chemical reactor design - seoul national university · 2018. 1. 30. · 5.2 batch reactor data...

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

Seoul National University

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.



★ 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.



★ 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)


5.1 Algorithm for Data Analysis

Steps in Analyzing Rate Data


A. Differential Analysis


B. Integral Analysis



tt


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


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)


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)


Differential Method to determine α and k

AdC

AAA Ck

dtdC lnlnln

ln ln

slope=P

P

(CAp)

kA=

lnCA

lnCApCA CAp


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


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


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


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)


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


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.



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)



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)



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



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.



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



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



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



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



Step 5A.1c Polynomial method

41239264 106973104853103431109782049990C 41239264 10697.310485.310343.110978.204999.0 ttttCA



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



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



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



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


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.


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


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.


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



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)



BAA CCr 223 min//mol)(dm244.0Note that differential method gives



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



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

( ) ( )


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.


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.


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.



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


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



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)


Nonlinear Regression (Nonlinear LeastNonlinear Regression (Nonlinear Least--Squares Analysis)Squares Analysis)


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


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



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


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)



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 )(


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.


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



Solution


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


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


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)


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)



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.



(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


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



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



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)



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



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



@ high PH2

1βPr ~@ low PH2 1

24

βHCH Pr ~

1

2

βHP

@ H2

12 2

2

2

4 βH

HCH bP1

r

~



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


(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


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.


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


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

chemical reactor design - seoul national university · 2018. 1. 30. · 5.2 batch reactor data...

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