Chemical Reactor DesignChemical Reactor Design
Y W LYoun-Woo LeeSchool of Chemical and Biological Engineering
Seoul National Universityy155-741, 599 Gwanangro, Gwanak-gu, Seoul, Korea [email protected] http://sfpl.snu.ac.kr
Chapter 4Chapter 4
Isothermal Reactor DesignIsothermal Reactor Design
Chemical Reactor DesignChemical Reactor DesignChemical Reactor DesignChemical Reactor Design
化學反應裝置設計化學反應裝置設計化學反應裝置設計化學反應裝置設計
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Objectives
ib h l i h h ll h d l h i l
j
• Describe the algorithm that allows the reader to solve chemical reaction engineering problems through logic rather than memorization.
• Size batch reactors, semibatch reactors, CSTRs, PFRs, and PBRs for isothermal operation given the rate law and feed conditions.
• Studying a liquid-phase batch reactor to determine the specific Study g a qu d p ase batc eacto to dete e t e spec creaction rate constant needed for the design of a CSTR.
• Design of a tubular reactor for a gas phase pyrolysis reaction• Design of a tubular reactor for a gas-phase pyrolysis reaction.
• Account for the effects of pressure drop on conversion in packed b d b l d i k d b d h i lbed tubular reactors and in packed bed spherical reactors.
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Fig. 4-1 Isothermal Reaction Design AlgorithmAlgorithm for Conversion
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Algorithm for isothermal reactor design
1. Mole balance and design equation1. Mole balance and design equation2. Rate law3 Stoichiometry3. Stoichiometry4. Combine5 E al ate5. Evaluate
We can solve the equations in the We can solve the equations in the combine step eithercombine step eithercombine step eithercombine step either
A. Graphically (Chapter 2)
B N i l (A di A4)B. Numerical (Appendix A4)
C. Analytical (Appendix A1)
D. Software packages (polymath)Seoul National University
French Menu Analogy
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Algorithm for isothermal reactorsFrench Menu
Analogy
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4.2.1 Batch Operationp
For constant volume batch reactor, the mole balance can be written
dN 1 dCVNddNdN /11
,in terms of concentration
AA r
dtdN
V
1
AAAAA r
dtdC
dtVNd
dtdN
VdtdN
V 0
0
/11
Generally, when analyzing laboratory experiments, it is best toprocess the data in terms of the measured variable. Becauseconcentration is the measured variable for most liquid-phaseconcentration is the measured variable for most liquid phasereactions, the general mole balance equation applied to reactionsin which there is no volume change becomes
AA r
dtdC
This is the form we will use in analyzing reaction rate data in Chap 5.Seoul National University
Reaction time in Batch Operation
Algorithm for isothermal reactor designAlgorithm for isothermal reactor design BA
AA VrdtdXN 00
1. Mole balance
& Design equation
AA
XCC
AinorderndleirreversibkCr 2
)1(
2,2. Rate law
AA
XkCdX
XCC
2
0
)1(
)1(
4 C bi ti
3. Stoichiometry
A
dtkCdX
XkCdt 0 )1(
2nd orderIsothermal
Liquid-phaseBatch reaction
4. Combination
Xt
A
dX
dtkCX 02
1
)1(
X1
Batch reaction
5. Analytical Evaluation
X
A
t
XdX
kCdt
0 200 )1(
1
X
XkC
tA 1
1
0Seoul National University
Reaction time in Batch Operation
Typical cycle times for a batch polymerization processt = tf + t + t + tActivity Time (h)
tt tf + te + tR + tc
1. Charge feed to the reactor and agitate, tf2. Heat to reaction temperature, te
1.5-3.00.2-2.0V i3. Carry out reaction, tR
4. Empty and clean reactor, tc
Varies0.5-1.0
Total cycle time excluding reaction 3 0 6 0Total cycle time excluding reaction 3.0-6.0
Batch polymerization reaction times may vary between 5 and 60 hoursBatch polymerization reaction times may vary between 5 and 60 hours.Decreasing the reaction time with a 60-h reaction is a critical problem. As thereaction time is reduced (e.g. 2.5 h for a 2nd-order reaction with X=0.9 andkC 10-3 s-1) it becomes important to se large lines and p mps to achie ekCA0=10-3 s-1), it becomes important to use large lines and pumps to achieverapid transfer and to utilize efficient sequencing to minimize the cycle time.
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Batch Reaction Times
BA
Mole balance VN
rdtdX A
0Ndt A
Rate lawkCr AA
order-First order Second2kCr AA
)1(00
XCVNC A
AA
kCr AA kCr AA
Stoichiometry (V=V0)
XkdtdX
)1( )1( 20 XkC
dtdX
A Combine
Xkt
11ln1
)1(0 XkCXt
A Integration
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Batch Reaction Times
t
s
1ln1
)10k 0.9,(Xorder -1st 1-4
Xt
s
R
)10kC 0.9,(Xorder 2nd 13A0
XktR
1ln11
ln
XkCt
AR
9.0
)1(0
k3.2
9.01ln
kCA
9
)9.01(0
k3.2
14
kCA
90
ssec000,23
10 14
ssec000,9
10 13
hr4.6 hr5.2Seoul National University
Batch Reaction Times
The order of magnitude of time hi 90% i
Table 4-3
to achieve 90% conversionFor first- and second-order irreversible batch reactions
1st-orderk (s-1)
2nd-orderkCA0 (s-1)
Reaction timetR
10-4 10-3 Hours
10 2 10 1 Mi t10-2 10-1 Minutes
1 10 Seconds1 10 Seconds
1,000 10,000 Milliseconds
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Design a Reactor to Produce of ethylene glycol
Design a CSTR to produce 200 million pounds of ethylene glycol per yearby hydrolyzing ethylene oxide. However, before the design can be carriedout, it is necessary to perform and analyze a batch reactor experiment todetermine the specific reaction rate constant (kA). Since the reaction will becarried out isothermally k will need to be determined only at the reactioncarried out isothermally, kA will need to be determined only at the reactiontemperature of the CSTR. At high temperature there is a significant by-product formation, while at temperature below 40oC the reaction does notproceed at a significant rate; consequently, a temperature of 55oC has beenchosen. Since the water is usually present in excess, its concentration maybe considered constant during the course of the reaction The reaction isbe considered constant during the course of the reaction. The reaction isfirst-order in ethylene oxide.
O CH OHO CH2-OH
CH2-CH2 + H2O CH2-OHH2SO4
A + B C Catalyst
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Example 4-1 Determining k from Batch Data
In the lab experiment, 500mL of a 2 M solution (2 kmol/m3) of EO inwater was mixed with 500mL of water containing 0 9 wt % sulfuric acidwater was mixed with 500mL of water containing 0.9 wt % sulfuric acidcatalyst. At T=55oC, the CEG was recorded with time. Determine thespecific reaction rate at 55oC
Time Concentration of EG
specific reaction rate at 55 C.
EO + H2O → EG(min) (kmol/m3)0.0 0.0000.5 0.145
A + B → C
1.0 0.2701.5 0.3762.0 0.4672.0 0.4673.0 0.6104.0 0.7156 0 0 8486.0 0.848
10.0 0.957Seoul National University
Problem Solving AlgorithmExample 4-1 Determining k from Batch Data
A. Problem statement. Determine the kAB Sketch
D. Assumptions and approximations:AssumptionsB. Sketch
C. IdentifyC1. Relevant theories
Rate law: AA Ckr
Assumptions1. Well mixed2. All reactants enter at the same time3. No side reactions
batchA, B, C
Rate law:
Mole balance:C2. Variables
AAA Ckr
Vrdt
dNA
A
3. No side reactions4. Negligible filling and emptying time5. Isothermal operationApproximations
Dependent: concentrations, CiIndependent: time, t
C3. Knowns and unknowns
pp1. Water in excess (CH2O~constant)
CB~ CBOE. Specification. The problem is neither
Knowns: CEG = f(time)Unknowns:
1. CEO = f(time)
overspecified nor underspecified.
F. Related material. This problem uses the2. kA3. Reactor volume
C4. Inputs and outputs: reactant fedll b h
mole balances developed in Chap. 1for a batch reactor and thestoichiometry and rate laws developedi Ch 3all at once a batch reactor
C5. Missing information: Nonein Chap. 3.
G. Use an Algorithm.(figs 4-1 & 4-2)Seoul National University
Problem Solving AlgorithmExample 4-1 Determining k from Batch Data
AA r
dtdN
V
11. MOLE BALANCE Batch reactor that is well-mixeddtV
2 RATE LAW kCrSince water is present in such excess, theconcentration of water at any time t is virtuallythe same as the initial concentration and the rate2. RATE LAW AA kCr the same as the initial concentration and the ratelaw is independent of the concentration of H2O.(CB~CB0)
3. STOICHIOMETRY
Species symbol Initial Change Remaining Concentration
CH2CH2O A NA0 - NA0X NA=NA0(1-X) CA=CA0(1-X)H2O B ΘBNA0 - NA0X NB=NA0(ΘB-X) CB=CA0(ΘB-X)H2O B ΘBNA0 NA0X NB NA0(ΘB X) CB CA0(ΘB X)
CB~ CA0 ΘB = CB0
(CH2OH)2 C 0 NA0X NC =NA0X CC=CA0X
NT0 NT =NT0 - NA0X
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Problem Solving AlgorithmExample 4-1 Determining k from Batch Data
AA r
dtdN
V
1
A rdC kCr 4 COMBINING
A kCdCAr
dt AA kCr 4. COMBINING
Mole balance Rate law
AkCdt
,
5. EVALUATE
For isothermal operation k is constant:For isothermal operation, k is constant:
At tC
A ktC
dtkkdtdCA 0ln
kt
AC A
eCC
ktC
dtkkdtCA
0 0ln
0
AA eCC 0
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Problem Solving AlgorithmExample 4-1 Determining k from Batch Data
The concentration of EG at any time t can be obtained from the reaction stoichiometry
A + B C
NNXNN
)1(00
00
ktAAA
CCC
AAAC
eCCCNN
C
NNXNN
)(000
AAAC VV
)1(0kt
AC eCC
kCC CA 0l kt
CA
CA 0
0lnSeoul National University
Example 4-1 Determining k from Batch Data
Rearranging and taking the logarithmof both side yields
CCAA Cr )min311.0( 1
We see that a plot ln[(CA0-CC)/CA0] as
ktC
CC
A
CA
0
0ln The rate law can now be used in the designof an industrial CSTR. Note that this ratelaw was obtained from the lab-scale batchWe see that a plot ln[(CA0 CC)/CA0] as
a function of t will be a straight line witha slope –k.
1
3210ln
reactor (1000 mL).
Time 0 CA CC
CA0
0.61
12
min311.055.195.8
3.210ln
tt
k
(min)0.0 1.0000.5 0.8551 0 0 730
0AC0.1
(CA0-C
C)/
CA
0.061.0 0.7301.5 0.6242.0 0.5333.0 0.390
(
4.0 0.2856.0 0.152
10.0 0.043
0.01
0 2 4 6 8 10 12
t (min)1.55 8.95Seoul National University
4.3 Design of CSTRDesign Equation for a CSTR
A XFV 0Mole balance
AA
exitA
XCVXCvV
rV
000
)(Mole balance
the space time000 AA CvF
AA rvr 0
For a 1st-order irreversible reaction, the rate law is
the space time
A
AA
XCX
kCr
01
Rate law
C biA
A
kCXk0
1
Combine
Rearrangingg g
kkX
1
CSTR relationship between space time and
conversion for a 1st-order k1liquid-phase rxn
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4.3.1 A single CSTR
We could also combine Eq (3-29) and (4-8) to find the exit t ti f A C
kCXCC AAA
1)1( 00
concentration of A, CA:
kX
kC
kkkC
kCXCC
AA
AAA
111
11)1(
00
00 k1
CC
kk
AA
11
0
kCA 1
kC
C AA
10exit concentration of Ak1
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4.3.2 CSTRs in Series
CA1, X1CA0
v0
CA2, X2
v0
-rA1, V1 -rA2, V2
For 1st-order irreversible reaction with no volume change (v=v0) is carried out in two CSTRs placed in series. The effluent concentration of A from reactor 1 is
11
01 1 k
CC A
A
From a mole balance on reactor 2,
21021 AAAA CCvFF 22
210
2
212
A
AA
A
AA
CkCCv
rFFV
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CSTRs in Series
Solving for CA2, the concentration exiting the second reactor, we get
1122
0
22
12 111 kk
Ck
CC AAA
11
01 1 k
CC A
A
If instead of two CSTRs in series we had n equal-sized CSTRs connected in series (1 = 2 = … = n = i = (Vi/v0)) operating at the same temperature (k1 = k k k) th t ti l i th l t t ld bk2 = … = kn = k), the concentration leaving the last reactor would be
AA CCC 00
nnAnk
CDa11
The conversion and the rate of disappearance of A for these n tank reactors in series would be
X 11 AkCkC 0
nkX
111
nA
AnAnk
kCr
1
0
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Conversion as a function of reactors in series for different Damköhler numbers for a first-order reaction
k=1
k=0.1
k=0.5
Da 1, 90% conversion is achieved in two or three reactors; 0
0
A
A
FVr
Da
, ;thus the cost of adding subsequent reactors might not be justified
Da ~0.1, the conversion continues to increase significantly with each reactor added Seoul National University
Reaction Damköhler number
rate"reaction a"Entranceat Reaction of Rate0
A VrDa
rate" convection a"A of Rate Flow Entering0
AFDa
The Damköhler is a dimensionless number that can give us a quickestimate of the degree of conversion that can be achieved in continuous-flow reactor.
For 1st-order irreversible reaction kCv
VkCF
VrDa
A
A
A
A
00
0
0
0
For 2nd-order irreversible reaction 000
20
0
0A
A
A
A
A kCCv
VkCF
VrDa
RULE OF THUMB
000 AA CvF
Da 0.1 will usually give less than 10% conversion.Da 10.0 will usually give greater than 90% conversion.
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4.3.4 A Second-Order Reaction in a CSTR
For a 2nd-order liquid-phase reaction We solve the above eq. for X:being carried out in a CSTR, thecombination of the rate law and thedesign equation yields
We solve the above eq. for X:
22121 20
200 AAA kCkCkC
X
200
kCXF
rXF
V AA
2
41212
00
0
000
AA
A
AAA
kCkCkC
kCX
(4-14)2kCr AA
For const density v=v0, FA0X=v0(CA0-CA)
Da2Da41Da21
2 0
AkC
(4-16)
20
0 kCCC
vV
A
AA The minus sign must be chosen in the
quadratic equation because X cannot
X
Using our definition of conversion, we have
q qbe greater than 1.
(4 15) Da41Da21 2
0 )1( XkCA (4-15)
Da2Da41Da21
X
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A Second-Order Reaction in a CSTR
V
0
0
A
A
FVr
Da
0.670.88
At high conversion, a 10-foldincrease in Da will increase theincrease in Da will increase theconversion only to 88% due tolowest value of reactantconcentration in CSTR
606concentration in CSTR.
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Example 4-2: Producing 200,000,000 lb/yr in a CSTR~91 ton/yr
It is desired to produce 200 x 106 pounds per year
CA0=1 vA0=vB0
p p p yof EG. The reactor is to operated isothermally.A 1 lb mol/ft3 solution of ethylene oxide (EO) inwater is fed to the reactor together with an equalwater is fed to the reactor together with an equalvolumetric solution of water containing 0.9 wt%of the catalyst H2SO4. The specific reaction rateconstant is 0 311 min-1 as determined in Ex 4-1constant is 0.311 min as determined in Ex 4-1.
(a) If 80% conversion is to be achieved, determineth CSTR lthe necessary CSTR volume.
(b) If two 800-gal reactors were arranged in parallel,what is the corresponding conversion?
(c) If two 800-gal reactors were arranged in series,what is the corresponding conversion?
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M.W. of EG=62
M.W. of EO=58
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v0 = vA0 + vB0
=1
10’3m
5’1.5m
~2 gps
1 gal ~3.78 L
5 gal~19.8L
ga 3 8
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Producing 200,000,000 lb/yr of EG in a CSTR
two equal-sized CSTRs in series two equal-sized CSTRs in parallel one CSTR
800gal
X=0.81
800gal
800gal 800gal
800gal
X=0.81800gal 800gal
X1=0.68 X2=0.901480galX=0.8
Conversion in the series arrangement is greater than in parallel for CSTRs.The two equal-sized CSTRs in series will give a higher conversion thanThe two equal sized CSTRs in series will give a higher conversion thantwo CSTRs in parallel of the same size when the reaction order is greater than zero.
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4.4 Tubular Reactors
-2nd-order liquid-phase rxn(AProducts)
Rate law: 2AA kCr
(AProducts)-Turbulent,- No dispersion- No radial gradients in T, u, or C
X
AA kC
dXFV0 20
St i hi t f li hPLUG-FLOW REACTOR
PFR mole balance
Stoichiometry for liq. phase rxnT & P = constant
)1( XCCPFR mole balance
AA rdVdXF 0
)1(0 XCC AA
Combination
The differential form of PFR designequation must be used when there is aP or heat exchange between PFR &th d I th b f P
XX
kCvdX
XkCFV
A
X
A
A
1)1(1
0
00 22
0
0
the surrounds. In the absence of P orheat exchange, the integral form of thePFR design equation is used.
2
2
0
0
11 DaDa
kCkCX
A
A
X
AA r
dXFV0
0
20 11 DakCA Da2 is the Damkohler numberfor a second-order reactionSeoul National University
4.4 Tubular Reactors
-2nd-order gas-phase rxn(AProducts)
Rate law: 2AA kCr
(AProducts)-Turbulent,- No dispersion- No radial gradients in T, u, or C
XA kC
dXFV0 20
PLUG-FLOW REACTOR
PFR mole balance
AkC0 2
Stoichiometry for gas phase rxnPFR mole balance
AA rdVdXF 0
Stoichiometry for gas phase rxnT & P = constant
)1()1(0 XCXFFC AA
The differential form of PFR designequation must be used when there is aP or heat exchange between PFR &th d I th b f P
)1()1( 00 X
CXvv
C AA
Combinationthe surrounds. In the absence of P orheat exchange, the integral form of thePFR design equation is used.
Combination
dXXFVX
A
22
2
0)1(
X
AA r
dXFV0
0XkCA
A 0 220
0 )1(Seoul National University
4.3 Tubular Reactors
dXXFVX
2)1( dX
XkCFV
AA
0 22
00 )1(
)(
CA0 is not function of X; k=constant (isothermal)
dXXX
kCF
VX
A
0 2
2
20
)1()1(
A0 ; ( )
000 vCF AA XkCA
00 )1(
2
Integration yields (see Appendix A.1 Eq. (A-7) @ page1009)
X
XXXkC
vV
A 1)1()1ln()1(2
22
0
0
XXX
vL )1()1l ()1(2
220
Reactor length will be
LAV c
X
XXAkC
LcA 1
)()1ln()1(2 2
0
0
Cross sectional area
V c
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Conversion as a function of distance down the reactor
1
1.2A 0.5B (=-0.5)
A B ( 0)
0.8
on (
X) A B (=0)
A 2B (=1)
A 3B (=2)
0.4
0.6
onve
rsi
0.2
Co
3
0
0 0.2 dmkC
v
A
0
0 2 4 6 8 10 12 14
L (m)L (m)
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The reaction that has a decrease in the total number of moles will have the highest conversion for a fixed reactor lengthwill have the highest conversion for a fixed reactor length.
0)5.01( vXv the reactant spends more timethe reactant spends more time
0)21( vXv the reactant spends less timethe reactant spends less time
The volumetric flow rate decreases with increasing conversionThe volumetric flow rate decreases with increasing conversion,and the reactant spends more time in the reactorthan reactants that produce no net change in the total number of moles.
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Change in Gas-Phase Volumetric Flow Rate Down the Reactor
v=vo(1+X)
X
XXXkC
vVA 1
)1()1ln()1(22
2
0
0
=1 : (A→2B)v = vo(1+X)o( )
= 0 : (A→B)( )v = vo
=0.5 : (2A→B)v = vo(1-0.5X)Complete
conversion
When there is a decrease in the number of moles in the gase phase,the volumetric flow rate decreases with increasing conversionthe volumetric flow rate decreases with increasing conversion,and the reactant spends more time in the reactorthan reactants that produce no net change in the total number of moles.
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Example 4-3: Determination of a PFR Volume
Determine the PFR volume necessary to produce 300 milliond f h l f ki f d fpounds of ethylene a year from cracking a feed stream of pure
ethane. The reaction is irreversible and follows an elementary ratelaw We want to achieve 80% conversion of ethane operating thelaw. We want to achieve 80% conversion of ethane, operating thereactor isothermally at 1100K at a pressure of 6 atm.
C2H6 C2H4 + H2
A B + C
F 300x106 lb/ 0 340 lb l/
A B + C
FB = 300x106 lb/year = 0.340 lb-mol/secFB = FAoXFAo = FB/X = 0.340/0.8 = 0.425 lb-mol/sec
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.2)
Ind. Eng. Chem. Process Des. Dev., 14, 218 (1975)Ind. Eng. Chem., 59(5), 70 (1967)
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C2H6 C2H4 + H2
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4.5 Pressure Drop in Reactors
In liquid-phase reaction In liquid phase reaction- the concentration of reactants is insignificantly affected by evenrelatively large change in the total pressure
- ignore the effect of pressure drop on the rate of reactionwhen sizing liquid-phase chemical reactorsth t i d i i d f li id h ki ti l l ti- that is, pressure drop is ignored for liquid-phase kinetics calculations
In gas phase reaction In gas-phase reaction- the concentration of the reacting species is proportional to total pressure- the effects of pressure drop on the reaction system are a key factorp p y yin the success or failure of the reactor operation
- that is, pressure drop may be very important for gas-phase reactions(Micro-reactors packed with solid catalyst)
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Pressure drop and the rate lawp
• for an ideal gas 기상반응에서는 반응 성분의 농도가 반응압력에 비례 for an ideal gas,
)/)(/)(1( 000
0
TTPPXvXvF
vF
C iiAii
의 농도가 반응압력에 비례하므로 압력강하에 대한
고려가 필수적이다.
• For isothermal operation
0 1 PP
XXv
CC iiAi
0ii
F 0Ay
01 PX 0A
i F 0Ay
abBA /,1
- determine the ratio P/P0 as a function of V or W
- combine the concentration, rate law, and design equation
- the differential form of the mole balance (design equation) must be used
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Pressure drop and the rate lawp
• For example• For example,
- the second order isomerization reaction in a packed-bed reactor
2A B C2A B + C
-the mole balance (differential form)
mincatalyst g
gmoles0 AA r
dWdXF
The differential form of PFR design equationmust be used when
there is a P- rate law
2AA kCr
there is a P
- stoichiometry for gas-phase reactions
TPXCC 01
TPXCC AA
0
00 1
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Pressure drop and the rate lawp
• Then, the rate law2
0
00 1
1
TT
PP
XXCkr AA (4-20)
- the larger the pressure drop from frictional losses, the smaller the reaction rate
C bi i ith th l b l ( i i th l ti T T )• Combining with the mole balance (assuming isothermal operation: T=T0)
220 )1(
PXCdX A
• Dividing by FA0(=v0CA0)0
00 1
)1(
PP
XXC
kdWdXF A
A
2
0
2
0
0
11
PP
XX
vkC
dWdX A
00 1 PXvdW
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Pressure drop and the rate lawp
22 2
0
2
0
0
11
PP
XX
vkC
dWdX A
-The right-hand side is a function of only conversion and pressure
),( PXfdWdX
(4-21)
-Another equation is needed to determine the conversion as a function of
l i h h i d l h d h lcatalyst weight: that is, we need to relate the pressure drop to the catalyst
weight)(WfP dP
We need)(WfP dW
We need
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Flow through a packed bedg p
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Flow through a packed bed dWdP
We need g p• The majority of gas-phase reactions are catalyzed by passing the reactant through a packed of catalyst particles
dW
through a packed of catalyst particles
• Ergun equation: to calculate pressure drop in a packed porous bed
d )(
G
DDgG
dzdP
ppc75.1)1(1501
3 (4-22)
laminar turbulent
G=u=superficial mass velocity [kg/m2s]; u=superficial velocity [m/s]; Dp=diameter of particlein the bed [m]; =porosity=volume of void/total bed volume; 1 =volume of solid/total bed
• The gas density () is the only parameter that varies with pressure on the right
in the bed [m]; =porosity=volume of void/total bed volume; 1- =volume of solid/total bedvolume, gc=1.0 [m/s2]; =viscosity [kg/ms]
• The gas density () is the only parameter that varies with pressure on the right-hand side. We are now going to calculate the pressure drop through the bed.
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Flow through a packed bedg p
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Flow through a packed bed dWdP
We need g p
• Equation of continuity
dW
vv
mm
00
0
- The reactor is operated at steady state, the mass flow rate at any point is equal to the entering mass flow rate
• Gas-phase volumetric flow rate
FTP
v
• Then,00
00
T
T
FF
TT
PP
vv
vv0
0
T
FF
TT
PP
vv 00
00
0
TFTPv 0
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Pressure drop in a packed bed reactor
• then, Ergun equation
00
03
075.1)1(150)1(
T
T
ppc FF
TT
PP
GDDg
GdzdP
• Simplifying
00
00
T
T
FF
TT
PP
dzdP
(4-24)
00 T
GD
G 75.1)1(150)1(30dP (4-25)
DDg ppc3
00
dWdP
We need
( )
Ac
dWdz
dzdP
dWdP
• The catalyst weight, cc zAW )1(
Volume of Density of
z
solid solid catalyst
dzAdW cc )1( (4-26)Seoul National University
Pressure drop in a packed bed reactor
00 TFTPdP
dzdPdP cc AdW
dz)1(
1
00
00
)1( T
T
cc FTPAdW
dWdz
dzdP
dWdP
cc )(
00
00
T
T
FF
TT
PP
dzdP
• Simplifying
00
0
0 /2 T
T
FF
PPP
TT
dWdP
(4-28)
0
)1(2
PA
(4-29)0)1( PA cc
XFFXFFF A
TATT 0000 1 XFT 1 0
0A
AF
y
X
FFXFFF
TTATT
0000 1 X
FT1
0
00
TA F
y
0PTdP (4 30))1(/2 0
0
0X
PPP
TT
dWdP
(4-30)
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Pressure drop in a packed bed reactorp p
)1(0 XPTdP
(4-30)
< 0 th d (P) ill b l th 0
)1(/2 00
XPPTdW
(4 30)
ε < 0, the pressure drop (P) will be less than ε = 0
ε > 0, the pressure drop (P) will be greater than ε = 0
• For isothermal operation
),( PXfdWdP
),( PXfdWdX
and (4-31))(fdW
),(fdW
• The two expressions are coupled ordinary differential equations.
We can solve them simultaneously using an ODE solver such as Polymath.
• For the isothermal operation and ε = 0, we can obtain an analytical solution.
• Polymath will combine the mole balance, rate law and stoichiometry
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Pressure drop in a packed bed reactorp p
PTdP
01For the isothermal operation and ε = 0,
Analytical Solution
)1(/2 0
0
0X
PPP
TT
dWdP
(4-30)
If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermal operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes
)/(2 0
0
PPP
dWdP
Isothermal withε = 0
Rearranging gives us
dW
PPdPP )/(
2 0
0
dWPPd 2
0 )/(
Taking P/P0 inside the derivative(4-32)
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Pressure Drop in a Packed Bed Reactorp
PPd 2)/(
dWPPd 2
0 )/(
Integrating with P=P @ W=0
WPP 1)/( 20
Integrating with P=P0 @ W=0
WP 10 WP
10
(4-33)ε = 0T = T0
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Pressure drop in a packed bed reactorp p
If ε = 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4-30) for isothermalIf ε 0 or ε X ≪ 1, we can obtain an analytical solution to Eq. (4 30) for isothermal operation (i.e., T=T0). For isothermal operation with ε = 0, Eq. (4-30) becomes
Pressure ratio WP 1 (4-33)
only for ε = 0 W
P 1
0(4-33)
0
0
)1(2
PA cc
02 zW
cc zAW )1( 0PW
021
zP Pressure as a
GDDg
G
ppc75.1)1(150)1(
30
0
0
0
01
PP
(4-34)function of reactor length, z
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4.5.3 Pressure Drop in Pipes
Pressure drop for gases along the length of the pipe w/o packing
DfG
dLduG
dLdP
22 02 2
2
00
DfG
PdLdPG
dLdP
PP/Gu
0 PP
Integrating with P=P0 at L=0, and assuming that f = constant
0
0P
PP
DLf
PG
PP 0
0
0222
0 ln22
0
0
Rearranging, we get
fGVP 2
0 41DAP
fGVP c
pp00
0 41
Example 4 4: 1½” schedule 40 x1000 ft L ( =0 018) P<10%Example 4-4: 1½ schedule 40 x1000-ft L (p=0.018), P<10%However, for high volumetric flow rates through microreactors, the P may be significant. Example 4-4 Seoul National University
4.5.4 Analytical Solution for Reaction with Pressure Drop
For gas phase reactions, as the pressure drop increases, the concentration decreases,resulting in a decreased rate of reaction hence a lower conversion when compared toresulting in a decreased rate of reaction, hence a lower conversion when compared toa reactor without a pressure drop.
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Reaction with Pressure DropConversion as a function of catalyst weighty g
A B
2nd-order isothermal reaction
Combining
dX
A B
Mole balance: 22/122
0 1)1( WXF
kCdWdX A
Combining
AA rdWdXF 0
Rate law:
0FdW A
Separating variable and Integrating
2AA kCr
S
1)1( 22
0
0 dWWX
dXkCF
A
A
0 )1(PPXCC AA
Stoichiometry: Gas-phase isothermal with =0
1
0,0
0
000
WWXv
vCFandWXat AA
0P
WPP
10
2
110
0 WXkCA
0
2/10 )1)(1( WXCC AA (4-37)
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Reaction with Pressure DropConversion as a function of catalyst weightConversion as a function of catalyst weight
Conversion for 2nd-order isothermal reaction in PFR with P
210 WWkCA
211
2
0
0
0
Wv
WkCv
XA
(4-38)
20v
Catalyst weight for 2nd-order isothermal reaction in PFR with P
2/1
2/100 )1/(/)2(11 XXkCv
W A (4-39)
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The Optimum Catalyst Particle Diameter
Internal diffusion inside catalyst
dominant
Pressure dropdominant
Conversion k ~ 1/DpdominantX
D t
Particle Diameter, Dp
Dp, opt
Problems with large diameter tubes:Why not pack the catalyst g(1) Bypassing or Channeling(2) Little efficient of heat transfer rate
y p yinto a large diameter
to reduce P?Seoul National University
Example 4-6 Production of ethylene glycol
H2, C2H4402 million lbC2H6 /yr
O C H N C H O
2
C2H6separator
C2H6 C2H4 + H21
V=81 ft3, X=0.8
lbC2H6 /yr
O2, C2H4, N2, C2H4O
3C2H4
C2H4+ ½ O2 C2H4OAg
3/23/1BAA PkPr C
hcatlbatmlbmolk [email protected]
56
separator
H O C H O
3
4 Air
W=45,440 lb, X=0.6
260oC, 10bar
C H O( )
7 8 H2O, 0.9wt% H2SO4
H2O C2H4O
C2H4O(aq)200 millionlb EG/yr9
V=197 ft3 X=0 8V 197 ft , X 0.8
CH2OHC2H4O + H2O
CH2OHCat.
absorber
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Example 4-6 Calculating X in a Reactor with Pressure Drop
yXX
Fk
dWdX
A 11
0
Xdy
A
)1(
0
ydW 2
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Example 4-6 Calculating X in a Reactor with Pressure Drop
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Example 4-6 Calculating X in a Reactor with Pressure Drop
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Example 4-6 Calculating X in a Reactor with Pressure Drop
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Example 4-6 Calculating X in a Reactor with Pressure Drop
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Negligible Pressure Drop in Pipes
Pressure drop along the length of the pipe
DfG
dLdG
dLdP
22 02 22
00
DfG
PdLdPG
dLdP
PP
Integrating with P=P0 at L=0, and assuming that f = constant
0
PP
DLf
PG
PP 0
0
0222
0 ln22
0
0
Rearranging, we get2
DAPfGV
PP
cpp
00
20 41
Example 4-4: 1½” sch. 40 x1000-ft L (p=0.018), P<10%Seoul National University
4.6 Synthesizing a Chemical Planty g
Always challenge the assumptions, constraints, andboundaries of the problem
The profit from a chemical plant will be the differencebetween income from sales and the cost to produce thechemicalchemical
Profit = value of products cost of reactantsProfit = value of products – cost of reactants
– operating costs – separation costs
The operating cost: energy, labor, overhead, anddepreciation of equipmentdepreciation of equipment
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Production of ethylene glycol
H2, C2H4402 million lbC2H6 /yr
O C H N C H O
2
C2H6separator
C2H6 C2H4 + H21
V=81 ft3, X=0.8
lbC2H6 /yr
O2, C2H4, N2, C2H4O
3C2H4
C2H4+ ½ O2 C2H4OAg
56
separator
H O C H O
3
4 Air
W=45,440 lb, X=0.6
C H O( )
7 8 H2O, 0.9wt% H2SO4
H2O C2H4O
C2H4O(aq)200 millionlb EG/yr9
V=197 ft3 X=0 8V 197 ft , X 0.8
CH2OHC2H4O + H2O
CH2OHCat.
absorber
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Production of ethylene glycol
bp(oC) 1 2 3 4 5 6 7 8 9lb mol M.W. $/lbbp( C) 3 5 6 8 9s0.425C2H6
C2H4
H2
$/ b
0.040
O2
N2
EO0.102EG 0.380
-H2SO4
H2O0.425 0.102Total
0.043
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Synthesizing a Chemical Planty g
Ethylene glycol = $0 38/lb (2x108 lb/yr) Ethylene glycol = $0.38/lb (2x10 lb/yr)Ethane = $0.04/lb (4x106 lb/yr)S lf i id $0 043/lb (2 26 108 lb/ )Sulfuric acid = $0.043/lb (2.26x108 lb/yr)Operating cost = $8x106/yr
Profit = $0.38/lb x 2x108 lb/yr - $0.04/lb x 4x108 lb/yr-$0.043/lb x 2.26x106 lb/yr - $8x106/yr
= $52 million$
How the profit will be affected by conversion separation recycle How the profit will be affected by conversion, separation, recyclestream, and operating costs?
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Fig. 4-1 Isothermal Reaction Design AlgorithmAlgorithm for Conversion
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4.11 The Practical Side
A practical guidelines for the operation of chemical reactors have beenA practical guidelines for the operation of chemical reactors have beenpresented over the years, and tables and some of these descriptions aresummarized and presented on the CD-ROM and web. The articles arelisted in Table 4-7.s ed ab e
For example, Mukesh gives relationships between the CSTR tank diameter,T, impeller size diameter, D, tank height, H, and the liquid level, l. To scalep g qup a pilot plant (1) to a full scale plant (2), the following guidelines are given
RHTD 2222
And the rotational speed, N2, is
RHTD
1111
Wh l f f diff t i iti d F d b
nRNN 12
Where values of n for different pumping capacities and Froude numbers aregiven in Mukesh’s article.
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Closure
After completing this chapter, you should be able to apply the algorithm building blocks
Evaluation
Stoichiometry
Combine
Mole Balance
Rate Law
Mole Balance
To batch reactor, CSTR, PFR, PBR., , ,
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