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Conversionand
Reactor Sizing
Suchithra Thangalazhy Gopakumar
H83RED Reactor
Design
Autumn Semester 2015
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Outline
Characterisation of reactor feed
Identification of Limiting reactant
Definition of the conversion
Conversion as a measurable design parameter
Design equations in terms of conversion
Levenspiel Plot
Levenspiel Plot as a reactor sizing tool
Sizing reactors in different configurations to
achieve a given conversion
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Limiting reactant is the one that get consumed faster in areaction
Excess reactant is the surplus amount of a reactant overthe stoichiometric amount required for the reaction tocomplete.
Characterisation of the Reactor Feed
+ +
=
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Characterisation of the Reactor Feed: Example
The gas phase reaction + 2 takes place inside a constant volume batchreactorat isothermal conditions. Initially, the reactor contains 1.5 kmol of CO, 1 kmol of O2and 1 kmol ofCO2 and 0.5 kmol ofN2at a pressure 5 atm.
(a) Identify the limiting reactant
(b) Determine the excess amount of the other reactant.
Solution
For reactants calculate theratios
= 1.52
= 0.75 = 1.01 = 1.0
Values of the ratiosshow
Therefore CO is the limitingreactant
One mole of CO2reacts with 0.5 moles of O2
Stoichiometric amount ofO2required
= 0.5 1.5 = 0.75
= .
. =.
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Characterisation of the Reactor Feed
When we consider the reactions taking place inside a reactor,the stoichiometric equation should be considered.
Consider following reaction
The stoichiometric equation is
Consider a chemical reaction as shown below
-a, -b, cand dare called stoichiometric coefficients. Negative and
positive signs indicate the consumption and generation ofmaterials respectively.
This can be rewritten in the form
Stoichiometric coefficients
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Conversion of species A is defined as the number of moles A that have
reacted per mole of A fed into the system:
fedAofmoles
reactedAofmolesAX (1) XA(max)
= 1 for irreversible reaction
=XAeqfor reversible reaction
In general, for reaction of any complexity we are able to follow conversion of any
substance via stoichiometry. For example, in case of the following reaction:
aA + bB cC + dD
every quantity could be put on a per mole A basis
A + B C + Da
b
a
c
a
d
Conversion of reactants correlates with product generation via stoichiometry
Conversion
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Conversion in a Batch Reactor
Batch reactors are operated for a given period of time. Consider ageneral form of the reaction within a batch reactor as
At timet= 0 the amount of A (the limiting reactant) charged into the batchreactor inNA0. After timet= tthe amount of A remaining in the reactor is NA.
Amount of A reacted during the period t
The number of moles of A that remain in the reactor after a time t:
[2]
[3]
[A]
XNA
0
reacted
Aofmoles
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Conversion in a Batch Reactor
The design equation for the batch reactor is (in which reaction [A] takes
place)
=
[4]
Substituting equation [3] in [4] =
=
The design equation indifferential formin term ofXA is
=
We call eq [5] the design equationfor a batch reactor, because we
have written the mole balance in terms of conversionXA .
[5]
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Conversion in a Batch Reactor
=
The design equation in differential form in term of XA is
= (
)
Integration with the limits at reaction start: t = 0,X = 0
= (
)
The design equation inintegral formin term of XA is
[6]
[7]
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Levenspiel* Plot
* Named after Octave Levenspiel, Oregon State University, who suggested this graphical method first
1
When we inspect the equations derived above, it is clear that wehave to carry out an integration of the form
It is clear from the rate lawthat
The above integral can be estimated graphically if we know the values of
or for various values of conversion X.
ConversionX
1.00
The rate of reaction for variousconversions of a limiting reactantcan be measure in a series ofcontrolled experiments.
Curve that shows
against
conversion x is called the LEVENSPIELPLOT
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Levenspiel Plot as a Reactor Sizing Tool
Batch Systems Graphical analysis
=
Consider a batch reactor
Therefore
As a result lowest at
Nearer to the end of the
reaction
have higher value that at()
Area=
1.00
Levenspiel Plot
ConversionX
At the start of the reaction (t=0)
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Levenspiel Plot as a Reactor Sizing Tool
=
When we consider aconstant volume batchreactor
=
=
= 1
=
= 1
= 1
and
= 1
()
Area=
Levenspiel Plot
Concentration
[8]
[9]
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Conversion: Example
A liquid phase reaction A + B C + D takes place in a constant volumeBatchReactor. The initial concentration of A, the limiting reactant, is 7 mol dm-3. Thereaction is non-elementary so that rA=k CAwith the specific rate constant k = 0.06
min-1. what is the conversion of A if the reactor has been operational for 30 minutes?
Solution
The reaction is
+ + With =
For a Batch reactor =
=
Reason out how the concentration is
included
Integrating gives
=
Reason out the limitsuses
=
= 1
=
= =
=
= 1
= 1
= 1
= .Therefore
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Conversion: Example
A liquid phase reaction A C + D takes place in a constant volumeBatch Reactor. The initialamount of A, the limiting reactant, fed to the reactor is 20 mol. The reaction is elementary so thatrA=k CAwith the specific rate constant k = 0.06 min
-1.
How long would it take to achieve 90% conversion?
How long would it take to achieve 99 % conversion?
Solution The reaction is + With = = 0.06
= 1 11
For a Batch reactor =
Fill the missing steps= ()
Substituting forNA = 1 ()
=
Therefore when XA=0.9
=1
0.06
1
1 0.9 min =
2.3
0.06 min
Therefore when XA=0.99
= 10.06
1
1 0.99 m in =
2.3
0.06 min
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Conversion in Flow Reactors
In batch reactors:X = f(t). In flow systems, reaction time increaseswith increasing reactor volume, therefore X = f(V)
Conversion is defined as
=
gives the molar rate at whichspecies A is reacting within the entiresystem at steady state
=
Molar flowrate
of A leaving
the system
=Molar
flowrate
of A fed
[10]
[11]
Rearranging equation [10]
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Flow Systems
The entering molar flow rate of species A, FAO(mol/s) is the product of the entering
concentration, CAO(mol/dm3) and the entering volumetric flow rate,v0 (dm3/s):
000 vCF AA
Liquid phase
CAOis given in term of molarity. Example: CAO =2 mol/dm3
Gas phase
CAOand FAOcan be calculated by the entering temperature and pressure using the
ideal gas law:
0
00
0
0
0 RT
Py
RT
PC AA
A
0
000000
RT
PyvCvF AAA
where NA0= entering number of mole of A
yA0 = entering mole fraction of AP0 = entering total pressure, e.g., kPa
PA0= yA0 P0 = entering partial pressure of A, e.g., kPa
T0 = entering temperature, K
R = ideal gas constant (e.g., R = 8.314 )mol.K
kPa.dm3
0000 RTNVP AA
Conversion in Flow Reactors
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Conversion in Flow Reactors
Since the exit composition fromthe reactor is identical to thecomposition inside the reactor(perfect mixing), the rate ofreaction is evaluated at the exitconditions.
Continuous Stirred Tank Reactor (CSTR)
Following equation was derived from molebalance
[12]
[11]
For flow reactor systems, from equation[11]
Substituting eq [11] in eq [12]
[13]
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Conversion in Flow Reactors
Plug Flow Reactor (PFR)
The mole balance applied toPFR provided
[11]
[14]
Substituting eq [11] in eq [14]
Packed Bed Reactor (PBR)
The mole balance applied toPFR provided
[11]
[16]
Substituting eq [13] in eq[18]
[15] [17]
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Conversion: Example
A liquid phase reaction A C + D takes place in a Packed Bed Reactor. Thereaction is elementary so thatrA=k CAwith the specific rate constant k = 0.005 dm
3
min-1 [kg(cat)]-1 . The expected conversion is 80%.
a. Find the mass of the catalyst as a function of the inlet flowrate.
Solution The reaction is + With = = 0.005 ( )
= 11
For a PBR =
Fill the missing steps=
()
Substituting forNA = 1
()
=
Therefore when XA=0.8 =
1
0.005
1
1 0.8 = 200 1.61
=
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Levenspiel Plot as a Reactor Sizing Tool
CSTR Graphical analysis
Rearranging the design equation
1.00
()
Area=
Levenspiel Plot
ConversionX
The area of the rectangleshown in the figure gives
the ratio.
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Levenspiel Plot as a Reactor Sizing Tool
Tubular Flow Reactors
Plug Flow Reactor Packed Bed Reactor
=
=
1.00
()
Area=
Levenspiel Plot
ConversionX
1.00
()
Area=
Levenspiel Plot
ConversionX
= 1
= 1
[16] [17]
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Space Time and Space Velocity
Thespace time(theholding timeormean residence time) is the time
necessary to process one reactor volume of feed fluid based on entranceconditions.
[18] Vthe reactor volume
0the volumetric flow rate entering the reactor
20 m20 m
Reactor
Consider the tubular reactor, which is 20 m long and 0.2 m 3 in volume. The dashed
line represents 0.2 m3 of fluid directly upstream of the reactor The time it takes for thisvolume to enter the reactor completely is the space time. For instance, if the space
time is 5 min, it means that every 5 minutes one reactor volume of feed at specified
conditions is being treated by the reactor.
Thespace velocity(SV) is defined as:
1so0 SV
VSV [19]
SVand
are not just the reciprocal parameters since they often refer to different
conditions:
to the entrance conditions; SV might be referred to the conditions at
specific location in the reactor
0
V
Levenspiel Plot as a Reactor Sizing Tool
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CSTR PFR
Design equations exit
0
)( A
A
r
XF
V
[20]
X
AA r
dX
FV0
0 [21]
Divide both sides of theequations by the entering
volumetric flow rate
exit0
0
0 )( A
A
r
XFV
X
A
A
r
dXFV
00
0
0
Put the left-hand sidesin terms of space times exit
0
)r(
XC
A
A
X
AA r
dX
C 00
[22] [23]
Area=/CA0=
XAX
ArX
1Ar
1
exitAr1
XAX
Area =/CA0=A
r
1
X
Ar
dX
0
Levenspiel Plot as a Reactor Sizing Tool
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Mole Balance for Different Reactors
Reactor Type Differential Algebraic Integral
Batch
CSTR
PFR
PBR
Vrdt
dXN AA 0
)(
0
0
tX
A
A
A
Vr
dXNt
exit
0
)( A
A
r
XFV
AA rdV
dXF0
out
in
X
X A
Ar
dXFV 0
AA rdW
dXF0
out
in
X
X A
Ar
dXFW 0
exit)r(
XC
A
A
0
out
in
X
X A
Ar
dXC 0
Conversion: Summary
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Constant Density Systems
Since the reaction volume does not change in the reactor (q= q0), molar flux
of species A into the reactor:
Lets express conversion from the equation (13):
In CSTR operated at steady-state
conditions, conversion does not change
through the reactor volume:
000 AA CF
)1(0 XFF AA 0
0
A
AA
F
FFX
0
0
00
000
0
0
A
AA
A
AA
A
AA
C
CC
C
CC
F
FFX
In PFR systems conversion changes
depending on location. The conversion at
specific point can be expressed viadifferentiation:
0
0
A
AA
C
CCX
0A
A
C
dCdX
Levenspiel Plot as a Reactor Sizing Tool
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CSTR PFR
Design equations
Replace conversionusing its relation withconcentrations
Design equations viaconcentration term
exit
0
)r(
XC
A
A
X
AA r
dX
C 00[26] [27]
Ar
1
A
r
1
0
00
A
A
A
A
C
CC
r
C
A
A
C
C A
A
A
AC
dC
rC
0 0
0
1
A
AA
r
CC
0
0A
A
C
C A
A
r
dC
[28] [29]
CACA(t)
Area ==
CA0
Constant densityonly
CACA(t)
Area ==
CA0
Constant densityonly
r
CC AA
0
0A
A
C
C A
A
r
dC
Levenspiel Plot as a Reactor Sizing Tool
Finding Space Time
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Methodology
If the rate of reaction is available as a function of conversion, -rA=f(X) (or in
some specific cases as a function of concentration, rA=f(CA)), or if it can begenerated by some intermediate calculation, one can size the reactor usingthe set of design equations by determining reactor volume or space time.
Task: The laboratory measurements of reaction rate for the gas phasereaction
A B + Cgave the following result - rates are in mol/(dm3 s)
0.0010.001250.00180.00250.00330.00400.00450.00500.00520.0053-rA
0.90.80.70.60.50.40.30.20.10.0X
T = 422.2 K, total pressure = 1013 kPa, initial charge 50 % inerts, 50 % A.Compare the volumes of CSTR and PFR required for the same conversion
of 60%, if the volumetric flow rate is 34.65 dm3
/s.
Reactor Sizing: Graphical Method
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X 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
-rA 0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001
Solution
mol
sdm24060400
1 3
0
.X
rF
V
AA
The CSTR volume necessary to
achieve 60% conversion is:
33
dm1200mol
sdm240
s
mol5
V
is the area of rectangle withvertices0A
F
V
(0, 0), (0, 400), (0.6,400), and(0.6, 0)
Reactor Sizing: Graphical Method -Example
0.80.60.40.2
200
400
600
0
ConversionX
Ar
1
CSTR
exit
0
)( A
A
r
XFV
exit)r(
X
F
V
AA
0
-1/rA
189 192 200 222 250 303 400 556 800 1000
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A BPFR
Solution (continuation)
X 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
-rA 0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001
-1/rA
189 192 200 222 250 303 400 556 800 1000
X
AA r
dX
F
V
00
).(r).(r)(r
X
r
dX
AAA
.
A 60
1
30
4
0
1
3
60
0
400]2224[1893
3.0
mol
sdm148
3
numerical evaluation of integralbased on Simpson's One-ThirdRule
The PFR volume necessary toachieve 60% conversion is:
33
dm740mol
sdm148
s
mol5
V
0.80.60.40.2
200
400
600
0
ConversionX
Ar
1
Reactor Sizing: Graphical Method -Example
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A B
Graphical Analysis
Solution (continuation)
X 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
-rA 0.0053 0.0052 0.0050 0.0045 0.0040 0.0033 0.0025 0.0018 0.00125 0.001
-1/rA
189 192 200 222 250 303 400 556 800 1000
0.80.60.40.2
200
400
600
0
ConversionX
Differencebetween
CSTR and PFRPFRAr
1
For isothermal reactions ofgreater than zero order, the PFRwill always require a smallervolume that the CSTR toachieve the same conversion
Reactor Sizing: Graphical Method -Example
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Reactor in Series
The same conversion may be reached in two stages, for example, one reactor will
provide conversion to 30% while second one will increase the conversion from 30%to 60%:
Two CSTR
X2=0.6
FA2
FA0
X1=0.3
X1=0.3
X2=0.6
FA0
Two PFR
FA1
V1
V2
V1
V2
FA1
FA2
Xi = Total moles of A reacted up to point i
Moles of A fed to the first reactor
i=1 i= 1
i= 2
i= 2
-rA1
-rA2
-rA1
-rA2
Conversion is defined in terms of location at a point downstream rather than with
respect to any single reactor
Reactor Sizing: Different Configurations
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Two CSTR
X2=0.6
FA2
FA0
X1=0.3
FA1
V1
V2
i=1
i= 2-rA1
-rA2
V = V1+ V2
?V
Xr
FV
2
1
A1
A01
A mole balance on the second reactor:
In Out + Generation = 0
Reactor 2:
A2
A2A12
2A2A2A1
r-FF
rFF
V
0V
A2
12AO2
A2
2AO1AO2
A2
2AOAO1AOAO2
r-
)X( XFV
r-XFXF-V
r-
)XF-( F)XF-( FV
2AOAOA2
1AOAOA1
XF-FF
XF-FF
Reactor Sizing: Different Configurations
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Two CSTR in series Two PFR in series
122
01
1
021 XX
r
FX
r
FVVV
A
A
A
A
0.80.60.40.2
200
400
600
0
ConversionX
CSTR 1
CSTR 2Ar
1
0.80.60.40.2
200
400
600
0
ConversionX
PFR 1
PFR 2Ar
1
Volume of two CSTR in series smaller
than the volume of one CSTR
Volume of two PFR in series is
identical to that of one PFR
The difference between overall volume of two CSTR in series and two (or one)
PFR decreases
2
1
1
0
021
X
X A
X
A
Ar
dX
r
dXFVVV
Reactor Sizing: Different Configurations
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1) CSTR-PFR 2) PFR-CSTR
Ar
1
0.80.60.40.2
200
400
600
0
ConversionX
CSTR
PFR Ar
1
0.80.60.40.2
200
400
600
0
Conversion X
CSTR
PFR
Reactor size can be optimised by the selection of number of stages, type of
reactors, and their combinations
2
1
01
1
021
X
X A
A
A
A
rdXFX
rFVVV 12
2
0
0
021
1
XXr
F
r
dXFVVV
A
AX
A
A
Reactor Sizing: Different Configurations
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3) PFR - CSTR - PFR
X1
X=0
V1
V2
V3
X2
X3
FA0
Ar
1
200
400
600
Conversion X
CSTR
PFR
V1
/ FA0
X1
X2
X3
V2
/ FA0
V3
/ FA0
122
02 XX
r
FV
A
A
1
0
01
X
A
Ar
dXFV
3
2
03
X
X A
Ar
dXFV
Reactor Sizing: Different Configurations
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Sequencing of reactors
Question:Which arrangement is the best?
Answer:It depends on the:
a) Shape of the Levenspiel plots
b) Relative reactor sizes
For autocatalytic reactions:
CSTR is more efficient at low conversions,
PFR is more efficient at high conversions
Reactor Sizing: Different Configurations
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Reactor Staging
Lets analyse multistage reactor system, say 5 stages of similar reactor:
A PFR can modelled as a
number of CSTR in series
1 2 3 4 5
31 2 4 5
X4X3X2X10 X5
V1V2
V3
V4
V5A
A
r
F
0 As we make the volume of each
CSTR smaller and increase the
number of CSTRs,
Total number of the CSTRs in series
Volume of the PFR
Reactor Sizing: Different Configurations
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Consider the reaction A B . It is carried out adiabatically in the liquidphase and following data were obtained.
X 0.0 0.2 0.4 0.6 0.65
-rA (kmol/m3
h)39 53 59 38 25
The reactor scheme is shown below
X2=0.6
FAex
FA0
X1=0.2
X2=0.65
FA1FA0
V1V2
V3
Calculate the volume of each of the reactors for an entering molar flowrate ofA of 50 kmol/h
Reactor Sizing: Different Configurations -Example
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Solution
For CSTR 1
X2
=0.6
FAex
FA0
X1
=0.2
X3=0.65
FA1FA0
V1V2
V3
CSTR 1
CSTR 2
PFR 1
10
1 Xr
FV
A
A
For PFR 1
dXr
FV
X
X A
A
2
1
02
For CSTR 2
230
3 XXr
FV
A
A
X 0.0 0.2 0.4 0.6 0.65
-rA (kmol/m3
h)39 53 59 38 25
[FA0/-rA] (m3) 1.28 0.94 0.85 1.32 2.0
hmolFA /50AofflowrateMolar 0
Reactor Sizing: Different Configurations -Example
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1.28
0.940.85
1.32
2
0
0.5
1
1.5
2
2.5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
310
1 188020940 m...Xr
FV
A
A
dXr
FV.
. A
A
60
20
02
60
0
40
0
20
02 4
3.xA
A
.xA
A
.xA
A
r
F
r
F
r
FXV
3
2 380 m.V
3230
3 106065002 m....XXr
FV
A
A
X 0.0 0.2 0.4 0.6 0.65
-rA(kmol/m3 h)
39 53 59 38 25
[FA0/-rA](m3)
1.28 0.94 0.85 1.32 2.0 CSTR 1
PFR
CSTR 2
Reactor Sizing: Different Configurations -Example
X2=0.6
FAex
FA0
X1=0.2
X3=0.65
FA1FA0
V1V2
V3
CSTR1
CSTR
2
PFR 1
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Summary