FA2
V3
X3
FA3
The relationships between conversion and molar flow rate
FA1 = FA0 - FA0 X1
FA2 = FA0 - FA0 X2
FA3 = FA0 - FA0 X3
where
total m olesof A reacted up to point 2
X2 
m olesof A fed to first reactor
similar definitions
exist for X1 and X3
Reactor 1:
V1  FA0 
X1
0
X=0
FA0
dX
 rA
V1
Reactor 3 :
dX
X2  r
A
V3  FA0 
X1
FA1
V2
X2
FA2
V3
X3
X3
FA3
Reactor 2 :
in  out  gen.  0
FA1  FA2  rA2V2  0
FA1 = FA0 - FA0 X1
FA2 = FA0 - FA0 X2
FA3 = FA0 - FA0 X3
V2 
F A0( X 2  X 1 )
 rA2
-rA2 is evaluated
at X2 for the CSTR
In this series
arrangement
Different Schemes of Reactors in Series
FA0
Two CSTRs in series
X1=0.4
FAe
X2=0.8
Two PFRs in series
FA0
X1=0.4
FA0
FAe
X2=0.8
X1=0.5
a PFR and a CSTR in series
FAe
X2=0.8
FA0
a CSTR and a PFR in series
X1=0.5
FAe
X2=0.8
Two CSTRs in Series
FA0
X1=0.4
FAe
X2=0.8
Reactor 1
FAo (X1-Xo)
 1 
 X 1 =
V1  FA0 
-rA1 0
  rA1 
Reactor 2
FA0 ( X 2  X 1 )
V2 
 rA2
Example 2-5: Two CSTRs in Series
What is the volume of each of
Two reactors?
FA0
X1=0.4
FAe
X2=0.8
XA
[FAo/-rA] (m3)
0.0 0.1 0.2 0.4 0.6 0.7 0.8
0.89 1.09 1.33 2.05 3.54 5.06 8.0
Reactor 1
[FAo/-rA]x=0.4=2.05 m3
V1=([FAo/-rA]x=0.4)(X1-X0)=(2.05)(0.4-0)=0.82 m3
Reactor 2
[FAo/-rA]x=0.8=8.0 m3
V1=([FAo/-rA]x=0.8)(X2-X1)=(8.0)(0.8-0.4)=3.2 m3
Example 2-5: Two CSTRs in Series
Therefore, V1+V2=0.82+3.2=4.02 m3
What is the reactor volume to achieve 80%
Conversion in a single CSTR?
[FAo/-rA]x=0.8=8.0 m3
V1=([FAo/-rA]x=0.8)(X1-X0)
=(8.0)(0.8-0)=6.4 m3
The sum of the two CSTR reactor volumes (4.02 m3) in series is less than the
volume of one CSTR (6.4 m3) to achieve the same conversion (X=0.8)
Example 2-5: Two CSTRs in Series
FA0
X=0.8
FA
Vtotal = 6.4 m3
[FAo/-rA]
(m3)
FA0
X1=0.4
FAe
X2=0.8
Vtotal = 4.02 m3
A PFR by a Large Number of CSTRs in Series
Approximating a PFR with a number of small, equal-volume CSTRs of Vi in series
1
1
2
2
3
4
3
5
4
5
A PFR by a Large Number of CSTRs in Series
1 2 3 4 5
1
2
3
4
5
80
FA0
 rA
60
40
20
.35
X
.53
.65 .74
.8
As we make the volume of each CSTR smaller and increase the number of CSTRs,
the total volume of the CSTRs and the PFR will become identical. The performance
of a PFR is equal to that of a number of (N) CSTRs in Series.
Can you verify this mathematically?
Two PFRs in Series
FA0
Reactor 1
Reactor 2
X1=0.4
V1  FA0 
X1
0
V2  FA0 
X2
X1
dX
 rA
dX
 rA
FAe
X2=0.8
Two PFRs in Series
VTotal= V1 + V2=

X1
0
FA0
-rA
dX +

X2
X1
FA0
-rA
dX =

X2
0
FA0
-rA
Sizing PFR in Series
What is the volume of each of
Two reactors?
FA0
X1=0.4
Molar flow rate of A is 0.4 mol/sec.
XA
[FAo/-rA] (m3)
FAe
X2=0.8
0.0 0.1 0.2 0.4 0.6 0.7 0.8
0.89 1.09 1.33 2.05 3.54 5.06 8.0
Reactor 1
By applying Simpson’s rule in Appendix A.4 (Text page 60),
V1=
0.2
( 3 ) [0.89+4(1.33)+2.05] =0.551 m =551 liter
3
Reactor 2
By applying Simpson’s rule in Appendix A.4 (Text page 60),
V2=
0.2
( 3 ) [2.05+4(3.54)+8.0] =1.614 m =1614 liter
3
Therefore, V1 + V2=0.551 m3 + 1.614 m3=2.165 m3 < 4.02 m3 (Two CSTR in Series)
Combination of CSTR and PFR in Series
Reactor 1
FAo (X1-Xo)
V1 =
-rA1
0
Reactor 2
FAo (X2-X1)
V2 =
-rA2
Reactor 3
V3 =

X3
X2
FA0
-rA
dX
Example 2-7: Liquid-Phase Isomerization
n-C4H10
i-C4H10
X
0.0
0.2 0.4 0.6 0.65
-rA (kmol/m3-h)
39
53
59
38
25
Calculate the volume of each of the three reactors for an entering molar
flow rate n-butene of 50 kmol/h.
Example 2-7: Liquid-Phase Isomerization
FAo = 50 kmol/h
X
0.0
0.2 0.4 0.6 0.65
-rA (kmol/m3-h)
39
53
[FAo/-rA](m3)
59
38
25
1.28 0.94 0.85 1.32 2.0
(a) Reactor 1 (X1=0.2)
FAo (X1-Xo)
V1 =
-rA1
0
(b) Reactor 2 (X2=0.6)
V2 =

0.6
FA0
0.2 -rA
dX
= (0.94)(0.2)=0.188 m3
Example 2-7: Liquid-Phase Isomerization
By applying Simpson’s three point formula in Appendix A.4 (Text page 64),
V2=
0.2
( 3 ) [0.94+4(0.85)+1.32] =0.38 m
3
(c) Reactor 3 (X3=0.65)
V3 =
FAo (X3-X2)
-rA3
= (2.0)(0.65-0.6)=0.1 m3
Reactor Sequence
FA0
X1=0.5
FAe
X2=0.8
X12
dX
 rA
Reactor 1
V  FA0 
Reactor 2
FA0 X
V 
 rA exit
X 10
Reactor Sequence
Scheme A
FA0
X1=0.5
FAe
X2=0.8
PFR
CSTR
Total volume= Vtotal=V1+V2= 305 dm3
Reactor Sequence
Scheme B
FA0
X1=0.5
CSTR
PFR
Total volume= Vtotal=V1+V2= 262.3 dm3
FAe
X2=0.8
Reactor Sequence
Scheme A
FA0
X1=0.5
FAe
X2=0.8
Scheme B will give the smaller
total volume for an intermediate
conversion of 50%.
However, the relative sizes of
the reactors depend on the
intermediate conversion.
Vtotal=V1+V2= 305 dm3
What if zero order reaction?
FA0
Scheme B
X1=0.5
FAe
X2=0.8
Vtotal=V1+V2= 262.3 dm3
Choosing Reactor Sequence
[FAo/-rA]
PFR
MFR
0
X
Space Time
Space-time :
The time necessary to process one reactor volume of fluid based on
entrance conditions. Also called the holding time or mean residence time.
 tim e required to process one 

V 
t
  reactor volum e of feed m easured  tim e
v0 

at
specified
condition


A space-time of 2 min means that every 2 min one reactor volume of
feed at specified condition is being treated by the reactor.
FA0
V
t 
v0
v0

X
0
X dX
dX
 C A0 
0 r
 rA
A
Space Time
Space-time :
The time necessary to process one reactor volume of fluid based on
entrance conditions. Also called the holding time or mean residence time.
20m
20m
Consider the tubular reactor, which is 20m long and 0.2 m3 in volume.
The dashed line represents 0.2 m3 of fluid directly upstream of the
reactor. The time it takes for this fluid to enter the reactor completely
is the space time.
Space Velocity
Definition of Space-velocity
 num ber of reactor volum es of 

v0 1 
SV     feed at specified condition which  tim e1
V t 

can
be
treated
in
unit
volum
e




• A space-velocity of 5 hr-1 means that five reactor volumes of feed at specified
condition are being fed into the reactor per hour.
• Difference in the definitions of SV and t
- space time : the entering volumetric flow rate is measured at the entrance condition
- space velocity : other conditions are often used
LHSV and GHSV
• LHSV ( liquid hour space velocity)
- v0 is frequently measured as that of a liquid at 60 or 75 0F,
even though the feed to the reactor may be a vapor at some
higher temperature.
• GHSV ( gas hour space velocity)
- v0 is normally measured at standard temperature and pressure
For reaction rate depending only on the concentration
• It is usually convenient to report –rA as a function of concentration rather
than conversion.
- PFR
X dX
design equation : V  FA0 0
 rA
molar flow rate :
FA0  v0CA0
flow system conversion :
X
FA0  FA
FA0
• For the special case when v = v0
X 
FA0  FA C A0 v0  C Av C A0  C A


FA0
C A0v0
C A0
when X  0, C A  C A0
when X  X , C A  C A
For reaction rate depending only on the concentration
• Differentiating yields
 dCA
dX 
C A0
V  v0 
C A0
CA
t 
C A0
CA
dCA
 rA
dCA
 rA
• A typical curve for determining
the space time, t
Homework
P2-7B
P2-8B
P2-9B
Due Date: Next Week