chemical reactor design - seoul national university · 2018. 1. 30. · 5.2 batch reactor data...
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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
第5章
Collection and Analysis of Rate Data
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
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Objective
★ Determine the reaction order and specific reaction rate from★ Determine the reaction order and specific reaction rate from experimental data obtained from either batch or flow reactors.
★ Describe how to use equal-area differentiation, polynomial fitting, numerical difference formulas and regression to analyze g, g yexperimental data to determine the rate law.
★ Describe how the methods of half lives, and of initial rate, are used to analyze rate data.
★ Describe two or more types of laboratory reactors used to obtain rate law data along with their advantages and disadvantages.
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Collection and Analysis of Rate Data
★ Two common type reactors for obtaining rate data:Two common type reactors for obtaining rate data:
(1) Batch reactor : Conc. vs. time(1) Batch reactor : Conc. vs. time( )( )- homogeneous reaction during transient operation - Concentration (or pressure) are usually measured andConcentration (or pressure) are usually measured and
recorded at different times during the course of reaction.
(2) Differential reactor : Conc. @ steady state(2) Differential reactor : Conc. @ steady stateSolid fluid heterogeneous reactions- Solid-fluid heterogeneous reactions
- Product concentration is usually monitoredf diff t t f f d ditifor different sets of feed conditions.
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Collection and Analysis of Rate Data
★ Six different mSix different methods of analyzing the data collectedethods of analyzing the data collected★ Six different mSix different methods of analyzing the data collectedethods of analyzing the data collected
(1) the differential method(1) the differential method(2) the integral method
primarily in analyzingbatch reactor data
(3) the method of half-lives(4) method of initial rates(4) method of initial rates(5) linear regression(6) nonlinear regression
(least squares analysis)(least squares analysis)
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5.1 Algorithm for Data Analysis
Steps in Analyzing Rate Data
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5.1 Algorithm for Data Analysis
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A. Differential Analysis
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B. Integral Analysis
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5.1 Algorithm for Data Analysis
tt
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5. 2 Batch Reactor Data
Measuring concentrationt CA0 50 0 Measuring concentration
as a function of time0 50.0
50 38.0100 30.6
Differential, integralData
......
integralor nonlinear regression methodanalysis
D i i d kDetermining and kin -rA = kCA
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5.2 Batch Reactor Data
Irreversible reactionIrreversible reactionDetermine and k-Determine and kby either nonlinear regression or by numerically diff i i i i ddifferentiating concentration versus time data
For example-for decomposition reaction (only one reactant)
A productsAssuming that therate la is of the A products
-rA = kACA
rate law is of theform
-rA = kACA
(5-1)AAA Ckr
then differential method may be used.Seoul National University
5.2 Batch Reactor Data
Consider the irreversible reaction : A + B products
Excess Experiments p
BAAA CCkr (5-2)
Excess A experiments:CA remains unchanged
Excess B experiments:CB remains unchanged
during the reaction (CAo)BBAABAAA CkCCkCCkr 0
during the reaction (CBo)AABABAAA CkCCkCCkr 0
(5-4)(5-3)
O d d t i d k b l l t d f th t f
moldmr 13 /
Once and are determined, kA can be calculated from the measurement of -rA
at known concentration of A and B
s
moldmCCrk
BA
AA
/ (5-5)
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5.2.1 Differential Method of Analysis
Consider a reaction carried out isothermally in a constant-volume batch reactor and the concentration recorded as a function of time. By combining the mole balance with the rate law given by Equation (5-1), we obtain
AA
A Ckdt
dC (5-6)
After taking the logarithm of both sides of Eq. (5-6)
dC AA
A Ckdt
dC lnlnln
(5-7)
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Differential Method to determine α and k
AdC
AAA Ck
dtdC lnlnln
ln ln
slope=P
P
(CAp)
kA=
lnCA
lnCApCA CAp
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How to get How to get ––dCdCAA/dt ?/dt ?
T b i h d i i dCA/d d i hi ldCATo obtain the derivative –dCA/dt used in this plot, we must differentiate the concentration-time data
dtdCA
either numerically and graphically. These methods are:
★ Graphical differentiation★ Numerical differentiation formulas★ Numerical differentiation formulas★ Differentiation of a polynomial fit to the data
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How to get How to get ––dCdCAA/dt ? (Graphical Method)/dt ? (Graphical Method)
Time (min) t0 t1 t2 t3 t4 t5C t ti ( l/d 3) C C C C C C1 Tabulate the (t C ) Concentration (mol/dm3) C0 C1 C2 C3 C4 C5
1. Tabulate the (ti, Ci)
2. For each interval, calculatet and C
ti
t
Ci
C
t and C
3. Calculate C/t as an estimateof the average slope in an interval.
t C dC/dt
(dC/d )
C/tt0
t1
C0
C1
g p
4. Plot these values(C/t) as a histogram versus ti.
t1-t0
t t
C1-C0
C C
(dC/dt)0
(dC/dt)1
C/t)1
C/t)t2
t3
C2
C3
5. Next draw in the smooth curvethat best approximates the area
d h hi
t2-t1
t3-t2
C2-C1
C3-C2
(dC/dt)2
(dC/dt)3
C/t)2
C/t)33
t4
3
C4
under the histogram.
6. Read estimates of the dC/dtfrom this curve at the data points
t4-t3
t5-t4
C4-C3
C5-C4
(dC/dt)3
(dC/dt)4
C/t)4
C/t)5t5 C5from this curve at the data points
t1, t2, … and complete the table.t5 t4 C5 C4
(dC/dt)5
C/t)5
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How to Get How to Get ––dCdCAA/dt ? (Graphical Method)/dt ? (Graphical Method)
★ Graphical Method (Equal-Area Graphical Differentiation)
Draw smooth curvethat best approximatesthe area under histogram
ddCA
1
tCA
0
dt
dCA
2
tCAA
B
1
dtdCA
dtdCA
4
tCA
CA
3
tCA2 dt
ddCA
0 1 2 3 4 5
4
5
tCA
4 dt
0 1 2 3 4 5
tSeoul National University
How to Get How to Get ––dCdCAA/dt (Numerical Method)/dt (Numerical Method)
Time (min) t0 t1 t2 t3 t4 t5Concentration (mol/dm3) C C C C C CConcentration (mol/dm ) CA0 CA1 CA2 CA3 CA4 CA5
★ Numerical Method (Independent variables are equally spaced)★ Numerical Method (Independent variables are equally spaced)
tCCC
dtdCpoInitial AAA
t
A
243:int 210
0
CCdCtCC
dtdCspoInterior
tdt
AA
t
A
t
2:int
2
02
1
0
CCd
dCtCC
dtdC
AAA
AA
t
A
2
2
24
13
2 )1()1(21
iAiAt
A CCtdt
dC
i
The three-pointdifferentiationformulas
tCC
dtdC
tdt
AA
t
A
t
2
2
35
4
3i
tCCC
dtdCpoLast AAA
t
A
234:int 543
5
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How to Get How to Get ––dCdCAA/dt ? (Polynomial Fit)/dt ? (Polynomial Fit)
★ Polynomial Fity
Time (min) t0 t1 t2 t3 t4 t5Concentration (mol/dm3) CA0 CA1 CA2 CA3 CA4 CA5
Polynomial fit with software program to get best value of ai
nn--thth order polynomialorder polynomial2
210 ... nnA tatataaC (5-11)
12321 32 nA tnatataadC
Differential equationDifferential equation
(5-12)321 ...32 ntnatataadt
(5-12)
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How to Get How to Get ––dCdCAA/dt (Polynomial Fit)/dt (Polynomial Fit)
3rd order 5th order N i3rd-orderpolynomial
5th-orderpolynomial
Negativederivative
t5
Care must be taken in choosing the order of the polynomial. If the order is toolow, the polynomial fit will not capture the trends in the data. If too large anorder is chosen the fitted curve can have peaks and valleys as it goes throughorder is chosen, the fitted curve can have peaks and valleys as it goes throughmost all of the data points, thereby producing significant errors when thederivatives, dCA/dt, are generated at the various points. Seoul National University
Finding the Rate Law ParameterFinding the Rate Law Parameter
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Finding the Rate Law ParameterFinding the Rate Law Parameter
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
The reaction of triphenyl methyl chloride (trityl) (A) and methanol (B) wascarried out in a solution of benzene and pyridine at 25oC. Pyridine reacts withHCl h h i i idi h d hl id h b ki h iHCl that then precipitates as pyridine hydrochloride thereby making the reactionirreversible. The concentration-time data was obtained in a batch reactor. Theinitial concentration of methanol was 0.5 mol/dm3.
(C6H5)3CCl (A) + CH3OH (B) (C6H5)3COCH3 (C) + HCl (D)
Time (min) 0 50 100 150 200 250 300CA (mol/dm3) x 103 50 38 30 6 25 6 22 2 19 5 17 4CA (mol/dm ) x 10 50 38 30.6 25.6 22.2 19.5 17.4
CAo=0.05 MCBo=0.5 M
Part (1) Determine the reaction order with respect to trityl (A)Part (2) In a separate set of experiments, the reaction order with respect to
methanol was found to be first order. Determine the specific reactionprate constant.
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
Concentration of triphenyl methyl chloridef ti f ti
55
as a function of time
/dm
3 )
40
45
5010
3 (mol
30
35
40
CA x
1
20
25
30
t (min)0 50 100 150 200 250 300
15
t (min)
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
SolutionPart (1) Determine the reaction order with respect to trityl (A)
Step 1 Postulate a rate lawBAAA CCkr
Step 2 Process your data in terms of the measured variable which is this case is CA
(E5-1.1)
Step 2 Process your data in terms of the measured variable, which is this case is CA.
Step 3 Look for simplifications. Because concentration of methanol is 10 timesthe initial concentration of triphenyl methyl chloride, its concentration isp y y ,essentially constant
0BB CC (E5-1.2)
Substituting for CB in Equation (E5-1.1)
CkCCkr (E5 1 3)AABAA CkCCkr 0 (E5-1.3)
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
Step 4 Apply the CRE algorithm
(E5-1.4)Mole Balance Vrdt
dNA
A dt
Rate law AA Ckr (E5-1.3)
0VV Stoichiometry: Liquid
0
VNC A
A
(E5-1 5)
0V
Combine A CkdC (E5-1.5)Combine ACkdt
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
Taking the natural log of both sides of Equation (E5-1.5)
AA Ck
dtdC lnlnln
(E5-1.6)
The slope of plot of versus will yield the reaction order
with respective to triphenyl methyl chloride (A)
dtdCAln
ACln
Step 5 Find as a function of CA from concentration-time data.
with respective to triphenyl methyl chloride (A).
dCAStep 5 Find as a function of CA from concentration time data.dt
We will find by each of the three methods just discussed, (1) the graphical
(2) finite difference, and (3) polynomial methods.dt
dCA
(2) finite difference, and (3) polynomial methods.
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
Step 5A.1a Graphical method
(min)t
)/(
103
3
dmmol
CA min)/(
10/3
4
dmmol
dtdCA min)/(
10/3
4
dmmol
tCA
0
50
50
38
3.0
1.862.40
5038
tC
100 30.6 1.21.48
1.004
3
104.2
10050
5038
150
200
25.6
22.2
0.8
0.50.68
200
250
22.2
19.5
0.5
0.470.54
0 42300 17.4
0.42
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
Step 5A.1a Graphical method
3.03.00
dtdCA
2.0
2.41
tCA
dCA 2.0
1.482
tCA
C
1.86
1 2
1
dt
ddCA
1.0 1.00
0.680.54
4
tCA
tCA
3
tCA
1.2
0.80.50 47
2 dt
4
dtdCA
0 50 100 150 200 250 3000.0
0.540.42 5 t0.47
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
Step 5A.1b Finite Differential Method
3
43
210
0
1086.2)50(2
10]6.30)38(4)50(3[2430
tCCC
dtdCt AAA
t
A
43
13
43
02
1
1024110]386.25[100
1094.1)50(2
10]506.30[2
50
CCdCt
tCC
dtdCt
AAA
AA
t
A
43
24
3
2
1084.0)50(2
10]6.302.22[2
150
1024.1)50(22
100
tCC
dtdCt
tdtt
AA
t
A
t
3
43
35
4
3
10]222417[
1061.0)50(2
10]6.255.19[2
200
)50(22
CCdC
tCC
dtdCt
tdt
AA
t
A
t
43
654
43
46
5
1036010)]4.17(3)5.19(42.22[34300
1048.0)50(2
10]2.224.17[2
250
CCCdCt
tCC
dtdCt
AAAA
AA
t
A
6
1036.0)50(22
300
tdt
tt
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
Step 5A.1c Polynomial methodt CA dCA/dt
Another method to determine (dCA/dt) is to fitthe concentration of A to a polynomial in timeand then to differentiate the resulting
l i l W fi h h f h
0 50 -0.299
50 38 -0.189 polynomial. We first choose the fourthpolynomial degree
2 ntatataaC
100 30.6 -0.120
150 25.6 -0.081
1221
210
...32
...
nn
A
nA
tnataatadt
dCtatataaC 200 22.2 -0.061
250 19.5 -0.049
300 17 4 0 034
41239264
dt 300 17.4 -0.034
41239264 10697.310485.310343.110978.204999.0 ttttCA
38253 1047911004510026860978210 tttdC A 10479.110045.1002686.0978.210 tttdt
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
Step 5A.1c Polynomial method
41239264 106973104853103431109782049990C 41239264 10697.310485.310343.110978.204999.0 ttttCA
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
SUMMARY: There is quite a close agreement betweenthe graphical technique finite difference and polynomial methodsthe graphical technique, finite difference, and polynomial methods
min mol/dm3 Graphical Finite Difference Polynominal
t CAx103 -dCA/dt -dCA/dt -dCA/dtt CAx10 dCA/dt dCA/dt dCA/dt0 50.0 3.00 2.86 2.98
50 38.0 1.86 1.94 1.88
100 30.6 1.20 1.24 1.19
150 25.6 0.80 0.84 0.80
200 22.2 0.68 0.61 0.60
250 19.5 0.54 0.48 0.48
300 17.4 0.42 0.36 0.33
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
From Figure E5-1.3, we found the slope to be 1.99 so that thereaction is said to be second order w r t triphenyl methyl chloridereaction is said to be second order w.r.t. triphenyl methyl chloride.To evaluate k’, we can evaluate the derivative and CAp=20x10-3
mol/dm3, which is,
minmol/dm1050 34
AdC minmol/dm105.0
pdt
then
dil/d1050 334
A
dtdC
minmoldm125.0
mol/dm1020minmol/dm105.0 3
233
34
2
Ap
p
Cdt
k
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
Excel plot to determine and kCoefficients:
10.00
99.10013.0 AA C
dtdC
b[0] -6.6685984257b[1] 1.9958592737r ² 0.9942392165
104dCA
99.11.00
mindmmol
10
3
4
dtA
min/moldm13.0 3 k
Graphical
Finite
Polynomial
mindm0.5
Regression again
0.10
10 100
mol220
3
3
dmmol10AC min/moldm122.0 3 k
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
Part (2) The reaction was said to be first order wrt methanol, =1,
kCkCk BB 00
Assuming CB0 is constant at 0.5 mol/dm3 and solving for k yields
dm12203
min/moldm244.0
dmol5.0
minmoldm122.0 23
30
BC
kk
dm3
The rate law is
BAA CCr 223 min//mol)(dm244.0
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Integral MethodIntegral Method
★ The integral method uses a trial-and-error procedure to find rxn order:We guess the reaction order and integrate the differential equation used tog g qmodel the batch system. If the order we assume is correct, the appropriateplot of the concentration-time data should be linear.
★ It is important to know how to generate linear plots of functions of CA
versus t for zero-, first-, and second-order reactions.
★ The integral method is used most often when the reaction order isknown and it is desired to evaluate the specific reaction rate constantst diff t t t t d t i th ti tiat different temperatures to determine the activation energy.
★ Finally we should also use the formula to plot reaction rate data in★ Finally we should also use the formula to plot reaction rate data interms of concentration vs. time for 0, 1st, and 2nd order reactions.
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zero orderzero order first orderfirst order second ordersecond order
kdt
dCA kCdt
dCA
A kCdt
dCA
A 2
CCt AA 0,0@ CCt AA 0,0@ CCt AA 0,0@
ktCC AA 0 ktCC
A
A 0
ln ktCC AA
0
11
slope = -k
0AC
1AC
slope k
0
lnA
A
CC
slope = k
AC1
slope = k
t t tThe rxns are zero, first, and second order respectively since the plots are linear.
t t t
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second ordersecond orderkC
dtdC
AA 2
second ordersecond order
CCt AA 0,0@AC
1
non linear
ktCC AA
0
11
A non linear
AA 0
t
If the plots of concentration data versus time hadturned out not to be linear, we would say that theproposed reaction order did not second order.
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Example 5Example 5--22
Use the integral method to confirm that the reaction is second order with respect to trityl(A) as described in example 5-1 and to calculate the specific reaction rate k . '
Trityl(A) + Methanol (B) Products
CkCkdt
dCAA
A 2
CCt AA 0,0@Integrating with
tkCC
11
CC AA 0
Time (min) 0 50 100 150 200 250 300C ( l/d 3) 0 05 0 038 0 0306 0 0256 0 0222 0 0195 0 0174CA (mol/dm3) 0.05 0.038 0.0306 0.0256 0.0222 0.0195 0.01741/CA (dm3/mol) 20 26.3 32.7 39.1 45 51.3 57.5
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Example 5Example 5--22
12.2012.01 t
Csecond ordersecond order
60
CA
ol) 50
Experiments
Regression
(dm
3 /mo
40
1/C
A
30 tkCC AA
0
11
0 50 100 150 200 250 300 35020
t (min)
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Example 5Example 5--22
BAA CCr 223 min//mol)(dm244.0Note that differential method gives
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Example 5Example 5--22
Linear Regression Reportby SigmaPlot
Function values:x f(x)
Function values:x f(x)
by SigmaPlot
Coefficients:a0 20.1142857143
x f(x)0 20.11428571436 20.863428571412 21.612571428618 22.361714285724 23.110857142930 23 86
x f(x)156 39.592162 40.3411428571168 41.0902857143174 41.8394285714180 42.5885714286186 43 33 1428
0
k’ 0.1248571429r ² 0 999903135
30 23.8636 24.609142857142 25.358285714348 26.107428571454 26.856571428660 27.6057142857
186 43.3377142857192 44.0868571429198 44.836204 45.5851428571210 46.3342857143216 47.0834285714r 0.999903135 66 28.3548571429
72 29.10478 29.853142857184 30.602285714390 31.351428571496 32 1005714286
222 47.8325714286228 48.5817142857234 49.3308571429240 50.08246 50.8291428571252 51 578285714396 32.1005714286
102 32.8497142857108 33.5988571429114 34.348120 35.0971428571126 35.8462857143
252 51.5782857143258 52.3274285714264 53.0765714286270 53.8257142857276 54.5748571429282 55.324k=0 25 dm3/mol/min 132 36.5954285714
138 37.3445714286144 38.0937142857150 38.8428571429
288 56.0731428571294 56.8222857143300 57.5714285714
k=0.25 dm3/mol/min
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Example 5Example 5--1: Determining the Rate Law1: Determining the Rate Law
zero orderzero order first orderfirst orderktCC AA 0 kt
CC
A
A 0
ln
3 )
4
50
55
1.0
1.2
A0
3 (mol
/dm
35
40
45
CA/
CA0
)
0.6
0.8
CA x
103
20
25
30 ln (C
0.2
0.4
t (min)0 50 100 150 200 250 300
15
20
t (min)0 50 100 150 200 250 300
0.0
( ) ( )
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ComparisonComparison
The differential method tends to accentuate theuncertainties in the data while the integral methoduncertainties in the data, while the integral methodtends to smooth the data, thereby disguising theuncertainties in it.uncertainties in it.
In most analyses, it is imperative that the engineery , p gknow the limits and uncertainties in the data.
This prior knowledge is necessary to provide for asafety factor when scaling up a process froml b t i t t d i ith il t l tlaboratory experiments to design either a pilot plantor full-scale industrial plant
Accentuate : 강조하다 . 두드러지게하다.Disguise: 위장하다. 변장하다.
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5.2.3 Nonlinear Regression5.2.3 Nonlinear Regression
★ In nonlinear regression analysis we search for those parameter★ In nonlinear regression analysis, we search for those parametervalues that minimize the sum of squares of the differences between themeasured values and the calculated values Not only can nonlinearmeasured values and the calculated values. Not only can nonlinearregression find the best estimates of parameter values, it can also be usedto discriminate between different rate law models, such as Langmuir-to discriminate between different rate law models, such as LangmuirHinshelwood models.
★Many software programs are available to find these parameter valuesso that all one has to do is enter the data In order to carry out the searchso that all one has to do is enter the data. In order to carry out the searchefficiently, in some cases one has to enter initial estimates of theparameter values close to the actual values. These estimates can beparameter values close to the actual values. These estimates can beobtained using the linear-least-squares technique.
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5.2.3 Nonlinear Regression5.2.3 Nonlinear Regression
★ We will now apply nonlinear least-squares analysis to reaction rate data to
d t i th t l t (k ) W th h f th ldetermine the rate law parameters (k, , ..). We then search for those values
that will minimize the sum of the squared differences of the measured reaction
t d th l l t d ti t Th t i t th f (rates, rm, and the calculated reaction rates, rc. That is, we want the sum of (rm-
rc)2 for all data points to be minimum. If we carried out N experiments, we
ld t fi d th t l th t i i i th titwould want find the parameter values that minimize the quantity
N rrs 222 )(
N n mber of r ns
i
icim
KNrr
KNs
1
2 )( (5-34)
N = number of runs
K = number of parameters to be determined
r measured reaction rate for run i (i e r )rim = measured reaction rate for run i (i.e., -rAim)
ric = calculated reaction rate for run i (i.e., -rAic)Seoul National University
Nonlinear RegressionNonlinear Regression
To illustrate this technique, let’s consider the first-order reaction
A Product
for which we want to learn the reaction order, , and the specific reaction rate, k,
kCAkCr
Th ti t ill b d t b f diff t t ti WThe reaction rate will be measured at a number of different concentrations. Wenow choose values of k and and calculate the rate of reaction (ric) at eachconcentration at which an experimental point was taken We then subtract theconcentration at which an experimental point was taken. We then subtract thecalculated value (ric) from the measured value (rim), square the result, and sumthe squares for all the runs for the values of k and we have chosen.the squares for all the runs for the values of k and we have chosen.
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Nonlinear RegressionNonlinear Regression
This procedure is continued by further varying and k until we find their best
values, that is, those values that minimize the sum of the squares. Many well-, , q y
known searching techniques are available to obtain the minimum value .
Figure 5-7 shows a hypothetical plot of sum of the squares as a function of the
2min
g yp p q
parameters and k:2 572
/ ld050.2
3
k 25.132
2
5
s/moldm0.5 3 k 3.72 85.12
)(2 kf2
045.02min
),(2 kf
k Seoul National University
Nonlinear RegressionNonlinear Regression
In searching to find the parameter values that give minimum of thesum of squares 2, one can use a number of optimizationtechniques or software packages.
A number of software packages are available to carry out theprocedure to determine the best estimates of the parameter valuesand corresponding confidence limits.
All on has to do is to type the experimental values in the computer,specify the model, enter the initial guesses of the parameter valuesalong with 95% confidence limit appear.g pp
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Nonlinear RegressionNonlinear Regression
Concentration-time data. We will now use nonlinear regression todetermine the rate law parameters from concentration-time data obtained inbatch experiments. We recall that the combined rate law-stoichiometry-molebalance for a constant-volume batch reactor is
dC kCdt
dCA
A (5-6)
We now integrate Eq (5 6) to give
11
0 )1( ktCC AA
We now integrate Eq. (5-6) to give
)1/(110 )1(
ktCC AA(5-18)
W t th l f d k th t ill k s2 i i
2)1/(1122 )1()( NN
ktCCCC
We want the value of and k that will make s2 a minimum.
10
1
2
1
2 )1()(
iAi
AmiAcii
Ami ktCCCCs (5-19)
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Nonlinear Regression (Nonlinear LeastNonlinear Regression (Nonlinear Least--Squares Analysis)Squares Analysis)
Seoul National University
Example 5Example 5--3 Use of Regression to find the Rate Law Parameter3 Use of Regression to find the Rate Law Parameter
We shall use the reaction and data in Ex. 5-1 to illustrate how to use regression to find and k’
(C6H5)3CCl (A) + CH3OH (B) (C6H5)3COCH3 (C) + HCl (D)
A CkdC A
A Ckdt
Integrating with the initial condition when t=0 and CA=CA0 for ≠1.0
)1(1 )1()1(
0
AA CC
kt
)1( k
Substituting for the initial concentration CA0=0.05 mol/dm3
)1()05.0(1 )1()1(
ACk
t)(
Let’s do a few calculations by hand to illustrate regression.Seoul National University
Example 5Example 5--3 Use of Regression to find the Rate Law Parameter3 Use of Regression to find the Rate Law Parameter
We now first assume a value of and k and then calculate t for theconcentrations of A given in Table E5-1.1 (pp 261). We will then calculate the sum
NN CC
211
concentrations of A given in Table E5 1.1 (pp 261). We will then calculate the sumof the squares of the difference between the measured time, tm and the calculatedtimes (i.e., s2). For N measurements,
t
AAmi
tcimi k
CCttts1
0
1
22
)1()(
Our first guess is going to be = 3 and k’ = 5, with CA0 = 0.05. Equation (E5-3.2) becomes )05.0(1 )1()1(
ACt
4001
10111
21
2221c CCCkt
)1(
kt
102 0 AAA CCCk
We now make the calculations for each measurement of concentration and fill in column 3 and 4 of Table E5-3.1. For example, when CA=0.038 mol/dm3 then
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Example 5Example 5--3 Use of Regression to find the Rate Law Parameter3 Use of Regression to find the Rate Law Parameter
11111 min2.29400
)038.0(1
10111
21
220
21
AAc CCk
t
Which is shown in Table E5-3.1 on line 2 for guess 1. We next calculate the squaresof difference (tm1-tc1)2=(50-29.2)2=433. We continue in this manner for points 2, 3, and4 to calculate the sum s2=2916.
After calculating s2 for = 3 and k = 5, we make a second guess for and k’. Forour second guess we choose = 2 and k = 5; Equation (E5-3.2) becomes
2011111t (E5-3.2)
W d i h d fi d h f ( )2 b 2 49 895
2050 AAA
c CCCkt ( )
We now proceed with our second guess to find the sum of (tm1-tc1)2 to be s2=49,895, which is far worse than our first guess. So we continue to make more guesses of and k and find s2. Let’s stop and take a look at tc for guesses 3 and 4.Seoul National University
Example 5Example 5--3 Use of Regression to find the Rate Law Parameter3 Use of Regression to find the Rate Law Parameter
G 1 G 2 G 3 G 4O i i l D t
Table E5-3.1. Regression of data
Guess 1 Guess 2 Guess 3 Guess 4
=3k’=5
=2k’=5
=2k’=0.2
=2k’=0.1
Original Data
t CA x 103
tc (tm-tc)2 tc (tm-tc)2 tc (tm-tc)2 tc (tm-tc)2
0 0 0 0 0 0 0 00 50
A(min) (mol/L)
129.2 433 1.26 2,375 31.6 339 63.2 17450 38
66.7 1,109 2.5 9,499 63.4 1,340 126.8 718100 30.62
3163 1,375 5.0 38,622 125.2 5,591 250 2,540200 22.2
s2 = 2,916 s2 = 49,895 s2 = 7,270 s2 = 3,4324
We see that (k’=0.2 dm3/mol·min) underpredicts the time (e.g., 31.6 min versus 50 minutes), while (k’=0.1 dm3/molmin) overpredicts the time (e.g., 63 min versus 50 minute)
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Example 5Example 5--3 Use of Regression to find the Rate Law Parameter3 Use of Regression to find the Rate Law Parameter
04.2
min/moldm147.004.2
3 k
Regression again
0.23
min/moldm125.0 3 k
BAA CCr 223 min//mol)(dm25.0 BAA )(
Seoul National University
5.3 Method of Initial Rates5.3 Method of Initial Rates
The use of the differential method of data analysis to determine reactionorders and specific reaction rates is clearly one of the easiest since itorders and specific reaction rates is clearly one of the easiest, since itrequires only one experiment. However, other effects, such as thepresence of a significant reverse reaction could render the differentialpresence of a significant reverse reaction, could render the differentialmethod ineffective.
In these cases, the method of initial rates could be used to determine thereaction order and the specific rate constant. Here, a series ofexperiments is carried out at different initial concentrations, CA0, and theinitial rate of reaction, -rA0, is determined for each run.
The initial rate of reaction –rA0 can be found by differentiating the data and extrapolating to zero time.
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5.3 Method of Initial Rates5.3 Method of Initial Rates
By various plotting or numerical analysis techniques relating –rA0 to C bt i th i t t lCA0, we can obtain the appropriate rate law.
kCr 00 AA kCr
the slope of the plot of ln(-rA0) versus lnCA0 is the reaction order
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Example 5Example 5--44
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Example 5Example 5--44
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Example 5Example 5--44
Solution
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Example 5Example 5--44
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Example 5Example 5--44
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Example 5Example 5--44
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Method of HalfMethod of Half--LivesLives
The half-life of a reaction
A products (irreversible) The half life of a reaction= the time it takes for the
concentration of the reactantt f ll t h lf f it i iti l l
AAA kCr
dtdC
to fall to half of its initial value
By determining the half-life of a 1
0111 11111 ACt
dt
reaction as a function of the initial concentration, the reaction order and specific reaction rate can be determined.
10
10
1
1
1)1()1( AAAA CkCCCk
t
p
2/1ln t Slope=1-1
02/1 21
AA CCwhentt
10
1
2/11
)1(12
ACkt
0ln AC 0
1
2/1 ln)1()1(12lnln AC
kt
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Differential ReactorsDifferential Reactors
• Most commonly used catalytic reactor to obtain experimental data
use to determine the rate of reaction as a function of either- use to determine the rate of reaction as a function of eitherconcentration or partial pressure
- the conversion of the reactants in the bed is very smallthe conversion of the reactants in the bed is very small- reactant concentration is constant: gradientless (~CSTR)- reaction rate is spatially uniform (~CSTR)p y ( )
A P
FA0 FAeA0 Ae
Inert fillingL Inert fillingLcatalyst
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Steady state mole balance on reactant A (CSTR)Steady state mole balance on reactant A (CSTR)
Differential ReactorsDifferential ReactorsSteady state mole balance on reactant A (CSTR)Steady state mole balance on reactant A (CSTR)
L
ofrateofratet
flowt
flow
FA0 FAe
naccumuatiof
generationf
outrate
inrate
CA0 FP
CP
catofmasscatofmass
reactionofrateFF AeA
0 0.)(
.][][
W
FFr
WrFF
AeA
AAeA
0'
'0 0))((
WrA
CCr AeA
00' (5-27)
A P
WrA
FXFReactor Design EquationReactor Design Equation
(5 27)
( )WF
WXFr PA
A 0'
(5-28)
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Differential ReactorsDifferential Reactors
CCCCC )(
For constant volumetric flowFor constant volumetric flow knownMeasuringthe productCan be determined
WC
WCC
WCCr PAeAAeA
A00000' )(
concentration
known
- using very little catalyst and large volumetric flow rates 0~)( 0 AeA CC
where CAb the concentration of A within the catalyst bed
0 AeA
)(''AbAA Crr
where CAb the concentration of A within the catalyst bed
th ith ti f th i l t d tl t t ti 0 AeA CCC
- the arithmetic mean of the inlet and outlet concentration:
- very little reaction takes place within bed: 2
0 AeAAbC
0~ AAb CC
)( 0''
AAA Crr Seoul National University
Example 5Example 5--5 Differential Reactors5 Differential Reactors
The synthesis of CH4 from CO and H2 using a nickel catalyst was carried out at 500oF in a differential reactor where the effluent concentration of CH4 was measured.
OHCHCOH 242 23
in a differential reactor where the effluent concentration of CH4 was measured.
a. Relate the rate of reaction to the exit methane concentration.
b. The reaction rate law is assumed to be the product of a function of the partial p ppressure of CO and a function of the partial pressure of H2:
)()( HgCOfr (E5 5 1)
Determine the reaction order w.r.t. CO, using the data in Table E5-5.1.
)()( 24HgCOfrCH (E5-5.1)
Assume that the functional dependence of on is of the form4CHr COP
Pr (E5-5 2)COCH Pr ~
4(E5-5.2)
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Example 5Example 5--5 Differential Reactors5 Differential Reactors
Table E5-5.1. Raw Data
Run PCO (atm) PH2 (atm) CCH4 (mol/dm3)1 1.0 1.0 1.73x10-4
2 1.8 1.0 4.40x10-4
3 4.08 1.0 10.0x10-4
4 1 0 0 1 1 65 10 44 1.0 0.1 1.65x10-4
5 1.0 0.5 2.47x10-4
6 1 0 4 0 1 75x10-46 1.0 4.0 1.75x10
The exit volumetric flow rate a differential packed bed containing 10 g ofThe exit volumetric flow rate a differential packed bed containing 10 g ofcatalyst was maintained at 300 dm3/min for each run. The partial pressure of H2
and CO were determined at the entrance to the reactor, and the methaneconcentration was measured at the reactor exit.
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Example 5Example 5--5 Differential Reactors5 Differential Reactors
(a) In this example the product composition, rather than the reactant concentration, isbeing monitored. r´CH4 can be written in terms of flow rate of methane from the
ireaction,
CHmoldmmolnmidmCF CHCH
45343
0 10251073130044 )/.)(/(nmicatg
CHmolcatg
dmmolnmidmWW
rr CHCHCHCO
450 102510
1073130044
4.)/.)(/(
Table E5-5.1. Raw and Calculated Data
R P P C ´Run PCO (atm) PH2 (atm) CCH4 (mol/dm3) r´CH4 (mol CH4/cat·min)1 1.0 1.0 1.73x10-4 5.2x10-3
2 1 8 1 0 4 40x10-4 13 2x10-32 1.8 1.0 4.40x10 13.2x103 4.08 1.0 10.0x10-4 30.0x10-3
4 1.0 0.1 1.65x10-4 4.95x10-3
5 1.0 0.5 2.47x10-4 7.42x10-3
6 1.0 4.0 1.75x10-4 5.25x10-3Seoul National University
Example 5Example 5--5 Differential Reactors5 Differential Reactors
Determining the rate law dependence in CO
For constant hydrogen concentration, the rate law can be written as
COHCOCH PkPgkPr )(
24(E5-5.4)
Taking the log
)()()( Pkr lll )()()( COCH Pkr4
lnlnln
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Example 5Example 5--5 Differential Reactors5 Differential Reactors
PlWe now plot versus for runs 1 2 and 3)( l COPlnWe now plot versus for runs 1, 2, and 3.)(4CHr ln
Table E5-5.1. Raw and Calculated Data
Run PCO (atm) PH2 (atm) CCH4 (mol/dm3) r´CH4 (mol CH4/cat·min)1 1.0 1.0 1.73x10-4 5.2x10-3
2 1 8 1 0 4 40x10-4 13 2x10-32 1.8 1.0 4.40x10 13.2x103 4.08 1.0 10.0x10-4 30.0x10-3
4 1.0 0.1 1.65x10-4 4.95x10-3
5 1.0 0.5 2.47x10-4 7.42x10-3
6 1.0 4.0 1.75x10-4 5.25x10-3
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Example 5Example 5--5 Differential Reactors5 Differential Reactors
0.1ca
t min
)
221COCH P00560r
4
..
0.01(mol
/gc
n R
ate
(
0.001
Rxn
0.1 1 10
Pressure of CO (atm)Pressure of CO (atm)
Seoul National University
Example 5Example 5--5 Differential Reactors5 Differential Reactors
Determining the rate law dependence in CO
Had we include more points, we would have found the reaction rate is essentially first order with =1, that is,
COCH Pkr (E5-5 5)221
COCH P00560r4
..
=1.22 → 1
COCH Pkr4
(E5-5.5)4
)f(PPkr24 HCOCH
Seoul National University
Example 5Example 5--5 Differential Reactors5 Differential Reactors
Determining the rate law dependence in H2
’From Table E5-5.2, it appears that the dependence of r’CH4 on PH2 cannot berepresented by a power law. The reaction rate first increases with increasingpartial pressure of hydrogen, and subsequently decreases with increasing partialpressure of hydrogen. That is, there appears to be a concentration of hydrogen atwhich the rate is maximum.
Run PCO (atm) PH2 (atm) CCH4 (mol/dm3) r´CH4 (mol CH4/cat·min)1 1.0 1.0 1.73x10-4 5.2x10-3
2 1.8 1.0 4.40x10-4 13.2x10-3
3 4.08 1.0 10.0x10-4 30.0x10-3
4 34 1.0 0.1 1.65x10-4 4.95x10-3
5 1.0 0.5 2.47x10-4 7.42x10-3
6 1 0 4 0 1 75x10-4 5 25x10-36 1.0 4.0 1.75x10 5.25x10
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Example 5Example 5--5 Differential Reactors5 Differential Reactors
@ high PH2
1βPr ~@ low PH2 1
24
βHCH Pr ~
1
2
βHP
@ H2
12 2
2
2
4 βH
HCH bP1
r
~
Seoul National University
Example 5Example 5--5 Differential Reactors5 Differential Reactors
1
2
βHCOPaP
r 2
2
4 βH
CH bP1r
(E5-5.11)
PolymathRegression
( )
2
0.61HCOP0.025P
2
2
4H
HCOCH 2.49P1
r
(E5-5.12)Hydrogen undergo
dissociate adsorption ( ) pon the catalyst surface
→ Dependence of H2 to the ½ powerPolymath
2
2
4H
0.5HCO
CH 1.49P1P0.018P
r
Regression
again
1, 221
1 Seoul National University
Example 5Example 5--5 Differential Reactors5 Differential Reactors
(E5 5 13)2
0.5HCOP0.018P
r 2
0.5HCO P
PP b1rearranging (E5-5.13)
2
2
4H
CH 1.49P1r
2
4
2H
CH
CO Pr aa
rearranging
This plot should be a straight line with an intercept of 1/a and a slope b/a.400
2
0.5HCOPP
3002
2
4H
0.5HCO
CH 1.49P1P0.018P
r
4CHr 200
Rate law is indeed100
0
indeed consistent with
the rate law data
)(atm2HP
0 1 2 3 40 data
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5.7 Evaluation of Laboratory Reactor5.7 Evaluation of Laboratory Reactor
The successful design of industrial reactors lies primarily with theg p yreliability of the experimentally determined parameters used in thescale-up. Consequently, it is imperative to design equipment andp q y, p g q pexperiments that will generate accurate and meaningful data.
Unfortunately, there is usually no single comprehensive laboratoryreactor that could be used for all types of reactions and catalysts. Inyp ythis section, we discuss the various types of reactors that can be chosento obtain the kinetics parameters for a specific reaction system.to obtain the kinetics parameters for a specific reaction system.
We closely follow the excellent strategy presented in the article by V.W.We closely follow the excellent strategy presented in the article by V.W.Weekman of ExxonMobil.
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5.7 Evaluation of Laboratory Reactor5.7 Evaluation of Laboratory Reactor
Criteria used to evaluate laboratory reactory
1. Ease of sampling and product analysis2. Degree of isothermalityg y3. Effectiveness of contact between catalyst and reactant4 Handling of catalyst decay4. Handling of catalyst decay5. Reactor cost and ease of construction
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Evaluation of Laboratory Reactor (Types of Reactors)Evaluation of Laboratory Reactor (Types of Reactors)
Diff ti l t Integral reactor(Fixed bed)
Stirred-Batch Reactor
Differential reactor(Fixed bed)
catalyst slurry
Stirred ContainedSolids Reactor
(SCSR)Straight-through transport reactor
Recirculatingtransport reactor
Continuous-Stirred Tank Reactor
(CSTR) (SCSR) transport reactor transport reactor(CSTR)
Figure 5-12 Type of ReactorsSeoul National University
Evaluation of Laboratory Reactor (Reactor Ratings)Evaluation of Laboratory Reactor (Reactor Ratings)Reactor type Sampling Isothermality F-S contact Decaying Catalyst Ease of construction
Differential P-F F-G F P G
Fixed bed G P F F P GFixed bed G P-F F P G
Stirred batch F G G P G
Stirred-contained solids G G F-G P F-GStirred contained solids G G F G P F G
Continuous-stirred tank F G F-G F-G P-F
Straight-through transport F-G P-F F-G G F-G
Recirculating transport F-G G G F-G P-F
Pulse G F-G P F-G GG=Good; F=Fair; P=Poor
CSTR and recirculating transport reactor appear to be the best choice, because they aresatisfactory in every category except for construction. However, if the catalyst understudy does not decay, the stirred batch and contained solid reactors appear to be bestchoices. If the system is not limited by internal diffusion in the catalyst pellet, largerpellets could be used, and the stirred-contained solids is the best choice. If the catalystis nondecaying and heat effects are negligible the fixed-bed (integral) reactor would beis nondecaying and heat effects are negligible, the fixed-bed (integral) reactor would bethe top choice, owing to its ease of construction and operation. However, in practice,usually more than one reactor type is used in determining the reaction rate lawparameters. Seoul National University