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Elements of Chenzicn1 Reaction Engineering
PRENTICE HALL PTR INTERNATIONAL SERIES IN THE PHYSICAL AND CHEMICAL ENGINEERING SCIENCES NEALR. AMUNDSON, SERIES EDITOR, University of Houston
ANDREAS AcRI\~OS, Stanford University JOHN DAALER, University of Minnesota H.SCOTTF%LER, University of Michigan THOMAS J. H A N R A University ~, of Illinois JOHN M. PRAUSNITZ. University of California L.E.S C R ~ I ~University EN, ofMinnewta
Chemical Engineering Thermodynamics BALZHISER, SAMUEIS, AND EWASSEN BEQUETTEProcess Controi: Modeling, Design, and Simulation BEQUETTE Process Dynamics BIEGLER, GROSSIWA?~~. AND WESTERBERGSystematic Methods of Chemical Process Design RRosILow A N 5 JOSEPH Techniques of Modelbased Control CQNSTAN~NXDES AND MOSTOUFI Numerical Methods for Chemical Engineers with MATLAB Applications CROWLAND LOUVAR Chemical Process Safety: Fundamentals with Applications, 2nd edition Problem Solving in Chemical Engineering with Numerical CUTLIP AND SHACHAM Methods DENY Process Fluid Mechanics ELLIOT AND LIRA Introductory Chemical Engineering Thermody narnics F ~ G L E RElements of Chemical Reaction Engineering, 4th edition HEMMELBLGUAND RIGGS Basic Principles and CalcuIations in Chemical Engineering, 7th edition H J N EAND ~ MADDOX Mass Transfer: Fundamentals and Applications PRAUSNITZ, LICHTENTHALER, AND DE AZEVEDO Molecular Thermodynamics of FluidPhase Equilibria, 3rd edition PRENTICEEIectrochernical Engineering Principles SHULER ASD KARGI Bioprocess Engineering, 2nd edition STEPHANOPOUU~S Chemical Process Control TESTERAND MODELL Thermodynamics and Its Applications, 3rd edition TURTON, BAILIE,WHITING.AND SHAEIWITZAnalysis, Synthesis, and Design of Chemical Processes, 2nd edition WII.KES Fluid Mechanics for Chemical Engineers, 2nd edition
Elements of Chemical Reaction Engineering Fourth Edition
H. SCOTT FOGLER Arne and Catherine Vennema Professor of Chemical Engineering The University of Michigan, Ann Arbor
Prentice Hall Professional Technical Reference PRENTICE
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FogIer. H. Scott. Elements of chemical reaction engineering I A. Scott Fogler4th ed.
p. cm. Includes bibliographical references and index. ISBN 23130473944 (alk. paper) 1. Chemical reactors. I. Title.
Copyright O 2006 Pearson Education, Inc. All rights reserved. Printed in the United States of Amwics This publication is pmtected by copyright. and permission must k obtained from the publisher prior to any prohibited reproductiod, storage in a retrieval system. or transmission in any form or by any means, electronic, mechanical. photocopying, recording, or likewise. For information r~gard~ng permissions. write to:
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ISBN 0130473944 Text printed in the United States on recycled paper at Courier in Westford. Massachusetts. First printing. August 1005
Dedicated fo rke tIrerno9 of Professors
Gi useppe Parravano Joseph J. Martin Donald L, Katz
of the University of Michigan whose standards and lifelong achievements serve to inspire us
Contents
PREFACE
1.I 1.2 1.3 J .4
1.5
The Rate of Reaction, 4 The Genera1 Mole Balance Equation 8 Batch Reactors 10 ContinuousRow Reactors 12 1.4.J ContinuousStirwd Tank Reactor, 1.4.2 Tubular R~aclor 14 1.4.3 PackedBed Reacror 17 Industrial Reactors 21 Summary 25 CDROM Material 26 Questions and Problems 29 Supplementary Reading 35
12
2 CONVERSION AND REACTOR SIZING 2.1
2.2 2.3
2.4
Definition of Conversion 38 Batch Reactor Design Equations 38 Design Equations for Flow Reactors 48 2.3.1 CSTR (alsr~known as a Backmix Reactor or Vat) 43 2.2.2 Tubular Flow R ~ Q C I O {PFR) P 44 2.3.3 PackedBed Rearm 45 Applications of the Design Equations for ContinuousFlow Reactors 45
Contents
2.5
2.6
in Series 54 CSTRs in Series 55 PFRs i n Series 58 Cr~tnbinarionsof CSTRs and PFRs in Series 60 Comparing rhe CSTR alid PFR Reuctor filrrmes ~ m l i Reactor Seqitencitlg 64 Some Further Definitions 66 2.6.1 Spacelime 66 2.6.2 Space Vekocic 68
Reactors 2.5.1 2.5.2 2.5.3 2.5.4
Summary 69 CDROM Materials 71 Questions and Problems 72 Supplementary Reading 77
PART 1 Rate Laws
80 Basic Definitions 80 3.1. I Relative Rates of Reaction 81 3.2 The Reaction Order and the Rate Law 82 3.2.1 Power Lnw Models nnd Elementnry Rate Ln~w 82 3.2.2 Nonelemenrav Rate Lnws 85 32.3 Reversible Reacrions 88 3.3 The Reaction Rate Constant 91 3.4 Present Status of Our Approach to Reactor Sizing and Design 98 PART2 Stoichiometry 99 3.5 Batch Systems 100 3.5.i Equations fur Batch Concentrations 102 3.5.2 Constant Volcfme Bnfch Reaction Systems 103 3.6 Flaw Systems 106 Eqttations for Concenfrarions in Flow 3.6.1 Systems 107 3.6.2 LiquidPhase Concmtmtions 108 3.6.3 Change in the Total Number of Moles with R~enctionin rhe Gas Phase 108 Summary 124 CDROM Material 126 Questions and Problems 131 Supplementary Reading 141 3.1
4 ISOTHERMAL REACTOR DESIGN PARTI
Mole Balances in Terms of Conversion 144 Design Structure for Isorherma! Reactors 144 3.1 ScaleUp of LiquidPhase Batch Reactor Data to the Design o f a CSTR 148 4.2.1 Batch Opemfion 148 4.3 Design of Contincous Stirred Tank Reactors (CSTRs) 156 4.3,J A Single CSTR 157 4.3.2 CSTRs in Series 158 4.3.3 CSTRs in PrrroIIeI 160 4.3.4 4 SecondOrder Reoctiott irt n CSTR 162 4.4 Tubular Reactors 168 4.5 Pressure Drop in Reactors 175 4.5.1 Presslire Drop a~rdthe Rate Law 175 4.5.2 Flow Throirgh a Packed Bed 177 4.5.3 Pressure Drop in Pipes 182 4.5.4 Analvfica! Solution for Reaction with Presstire Drop 185 4.5.5 Spherical PackedBed Reactors 196 4.6 Synthesizing the Design of a Chemical Plant 196 PART2 Mole Balances Written i n Terms of Concentration and Molar Flow Rate 198 4.7 Mole Balances on CSTRs, PFRs, PBRs. and Batch Reactors 200 4.7.1 Liquid Phase 300 4.7.2 Gas Pi~nse 200 4.8 Microreactors 201 4.9 Membrane Reactors 207 4.10 UnsteadyState Operation of Stirred Reactors 215 4.10.1 Startup of a CSTR 216 4.10.2 Sernibcrrct?Reactors 217 4.10.3 Writing the Semibatch Reactnr Equa [ionsin Terms qf Cancentrntions 219 4.10.4 Wriring the Semibnrch Reacror Equations in Terns of Conversion 223 4.1 1 The Practical Side 226 Summary 227 ODE Solver Algorithm 230 CDROM Material 231 Questions and Problems 234 Some Thoughts on Critiquing What You read 249 Journal Critique Problems 249 Supplementary Reading 253
4.1
Contents
5
COLLECTION AND ANALYSIS OF RATE DATA 5.1
5.2
5.3 5.4 5.5 5.6 5.7
The Algorithm for Data Analysis 254 Batch Reactor Data 256 5.2.1 Differential Method ofAna!ysis 257 5.2.2 bztegral Mefhod 267 5.2.3 Nonlinear Regression 271 Method of Initial Rates 277 Method of HalfLives 280 Differential Reactors 281 Experimental Planning 289 Evaluation of Laboratory Reactors 289 5.7. i Criteria 289 5.7.2 Types of Reacrors 290 57.3 Su171mayqf Reactor Ratings 290 Summary 291 CDROM Material 293 Questions and Proble~ns 294 Journal Critique Problems 302 Supplementary Reading 303
Definitions 305 6.I. I Types ?f Renctio~u 305 Parallel Reactions 310 6.2.1 Moxilni: b g rhe Desired Product,for Oize Renciant 311 6.2.2 Reartor Selection n11d Opemfing Corrdiflirfons 31 7 Maximizing the Desired Product in Series Reactions 320 Algorithm for Solution of Complex Reactions 327 6.4.1 Mole Boln~lces 327 6.4.2 Npt Rures ?f Reaction 329 6.4.3 Stnrclrinmerp: Co~~r~oerr~rurio~~s 334 Multiple Reactions in a PFWPBR 335 Multiple Reactions in a CSTR 343 Membrane Reactors to Improve Selectivity in Multiple Reactions 347 Complex Reactions of Ammonia Oxidation 351 Sorting Jt All Out 356 The Fun Part 356 Summary 357 CDROM Material 359 Questions and Problems 361 Journal Critique Problems 372 Supplementary Reading 375
253
Contents
7 R EA CTION MECHANISMS, PA THWAYS, BIOREACTZONS, AND BIOREACTORS 7.1
7.2
7.3
7.4
7.5
Active Intermediates and Nonelementary Rate Laws 377 7.1.1 PseudoSreadyState Hypothesis (PSSH) 379 7.1.2 Searching for a Mechanism 383 7.1.3 Chain Reactions 386 7.1.4 Reaction Pathways 391 Enzymatic Reaction Fundamentals 394 7.2.1 Et~z~rneSuhsrrate Complex 395 7.2.2 Mechanisms 397 7.2.3 MicheelisMentenEquation 399 7.2.4 Batch Reactor Calcularionsfor Enzyme Reactions 404 lnhibi tion of Enzyme Reactions 404 7.3.1 Comperirive Inhibirion 410 7.3.2 Uncomperitive Inhibition 412 7 3.3 Noncclmpetitive Inhibition (Mixed Inlzibi~ion) 41 4 7.3.4 Substrate Inhibition 416 7.3.5 Multiple Enzyme and Substrare Systen~s 417 Bioreactors 418 7.4.1 CelI Growth 422 7.4.2 Rare Laws 423 7.4.3 Stoichiometiy 426 7.4.4 Mass Balances 431. 7.4.5 Chemosrafs 434 7.4.6 Design Equations 435 7.4.7 Washout 436 7.4.8 0.rqgenLimited Growth 438 7.4.9 Scaleup 439 Physiologically Based Pharmacokinetic (PBPK) Models 439 Summary 447 CDROM Material 449 Questions and Problems 454 Journal Critique Problems 468 Supplementary Reading 469
8 STEA DYSTATE NONISOTHERMA L REACTOR DESIGN 8.1 8.2
Rationale 472 The Energy Balance 473 8.2.1 First Law of Tl~erpnodynamics 473 8.2.2 E\3aluarit~gthe Work Tern 474 8.2.3 O\?en,ien4 of Ellelgy BaIa~~ces 476 8.2.4 Dissecti?t,q the Stead!Srate Molar Flow Riir~s to Okrain !he Hear of Reaction 479 8.2.5 Dissecring h e Enrhalpies 48 1 8.2.6 Relating AHR,IT1, AHOR, (TR1 and AC, 483
Adiabatic Operation 486 8.3.1 Adiabafic Energy B~lance 486 8.3.2 Adiabatic Tirbular Reactor 487 SteadyState Tubular Reactor with Heat Exchange 495 8.4.1 Deriving the Dtergv Balance for a PFR 495 8.4.2 Balance on tire CovInnt Heat Transfer Fl~rirl 499 Equilibrium Conversion 511 8.5.1 Adiabatic Temperature and Equilibrilam Conversion 5t2 8.5.2 Optimum Feed Temperature 520 CSTR with Heat Effects 522 8.6.1 Hear Added to the Reactot; 6 522 Multiple Steady States 533 8.7.1 HeatRemoved Term, R(TI 534 8.7.2 Heat of Generation, G ( T ) 534 8.7.3 Ignition Extinction Curve 536 8.7.4 Runaway Reactions in a CSTR 540 Nonisotherrnal Multiple Chemical Reactions 543 8.8.1 Energy Balance for Mulriple Reactiorls in PlrcgFlow Rencdors 544 8.8.2 Energy Balance for Multiple Reactions in CSTR 548 Radial and Axial Variations in a Tubular Reactor 551 The Practical Side 561 Summary 563 CDROMMaterial 566 Questions and Problems 568 Journal Critique Problems 589 Supplementary Reading 589
9
UiVSTEADYSTRTE NOAXSOTHERMAL REA CTOR DESIGN 4.1
9.2
9.3 9.4
9.5 9.6
The UnsteadyState Energy Balance 591 Energy Balance on Batch Reactors 594 9.2.1 Adinbatic Operation of a Batch Reactor 594 9.2.2 Batch Reactor with Intermpred Isothermal Operation 599 9.2.3 Reactor Safety: The Use ofthe ARSST ro Find AH,,, E 605 and to Size Pressure Relief Valves Semibatch Reactors with a Heat Exchanger 614 Unsteady Operation of a CSTR 619 9.4.1 Startup 619 9.4.2 Falling Offthe Steady State 623 Nonisothermal Multiple Reactions 625 Unsteady Operation of PlugFlow Reactors 628
xiii
Summary 629 CDROM Material 630 Questions and Problems 633 Supplementary Reading 614
645 D.cjniri017s 646 10.1.2 Cnrctl.;st Properties 648 10.I . 3 CdassiJicationof Crrml?r.sts 652 Steps in n Catalytic Reaction 455 10.2.1 Step I Overview: Difllsion fram tile Btdk to the External Transport 658 10.2.2 Step 2 Overview: Int~rnalDiffiision 660 10.2.3 Adsorption Isotherms 661 10.2.4 Surfnce Reection 646 10.2.5 Desorption 668 10.2.6 The RateLitniring Step 669 Synthesizing a Rate Law, Mechanism, and MeLimiting Step 671 10.3.I Is the ddsorprion of Curn.me RateLimi~ing? 674 10.3.2 Is the Scttface Reaction RateLimiting? 677 10.3.3 IS the De.wrprion of Benzene Rate Limiting? 678 10.3.4 Summary of the C~rmeneDecomposition 680 10.3.5 Reforming C ~ ~ t a l y , ~ r s681 10.3.6 Rate Lnws Derived from the PseudoSteady S r ~ f Hypothesis e 684 10.3.7 Terrtperuture Dependence of the Rare Caw 687 Heterogeneous Data Analysis for Reactor Design 688 10.4.1 Dediicing a Rare h w f m r n the E~perirnentulDara 689 10.4.2 Finding n Mechanism Consistent with Experimental Observations 691 10.4.3 Evnluation of the Rare Law Pammeters 692 10.4.4 Reactor Design 694
10.1 CataIysts 10.1.1
0
10.3
10.4
10.5
10.6
10.7
Reaction Engineering in Microelectronic Fabrication 698 IQ5.I Overview 698 10.5.2 Etching 700 10.5.3 Chemical Vapor Deposition 701 Model Discrimination 704 Catalyst Deactivation 707 10.7.1 Types of Catalyst Deactivation 709 10.7.2 TenaperatlireErne Trajectories 721 10.7.3 MovingBed Reactors 722 10.7.4 sf mightTElmugh Tmnsporr Reactors (STTR) 728
Contents
Summary 733 ODE Solver Algorithm 736 CDROM Material 736 Questions and Problems 738 Journal Critique Problems 753 Supplementq Reading 755
11 EXTERNAL DIFFUSION EFFECTS ON HETEROGENEOUS REACTIOM 11.1 Diffusion Fundamentals 758 1I . 1. I Defilitions 758 11.1.2 Molar Flux 759 11.1.3 Fick'sFirsrLaw 760 1 1.2 Binary Diffusion 761 J 1.2.1 Eiralrtatirag The Molar F l u 761 11.2.2 Boundary Corlditions 765 11.2.3 Modeling Difusion Withorcr Reaction 766 1 I . 2.4 Temperature art$ Pres.ture Depend~nce of DAB 770 I J.2.S Modeling Difision with Chemical Reaction 771 11.3 External Resistance to Mass Transfer 771 1 1.3.1 TIne Mass Transfer Coeficient 771 11.3.2 Mass Transfer Coeficient 773 11.3.3 Correlario~~s for fhe Mass Transfer Co~firienr 774 11.3.4 Mass Transfer to a Single Particle 776 11.3.5 Mnss TransferLimited Reactions in Packed Beds 780 11.3.6 Robert the Worrier 783 1 1.4 What If. . . ? (Parameter Sensitivity) 788 1 1.5 The Shrinking Core Model 792 11 5.1 Cara!\.sr Regenerarion 793 11.5.2 Phanl~acokineticsDissoIufinn qf Monodispers~d SoIid Particles 798 Summary 800 CDROM Material 801 Questions and ProbIems 802 Supplementary Reading 810
12 DIFFUSION AND REACTION 12.1 Diffusion and Reaction in Spherical Catalyst Pellets 814 12.1.I Efccti~,eD~fif~rrsil'iry 814 12.1.2 Deri~nfionqf rhe D$fer@nrialEquatinn D~scribing Diffusinrr artd R~ucrion 816 12.1.3 Wririrrg the Equarion in Dimensionless f i m n 819
757
Contents
12.1.4
12.2 12.3 1 2,4 12.5
12.6 12.7
12.8
12.9 12.10
Solution to the Dlferential Equation for a FirstOrder Reaction 822 Internal Effectiveness Factor 827 Falsified Kinetics 833 Overall Effectiveness Factor 835 Estimation of Diffusion and ReactionLimited Regimes 838 12.5.1 wei.TiPrsate Crilerion for I~tterna!Diffusion 839 12.5.2 Mearx' Crirerion for External Difusion 841 Mass Transfer and Reaction in a Packed Bed 842 Determination of Limiting Siruations from Reaction Data 848 Multiphase Reactors 849 12.8.1 SlurnRenctors 850 12.8.2 Trickk Bed Reactors 850 Fluidized Bed Reactors 851 Chemical Vapor Depwi tion (CVD) 851 Summary 853 CDROM Material 852 Questions and Problems 855 Journal Article Problems 863 Journal Cririque Problems 863 Supplementary Reading 865
13 DTSTRIBUTZOM OF RESIDENCE TIMES FOR CHEMICAL REACTORS 1 3.1
868 PARTI Characterislics and Diagnostics 868 13.I. J Reside~rceTirneDidribulion (RTD)Functior? 870 13.2 Measurement o f the RTD 871 13.2.1 Puhe Irrput E~prrit~ient 871 13.2.2 Step Tracer E.rl?erinzenr 876 13.3 Characteristics of the RTD 878 13.3.S Jniegra1R~lnrir~tlshil~s 838 13.3.2 Mearr Residenre Tinw 879 13.3.3 Orher Mor~lerrtsof the RTD 881 13,3.4 Not.rlla/ted RTD F~o~crion. E(O) 884 13.3.5 I ~ ~ r e ~ n o l  A Di.~~rihuriorr, ge I(a) 885 13.3 RTD in Ideal Reactors 885 13.4.1 RTDs i ~ Batch t and PlugFlow RPCICIOKT 885 13.4.2 SingleCSTR RTD 887 13.4.3 Lcrrlri~lc~r FICJM* Reocror ( L F R ) 888 13.5 Diagnoqtics and Troubleshnoting 891 13.5.1 Gniewl Cnn~rlrenrs 891 12.5.2 Si171plcDiog\~os~ic.~ o ~ Tt~~~lhlesho(~fit7g d U S ~ I the FR KTD for kIenl Rericrors 892 1.q.5.3 PFR/CSTRSeriesRTD 897 General CIlaracterislics
867
PART2 Predicting Conversion and Exit Concentration 902 1 3.6 Reactor Modeling Using the RTD 902 3 3.7 ZeroParameter Models 904 13.7.I Segr~gnrionM o d ~ l 904 13.7.2 bfLL~humMi.redne.7~Mode/ 915 ~~l 13.7.3 Comparirzg Segregarion and M o x i r n ~ / iWi.xedness Predictions 922 I 3.8 Using Software Packages 923 13.8 1 Heot Eflect.? 927 1 3.9 RTD and Multiple Reactions 927 13.9.1 Segregafion Model 927 13.9.2 Ma~itnurnbIixedizess 928 Summary 933 CDROM Material 934 Questions and Probkms 936 Supplementary Reading 944
14.1
14.2 14.3 14.4
14.5 14.6 14.7
14.8
14.9
14.10
Some Guidelines 946 14.1.I OnePornmeter Models
947 14.1.2 TwoParnmer~rModels 948 TanksinSeries (T151 Model 948 Dispersion Model 955 Flow. Reaction, and Dispersion 957 14.4.1 Balance Eqlrnrinns 957 14.4.2 Bouadcd~Conditions 958 14.4.3 Finding D, and the Pecler Number 962 14.4.4 Dispersion in a Ethular Reactor with Laminar Florv 962 14.4.5 Correlationsfor D, 964 14.4.6 Experimental Determination of D, 966 14.4.7 Slopp?: Tracer Inputs 970 TanksinSeries Model Versus Dispersion Model 974 Numerical Solutions to Flows with Dispersion and Reaction 975 TWOParameterModelsModeling Real Reactors with Combinations of Ideal Reactors 979 147.1 Real CSTR Modeled Using Bypassing and Deadspace 979 14.7.2 Real CSTR Modeled as Two CSTRJ with Interchange 985 Use of Software Packages to Determine the Model Parameters 988 Other Models of Nonideal Reactors Using CSTRs and PFRs 990 AppIications to Pharmacokinetic Modeling 991
Contents
Summary 993 CDROM ~Vnterial 994 Questions and Problems 996 Supplementary Reading 1005
Appendix A
NUMERICAL ECHIVIQ UES
Appendix B
IDEAL GAS CONSTANT AND COWERSION FA CTORS
Appendix C
THERMODYNAMIC R ELA TIOIVSHIPSZ W L VING THE EQUILIBRIUM CONSTANT
Appendix D
MEASUREMENT OF SLOPES ON SEMILOG PAPER
Appendix E
SOFTWARE PACKAGES
Appendix
F
NOMENCLATURE
Appendix G
RATE LAW DATA
Appendix H
OPENENDED PROBLEMS
Appendix I
HOW TO USE THE CDROM
Appendix J
USE OF COMPUTATIONAL CHEMISTRY SOFTWARE PA CKAGES
INDEX ABOUT THE CDROM
xvii
Preface
The man who has ceased to learn ought not to be allowed to wander around Ioose in these dangerous days.
M. M.Coady
A. The Audience This book and interactive CDROM is intended for use as both an undergraduatelevel and a graduatelevel text in chemical reaction engineering. The level will depend on the choice of chapters and CDROM Prufessionaf R@ference Shelf (PRS) material to be covered and the type and degree of difficulty of problems assigned.
B. The Goals B.4.
To Develop a Fundamental Understanding of Reaction Engineering
The first goal of this book is to enable the reader to develop a clear understanding of the fundamentals of chemical reaction engineering (CRE). This goal will be achieved by presenting a structure that allows the reader to solve reaction engineering problems through reasoning rather than through memorization and recall of numerous equations and the restrictions and conditions under which each equation applies. The algorithms presented in the text for reactor design provide this framework, and the homework problems will git~e practice at using the algorithms. The conventional home problems at the end 05 each chapter are designed to reinforce the principles in the chapter. These problems are about equally divided between those that can be solved with a
XX
Preface.
calculator and those that require a personal computer and a numerical snftware package such as Polymath, MATLAB, or FEMLAB. To give a reference point as to the level o f nnderst:inding o f CRE required in the profession. a number of reaction engineering problems from the California Board of Registration for Civil and Professional EngineersChernical Engineering Examinations (PECEE) are included in the text.] Typically. these problems should each require nppmximately 30 minutes to solve. Finally, the CDROM should greatly facilitate learning the fundamentals of CRE because it includes summary notes o f the chapters, added examples, expanded derivations, and self test>. A complete description of these knrnirrg resorrrces is given in the 'The Integration of the Text and the CDROM" section in this Preface. 8.2.
To Develop Critical Thinking Skills
A second goal is to enhance critical thinking skills. A number of home problems have been included that are designed for this purpose. Smratic questioning is at the h e m of critical thinking, and a number of homework problems draw from R. W.Paul's six types of Sacsatic questions2 shown in Table PI. 1 4 1) Q~resticmsjbfirr ck~r$uatiun: Why do you fay that7 Hoic does
1 (2)
th~qrelake
to our d i ~ u s s ~ o n ?
*'Are you going to include diffusion In )our mole balance equations?" Quasrionr rhnr pmhc nssrrmpnons: What could we assume instead? How can you verify or disprove that assumption?
"Why are you neglecting rddial diffusion and including only ~ ~ l diffu~ionT' a l
( ( 3 )Q~trsrionsrhar p m k
(
reasons and evirleucu: What would be an example?
"Do you think that diffusion is respnnsibIe For the Iower cnnvers~onr'
((4) Quesrions about viewpoinrs and perspctrle~:Whar would be an alternative?
"With all the bends in the pipe. from an industriallpracticaI rtShelf on the CDROM
A batch reactor has neither inflow nor outflow of reactants or products while the reac~ionis being carried out: F,o = F, = 0. The resulting general mole bal
ance on species j is
If the reaction mixture is perfectly mixed (Figure 1 5[b]) so that there is no variation in the rate af reaction throughout the reactor volume. we can take r, out of the integral, integrate. and write the mole balance in the form
Perfect mixing
Let's consider the isomerization of species A in a batch reactor
As the reaction proceeds. the number of moles of A decreases and the number of moles of B increases, as shown in Figure 16.
Mole Balances
0
ti
Figure 16
t
0
Chae. 1
1 '
Moletime rrajectories.
We might ask what time. t , , is necessary to reduce the initial number of moles from NAo to a final desired number N A I .Applying Equation (15) to the isomerization
rearranging.
and integrating with limits that at r = 0. then N, = N,, NA = we obtain
and at t = I , , then
This equation is the integral form of the mole balance on a batch reactor. It gives the time, r,. necessary to reduce the number o f moles from /VAoto N,, and also to form rYB1moles of B.
1.4 ContinuousFlow Reactors Continuous flow reactors are almost always operated at steady state. We wiil consider three types: the continuous stirred tank reactor (CSTR), the plug flow reactor (PFR), and the packed bed reactor (PBR). Detailed descriptions of these reactors can be found in both the Professional Reference Shelf IPRS) for Chapter 1 and in the Visrral Encyciopeclin of Equiprnenr on the CDROM, 1.41 ContinuousStirred Tank Reactor
What is a CSTR
uxd
A type of reactor used commonly in industrial processing is the stirred tank operated continuously (Figure 17). It is referred to as the continuo~ustirred tnrlk renrtor (CSTR)or vat, or backmix reactor; and i s used primarily for liquid
phase reactions. I t i s normafly operared a t steady state and I \ a\surned to be perfectly mixed: conrequenily, t hers is no ti me dependence or pokition dependence o f the temperature. [he concentration, or the reaction rnte inside rile CSTR. That is, every variable i> the wme at every point inside the reactor. Becat~sethe temperature md concenrration are identical everywhere u itliin the reaction vessel, they are the same at the exit point as they are elsewhere in the tank. Thus the temperature and concentralion in the exit stream are modeled a5 being the same as those jnside the reactor. In systems where mixing is highly nonideill, the weIImixed model i* inadequate and we must resort to other modeling techniques, such ns rebidencetime distributions, to obtain meaningful results. This topic o f nonideal mixing is discussed in Chapters 13 and 14.
What reac~ion systems use a CSTR?
The idtal CSTR i v assumed trr ht. perrec~lymixed.
Figure 17(a) CSTRharch reactor. [Courtesy of Pfnudlcr. Inc.1
Figure I71bk CSTR mixing patterns. Alw see the Vir~rnlEnr~cluped~n oJ Eqwp~nenton the CDROM.
When the general mole balance equation
is applied to a CSTR operated at steady state (i.e., conditions do not change with time),
14
Mole Balances
Chap. 1
in which there are no spatial variations in the rate of reaction (i.e., perfect
mixing),
it takes rhe familiar form known as the design eqrrarion for a CSTR: F~~
Al
The CSTR design equation gives the reactor volume I{necessary to reduce the entering flow rate of species j, from Fj, to the exit flow rate F,,when species j i s disappearing at a rate of rj. We note that the CSTR is modeled such that the conditions in the exit stream (e.g.. concentration, temperature) are identical to those in the tank. The molar flow rate 6is just the product of the concentralion of species j and the volumetric flow rate u :
I Fl = C j +v
& p
time

moles
volume
volume time
Consequently. we could combine Equations (17) and ( 18) to write a balance on species A as
1.4.2 Tubular Reactor
When is reactor mnsi u,ed?
In addition 10 the CSTR and batch reactors, another type of reactor commonly used in industry is the ruhlliar rmr.ror: It consists of a cylindrical pipe and is normally operated at steady state, as i~ the CSTR. Tubular reactors are used most often for gaspha~ereactions. A schematic and a photograph of industrial tubular reactors are shown in Figure 18. In the tubular reactor, the reactants are continually consumed as they flow down the length of the reactor. In modeliny the tubular reactor. we assume that the concentration varies continuoucly in the axial direction through the reactor. Con~equcntly.the reaction rate, which is a function of concentration for all but zeroorder reactions, will also vary axially. For the purposes of the material presented here. we consider systems in which the flow field may be modeled by that of a plug flow profile l e g . . uniform veIocity as in turbulent floa). as shewn rn Figure 19. Thar is. there i h no radial variation it1 reaction rate and the reactor is referred to as a plugfiow rcactor [PFR). (The laminar flow renctor is diwurced in Chapter 13.)
Sec. l.d
15
ContintlousFlow Reactors
ee PRS and Enr~cloprEqltip~~~~nt.
Figure 18(a) Tubular reoclrw wherna~ic Longitudinal tubular reactor. [Excerpted b! spctnl permission from CIINII.Ella, h3( 10). 2 1 1 (Oct. lq56). Copyright 1956 by McCrau Hill. Inc , Kcu York, Y 1' 10010.1
Figurc 18(bl Tuhulw iellclur photo Tubular reac~orfor p~vductlonof Dlmerwl G. [Photo Colrnecy of Editions Technvq Institute Frrrncois du Pc~rolj
Plug tlr>url~o radral iarinlionr in vclncity.
Reactants
Products Fiaure 19 Pluyflow
The general
t u l l ~ ~ l are;lctor. r
moIe balance equation i s
given by Equation ( 3 4 ) :
The equation
we will use to design PFRs at steady /,2200, Chapter 4. HIMMELBLAU.D. M,, and J. D. Riggs. Basic Prinripl~sand Calcirlariorts in Chen~icnlE~tgi?~eeriltg, 7th ed. Upper Saddle Rwer, N.J.: Prentice Hall. 2004. Chapters 2 and h. SANDERS. R . 1.. The Alrafninx ofSkzing, Denver. CO:Golden Bell Press, 1974. 2. A detailed explanation of a number a f topics in ?his chapter can be found in
H. S. FOCLEH,eds.. AlCItE Mndulular I~t.~rrucriorlSerz~rE. K ~ I I E Y ~VOIS. T S . 1 and 2. New York: AlChE, 1981.
C R Y ~ E R. S , L.. and
3. An excellent description o f the various types of commercial reactors used in industry is found in Chapter I I of WAIAS.S. M., Rencfiotl Kinerfc.sfor CCheicuI Brgil~eers.New Y
d McGnwHiU, 1959.
1. A discussion nf some of the most important industr~alprocesses i s presented by
MEYERS,R.A., Handbook of Chcnlicols Pmd~rcriott Processes. New York: McGrawH111. 1986 See also
MCKETTA. J. 1.. ~ d . .Etrryrloprriiu oj" Cl~e~rricnl Pmrrsse.~urld Design. New York: Marcel Dckker. 197h.
36
Mola Balances
Chap.
.4 simiIar book. which describes a larger nuniber o f processes, i s
C. T.. S l ~ r e ~ ~Clirrnicnl e'~ PTOCF.TF Indrrstrirs. 5th ed. New Yorb McGrawHill, 1984.
AL'STIY.
d LlnRr
5. The following journals may be useful in obtaining infomation on chemical reac tiun engineering: Internotional Jo~~rnul of CS~ernEcrlEKinetics, Joltrnnl of Car~ilyx Journal of Applied Catul~r.~is. AlChE Jolurr ma/. CEwmical Engineering Scie~c Canadian Journal of Chemical Etigineering, Chemical Drgzr~eering Comrnunica tro~ls,Joi~rnnlof Physical C.hettiisrr?: and Jndusrrrol nnd Engit~cerirrg Chemisrc Reseami1.
6. The price of chemicals ern be found in such journals as the Ch~ttiicoiMarketin! Reporter, Cherniml Weekly3and Chemical Engrr~eeringNews and on the ACS wet site hrrp://pubs. acs.or~/c~n.
Conversion and Reactor Sizing
2
Be more concerned with your character than with your
reputation, because character is what you redly are while reputation is merely what others think you are. John Wooden, coach, UCLA Bruins
Overview. In the first chapter, the general mole balance equation was detived and then applied to the four most common types of industrial reactors. A balance equation was developed for each reactor type and these equations are summarized in TabIe SI. In Chapter 2, we will evaluate these equations to size CSTRs and PFRs. To size these reactors we first define conversion, which is a measure of the reaction's progress toward completion, and then rewrite all the balance equations in terms of conversion. These equations are ofren referred to as the design equations. Next, we show how one may size a reactor line., determine the reactor volume necessary to achieve a specified conversion) once the relationship between the reaction rate,  r ~ and , conversion, X,is known. In addition to being abIe to size C S R s and PFRs once given r, =Am, another goal of this chapter is to compare CSTRs and PFRs and the overall conversions far reactors arranged in series. It is also important to arrive at the best arrangement of reactors in series. After completing this chapter you will be able to size CSTRs and PFRs given the rate of reaction as a function of conversion and to calculate the overall conversion and reactor volumes for reactors arranged in series.
38
Conversion and Reactor Sizing
Chap. 2
2.1 Definition of Conversion In defining conversion, we choose one of the reactants as the basis of calculation and then late the other species involved in the reaction to this basis. In virtually all instances it is best to choose the limiting reactant as the basis of calculation. We develop the stoichiometric relationships and design equations by considering the general reaction
The uppercase letters represent chemical species and the lowercase letters represent stoichiometric coefficients. Taking species A as our basis of c~lcularion, we divide the reaction expression through by the stoichiometric coefficient of species A. in order to arrange the reaction expression in the form
to put every quantity on a "per mole of A basis. our limiting reactant. Now we ask such questions as "How can we quantify how far a reaction [e.g., Equation (22)] proceeds to the right?" or "Mow many moles of C are formed for every mole A consumedr' A convenient way to answer these questions is to define a parameter called conversion. The conversion XA is the number of moles of A that have reacted per mole of A fed to the system:
of A reacted x* = Moles Moles af A fed
Definition of X
Because we arre defining conversion with respect to our basis of calculation [A in Equation (2211, we eliminate the subscript A for the sake of brevity and let X = X, . For irreversible reactions, the maximum conversion is 1.O, i.e., complete conversion. For reversible reactions, the maximum conversion is the equilibrium conversion & (i.e., X,, = X,).
2.2 Batch Reactor Design Equations In most batch reactars. the longer a reactant stays in the reactor, the more the reactant is converted 10 product until either equilibrium is reached or the reactant is exhausted, Consequently. in batch systems the conversion X is a function of the time the reactants spend in the reactor. If WAO i s the number of moles of A initiaIly in the react* then the total number of moles of A that have reacted after a time r is [h'A,lXj
I
Moles of A reacted r
7
Moles of A I reacted j
=
I.Wqul
[XI
(23)
Sac. 2.2
39
Batch Reactor Design Equations
Now, the number of moles of A that remain in the reactor after a time r, N,, can be expressed in terms of NAOand X:
Moles of A that
The number of moles of A in the reactor after a conversion X has been achieved 1s
When no spatial variations in reaction rate exist, the mole balance on species A for a batch system is given by the following equation [cf. Equation (15)J:
This equation is valid whether or not the reactor volunle is constant. In the general reaction, Equation (22). reactant A is disappearing: therefore, we rnultiply both sides of Equation (25) by 1 to obtain the mole balance for the hatch reactor in the form
The rate of disappearance of A. r,, in this reaction might be given by a rate law similar to Equation (12), such as  r , = kCACB. For batch reactors. we are interested in determining how long to leave the reactants in the reactor tn achieve a certain conversion X . To determine this length of time, we write the mole balance. Equation (25). in terms of conversion by differentiating Equation (14)with respect to time, remembering that NAo is the number of moles of A initially present and is thereforc a conqtant with respect to time.
Combining the above with Equation ( 2  5 ) yields
For n batch reactor. the dedgn equation in differentia! form is
40
Conversion and Reactor Sit~ng
Chap.
We call Equation (26) the differential form of the design equation for batch reactor because we have written the mole balance in terms of conversior The differential forms of the batch reactor mole balances. Equations (25) an1 (26). are often used in the interpretation of reaction rate data (Chapter 5 ) ant for reactors with heat effects (Chapter 9), respectiveIy. Batch reactors are fre quently used in industry for both gasphase and liquidphase reactions. Thl laboratory bomb calorimeter reactor is widely used for obtaining reaction rat1 data (see Section 9.3). Liquidphase reactions are frequently carried out ii batch reactors when smallscale productton is desired or operating difficuftie,
mle out the use of continuous flow systems. For a constantvolume batch reactor. V = V,,, Equation (25) can bc arranged into the form
Constantvolume batch reactor
As previously mentioned. the differential form of the mole balance, e.g.. Equa tion (27). is used for analyzing rate data jn a batch reactor as we will see ir Chapters 5 and 9. To determine the time to achieve a specified conversion X, we first separate the variables in Equation (26) as follows
Batch time t to achieve e conversion X
u Batch
Design Equation
This equation is now integrated with the limits that the reaction begins at time equal zero where there is no conversion initially (i.e., t = 0,X = 0). Carrying out the inteption, we obtain the time t necessary to achieve a conversion X in a batch reactor
The longer the reactants are left in the reactor, the greater will be h e conversion. Equation (26) is the differential Form of the design equation. and Equation (29) is the integral form of the design equation for a batch reactor.
2.3 Design Equations for Flow Reactors For a hatch reactor. we saw that conversion increases with time spent in the reactor. For continuousflow systems, this time usually increases with increasing
Sec. 2.3
41
Design Equatio~sbr F!ow Reactors
reactor volume. e.,a.. the biggert'lonper the reactor, the more time it will take the reactant5 to Row conipleteIy through the reactor and thus, the more time to react. Consequently, the conversion X is a Function of reactor volume V. If FA,, i s the molar flnw rate of specres A fed to a system operated at steady state. the molar rate at which species A is reacting rr3ithirrthe entire system will be F,d.
Moles of A fed, Moles of A reacted lime Mole o f A fed Moles of A reacted '4 = time
LF*ol.L.\1=
The molar feed rate of A ro the system minus the rate of reaction of A within the system eqltnls the moIar flow rate of A leaving the system FA. The preceding sentence can be written in the form of the following mathematical statement:
Molar flow rate fed to the system
Molar rate at
Molar flow rate
consumed within
the system
Rearranging gives
The entering molar flow rate of species A. FA, (mol/s), is just the product of the entering concentration, CAo(mol/dmf ), and the entering volumetric flow rate, u, (drn31s):
?%IJ = c~n uo
Liquid phase
For liquid systems, C ,,
is commonly given in terms of molarity, for example, CAO= 2 rnol/drn3
For gas sysrerns, CAocan be calculated from the entering temperature and pressure using the Ideal gas law or some other gas law. For an ideal gas (see Appendix B):
Gas phase
42
Conversion and Reactor Sizing
Chap. 2
The entering molar flow rate is
where C,, = entering concentration, mol /dm3 = entering mole fraction of A
y,
P, = entering total pressure, e.g., kPa PA, = .v,,Po
= entering partial pressure of A,
e.g., kPa
To = entering temperature, K R
= ideal gas constant
kPa ' mol * K
see Appendix
B
1
The size of the reactor will depend on the flow rate, reaction kinetics, reactor conditions, and desired conversion. Let's first calculate the entering molar flow rate. Exumpk 21
U~ittgthe Ideal Gas Law to Caicuhfe CAl and FA*
A gas of pure A at 830 kPa (8.2 atm) enters a reactor with a volumetric flow rate, v* of 2 dm% at 500 K. Calculate the enterlng concentration of A, C,,, and the entertng molar Bow rate.
rho.
Solurioil
U'e again recall that for an ideal pas:
where Po = 8.70 k% (8.3 atm) YM) = 1 . 0 ( P u ~ A ) To = in~traltemperature = 500K R = 8.3 14 dm3 1;Palniol , K (Appendix B ) 4
Substituting the given paraineter values inlo Equation (E21. I ) yields .c.40 =
mol I = 0.20dm3 (8.334 dm? kPdrnol . KJ(500K) (1)(830 kPa)
We could also solve for the partial pressure in terms of the concentration:
Sec. 2.3
Design Equations for Flow Readors
43
pure A enters, the total pressure and partial pressure entering are the same. The entering molar flowrate, FA,, is just the p d u c t of the entering concentration, C,,, and the entering volumetric flow rate, vo: However, since
I
FA, = CA,vo = (0.2 mal/dm3)(2 dm3Is) = (0.4 rnol/s)
This feed rate (FA, = 0.4 moYs) is in the range o f that which is necessary to form several million pounds of product per year. We will use this value of FA, together with either Table 22 or Figure 21 to size and evaluate a number of reactor schemes in Examples 22 through 25.
Now that we have a relationship [Equation (?lo)] between the molar flow rate and cenversion, it is possible to express the design equations (i.e., mole balances) in terms of conversion for the flow reactors examined in Chapter I .
2.3.1 CSTR (also known as a Backmix Reactor or Vat) Recall that the CSTR is modeled as being we11 mixed such that there are no spatial variations in the reactor. The CSTR mole balaoce, Equation ( I 7), when applied to species A in the reaction
can he arranged to
We now substitute for FA in terns of FAOand X
and then substitute Equation (2 12) into (21 1)
 F,d v= 5 0  (50  r* Simplifying, we see the CSTR volulne necessary to achieve a specified conversjon X is FA ruU)Im
Perfec~mixing
(2 13)
44 Evalrlate r,
at
CSTR exit.
Conversion and Reactor Sizing
Chap. .
Because the reactor is perfprrk mixer/, the exit composition from the reactor i identical to the composition inside the reactor, and the rate of reaction is eval uated at the exir conditions.
2.3.2 Tubular Flow Reactor (FFR) We model the tubular reactor as having the fluid flowing in plug flow. i.e.. nr radial gradients in concentration, temperature, or reaction rate.' As the reac [ant$ enter and flow axially down the reactor, [hey are consumed and the con version increases along the length of the reactor. To develop the PFR desigr equation we first multiply both sides of the tubular reactor design equatior (1 123 by  I . We then express the mole balance equation for species A in tht reaction as
For a flow system, FA has previously been given in terms of the entering molat Row rare FM and the conversion X
differentiating
dFA=  F A d X
and substituting into (7141, gives the differential form of the design equation for a plugflow reactor (PFR):
4
blgn
PFR cquntlon
b,
We now separate the variables and integrate with the limits V = 0 when X = 0 to obtain the plugflow reactor voluine necessary to achieve a specified conversion X:
To carry out the integrations in the batch and plugRow reactor design equations (29) and (216). as welI as to evaluate the CSTR design equation (223), we need to know how the reaction rate  r ~varies with the concentration (hence conversion) of the reacting species. This relationship between reaction rate and concentration is developed in Chapter 3.
This constraint will be removed when we extend our analysis to nonideal (industrial) reactors in
Chapters 13 rind
Id.
Sm. 2.4
Apolicatiors cl ths Design Equaticns for ContinuousFlow Fleactors
45
2.3.3 PackedBed Reactor
Packedbed reactors are tubular reactors filled with catalyst particles. The drrivation of the differential and integral forms of the design equations For packedbed reactors are analogous 10 those for a PFR {cf. Equations ( 2  15) and (2 1611. That is, substituting Equation (2 12) for FA in Equation ( 1  15) gives PBR design equation
The differential Form o f the design equation [i.e., Equation (217)J must be used when analyzing reactors that have a pressure drop along the length of the reactor. We discuss pressure drop in packedbed reactors in Chapter 4. I n the abserlce of pressure drop, i.e., AP = 0. we can integrate (2 17) with Iimits X = 0 at W = 0 to obtain
Equation (2181 can be used to determine the catalyst weight W necessary to X when the total pressure remains constant.
achieve a conversion
2.4 Applications of the Design Equations for ContinuousFlow Reactors In this section. we are going ro show how we can size CSTRs and PFRs (i.e., determine their reactor volumes) from knowledge of rbe rate of reaction. r,. as n function of conversion, X. The rate o f disappearance of A. r,. is almost aIways a function of the concentrations of the various species present. When only one reaction is occurring. each of the concentrations can be expressed as a function of the conversion X (see Chapter 3); consequently, r, can be expressed as a function of X. A particularly simple functional dependence, yet one that occurs often, is the firstorder dependence
Here. k is the specific reaction rate and is a function only of temperature, and CA0is the entering concentration. We note in Equations (213) and (216) the reactor volume in a function of the reciprocal of r,. For this firstorder dependence, a plot of the reciprocal rate of reaction (I/r,) as a function of conversion yields a curve similar to the one shown in Figure 21, where
46
Conversion and Reactor Sizing
Chap. 2
To illustrate the design of a series of reactors, we consider the isotherrnaI gasphase isomerization AB We are going: to the laboratory to determine the rate of chemical reaction as a function of the conversion of reactant A. The laboratory measurements given in Table 21 show the chemical reaction raze as a function of conversion. The temperature was 500 K.(440"FI. the total pressure was 830 kPa (8.2 atm), and the initial charge to the reactor was pure A.
If we know r, as a function of X,we can size any isothermal mcrion system.
Recalling the CSTR and PFR design equations, (213) and (2I&}, we see that the reactor volume varies with the reciprocal of r,, (I/r,4f. e.g., V=
(%)(F*&').
Consequently, to size reactors, we conven the rate data in
Table 2 I to reciprocal rates, ( 1 IrA). in Table 22.
These data are used to arrive at a plot of (I/r,) as a function of X. shown in Figure 2 1 . We can use chis figure to size Row reactors for different entering molar flow rates. Before sizing flow reactors let's first consider some insights. If a
Sec. 2.4
Appl~cat~ons of the Desrgn Equations for ContfnuousFlow Reactors
Figure 21 Processed data I.
reaction is carried out isothermally, the rate is usually greatest at the start of the reaction when the concentration of reactant is greatest (i.e., when there is negligible conversion SX E 03). Hence I 1/rAf will be small. Near the end of the reaction, when the reactant has been mostly used up and thus the concentration of A is small (i.e., the conversion is large), the reaction rate will be small. Consequently. (I/rA)is large. For all irreversible reactions of greater than zero order (see Chapter 3 for zeroorder reactions), as we approach complete conversion where all the limiting reactant is used up. i.e., X = 1. the reciprocal rate approaches infinity as does the reactor volume. i.e.
As X
A+BtC "To infinity and beyond" Buzz Lightyear
A#B+C
I +x + I .  I : , + 0 , thus, 
rA
therefore If
+
Consequently, we see that an infinite reactor volume is necessary to reach complete conversion, X = 1.0 For reversible reactions (e.~., B), the maximum conversion is the  A equilibrium conversion X,. At equilibrium. the reaction rate is zero ( r , s 0). Therefore. As
I + X + X,. 7, + 0 . thus. r,
and therefore 'I
+
~3
and we see that an infinite reactor volume would also be necessary to obtain the exact equilibrium conversion, X = X,. To size a number of reactors for the reaction we have been considering, uRewill use FA, = 0.4 moI/s (calculated in Example 2 1) to add another row to the processed data shown in Table 22 to obtain Table 13.
Sec. 2.4
Apal~caltronsof the Design Equat~onsfor Cont~nuousFfowReactors
49
I
(8)
Quation (213) gives the volume o f a CSTR as a function of FA,,.X,and r,,:
I n a CSTR, the composition, temperature, and conversion of the effluent stream an: rdentical ta that of the Ruid within the reactor, because perfect mixing is assumed. Therefore, we need to hiid the value of rA (or reciprocal thereof) at X = 0.8. From either Table 22 or Figure 1I , we see that when X = 0.8, then
Substitution into Equation (213) for an entering molar flow rate. FA* of 0.4 mol A/s and X = 0.8 gives
(b) Shade the area in Figure 22 that yields the CSTR vocllume. Rearnnging Equa
tion (2 13) gives
In Figwe E22.1, the volume is equal to the area of a rectangle with a height (X= 0.81. This rectangle is shaded in the figure.
(FAIjrA = 8 rn3) and a base
(E22.2) V = Levenspiel rectangle area = height x width
The CSTR volume necessary to achieve 80% conversion is 6.4 m7when operated at 500 K. 830 kPa (8.2 am), and with an entering molar flow rate ~f A of 0.4 rnolh. This volume corresponds to a reactor about 1.5 rn in diametervand 3.6 rn high. It's a large CSTR, but this is a pasphase reaction, and CSTRs ate normally not used for gasphase reactions. CSTRs are used primarily for liquidphase reactions.
Conversion and Reactor Siting
Chap. 2
Plots of Ilr* vs. X arc sometimes referred to a
\ perox~deper day? (Hirlr. Recall Table 4 I .1 (fl Assume that the reaction is reverrrblc with Kr = 0.025 rnol'ldmh. and calculate the equilibr~umconversion; lhen redo l a ! through (c) to achieve a con\erainn thal i\ 90% of the equilibrium conver4nn. (g) Membranc reactor. Repeat Part ( f ) for the care when C:H, flows out through the .side\ ol' lhe reactor and !he tr:in%porr coefficient kr = 0.08 \I.
Chap. 4
I
239
Questions and Problems
P488 ~ u b l e s h o o t i n g (a) A
J'Q
liquidphase isomerization A
B is carried out in a 1000gal
CSTR that has a single impeller located halfway down the reactor. The liquid enters at the top of the reactor and exits at the bottom. The reaction is second order. Experimental data taken in a hatch reactor predicted the CSTR conversjon should be 5056.However, the conversion measured in the actual CSTR was 579. Suggest reasons for the discrepancy and suggest something that wouid give closer agreement between the predicted and measured conversions. Back your suggestions with cnkulatioos. F.S. It was raining that day. (b) The firstorder gasphase isornerization reaction
Creative Thlnking
B with k = 5 min1
A
is to be carried out in a tubular reactor. For a feed of pure A of 5 drn31min. the expected conversion in a PFR is 63.2%.However, when the reactor was put in operation. the con\~ersionwas only 5R.h%. U'e should note that the straight tubular reactor would not fit in the available space. One engineer suggested that the reactor be cut in half and the two reactors he put side by side with equal feed to each. However. the chief engineer overrode this suggestion saying the tubular reactor had to be one piece so he bent the reactor in a U shape. The bend was not a good one. Brainstorm and make a list of things that could cause this offdesign specification. Chaose the most logical explanationlmodel, and carry out a calculation to show quantitatively that with your model the conversion I< 5X.6&. (An Ans: 57% of the total)
Mall c f Fame
(c) The liquidphase reaction
was carried out in a CSTR.For an entering concentration of 2 rnol/dm3. the conversion was 4 0 9 . For the qame reactor volume and entering conditions as the CSTR, the expected PFR conversion is 48.6%. However. the PFR conversion was amazingly 50% exactly. Bninqtonn reasons for the disparity. Quantitatively show how these conversions came about (i.e., the expected conversion and the actual conversion). (d) The pasphase reaction
AtB
P49,
C+D
is carried out in a packed bed reactor. When the panicle size was decreased hy 15%. the conversion remained unchanged. When the particle size was decreased by 20%. the conveaion drcreaced. When the original particle size was increaced by 15'3, the conver.If nece\s a y rewrlte your mole balance 11: terns of the rnea~uredvar~able(e.g.. P,) 4. L w k For simplifications. For example, ~f one of the reactants in excess. assume its concentration is conqtant. IT the gas phare mole fraction of reaclanl 1s cmall. ser E=O. 5. For a batch reactor. calculate 4, a5 a function of concentration C, to determine reaction order. A.
D ~ f f e r e n ~ i anal! a l math tutorial on regression with screen rhuts ih ~ h u w nin [he Ch;~p~et5 Srrmrllrrr? Nofcs on the CDROM and web 6. For differential PER calculate r', as a function of C, or PA Summary gofer
ri
C IW
=~
Pa
function of re.lctant concentration.
C,.
A.
Calculate
C.
Use nonlinear regressloo 10 Hnd the be\r model and model parsnjerers. See example on [he CDROM Slrmlnun WCI!(V using dafa fbm hctrmpe~~enus catal
:iq
y w . Ch;~p~cr10 7. Analyze your rate law mudcl Fur "gwdneu uf tit:'Calcula~e n correla~iolrcocftic~ent.
1
256
Collection and Analysis of Rate Data
Chap
5.2 Batch Reactor Data
Process data in terms o f the
measured variable
Batch reactors are used primarily to determine n t e law parameters for horr geneous reactions. This determination is u s u ~ l l yachieved by measuring cc centration as a function of time and then using either the differential, integr or nonlinear regression method of data analysis to determine the reacti order, a , and specific reaction rate constant, k. If some reaction parame other than concentration is monitored, such as pressure, the mote balance mt be rewritten in terms of the measured variable (e.g., pressure as shown in t example in Solved Pmblenls on the CD). When a reaction is irreversible, it is possible in many cases to determi the reaction order a and she specific rate constant by either nonlinear regrt sion or by numericaIly differentiating concenrrntion versus time data. This 1, ter method is most applicable when reaction conditions are such that the rr is essentialty a function of the concentration of only one reactant; for examp if, for the decomposition reaction.
Assume that the rate law is of the
form  r A = kACl
then the differential method may be used. However, by utilizing the method of excess, it is also possible to dett mine the relationship between r, and the concentration of other reactan That is, for the irreversible reaction
A + B + Products with the rate law
where rw and f3 are both unknown, the reaction could first be run in an exce of B so that C , remains essentially unchanged during the course of the rea
tion and
where Method of excms
After determining a , the reaction is carried out in an excess of A, f which the rate law is approximated as
where k" = kACi = kAC;*
Sec. 5.2
257
Batch Reactor Data
Once u and p are determined. A, can be calculated from the measurement of  r , at known concentrations of A and B:
Both a and P can be determined by using the method of excess, coupled with a differential analysis of data for batch systems. 5.2.1
Differentia! Method of Analysis
To outline the procedure used in the differential method of analysis. we consider a reaction carried out isothermally in a constantvolume batch reactor and the concentration recorded as a function of time. By combining [he mole b d once with the rate Inw given hy Eqtrarion (51). we obtain Conutarttvolume batch rcuctor
After taking the natural logarithm of both sides of Equation (561,
observe that the slope of a plot of In (dC,ldr) reaction order, a (Figure 52 ).
as a function of (la C,} is the
vesus ln C, ro find
a and k,
(a)
~ b )
Figure 51 Differential method to determine reaction order.
Figure 5l[a) shows a plot of [ (dCAJdt)]versus [CAIon loglog paper (or use Excel to make the plot} where the slope is equal to the reaction order a . The specific reaction rate, k,, can be found by first choosing a concentration
258
Cellectlon and Analysis of Rate Data
Chap. 5
in the plot, say CAP,and then finding the corresponding value of I (dC,ldt)] as shown in Figure 5l(b). After raising CAPto the a power, we divide it into [ (rdC,/dr),]
Methods for finding

2
to determine X., :
To obtain the derivative dCA/dt used in this plot, we must differentiate the concentrationtime data either numerically or graphically. We describe three methods to determine the derivative from data giving the concentration as a function of time. These methods are: Graphical differentiation Numerical differentiation formulas Differentiation of a polynomial fit to the data
from
concentntmntime data
5.2.1A Graphical Method
krI
gig
1
can be written as
Taking the log of Equation (ES5.4) gives us
InIr;,
= Ink'+a InPco
We now plot In(rmJ) versus In Pco for runs 1. 2, and 3 . (b) Runs 1, 2, and 3. for which the H2 concentration is constant, are plotted in Figure E55.1.We see from the Excel plot that a = 1.22. Had we included more p i n t s we would have found that the reaction is essentially first order with a = 1, that is,
rho = k'Pc0 F~nmrhe first three data points where the partial pressure of the rate is linear in partial pressure. of CO.
(E.55.5)
Hzis constant. we see
CoflMion and Analysis of Rate Data
Chap. 5
0.001 1
0.1
10
PCO latm) Figure E55.1 Reaction rate
as a function of concentration.
Now let's look at the hydrogen dependence
Determining rhe Rote Lnw Dependence on HI
Fmm Table E55.2it appears that the dependence of riH4on PHl,cannot be represented by a power law. Comparing run 4 with run 5 and run I w ~ t hrun 6. we see that the reaction rate first increases with increasing partial pressure of hydrogen. and subsequenrly decreases with increasing P,, . That is, there appears to be a concentration of hydrogen at which the rate is mhimum. One set of rate laws that is conslstent with these observations is: 1. At low HZ concentrations whcre rhH4increases as PHI increases, the rate law may be of the form
2. At high H2 concentrations w)here rbHddecreases as PHI increases,
We would like to find one rate law that is consistent with reaction rate data at both high and low hydrogen concentrations. Application of Chapter 10 material 5uggesrs Equations (E55.6) and tE55.7) can be combined into the form
We wiIT see in Chapter 10 that this combination and similar rare laws which have reactant concentrations for partial pressures) in the numerator and denominator are common in heremgeneous caralysis. Let's see if the resulting ratc law cE55.8) is qualitatively consistent with the rate observed. 1 . For condition 1: At low P H 2 .( b ( ~ ~ ~1) ) and ' ~Equation e (E55.8)reduces to
Equation (E55.9)is consistent with the trend in comparing mns 4 and 5. 2. For candirion 2: At high P H 2 ,b ( ( ~ ~ , ) ' "11) and Equation (E55.8) reduces to
where p, > P, . Equation (E55.10) 1s consistent with the lrends in comparing runs 5 and 6. Combining Equations (E55.8)and (65.5) Typical form of the rate law For heterogeneous
catalysis
We now uqe the Polymath regression program to find the parameter values n, b, Dl. and P2. The results are shown in Table E55.3.
I
Model: Rale = a'Pco'PM%lall(l*b'PWeZ)
I
Msx
Polymath rcshion tutanal 1s in the Chapter 5 Summuy .Vnres.
Nonlinear regrassion M n g s #
iteralions = 134
The corresponding rale law is
h'e could use the rate Ian, as given by Equation (E55.17)as is. hut there are only six dara points, and we should be concerned about extrapolating the ratc law over n wider range of partial pre3sures. We could lake more data, and/or w e could carry out a theomtical analysis of the type diqcus9ed in Chapter IO for heterogeneous reactions. If we nwume hydrogen underptxs disiociative ;~dsorption on the catalyst
288
Collection and Analvsis of Pate Data
Cht
surface one would expect a dependence of hydrogen to the % power. Because I 1s close to 0.5, we are going to regress the data again setting = I and 0: = The results are shown in Table E55.4.
D,
Modd: Rate = a'Pco'PhZ"O.SI{l+b'PhZ)
The rate law is now
mol/gcat . s and the partial pressure is in atm, also have set Dl = !+ and f12 = 1.0 and rearranged Equa (ES5.11)in the form where r'&, is in
We
>auld
Linearizing the rate law to determine the rate Iaw parameters
A plot of F ~ ~ Pas a~ function ~ ~ of IPH2 ~should , ~be a straight line with an ir cept of I l a and a sbpe of bla. From the plot in Figure E55.2. we see that the law i s indeed consistent with the rate law data.
I 0
I
t
2
3
PH*(arm)
Figure E552 Linearized plot of data.
4
Sec. 5.6
Expertmental Planning
5.6 Experimental Planning Four to six weeks in the lab can save you an hour in the library. G. C.Quarderer. Dow Chemical Co.
Reference Shelf
So f a , this chapter has presented various methods of anaiyting rate data. It is just as imponant to knaw in which circumsrances to use each method as it is to knaw the mechanics of these merhods. On the CDROM. we give a thumbnail sketch of a heuristic to plan experiments to generate the data necessary for reactor design. However. for a more thorough discussion, the reader is referred to the books and articles by Box and Huntern9
5.7 Evaluation of Laboratory Reactors
1 I
!i
I I
I
Reference Shelf
The successful design of industrial reactors ties primarily with the reliabi!ity of p c experimentally determined pnmmercrr w.ed in the scaleup. Consequently, rt i s imperative to design equipment and experiments that will generate accurate and meaningful data. Unfortunately, there is usually no single cornprehensive laboratory reactor that could be used for all types of reactions and catalysts. In this section, we discuss the various types of reactors that can be chosen to obtain the kinetic parameters for a specific reaction system. We closely follow the excellent strategy presented in the articIe by V. W. Weekman of Mobil Oil. now E x x ~ n M o b i I . ' ~
5.7.1 Criteria The criteria used to evaluate various types of laboratory reactors are listed in Table 52. I. Ewe of sampling and pmduct analysis 2. Degree of isothemality 3. Effectiveness of contact between catalyst and reactant 4. Handling of catalyst decay 5. Reactor cost and e s e of construction
Each type of reactor is examined with respect to these cdteria and given (F). or poor (P). What follows is a brief description of each of the laboratory reactors. The reasons for rating each reactor for each of the criteria are given in Pmfessiunal Reference Shelf R5.4 on the CDROM. a rating of good (G), fair
G. E. P.Box, W.G.Hunter, and J. S. Hunter. Srnrisficsfor Erperimenrers: A n Inrmdiiction to Design, Dara Analysis, and Model Building (New 'fork: Wiley, 1978). j0V. W. Weekman. AlChE J., 20,833 (1974).
290
Collection and Anatysis of Rate Data
Chap. 5
5.7.2 Types of Reactors
The criteria in Table 52 is applied to each of the reactors shown in Figure 512 and are also discussed on the CDROM in Pmfessional Rderence ShelfR5.4. Reference Shelf
(a) In~egraIreactor
(h~ Stirred batch reactor.
( c ) Stirred contained sol~ds
reacror.
(dl Sold\ in a CSTR.
(el Stra~ghtthrough transpn reactor
(0 Recirculating transport reactor.
Figure 512 [From V. Weekman, AICiw J , 20. 833 ( 1974) with perml~sionof the AIChE. Copyright 0 1974 AIChE. All rights reserved.]
5.7.3 Summary of Reactor Ratings The ratings of the various reactors are summarized in Table 53. From this table one notes that the CSTR and recirculating transport reactor appear to be the best choices because they are satisfactory in every category except for construction. However, if the catalyst under study does not decay, the stirred batch and contained solids reactors appear to be the best choices. If the system is not limited by internal diffusion in the catalyst pellet, larger pellets could be used. and the stirredcontained solids is rhe best choice. Ef the catalyst is nondecaying and heat effects are negligible. the fixedbed (integral) reactor would be the top choice. owing to it< ease of construction and operation. #o\vever. in practice, usually more rhnn nrle reactor type is used in determining the reaction rate law parameters.
Chap. 5
291
Summary
Reac~orfipe Differenrial Fixed bed S t ~ batch d Surredcontained solids Continuousmmed tank Straightthrough transpan
SOI~IP~IR~ and FltiidSolid Analyszs Ixothennali~ Contacr
Drcnx~ng Caratur
PF
F 4
F
G
PF
F
F
G G G
G
P P P
FG
P
FG
FG
P F
FG
G
G FF
G P
FG
G E
Reciruular~ngIranspon
FG FG
Pulre
G
FG
Enrr of Consrnrction
C G
G F 4 PF FG P F G
%, g d : F. fmr: P, poor.
Closure. After reading this chapter, the reader should be able to analyze data to determine the rate law and rate law parameters using the graphical and numerical techniques as well as software packages. Nonlinear regression is the easiest method to analyze rateconcentration data to determine the parameters, but the other techniques such as graphicd differentiation help one get a feel for the disparities in the data, The reader should be able ta describe the care that needs to be taken in using nonlinear regression to ensure you do not arrive on a false minimum for d . Consequently, it is advisable to use more than one method to analyze the data. Finally, the reader should be abIe to carry mit a meaningful discussion on reactor selection to determine the reaction kine~icsalong with how lo efficiently plan experiments.
SUMMARY
a. Plat bCA /ill as a function nf r. h. Determine dC,/dt from t h ~ sphr. c. T~kerhe In of both hides of (551) to get
292
Ccl!ection and Analysis of Rate Qa!a
Cha
Plot In[dCAldt) versus In C., The slope will be the renclion order a . could use finitedifference formutns or software packages to eval (dC,/df) as a functlon of time and concentration.
2. Ftltegml method
a. Guess the reaction order and inrgrate the mole baEance equation. b. CalcuEnte the resulting function of concentntion for the data and plot i a function o f time. Tf the resulting plot is linear, you have prob: guessed the correct reaction order. c. If the plot i s not linear, guess another order and repeat the procedure. 3. Nortlinear regression: Search for the parameters of the rate law that will n irnize the sum of the squares of the difference between the measured rat1 reaction and the rate of rtaction calculnted from the pnrameter values cho For N experimental suns and K parameters to be determined, use PoIyma
"',. 1"
[r,(measured)  ri(calculatedEl?
=
NK
(S:
4. Method of initial sates
In this method of analysis of rate data, the slope of a plot of In(rAo) ve, In CA0will be the reaction order. 5. Modeling fhe diflerentinl reacror: The rate of reaction i s calculated from the equation
In calculating the reaction order, a ,
the concentration of A is evaluated either at the entrance conditions or mean value between C,, and C,,
.
Chap. 5
CDROM Maler~al
CDROM MATERIAL Learning Resources f ummsrv No:cr
I. s l l l ? l ~ ~ lWoOt ~~ L , . ~ 3. Interactive Comprrrer Modrrles A . Ecology
interactive
f$
\
Computer Modulcs
Solved Prcblems
Q F L~ 5%
L;v~ngExamu!e Problerv
B. Reactor L a b ( u ~ ~ v ~ ~ : r r a ~rwt) ~ u See r i n hReactor Lab Chapter 4 and P53,. 4. Solvrd Ptvblem.~ A. Example Differential Method of Analysis of PressureTime Data B. Example Integral Method of Analysis OF PreswreErne Data C. Example Oxygenating Blood Living Example Problems I. Ernmplr 53 U1.r (fReL~rrssintito Find !he Rore Lcrrc Prrrurnerers FAQ [Frequently Asked Questions]Tn UpdateslFPIQ icon section Professional Reference Shelf RS. 1 Lmsr Sqlraies Annlni.7 r,f the Lrr~euri:edRrrte Lntv The CDROM describes how the rate law
i s linearized In(r,)
= In k + a In C,$+ p In CH
and put in the form Y=llo+aYt+pX1
I
and used to solve for a,p, and k . The etching of a semiconductor. MnO,. is used as an example to illwtrate this technique.
R5.2 A Discussiun of Weigl~tedLeast Sqltnres For the case when the error in measurement is not constant. we must use a weighted least squa~es. R5.3 Erperrrnent~ilP l i l n n i ~ ~ ~ A. Why perform the experiment? B. Are you choosing the correct parameters? C. What is the n n g e of your experimental variables'?
294
Collection and Analysis of Rate Data
Chap. 5
D. Can you repeat the measurement? (Precision) E. Milk your data for all it's worth. F. We don't believe an experiment untiI it's proven by theory. G. Tell someone about your result. R5.4 Evaluation or Lubornrov Reacfors (see Table 53)
QUESTIONS AND PROBLEMS The subscript to each of the problem numbers indicates the lwel of difficulty: A, least difficult; D. most difficult. A=.
uome'hto''~
P51,
B = l l C = f D=++
(a) Compare Table 53 on laboratory reaaors with a similar table on page 269 of Bisio and &el (see Supplementary Reading, listing 1 ). What are the similarities and differences? (b) Which of the ICM5s for Chapters 4 and 5 was the most fun? (c) Choose a FAQ from Chapters 4 and 5 and say why it was the most helpful. (d) Listen to the audios on the CD and pick one and say why it could be eliminated. (el Create an onginaI problem based on Chapter 5 material. (f) Design an experiment for the undergraduate laboratory that demonstrates the principles of chemlcal reaction engineering and wifl cost less than $500 in purchased parrs to build. {Fmm I998 AIChE National Student Chapter Competition) Rule? are provided on the CDROM. (g) Plant a number of seeds in different pots (corn works well). The plant and soil of each pot will be subjected to different condit~ons.Measure the height o f the plant as a function of time and fertilizer concentration. Other variables rn~ghtincrude lighting, pH, and room temperature. (Great Grade School or H ~ g hSchool Science Project)
a
Creative Thinking
(h) Example 51. Discurs the differences for finding
[$I
dC*
shown in
Table E53. I by the three techniques. Example  1 .Construct a table and plot similar to Table E57.1 and Figure ES2.1. aqsuming a zeroorder and a firstorder reactinn. Looking at the plotr. can e~therof there ordcrs possibly explain the data? Example 53. Explain why the reFr&sron had to be carried out twice to find k' and k. (k) Example 54. Use regreszion to analyze the data in Table E54.1. What do you iind for the renct~onorder? (I) Example 55. R e p e r 5 the data to lil the rate lab i
[n:erac?rve
$
4%@ Comou'er Modules
P52,
Whel is the d~fferencern the correlation and sumsofsquares coinpared with those given In Example 55? Why was it necessary tn regre55 the data twice. once 10 ohtaln Table E55.3 and once to obtain Table E55.41 h d h e Interactive Computer Mtdule IICM) Fmrn the CDROM. Run the module and then mord >our pcrfommce number f i r the d u l e which ~ndlcatesyour mastenng of h e malerial. Your pmfesm h x the key 10 decode your p d o r m m numkr. ~
ICM E c o l o ~ yPerformance #
Chap. 5
Visit Reactor Lab
PS3*
d Links
P54,
295
Questions and ProbCems
GO to Professor Herz's Reactor Lab on the CDROM or on [he web at wuwreactorlab.nc~.Do (a) one quiz, or (bl two quizzes from D~vision 1. When you first enter a lab. you see all input values and can vary them. In a lab, click on the Quiz button in the navrgation bar to enter the quiz for that lab. In a quiz, you cannot see some of the input values: you need to find those with *'???" hiding the values. In the quiz, perform experiments and analyze your data in order to determine the unknown values. See the bottom of the Example Quiz page at nw~lc:rear!orlnb.nerfor equations that relate E and k. Click on the "???" next to an input and supply your value. Your answer will be accepted if is within +?O% of the correct \glue. Scoring is done with imaginary dollars to emphasize that you should design your experimental study rather than do random experiments. Each time: you enter a quiz. new unknown values are assigned. To reenter an unfinished quiz at the same stage you left, click the [i] info button in the Directory for instruct!ons. Turn in copies of your data. your analysis work. and the Budget Repon. When arterial blood enters a tissue capillary. lr exchanger oxygen and carbon dioxide with its environment. as shown in this diagram.
The kinetics of this dtoxygenation of hemoglobin in blood was studied with the aid of a: tubular reactor by Xakamura and Siaub [J. Phl~iol.,173. 1611.
Although this i s a reversible reaction, measurements west made In the inilia1 phases of the decamposit~onso that the reverse reaction could be neglected. Consider a system similar to the one used by Naknmura and S~aub:the solution enters a tubular reactor (0.158 r m In diameter) that has oxygen electrodes placed at 5cm interralr dou)n the tube. The solution floq rate into he reactor is 19.6 cdk. Electmde Povtion Percent Decompotition ot HhO,
P55,
I
2
3
4
5
h
7
O.OO
1 .Y3
3.82
5.68
7.38
9.25
t 1.00
(a) Using the method of differential analysis af rate data. determine the reaction order and the forward specific rcactjon rate constant d for [he deaxypenation of hemoglobin. (b) Repeat using regression. The liquidphase irreverqible reaction
296
Collection and Analysis of Rate Data
Char
is carried out in a CSTR. To learn the nte lam the volunietric flow rate, I [hence t = V l u , , ) 1s varied and the elfluent concentratloris of qpecies recorded as a function of the space time t . Pure A enters the reactor at a cc centratton of 2 rnolldrn'. Steadystate conditlonh exist when the rnensu
ments are recorded.
PS6,,
(a) Determine the reaction order and \pecific reaction n t e . (b) If you were to repeat this experiment to determine the kinetics, wt would you do differently? Would you run at a higher, lower, or the sar temperature? If you were to rake more data. where would you place t measurements (r.g., r )'? (c) 11 is believed that the technician may have mnde a dilution factoroferror in one of the concentmtiun measurements. What do you think? Hc do your answers compare using regression (Polymath or other softwnr with those obtained by graphical methods? Nore: All measurements were taken at steadystnte conditions. The redchon
was carried out In a constantvolume batch renetor where the following co centration measurements were recorded as a function of time.
P57,
(a) Use nonlinear lrnst squares (i.e.. regression) and one other method determine the reaction order a and the specific reaction rate. (b) If you were to take more data. where would you place the points? Wh: (c) If you were to repeat this experiment to determine the kinetics, wh would you do differentty? Would you run at a higher, tower. or the san temperature? Take different data points? Explain. (d) Lt is believed that the technician mnde a dilution emr in the concentr tion measured at 60 mio. What do you think? How do your answe compare using regression tPoIyrnath or other software) with tho: obtained by graphical methods? The Iiquidphase reactlon of methanol and triphenyl takes place in n batc reactor at 25°C
For an equal molar feed the following concentrationtime data was obtaine for methanol:
Chap. 5
297
Questions and Problems
The following concentration time data was carried out for an initial methanol conuentmtion 0.01 and an initial tnpheoyl of 0.1 :
(a1 Detcrniine the rate law and rate law parameters. (b) if you were to take more data points, what would be the reawnable settlngfi (e.g.. C,a, When the reaction order of the undesired product Is greater than that of the desired product,
Case 2:
A & u
Let h = a,a, , where h is a positive number: then
For the ratio Idr,, to be high, the concentration of A should be as low ar possible. This low concentration inay be accomplished by diluting the feed with For rx,>al uqe inerts and running the reactor at low concentration^ of A. A CSTR should be a CSTR and dilute used because the concentrations of reactants are maintained at a low level. A the feed stream. recycle reactor in which the product stream acts as a diluent could be used to maintain the entering concentralions of A at a low value. Because the activation energies of the two reactions in cases 1 and 2 are nor given. i t cannot he determined whether the reaction should be run at high For a nurnkr of liquidphase reac~iclnr,the prcqxr choice of a soI~enIcan enhance selectiv~ty.See. for example. Ir~il.Et~g.Cl~rn~.. rST(9). I h E 1970). In pa 3C + D Given: rIA= kbACACB
We need to relate the rates of formation of the other species in Reaction 1to the given rate law.
Similarly for Reaction 2. Reaction (2):
A
+ 2C
1'" 3E
the rate of formation of species E in reaction 2, r ? ~is,
and the rate of formation of C in reaction 2 is T~~
species
reaction n~~rnber
2
Given: rz4=kzACACc
Multiple Reactions
332
Char
6.4.21) Combine Individual Rate Laws to Find the Net Rate
We now substitute the rate laws for each species in each reaction to obtain t net rate of reaction for that species. Again, considering only Reactions 1 anc Summary Rates
Relative Rntea Rate Laws Net Rates
!I)
(2)
A + B >
'IA
3C+D
k~~ A+2C + 3E
the net rates of reaction for species A, B. C,
D.and E are
Now that we have expanded step two of our aIgorirhm, let's consider an exal ple with real reactions. Example 6 5 Stoichiomchy and Rate Laws for Multiple Reactions Consider the following set of reactions:
Rate Law,?
5N; + 6 H z 0 rIbO= k l N O ~ N H , ~ h A (E65
Reaction I: 4NH,+6NO
Reaction 2: 2 N 0
NZ+ O?
Reaction 3: N,+20, + 2 N 0 2
2 r ~ " i : =k 2 ~ 2 C ~ ~
'ro,
(E65
~ ~ ! c N(E65 ~G
'
Write the rate law for each species in each reaction and then write the net rates formation of NO. 0:. and N,
.
A. Net Rates of Reaction In writing the net rates of reaction, we set the rates to zero For those spec that are not in a given reaction. For example, H,O is not involved in Reactic 2 and 3; therefore. = 0 and r,, = 0. The net rates are
From tortusirnetry data (1 1121200h).
Sec. 6.4
333
Algorithm br Solution of Complex Reactions
8.
Relative ~ a t e sof Reaction The rate laws for Reactions 1. 2, and 3 are given in terms of species NO, Nz,and Os, respectively. Therefore, we need to relate the rates of reactions of other species i n a chosen reacrlon to the given rate laws.
Recalling Equation 1618), the corresponding rate laws are relafed by
Reaction 1 : The rate Saw wrt NO is
The relative rates are Multiple reaction stoichiometty
Then the rate af disappearance of NH3 IS Net rate NH,
THIO
= r ~ H L=o
vo=k1n:oc,vHJccLB
2 N 0 + N,
Reaction 2:
+ O2
2
rzNl is given (i.e., r,, = kZN2CNO); therefore, 2
+:NO
= Z~ZN:= ~ ~ I H > ~ ? I O 'to,
= r2u2
(E65.13)
334
Multiple Reactions
Chap. 6
We now combine the individual rates and the rate taws for each reaction to find the net rate of reaction. Next. let us examine the nP! rate of fonnations. The net rate of formation of NO is
Net rate NO
Next consider N: 3
Net rate
N?
(E65.21 ) Finally 0:
= r201+r.ro: = r 2 ~ : + 1 ' 3 0 L
{E65.32)
6.4.3 Stoichiomefry: Concentrations
In this step. if the reactions are liquidphase reactions. we can go directly the combine step. RecaIl for liquidphase reactions, u = uo and Liquid phase
cI , =Fi ufl
to
Sec. 6.5
Multiple Reactions In a PFWPBR
335
If the reactions are gasphase reactions, we proceed as follows.
For ideal gases recall Equation (342): Gas phaw
where
and
For isothermal systems (T = To}with no pressure drop (P = Po) Gas phase
and we can express the net rates of disappearance of each species (e.8.. species 1 and species 2) as a function of the molar flow rates ( F , ,... ,F; ):
where,fn represents the functional dependence on concentration of the net rate of formation such as that given in Equation (E65.21) for N2.
6.5 Multiple Reactions in a PFRlPBR We now insert rate laws written in terms of molar flow rates [e.g., Equation (34211 i a o the mole balances (Table 61 1. After performing this operation for each species, we arrive at a coupled set of firstorder ordinav differential
equations to be solved for the molar Row rates as a function of reactor volume (i.e., distance along the length of the reactor). In liquidphase reactions, incorporating and soIving for total molar flow rate is not necessary at each step along the solution pathway because there i s no volume change with reaction.
336
Multiple Reactions
Chap
Combining mole balance, rate laws, and stoichiometsy for species through species j in the gas phase and for isothermal operation with no prr sure drop gives us Coupled ODES
For constantpressure batch systems we would simply substitute (P for F; the preceding equations. For constantvolume batch systems we would use co centrations:
We see that we have j coupled ordinary differential equations that mt be solved simultaneously with either a numerical package or by writing ODE solver. In fact. this procedure has been developed to take advantage the vast number of computation techniques now available on personal compt ers (Polymath, MATLAB). Example 6 6 Combining Male Balances, Rate Caws, and Stoichiomern Jor itiulfiple Reactions
Coosider again the reaction in Example 65. Write the mole bafances on a PFR terms of molar Row rates for each species. Reaction 1: 4NH, + bNO
+5N2+6H20 r,,,
Reaction 2 : 2N0
N,+02
Reaction 3:
N,+20, + 2NOz
IN: = kLN2 ' 3 0 r
SoIurion For gasphase reactions. the concentmtion of species j is
For no pressure drop and isothermal operation.
= k l w ~ y H (j ~ ~6
(EB5.
6: (E65,
=k302~N2
Sec. 6.5
Multiple Reactions in a PFRIPBR
337
I n combining the mole balance, rate laws, and stoichiometry, we will use our results fmrn Example 65.The total molar flow rate of all the gases is
We now rewrite mole balances on each species in the total molar flow rate. Using the results of Example h5
( 1) Mole balance on N
O
(2) Mole balance on NH,: 1.5 ~ dY A F N H= r w , = r ~ m 323 =  r l v o =  ~ k ~ w o c ~ ~ 3 c ~ o '
Combined Mole balance
Rate law
Stoichiornetry
(3) Mole balance on H,O : 4 4 0 = 1
dy
1.5
r~,o=r~~20=r~no=k~uoC~~,C~o
(4) Mole balance on
N, :
Multiple Reactions
Do we need the Combine step when we use Polymath or another ODE solver? (Answer:
Chap. 6
( 5 ) Mole balance on 0,:
No) (See
Table EB6.1)
(6) Mole balance on NO, :
The entering molar flow rates. F,", along with the entertng rernperatuE, T o , and 3 pressure. P,. (C, = P/RTn = 0.2 moUdm ) , are spcified as are the specific reaction rates k,, [e.g.. k,,, = 0.43 (dm~/mol)I5/s, k,, = 2.7 dm3fmol s, etc.]. Consequently. Equations (E66.1) fhrough (E66.8) can be solved simulraneously with an ODE solver (e.g.. Polymath, MATLAB). I n fact, with almost all ODE sdvers. the combine step can be eliminated as the ODE solver will do the work. I n th~scase, the ODE solver algorithm is shown in Table E66.1. TARLE E661. ODE SOLVER ALGORITMMFOR MULTIPLE REACTIONS 1
Note: Polymath will do all the sub~tlmt i m g for you Thank you. Polymath!
339
Multiple Reactions in a PFRlPBR
Sec. 6.5
TABLEE66.1. ODE SOLVERA
m FOR M U L ~REACTIONS E (CO~~ED)
m
=
(2 1)
T H ~ O TI H ~ O
(27) F T = FNO+FWH~+FN~+FO,+EH~~+FN~~
fZ2)
NO^ = r 3 ~ 0 2
(28)
(23) C,, = c ,%
(29) k,, = 0.43
PT
(24)
FN: CN,= Cm
(25)
CNN, =C m d

cm= 0.2
(30) kZN2= 2.7
Fr
FNH Fr
(31) kTOl=S.R
For
(26) Coz= Cm
FT
Summarizing to this point, we show in TabIe 62 the equations for species j and reaction i that are to be combined when we have q reactions and n species.
Mole haIances:
ELc dY
(626)
Rates: Relative wtm:
2= 3 = k 3 h,
Q,
c,
d,
Rate laws:
r,,=$,f?(cl.c,.C.l
Net laws:
r, =
(616)
4
The basic equations
z 1=
Y,,
(6 1 7)
I
Stoichiometry: (gag phase)
C , = C ,F~ P   To FTPOi
F,=
r.
Fl
(342)
(620)
,I= I
(liquid phase)
F
C, = A L'o
(619)
340
I I I
Multiple Reaeions
Chap
Example 6 7 HydrodeaIkylafion of Mesitylena in a PFR
The production of mxylene by the hydrodealkylation of rnesitylene over a Houc Detral catalystt involves the folfowing reactions:
mXylene can also undergo hydrodealkylation to form toluene:
The second reaction is undesinbie, because mxyIene selIs for a higher price th toluene: (S 1.32/lb vs. $0.3011b)? Thus we see that there is a significant incentive maximize the prduction of mxylene. The hydrodealkylation of meqitylene is to be carried out isothermally 15U0°R and 35 atm in a packedbed reactor in which the feed is 66.7 mol% hyd gen and 33.3 mol% mmesitylene. The volumetric feed rate is 476 ft'/h and the reac volume (i.e., V = W/p,) is 238 ft3. The rate laws for reactions 1 and 2 are, respectively.
A significant economic incentive
where the subscripts are: M = mesitylene, X = rnxylene. T = toluene, Me methane, and N = hydrogen (Hz). At 15W0R.the specific reaction rates are Reaction I: k, = 55.20 (ft3Ab r n ~ l ) ~ . ~ / h Reaction 2: k, = 30.20 (ft3Ab r n 0 1 ) ~ ~ ~ ~ l r
The bulk density of' the catalyst has been included in the specific reaction rate (i k , = k; 1. Plot the concentrations of hydrogen. mesitylene, and xylene as a function space time. Calculate the space time where the production of xylene is a maximi
5 6
Fnd Eng. Chem. Process Des. Dm.,4, 92 (1965); 5, 146 ( 1466). November 2004 prices, from Chemical Marker Reporter (Schnett Publishing Cc 265. 23 (May 17, 2004). Also see www.chemweek.com/ and www.icisloccom
Sec. 6.5
M o l e balance on each and even, species
Mul!iple React~onsin a PFRiPBR
Reaction 1 :
M+H
Reaction 2:
X+A

X+Me
w T+Me
1. Mole balances:
dF" = rH dl.'
Hydrogen:
Xy lene: Toluene:
_
~Fu, dp'  r,,
Methane:
2. Rate laws and net rates: Given Reaction 1:
rl, = klC ~ C ,
Reaction 2:
r:, = ~ , C ; ~ C ,
Relative rates: (1)
rlH =riM= rlMe = rlx
(2)
r~ = r2nt =  r 7 ~=  r 2 ~
Net rates: r ~ = r , ~ =  k1:~zCL1 C, 11: rn = rl,+r2, = rI,rzT =RICH C,k2CH1 1C,
rx = r,,+r,, = rIHr2,= '%
~,c;c,~,C;'C,
= rlMe+r~Mc =  r l ~ + T z T = k l ~ l $ +k2cVcX ~M IR
rs = rzr = k2CHCX
3. Stoichiometry The volumetric flaw rate is
342
Mullipla Reactions
Chap. 6
Because there is no pressure drop P = Po fie., y = 1) the reaction is carried out isothermally. T = To,and there is no change in the total number of moles: consequent1y, V = Uo
Flow rates:
4. Combining and substituting in terms of the spacetime yields
If we know C., C,, and C,. then C,, and C, can be calculated from the reaction stoichionetry. Consequently, we need only to sohe the forlowing three equations:
The emergence of userfriend1y ODE solwrs favors this approach over frsctional conversion.
5. Parameter evaluation: At To= 1.500" R and Po = 35 atm, the total concentration is
Sec. 6.6
MuRiple Reactions in a CSTA
We now solve these three equations, (E67.22) to (867.24).simultaneously using Polymath. The program and output in graphical form are shown in Table E67.1and Figure E67.1. respectively. However, I hasten to p i n t our that these equations can be solved analytically and the solution was given in the first edition of this text.
s(h) Figure EX7.1 Concentration profiles: in a PFR.
Dlffermtial quatione as entered by the mer [ 11 d(Ch)ld(tau)= r!h+flh 12 1 d(CmVd(tau)= r l m I3 I d(Cx)ld(tauE = rlxtRx Living Ewamnle Problem
Exp!lci! aquatiom as entered by the user [ 11 kl = 55.2 [ 2 ] k2 = 30.2 11 1 r l m = .kl' Crn'(Chn.5) 1 4 1 r2t = k2*Cx'(ChA.5) ~ 5 1r l h c r l m [ h i E!mr:Rt 171
rlxsrlm
[ 5 1 RW=.PI 1'31 QhrI% 
6.6 Multiple Reactions in a CSTR For a CSTR, a coupled set of algebraic equations analogous to PFR differential equations must be solved.
344
Multiple Reactions
Ck
Rearranging, yields
Recall that r, in Equation (617) is a function ($ ) of the species con
trations
After writing a mole balance on each species in the reaction set. we substi for concentrations in the respective rate laws. If there is no volume chi with reaction, we use concentra~ions.C,, as variables. 1f the reactions are phase and there is volume change, we use molar flow rates, as variat The total molar flow rate for n species is
5,
For q reactions occurring in the gas phase, where N different species present, we have the following set of algebraic equations:
We can use an equation solver in Polymath or a similar program to sc Equations (631) through (633).
I
Example &s Hydrodealkylntion of Mesitylene in a CSTR For the multiple reactions and conditions described in Example 67, calculate conversion of hydrogen and mesitylene along with the exiting concentrations mesitylene, hydrogen, and xylene in a CSTR.
Solution As in Example 67, we assume u = 0,: for example,
FA= vCA= uOCA,etc.
Sec. 5.8
1.
Multiple Reactions in a CSTR
345
CSTR Mole Bala~ces: Hydrogen:
uoCH, uoCH= r,V
Mtsitylene:
~OCM  VOCM O = r~ v
Xylene:
v,Cx = rxY
Toluene:
vDCT= rTY ~ O C=M he ~V
Methane:
(E68.5)
2. Net Rates The rate laws and net rates of reaction for these reactions were given by Equations (E67.12) through (E6716)in Example 67. 3. Stoichiornetry: Same as in Example 67. Combining Equations (E67.12) through (E67.16)with Equations (E68.1) through (E68.3) and after dividing by uo, (T = Vlvo), yields
Next, we put these equations in a form such that they can be readily solved using Polymath.
fl c, )
= 0 = (k,C , ! ~ C ,  ~ ~ C ~)~tCC, ,
(E68. I 1)
The Polymath program and solution lo equations (E68.9), (E68.101, and (E68.11) are shown in Table E68.1. The ptobfem was sotved For different values of r and the results are plotted in Figure E68.1. For a space time of T = 0.5, the exiting concentrations are C, = 0.0089, CM = 0.0029. and Cx = 0.0033. The overall conversion is
OveralI conversion
346
1
Multiple Reactions
Chap. 6
NLES Solutian
I
Variable
Ch Cm
CX tau
Value 0.0089436 0.0029085 0.0031266 0.5
f (x)
1.99%10 7.834E12 1.839E10
I n i Guess 0.006 0.0033 0.005
NLES Report (safenewt)
Living "ample Problem
Nonlinear equations , f(Ch) = Ch,021+[55.2'Cm*ChA.5+30.2*Cx*ChA.5)'tau= 0 , f(Crn) = Cm.0105+(55.2*Cm*ChA.5)'tau= O ; '$ : f/Cx) = {55.2*Crn'ChA.530.2CCx"Chh.5)*tauCx =0 '
Explicit equations : tau = 0.5 '
To vary p , , , one can v q either v, for a fixed V or very V for a fixed vo.
Figure E68.1 Concentrations as a function of space time We resolve Equations (E68.6) through (E68,Il) for different values of r la
arrive at Figure E68.1. The moles of hydrogen consumed in reaction 1 are equal to the moles of mesitylene consumed. Therefore, the conversion of hydrogen in reaction 1 IS
The conversion of hydrogen in reaction 2 is just the overall conversion minus the conversion in reactlon I : that is.
The vleld of xylene from meutklene ha3ed on molar flow a r e s exiting the CSTR lor t =
0.5 i q
Sec. 6.7
DvamIl seleclivity, i,and yield. ?.
Membrans Reactors to Improve Sefecffviin Multiple Reactions
11
347
mole mesitylene reacted
The ovemlI selectivity of xylene relative to toluene is
RecalI that for a CSTR the ovenll selectivity and yield are identical with the instantaneous selecrivity and yield.
6.7 Membrane Reactors to Improve Selectivity in Multiple Reactions In addition to using memhrane reactors to remove a reaction product in order to shift the equilibrium toward completion, we can use membrane reactors to increase selectivity in multiple reactions. This increase can he achieved by injecting one of the reactants along the length of the reactor. It is particularly effective in partial oxidation of hydrocarbons, chlorination, ethoxylation, hydrogenation, nitration, and suIfunation reactions to name a few.'
t
+ CH,

+ CH,
W.S . bsher. D.C. Bornberger. and D. L. Huestis. Eljaluatfon o f S R / k No~,elReactor Process Per'pmlixTM(New York:AIChE).
348
Multiple Reactions
Ck
In the top two reactions, the desired product is the intermediate (e.g., C2H,1 However, because there is oxygen present, the reactants and intermediates c be completely oxidized to form undesired products C 0 2 and water. The des product in the bottom reaction is xylene. By keeping one of the reactants at low concentration, we can enhance selectivity. By feeding a reactant thmug the sides of a membrane reactor, we can keep its concentration low.
In the solved example problem on the CDROM, we have used a n brane reactor (MR) to continue the hydrodealkylation of mesitylene reactic Examples 67 and 68. In some ways, this CD example parallels the u: MRs for partial oxidation reactions. We will now do an example for a diffc reaction to ilIustrate the advantages of an MR for certain types of reactiot Solved Problems
I
Example 69 Membrane Reactor to Improve Selretivi~in ~WultipbReoctio The reactions
take place in the gas phase. The overall selectivities, S m . are to be compare a membrane reactor (MR)and n conventional PFR. First, we use the instantar selectivity to determine which species should be fed through the membrane
We see that to maximize dm we need to keep the concentration of A high an concentration of B low: therefore, we feed B through the membrane. The moIar rate of A entering the reactor is 4 moYs and that of B entering through the I bnne is 4 molls as shown in figure E64.1.For the PFR,B enters along with
The reactor volume is 50 dm3 and the entering total c_oncentmtion is 0.8 moVc Plot the molar flow rates and the overall selectivity, SWTJ, as a function of rr volume for both the MR and PFR.
Sec. 6.7
Membrane Reactors to Improvs Selectivity In Multiple Reactions
Solirfiun
Mole Balances for both the PFR and the .MR
MR
PFR
Species B: (2)
5 =r dV
2
(E69.214)
= rB+RB
(E69.L[b])
Species C: (3)
Species D:(4)
1
Net Rates md Rate Laws
Transpnrt Law The volumetric flow rate through the membrane i s given by Darcy's Law (see Chap
ter 4):
where K is the membrane pemeab~lity( d s kPa) and P, (kPa) and P, (kPa) are the she[[ side and tube side pressures, and A, i s the membrane surface area m?.The flow rate through the membrane can he controlled by pressure drop across the membrane (P, P,). Recall from Equation 1443) that "a" i s the membrane surface area per unit volume of reactor, +
A, = nVI
The total molar flow rate of B through the sides of the reactor is
I
The molar flow rate of B per unit volume of reactor is
(E69.10)
350
Multiple Reactions
Chap. 6
Stoichiometry: Isothermal (T = To)and neglect pressure drop down the length of the reactor (P= Po, y = 1.0) For borh the PFR and MR for no pressure drop down the length of the rector and isothermal operation. the concentrations are Here T = Toand 3P = 0
Combine The Polymath Program will combine the mole balance, net rates, and stoichiornetric equations to solve for the molar flow rate and selectivity profiles for both the conventional PFR and the MR and also the selectivity profile. A note of caution on calculating the overall selectivity
Fool Polymath!
We have to fool Polymath because at the entrance of the reactor F,:= 0. Polymath will Eook at Equation (E69.17) and will not run be~avseit will say you are d~vidine by zero. Therefore. we need to add a very small number to the denominator. say 0.0001; that is,
(E69. IS) Skach the trends or results you expect before working out the details of the problem.
Table E69.1shows the Polymath Program and report sheet. WLYCIATH Results Err~nylr6.9 Membmn* Reactor ID trnprout hIKltvlq lo Mvlriplr Rwctlonr o s  l s . ? ~~, c v 5 . 13:
W.r#nUa apvauoneaa snlerafcymm usmr Varrabo !L I d ( F n Y ~ V J . 11 v 1: I ~ ; F O Y ~ Y ) .f b + ~ b .F 11 I diFdWrnV1
.
lu
[ I I I:FYII(I,Y]  m
Fb
P?
Inizih: valve
r n ~ n i z a lvalue
.mxima: u e l u c
3
3
:c
1
i.3513875 D 3
4
!:r.al sc
valu.
L J5I1815
1 :5:~n~s 1 15138'5
: 9299793
:9099'89
Ewtlcd s ~ a l m n aM e n w a d Oy h a umr I :, Fl = PbcFb+FO+Fy :2' C 1 0 1 3 8 i l l Xll.2 t i l *21=3 : 5 : Cb.Cm~w! l a r Ca k ttP'F& 181

111;
w50
: 7 1 I b rn .ktm'cd*Z'C&k2m'Cd.CM
ns I C I Cd CWFW! N I P ! C u r CWFWFt l t l l rd kla'Cs*I'Cb I .!: N rn *2aUa*Cb*2 [11' F b 0 . A
1::'
R l r f m
[:&I Sau . F W F u + W l )
We can e a ~ i l ymodif!. the program, Table Eh9 1. for the PFR simply by setting RH ~(l113~ to zeru (Rg= 0 ) and the initla1 cundi~ionfor B EO be 4.0.
Sec. 6.8
I
Complex Reactions of Ammonia Oxidation
351
Figures E69.2(a) and E69.2(b) show the molar flow rate profiles for the conventional PFR and MR, respectively.
la) PFR
Ibj
MR
Figure E692 Mnlar Row rates.
:Iectivities = 5 dm"
at
mm = 14 pFMDK = 0.65
(a)
tb) MR
PFR Figure E69.3 Selecrivity.
Figures W9.3(a) and Eh9.Xb) show the selectj\iry for the PFR and MR. One notlces the enormous enhancement in selectivity the MR has over the PFR. Be sure to load this liling example problem and play with the reactions and reactors. With minor modifications. you can explore reactions analogous to paaial oxidations.
]
where oxygen (3)is fed lhmugh the membrane. See Problems P69 and P619.
6.8 Complex Reactions of AmmonEa Oxidation In the two preceding examples. there was no volume change with reaction: consequently, we could use concentration as our dependenr banable. We now consider a gasphase reaction with volume change taking place in a PFR. Under these conditions. we must use the molar Row rates as our dependenr variables.
352
Multiple Reactions
Ch:
Example 410 Calculating Concentrations as Functions of Position for NH, Oxidation in a PFR The following gasphase reactions take place simuttaneously on a metal oxide: ported cataiyst: 1.
4NH3+5OZ+
2. 2NH3+E.502
3.
2N0+Q2
4.
4NH,+6NO
4N0+6H20

WL+3H20
+2N02 5N2+6H,O
Writing these equations in terms of symbls yieids Reaction 1:
Reaction 2: Reaction 3: Reaction 4: withs
414
+ 5B +
+ 1.5B
2A
2C + B 4.4
+ 6C
4C
4

+ 6D
r,, = ~,,c,c; 0261
E + 3D
r2, = k,,C,C,
2F
rJB=
5E I6D
r4,
tE61
k 3 B ~ l ~ (E61 B
= R,,c,c?
k,, = 5.0 (m3/kmol)2/min
k,, = 2.0 m 3 h o l . m i n
k,, = 10.0 (m3/km0l)~/min
k,c = 5.0 (rn3/kmol)m/min
(E6I
Note: We have convened the specific reaction rates to a per unit volume bask multiplying h e K on a per mass of catalyst basis by the bulk density of packed bed (i.e.. k = k'p,).
Determine the concentrations as s function of position (i.e., volume) in a PI Additional information: Feed rate = 10 drn3/min: volume of reactor = 10 dm3;
C,, = , C = 1 .O moVdm3, C, = 2.0 moVdm3 Solution
Mole balance:
Species B:
3 = rs dV
Reaction orders and rate constants were estimated from perjscosity measurements a bulk cataiyst density of 1.2 kgldm3.
Complex Reactions of Ammonia Oxidation
Sec. 6.8 Sr~lutiunsto ihc3e ellu:ltion5 are ~ i i r l > i
msily obtained with an ODE solwr

Species E:
[~FE d  r~
Species P:
5 = ri dV
Total:
FT = FA*FB+FC+FO+FE+FF
Rate laws: See above for r,,, r,,, r,,, and r,,. Stoichiametry: A.
Relative rutes
Reaction I : Reaction 2:

 X~~  r~~
~ I A r~~
4  5
4
6
r 2 ~ r ? ~ r~~  r
1.5
2
1
2 ~
3
(E610.13)
Reaction 3: Reaction 4:
B.
 r~~  r 4 ~
r 4 ~ r 4 ~
4

6
5
6
Concenrmtiorls: For isothermal operation and no pressure drop, the of the molar flow rates by
concentrations are given in terms
Next substitute for the concentration of each species in the rate laws. Writing the rate law for species A in reaction I in terms of the rate of formation, r , , , and molar flow rates, FA and F, , we obtain
Thus
Similarly for the other reactions,
354
Multiple Reactions
Chap. 6
Next, we determine the ne! rate of reaction for each species by using the appropriate stoichiometric coefficients and then summing the rates of the individual reactions. Net rates of formation:
Species A:
L r, = rlA+rza+~r,,
(E610.20)
Species B:
r, = 1.2Sr,,+0.75r2,+r,,
(E610.2 1)
SpeciesC:
rc=r,Af2r3s+rnC
(E610.22)
Species D:
rD = 1.5rIA 1.5r2Ar4c
(E6 10.231
Species
V?A
E:
S p c i e s F:
sE
=
5
T  :r4c
r F = 2rJR
(E6 1 0.24) (E610.25)
Combining: Rather than combining the concentrations, rate laws, and mole balances to write everything in terms of the molar flow rate as we did in the past. it IS mare convenient here to write our computer solution (either Polymath or our own program) using equations far r , , , F A , and so on. Consequently, we shall write Equations (E610. I61 through (E6 10.19) and (E610.5) through (E610.11) as individual lines and let the computer combine them to obta~na solution. The corresponding Polymath program written for this problem is shown in Table E610.1 and a plot of the output is shown in Figure E6 10.1. One notes that there is a maximum rn the concentration of NO (i.e.. C) at approximately 1.5 dd.
*b. TJ,'
+*
rJ/P
.
However, there is one fly in the ointment here: It may not be possible to determine the rate lau s for each of the reactions. In this case, it may be necess a y to work with the minimum number of reactions and hope that a rate law can he found for each reaction. That is. you need to find he number of Iinearly independent reactions in your reaction set. In Example 610. there are four reactions given [(E6 10.5) through (€6 10.811. However, only three of these reactions are independent. as the fourth can be formed from a linear cornbination of the other three. Techniques for determining rhe number of independent reactions are given by AsisnY .


' R.Aris,

E l r r n ~ ~ ~ rCllr~r~icr~l np Rrmr?ur At~aly.ri.r(Upper Saddle R~ver.N.J.: Prentlce Hall, 1969).
Suing Example Problcm
Sec. 6.8
Complox Reactions of Ammonia Oxidation
355
POLYMATH Rcgultr Fslmplt 6.10 Cllculmllw Carw4~0tloarna hdhdklllw br W Wdrtk.in a PPR
I
Variable
i n i t i a l value
minlmbl ualrre
V
c
0
PA
10
ao
1.501099 ~.&OPO?~P
PC
0
0
FD
0
0
PE
I
?I
C
0
Pt
10
I0
vn
0
XlA
5
5
r2A
2 0
I 0 5619376 0.1148551
r4C r3U CA
0
1
rA
rB
rC
I 7.15 5
0.1188767 7.73
3.2008313 1 2182361
rD
10 1
CL
1
O.Og3074q
rP
0
E
amwmlar s a u d m us e I .I
final valua 10
Q W M Y I = rA
m w ~ h LW s
12 1 d(FBW(Y) = IB J 1 d(FCVfl(VJ= rC !d f d,FDVO[V) m
:
I 5 I dlFEYdlW= rE 1 6 1 d{FFpd(V) e rF
V
Iddl
Figure E610.1 Molar Row rates proliles.
356
Multiple Reactions
Ct
6.9 Sorting It All Out In Example 69 we were given the rate laws and asked to caIculilte the produc
Gonllnear leartsquxes
tribution. The inverse of the problem descrihed in Example 69 must frequent solved. Specifically, the rate Iaws often must be determined from the variati the product distribution generated by changing the feed concentrations. In ! instances this determination may not be possible without carrying out indr dent experiments on some of the reactions in the sequence. The best strate, use to sort oirt all of the rate law parameters will vary from reaction sequen reaction sequence. Consequently. the strategy developed for one system ma be the best approach for other mu1tiplereaction systems. One general n~l start an analysis by lookins for species produced in only one reaction: next. ! the species involved in only two reactions, then three, and so on. When some of the intermediate products are free radicals, i t may nl possible to perform independent experiments to determine the rate law pa eters. Consequently, we must deduce the rate law parameters from chang the distribution of reaction products with feed conditions. Under these c i a stances. the analysis turns into an optimization problem to estimate the bes ues of the parameters that will minimize the sums of the squares betwee calculated variables and measured variables. This process i s basically the as that described in Section 5.2.3, but more complex. owing to the larger nu of parameters to be determined. We begin by estimating the parameter v using some of the methods just discussed. Next, we use our estimates tc nonlinear regression technrques to determine the best estimates of our parat values from the data for a11 of the experiments.In Software packages are be ing available for an analysis such as this one.
6.10 The Fun Part
I'm not talking about fun you can have at an amusement park. but CRE Now that we have an understanding on how to solve for the exit concentra of multiple reactions in a CSTR and how to plot the species concentra down the length of a PFR or PBR, we can address one of the most i m p and fun areas of chemical reaction engineering. This area. discussed in Se 6.2, is learning how to maximize the desired product and minimize the t sired product. It is this area that can make or break a chemical process f cially. It is also an area that requires creativity in designing the re schemes and feed conditions that will maximize profits. Here you can mi: match reactors, feed streams, and side streams as well as vary the ratios of concentration in order to maximize or minimize rhe selectivity of a parti species. Problems of this type are what I call digitnlage problemsH bet we normally need to use ODE solvers along with critical and creative thir skills to find the best answer. A number of problems at the end of this ch loSee. for example. Y. Bard, Nonlinaor Pornmeter Estrrnation
(San Diego, Calif.: demic Press. 19741. "H. Scott Fogltr. Teaching Cririca! Thinking. Creative Thinking, and Problem Sc in the Digitnl Age, Phillips Lecture (Stillwater. Okla.: OSU Press, 1997).
Chap. 6
357
Sdmmary
will allow you to practice thew critical and crci~tivethinking skills. These problems offer opportunity to explore many different solution alternatives to enhance selectisity and have fun doing it. However. lo carry CRE to the next level and t t r have a lot more fun solving multiple reaction problems, we will have to be patient a little longer. The reason is that in this chapter we consider only isothermal multiple reactions, and i t is nonisothermal multiple reactions where things really get interesting. Consequently, we will have to wait to carry out schemes to maximize the desired product in nonisothermal multiple reaclctions until we study heat effects in Chapters 8 and 9. After studying these chapters, we will add a new dimenMultiple Rsactlonc sion to muItiple reactions, as we now have another variable, temperature. that ~ ~ heat t hutTecrs is we may or may not be able to use to affect selectivity and yield. In one particunique tu thls h ~ k ularly interesting problem (P826). h e will study is the production of styrene from ethylbenzene in which two side reactions, one endothermic, and one exothermic, must be taken into account. Here we may vary a whole slew of variables. such as entering temperature. diluent rate, and observe optima, in the production of styrene. However, we will have to delay gratification of the styrene study until we have mastered Chapter 8.
Closure. After completing this chapter the reader should be able to describe the different types of multiple reactions (series, parallel, compIex, and independent) and to select a reaction system that maximizes the selectivity. The reader should be able to write down and use the algorithm for solving CRE probIems with rnultipIe reactions. The reader should also be able to point out the major differences in the CRB algorithm for the multiple reactions from that for the single reactions, and then discuss why care must be taken when writing the rate law and stoichiornetric steps to account for the sate laws for each reaction, the relative rates, and the net rates of reaction. Finally, the readers should feel a sense of accomplishment by knowing they have now reached a IeveI they can solve realistic CRE problems with complex kinetics.
SUMMARY 1. For the cornpetin5 reactions
Reaction 1:
A+B
% D
a [ $ (S61) rD= A e  ~ d ~A ~ C B
Reaction 2:
A+B
A O
s, = A
e~~~ a22? o ~C (562)
the instantaneous selectivity parameter is defined as
A
Multiple Reactions
Chap. 6
a. If ED>EU,the selectivity parameter SDnrwill increase with increasing tempwature. b. If a,>a, and PZ> the reaction should be carried out at high concentrations of A and low concentrations of B to maintain the selectivity parameter S,., at a high value. Use a semibatch reactor with pure A initially or a tubular reactor in which B is fed at d i f f e ~ n tlocat~onsdown the reactor. Other cases discussed in the text are (a,> a,,P, > P?). (a2> a,, B2 P,x and (a, > P I > B2)
Pi,
.
The overall selectivity, based on molar flow rates leaving the reactor, for the reactions gwen by Equations (S61) and (S62) is
2. The overall yield is the ratio of the number of moles of a product at the end of a reaction to the number of moles of the key reactant that have been consumed:
I . The algorithm:
Mole balances: Following the Algorithm
PFR
CSTR Batch
Membrane ("i" diffuses in)
Liquidsemi batch
Rate laws and net rates: Laws
Chap. 6
CDROM Material
Net rates
Relative rates Stoichiornetry:
F. P To r 55,: c.= C A_n ~ , p T, r n ~ , T ,
Cur p k n ~e
CDROM MATERIAL Learning Resources 1. Szt~lrrnunNorrs
2. Web module.^ A. Cobra Bites
B. Oscillating Reactions
1 2 .
,. ....... ,
l*
(S6 14)
360
4L
.H
Multiole Reactions
Ck
3, {trrpr.rrctir.c2 Cot)+pfr1riA b l o t l e l r ~1ICil.l, The Grzi~tRace
&[+
@ C o ~ g u t e Modules r
4. Reir~torLr~h.her) L L , L I ) . I #R~~~, TI ~O U I TLII Y I~lrrC rg"tlre~rit~rrrui?ri~,e rmrrlprit~r ulmersr:es.
I I (tf ~
C / ~ q l t ~4 'fur r dercri~
5. Solveil Pmhlernr A. Blood Coagulation B. Hydrodealkylation of Mes~tylrnein s Membrane Reactor C. "411Yotl Wanted to Knnw About Making Malic Anhydrtde and More 6. ClnrIficnrinn: PFR wirh frril srrealnr nk~rr,qrkr lrrtpth ofthe rmcror: Living Example Problems Solved Problemr ' I . Exotnple 62 Tr~rrrnbotcrRrrrctint~r 1 Ertrrnplr 67Hydn)deolk!Enrion r$bfcsih/etl~ in n PFR 3. Exat~~ple 68 H!8(/r'drrlr/~n/k~i~r!on qf !fMrsr!vlr~~e in n CSTR 4. Erntnple 159 Me~~hrrlrfr Rmctor lo Itliprove Selecririn it1 Mirlrrple Renct 5. E.mnrplr 6 10 Crrlclrlirri~lgCorrcr~~rnrrio~ls as rr F~rtlr.!imnof Porltior~for, 0ridatron in a PFR 6. E.wmple weh  Cohm Bite Problern Living Example Problem 7. E . ~ m ! p l eweb  0sc.illrrtmg Reacricvts Pmblern 8. , ? h : n / ~ l eCD Solved Pmhlemh  H ~ d r ( ~ c I ~ a l X ~ufiW~>.rit?.lenc brio~~ In n M bmne Reclrrer 9. Erfintple CD Sohrrl Plnblrrns  Blood Congrr/ano?~ FAQ [Frequently Asked Questions] In Updates/FAQ icon section Professional Reference Shelf R6,1 Attuir~nhls.Rginn Annlysis IARA) The ARA allowr one to find the optimum reaction system for certain type rate laws. The example used in one of modified van de Vusse kinetics ~eferenceShelf

to find the optimum wrt
Lints
B using a combrnation of PFRs and CSTRs
Chap F
Ques:ions 2nd Prob!ens
Q U E S T I O N S AND
PROBLEMS
The rubscr~ptto each of the problem n u m k r s indicates the level of diffictilty: A. least difficult: D, most difficult.
I n each of the following questions and problems. rather than just dnwing a box around yuur answer. write 3 sentence or two describ~nghow you solved the problem. the nssumptrons you made. the reasonableness of your answer. what you learned. and any other Facts that you want to include.
Clornewc~r"~sb!ems
P61
2nd solve an ongins1 problem tn illustrate the principles of this chapter. Ser Probleni PJl for guidelines. (b) Write a question based on the material rn this chapter that require5 critical thinking. Explain why your question rtquirer critical thinkinc. [Hint: See Preface section 33 2.1 Ic) Choose a FAQ from Chapter 6 to be eliminated and say why it should be elirn~nnted. (d) Listen to the audios , on the CD and then pick one and say why it
(a) Make up
,,
P62,
was the most helpfu1~~t (e) Which exmple on the CDROM kclure Note$for Chapter 6 w a s Iea5t helpful? (a) Example 62. ! I ) What wot~ldhave k e n nhe selectivity SBqY nnd conversion, X, if the reaction had been cartied out in a singlc PFR with the same volume as the CSTR'?(2) How would your answers change if the pressure were ~ncreasedby a factor of 100? (hl Exsmpla 63. Make a tableAist for each reactor shown i n Figure 63 identifying all the types of reactions that would he best carried out in this reactor. For enamplz, Figure 63(d) Semibatch: (I)high?) exothermic reactions and ( 2 ) selectivity. for example, to maintain concentntion A high and B Iow. (3) ta control convcr>KM).For example. turnover Turnover number for the decomposition H 2 0 2 by the enzyme catalase is 40 x 1 Oh s'. That IS. 40 million molecules of H,02 are decomposed every second on a ,,,k,r.,,, singleenzyme molecule saturated with Hz02. The constant K M (mol/dm') is called the Michaelis constant and for simple systems is a measure of the
400 M~chaelis conrtant Kq
Reaction Mechan~sms,Pathways. Bioreactions, and Biorsactors
Cha
attraction of the enzyme for its substrate. so it's also called the nfinio c stant. The Michaelir constant. K,, for the decomposition of H20z discus earlier is I . I M while that for chymotrypsin is 0.1 M.9 If, in addition, we let V,,, represent the maximum rate of reaction f( given total enzyme concentration, Ymex
= kcat(Et)
the MichaelisMenten equazian takes the familiar form MichaelisMenten equation
For a given enzyme concentration, a sketch of the rate of disappearance of
substrate i s shown as a function of the substrate concentration in Figure 7
Figure 76 MichaelisMenten plot identifying the parameters V,,,
and
K,.
A piot of this type is sometimes called a MichaelisMenten plot. At low s strate concentration, KM (S),
and the reaction is apparent firs.[ order in the substrate concentration. At h substrate concentrations, ( S ) 9 K,,
and she reaction is apparent zero order
rs = Y,,
13. L. Nelson and M. M. Cox, Lehninger Principles of Biochemistry, 3d ed. (E York: Worth Publishers, 2000).
40t
Enzymahc Reaclron Fundamentals
Sec. 7.2
Consider the case when the substrate concentration is such that the reaction rate is equal to onehalf the maximum rate,
then
Solving Equation (727) for the Michaelis constant yields Interpretation of
Michael~sconftant
The Michaelis constant is equal to the substrate concentration at wwhich the rate of reaction is equal to onehalf the maximum rate. The parameters V,, and KM characterize the enzymatic reactions that are described by MichaelisMenten kinetics. V,,, is dependent on total enzyme concentration, whereas KM is not. Two enzymes may have the same values for kc, but have different reaction rates because of different values of K,. One way to compare the catalytic efficiencies of different enzymes is to compare the ratio kcJKM. When this ratio approaches 108to lo9 (dm3/mol/s) the reaction rate approaches becoming diffusionlimited. That is. it takes a long time for the enzyme and substrate to find each other, but once tbey do tbey react immediately. We will discuss diffusionlimited reactions in Chapters 11 and 12. ExarnpIs 73 Evaluation oJMichaelisMenten Parameters V,,, and K,, Determine the MichaelisMenten parameters V,, Urea + Urease
k,
[Urea. Urease]'
tl
and KM for the reaction
& H*O
2NH,
+ CO, + Urease
The rate of reaction is given as a function of urea concentration in this table. Cu,,(kmollrns)
10.2
Sohtion
Inverting Equation (726) gives us
0.02
0.01
0005
0.002
402
Reactton Mechanisms, Pathways, Bmreactlons, and Bioreacfors
Chap 7
A plot of the reciprocal reaction rate versus the reciprocal urea concentration should be a straight line with an intercept llV,,, and slope K M / V  . This type of plot is called a LineweaverBurk plor. The data in Table E73.1 are presented in Figure E73.1 in the form of a LineweaverBurk plot. The intercept is 0.75, so
TABLEE?3.1.
b~AND PROCESSEDDATA
LineweaverBurk plot
l'uw
e
Figure E73.1 (a)MicheeIisWenten plot; (b) LineweaverBurk plot.
Therefore, the maximum rate of reaction is
Vma,= 133 krnolirn"~ = 1.33 molldm3s From the slope. which i s 0.02 s, we can calculate the Michaelis constant, Khl: For enzymatic reactions, the two
key ratelaw paramelers are V,,, and K,.
Sec. 7.2 1
Enzymatic Rsactbn Fundamentals
403
Substituting K M and V,,,',,,into Equation (726)gives us

where C,,, has units of kmol/m3 and r, has units of kmol/mLs. Levine and Lacourse suggest that the total concentration of urease, (E,),corresponding to the value of V,, above i s approximately 5 g/dm3, In addition ro the LlneweaverBurk plot, one can also use a HanesWookf pIot or an EadieIfofstee plot. Here S r Cum,and r, s r,,,. Equation (726)
can be rearranged i n the following forms. For the EadieMofstee f m , ZadieHofstee plot
For the HanesWoolf form, we have
pz&J I

 r ~
'max
'man
I
For the EadieHofstee model we plot rs as a function of (r,/S) and for the HanesWoolf model. we plot [(S)/r,l as a function of (dl. The EadieHofqtee plot does no! bias the points at low substrated concentrfltions, while the HanesWoolf plot gives a more accurate evatuation of V,,,. In Table E73.2, we add two columns to Table E73.1 to generate these plots (C,,, r S).
PIotting the data in Table E73.2, we arrive at Figures E73.2 and E73.3.
S
Figure E73.2 Hane+Waolf plc~t.
Figure E73.3 EadieHofstee plot.
404
I
Reaction Mechanisms, Pathways, Bioteactions, and B~oreactors
Cha
Regression Equation (726) was used in the regression program of Polymath with the follow results for V,, and Ku.
I
Madel: rate = Vrnax*Cursal(KmCurea)
Vmax Km
u.busEs 1
rmess 1.2057502
0.02
0.0233322
Nonlinear regression settings Max # iterations = 64 Precision R"2 RA2adj Rrrisd
V, K,t

95% 0.0598303 0.003295
= 1.2 mol/dm7 s = 0.0233 moVdm3
= 0.9990611 = 0.9987481
=
0.0047604
variance = 1.888E04
The ProductEnzyme Complex In many reactions the enzyme and product complex (E P) is formed direc from the enzyme substrate complex (E S) according te the sequence
Applying the PSSH to both (E S) and
(E P), we obtain
BriggsHaldane Rate Law
rs =
vm, (C,  C,/Kc3
(72
Cs+K,, + K P ~ P
which is often referred to as the BdggsHaldane Equation (see Problem P71 and the application of the PSSH to enzyme kinetics often called t BsiggsHaldane approximatian. 7.2.4 Batch Reactor Cakulations for Enzyme Reactions
A male balance on urea in the batch reactor gives Mole balance
Because this reaction is liquid phase, the mole baIance can be put in the fc lowing form:
Sec. 7.2
405
Enzymatic Reaction Fundamentals
The rate law for urea decomposition is
r u m 
Rate law
'mil3Cure3
KU +
cures
(73 1 )
Substituting Equation (73 1 ) into Equation (730) and then rearranging and integrating, we get
Integrate
/ =  I nKY Vmax
 Cure2
Cureno + Curco~
Cure,
ym,,
We can write Equation (73 1) in terms of conversion as Time to achieve a conversion X i n a batch enzymatic reaction
The parameters K, and V,, can readily be determined from batch reactor data by using the integral method of analysis. Dividing both sides of Equation (732) by fKM/Vmaxand rearranzing yields
We see that K, and V,,, can be determined from the slope and intercept of a pIot of l l t In[l/(l  X)] versus Xlt. We could also express the MichaelisMenten equation in terms o f the substrate concentration S:
where So is the initial concentration of substrate. In cases similar to Equation (733) where there is no possibility of confusion, we shall not bother to encEose the substrate or other species in parentheses to represent concentration [i.e., Cs = (S) E S]. The carresponding plot in terms of substrate concentration is shown in Figure 78.
Reaction Mechanisms, Pathways, Biomactions, and B~oreactors
Figure 77 Evaluating V,
Chap. 7
and K,.
&le 74 Batch Enwmtic Reactors
Calculate the time ~Bededto convert 99% of the urea to ammonia and carbon dioxide in a 0.5dm' batch rcactor. The initial concentration of urea is 0.1 mol/dm3, and the urease concentration is 0.001 g/drn3. The reaction is to be carried wt isotheimally at the same temperamre at which the data in Table E73.2 were obtained. Solution
We can use Equation (732),
KM = 0.0266 rnolldm~,X = 0.99, and CuWm = 0.1 molldm3,V,,, was 1.33 molldrn3.s. However. for the conditions in the batch reactor, the enzyme concentration is only 0.001 g/dm7compared with 5 g in Example 73. Because V, = E;b, V,, for the second enzyme concentration is where
I
K M = 0.0266 m o l l d m ~ a n d X = 0.99 Substituting into Equation (732)
! 1
=460s+380s
= 840 s (14 minutes)
Sec. 7.2
Enzymatic React~onFundamentals
407
Effect of Temperature
vmx
T
The effect of temperature on enzymatic reactions is very camplex. ~f the enzyme structure would remain unchanged as the temperature: i s increased, the rate would probably follow the Arrhenius temperature dependence. However. as the temperature increases, the enzyme can unfoId and/or become denatured and lose its catalytic activiry. Consequently, as the temperature increases, the reaction rate, rs, increases up to a maximum with increasing temperature and then decreases as the temperature is increased further. The descending part of this curve is called temperature inactivation or thermal denaturi~ing.~~ Figure 79 shows an example of this optimum in enzyme activity."
Figure 78 Catalytic breakdown rate of HID2 depending on temperature. Counecy of 5. Aha. A. E.Humphrey. and N. F. Mills. B t n r h ~ m i c E ~ ln ~ i n e t r r t r Academic ~. Press 11473).
'OM.L.Shuler and F. Kargi, B i o p r o c ~ Engittecring .~~
Basic Conc~prs,2nd ed. (Upper Saddle River. N.J : Prentrce Hall. 20011). p, 77. " S . Aiba, A. E. Humphrey, and N. F. Mills, Biochemical engineer in^ (Nm York: Academic Press, 1473). p. 47.
408
React~onMechanisms, Pathways, Bloreact~ons,and 81oreactors
CI
Side note: Labonachip. Enzymecatalyzed polymerization of nucleoti is a key step in DNA identification. The microfluidic device shown in I ure SN7.I is used to identify DNA strands. It was developed by Frofe! Mark Bums's group at the University of Michigan.
SAMRE LO*L*IVO
Figure SW7.1
GEL
z:g?&
M & g w MlXlM
GEL
L ~ OELECTROP~RESIS M
I
Micmfluidi device to identify DNA. Courtesy of Science. 282. 484 (1998).
In order to identify the DNA, its concentration must be raised to a level t can be easily quantified. T h i s increase is typically accompIished by replic ing the DNA in the following manner. After a biological sample (e.g., pl; fied saliva, blood) is injected into the micro device, it is heated and 1 hydrogen bonds connecting the DNA strands are broken. After breaking
primer attaches to the DNA to form a DNA primer complex. DNA*. , enzyme @ then attaches.to this pair forming the DNA* enzyme complc DNA* E. Once this complex is formed a polymerization reacrion oca as nucleotides (dNTPsdATP, dGTP, d m , and d m  N ) attach to t primer one molecule at a time as shown in Figure SN7.3,. The enzyme int~ acts with the DNA strand to add the proper nucleotide in the proper ord The addition continues as the enzyme moves down the strand attaching x nucleotides until the other end of the DNA strand is reached. At this poi the enzyme drops off the strand and a dupIicate, doublestranded DNA mt ecule is formed. The reaction sequence is DNA
1
k

I

.
&
_
*
*.A
.
 . 
+

DNA'
*aHeaf
*

[I flpf;
+ Pdmm.r
DPlAStrend
.@

DNA' Enzyme
Complex
DNA'
DNA Strand Prlmer Complex & r
Sm. 7.3
Inhibition of Enzyme Reactions
409
The schematic in Figure SN7.2 can be written in terms of singlestep reactions where N is one of the four nucleotides. Complex Formation:
DNA + Primer +DNA* Nucleotide additiodporyrnerization
The process then continues much like a zipper as the enzyme moves Bong the sttand to add more nucleatides to extend the primer. The addition of the last nucleotide is
where i is the number of nucleotide moIecuIes on the originaI DNA minus the nucleotides in the primer. Once a complete doublestranded DNA is formed, the poZyrnerizatian stops, the enzyme drops off, and separation occurs.
Here ?DNA strands really r'epresents one doublestranded DNA helix. Once replicated in the device, the length of the DNA molecules can be analyzed by electrophoresis to indicate relevant genetic information.
7.3 Inhibition of Enzyme Reactions In addition to temperature and solution pH, another factor that greatly influences the rates of enzymecatalyzed reactions is the presence of an inhibitor. inhibitors are species that interact with enzymes and render the enzyme inef
fective to cataiyte its specific reaction. The most dramatic consequences of enzyme inhibition are found in living organisms where the inhibition of any particular enzyme involved in a priman, rnerabolic pathway will render the entire pathway inoperative, resulting in either serious damage or death of the organism. For example, the inhibition of a single enzyme. cyrochrorne oxidnse, by cyanide will cause the aerobic oxidation process to stop; death occurs in a very few minutes. There are aIso beneficial inhibitors such as the ones used in the treatment of leukemia and other neoplastic diseases. Aspirin inhibits the enzyme that catalyzes the synthesis of prostaglandin involved in the painproducing process. The three most common types of reversible inhibition occurring in enzymatic reactions are competitive, uncompefitive, and noncomperfrive. The enzyme moIecule is analopous to a heterogeneous cataIyzic surface in that it contains active sites. When competitive inhibition eccurs. the substrate and
410
Reacfion Mechanisms, Pathways, BIoreaCtlons, end Bioreactors
Chap. 7
inhibitor are usually similar molecules that compete for the same sire a n the enzyme. Urzcomperirke inhibition occurs when the inhibitor deactivates the enzymesubstrate complex, sometimes by attaching itself to both the substrate and enzyme molecules of the complex. Noncompe~itiveinhibition occurs with enzymes containing at least two different types of sites. The substrate attaches only to one type of die, and !he inhibitor attaches only to the other to render the enzyme inactive.
7.3.t Competitive Inhibition Competitive inhibition is of particular importance in pharmacokinetics (drug therapy). If a patient were adminisrered two or more drugs that react simuitaneously within the body with a common enzyme, cofactor, or active species, this interaction could lead to competitive inhibition in the formation of the respective metabolites and produce serious consequences. In competitive inhibition another substance, I, competes with the substrate for the enzyme molecules to form an inhibitorenzyme complex, as shown here. Reaction Steps
Competitive inhibition pathway E+S+E4S
+ I
u
KI
t.1
4E + P
Competitive Inhibition Pathway Active
(1) (2)
E+S E.S
(3) (4)
EaS I +E
(5)
Em1
"
"
>E*S > EE+S
,P + E t'
"
EE.I (inactive)
r E+I
5 GO.
0:30 Inactive
(a) Competitive inhibition. Courtesy of D.L. Xelson and M. M. Cox, Lehn~nger Prinriples of Bioch~rnistty~ 3rd ed. (New York: Wonh Publishers, 20(M),p. 265.
In addition to the three MichaelisMenten reaction steps, there are two additional steps as the inhibitor reversely ties up the enzyme as shown in reaction steps 4 and 5. T h e rate law for the fodiation of product is the same [cf. Equation (718A)I as it was before in the absence of inhibitor
S6c.7.3
Inhibition of Enzyme Reactions
41 1
Applying the PSSH, the net rate of reaction of the enzymesubstrare complex is
The net rate of reaction of inhibitorsubstrate complex is also zero The total enzyme concentration is the sum of the bound and unbound enzyme concentrations
Combining Equations (733, (736), and (737) and soIvjng for (E and substituting in Equation (734) and simplifying
S)
Rate law for competitive inhitrit~on
V,,, and K,, are the same as before when no inhibitor is present, that is, Vmax= k3E, and

KM = kz i,+ k3 k,
and the inhibition constant K, (rnol/dm3) js
By letting K; = 4 ( 1 + I/K,II we crm we that the effect of a competitive inhibition is to increase the "apparent" Michaelis consmt. KM.A consequence of the larger "apparent" Mjchaelis constant K M is that a Iarger substrate concentration is needed for the rate of substrate decomposition, rs, to reach half its maximum rate. Rearranging in order to generate a LineweaverBurk plot, I
(739)
From the LineweaverBurk plot (Figure 710'), we see that as the inhibitor ( I ) concentration is increased the slope increases (i.e,, the rate decreases) while the intercept remains fixed.
412
React~onMechanrsms, Pathways. Bioreacttons, and Bioreactors
CP
Increasing inhibitor Concentration ( I ) I
Figure 7Ill
Line~,rilverBurk plot for compztitive inhibition.
Side note: Methanol Poisoning. An interesting and important example competitive substrate inhibifion is the enzyme alcohol dehydrogenase (AC in the presence of ethanol and methanol. If a person ingests methanol, A1 will convert it to formaldehyde and then formaze, which causes bIindnc Consequently, the treament involves intravenously injecting ethanol (wh is metabolized at a slower rate than methanol) at a controlled rate to tie ADH to slow the metabolism of melhmoltofamaldehydetoformatt sot the kidneys have time ro filter out the methanol which is then excreted in urine. With this treatment, blindness is avoided. For more on the met. nollethmol competitive inhibition, see Problem W25,. 7.3.2 Wncompetitive In hibition
Here the inhibitor has no affinity for the enzyme itself and thus does not ( pete with the substrate for the enzyme; instead it ties up the enzymesubs complex by forming an inhibitorenzymesubstrare complex. (I E S) w is inactive. In uocompetitive inhibition, the inhibitor reversibly ties enzymesubstrate complex oftfter it has been formed. As with competitive inhibition. two additional reaction steps are addt the MichaeIisMenten kinetics for uncomvtitive inhibition as shown in r tion steps 4 and 5 .
Sec. 7.3
413
Inhibition ol Enzyme Reactions
Reaction Steps
Uncornpetilive Prithway
Uncompetitive lnh~bitionpathway E+S'E.SE+P
E.S.1
(1)
E + S L$ Ems
12)
E*S
(3)
ES
(4)
I+E*S IE*S
(5)
Active
"'aE+S ''>P+E "
ila
+ 1eE.S (inactive)
k,
1 0 ~ ~ s Inactive
Rate law for Starting with equation for rate of formation of product, Equation (734). uncomperitive and then applying the pseudosteadystate hypothesis to the intermediate ~nhibirion
(I E S), we arrive at the rate law for uncompetitive inhibition
Rearranging
I
I
The LineweaverBuck plot is shown in Figure 71 1 for different inhibitor concentrations. The slope (KMIV,,,) remains the same as the inhibition ( I ) concentration is increased, while the intercept (1 + (flIKI) increases.
Increasing Inhibitor Concentration (I)
L
1 S
Figure 711 LineweaverBurk plot for uncornpetitive inhibition.
474
Reaction Mechanisms, Pathways, Bioreactions, and Biomctors
Chap. 7
7.3.3 Noncompetitive Inhibition (Mixed Inhibition)? In noncompetitive inhibition, also called mixed inhibition, the substrate and inhibitor molecules react with different types of sites on the enzyme molecule. Whenever the inhibitor is attached to the enzyme it is inactive and cannot form products. Consequently, the deactivating complex (I E S) can be formed by two reversible reaction paths. 1. After a substrate molecule anaches to the enzyme molecule at the substrate site. the inhibitor molecu~eattaches to the enzyme at the inhibitor site. 2. After an inhibitor molecule attaches ta the enzyme molecule at the inhibitor site, the substrate molecule attaches to rhe enzyme at the substrate site. These paths, along with the formation of the product, P, are shown here. In noncompetitive inhibition, the enzyme can be tied up in its inactive form either before or nfer forminy the enzyme substrate complex as shown in steps 2, 3, and 4. Reaction Steps
Mixed inhibition
Noncompetitive Pathway Active
E.l+sE.svI
( 2 ) E + I 2 1 E (inactive) (3) 1 + E * s I * E * s (inactive) (4) S t I E Z 1 E S (inactive)
z
(5)
ES

os(p(~ A
G
P
I!
Inactive
Summery Notes
Again starting with the rate law for the rate of formation of product and then applying rhe PSSH to the complexes (I E) and (I E S) we arrive at: the rate law for the noncompetitive inhibition
Rale l a w for noncompetitive inhibition
The derivation of the rate law is given in the S u r n m a ~Notea on the web and CDROM. Equation (742) is in the form of the rate law that is given for nn enzymatic reacflon exhiQiting noncompetitive in hibition. Heavy metal ions such a$ pb2'. ~ g ' . and Hg+. as well as inhibitors that react with the enzyme to form chemical derivatives. are typical examples of noncompetitive inhihitors. ' In some text>(KMc S), then
and we see that the rate decreases as the substrate concentration inc Consequently, the rate of reaction gives through a maximum in the su concentration as shown in Figure 714. We also see there is an optimul strate concentration at which to operate. This maximum is found by takj derivative of Equation (744) wrt S, to obtain
Inhibition of Enzyme Reactions
Sec. 7.3
D
5A. Sub.irrate ~nhlbitron
4s
S
S ma
Figure 714 Substrate reaction rate as a function of substrate concentration for subqtrdte inhlb~tinn,
When substrate inhibition is possible. a semibatch reactor called a f e d batch is often used as a CSTR to maximize the reaction rate and conversion.
7.3.5 Multiple Enzyme and Substrate Systems In the preceding section, we discussed how the addition af a second substrate, I, to enzymecatalyzed reactions could deactivate the enzyme and greatly inhibit the reaction. In the present section, we look not only at systems in which the addition of a second substrate is necessary to activate the enzyme, but also at other multipleenzyme and mulriplesubstrate systems in which cyclic regeneration of the activated enzyme occurs. Cell growth on multiple substrates is given in the Sirrn~nopNotes.
Enzyme Regeneration. The first example considered is the oxidation of glucose (S,) with the aid of the enzyme glucose oxidase (represented as either G.O. o r [E,lj) to give 8gluconolactone (PI: Glucose
+ G.O.
(Glucose . G.O.)
; ((6glucona'lactone  G.O.H,)
In this reaction. the reduced form of glucose oxidase (G.0.H2), which will be represented by E,, cannot catalyze further reactions until it is oxidized back to E,. This oxidation is usually carried out by adding molecular oxygen to the system so that glucose oxidase, E,, is regenerated. Hydrogen peroxide is also produced in this oxidation regeneration step:
G.Q.H2 + O2
G.O.+ HIOz
Overall, the reaction is written Glucose + OZ
plucosc ox,acc
+
> HzOl 8Gluconolactone
418
Reaction Mechanisms, Patftways, Bioreactions, and Bioreactors
Chap. 7
In biochemistry texts, reactions of this type involving regeneration are
usuaIly written in the form
Derivation of the rate laws for this reaction sequence is given on the CDROM.
Enzyme Cofactors. In many enzymatic reactions, and in particular biologicaI reactions, a second substrate line.,species) must he introduced to activate the enzyme. This substrate. which is referred ra as a cofaaor or coenzyme even though it is not an enzyme as such, attaches to the enzyme and is most often either reduced or oxidized during the course of the reaction. The enzymecofactor complex is referred to as a holoengfme.The inactive form of the enzymecofactor cnmpIex far a specific reaction and reaction direction is called an apoenzyme. An example of the type of system in which a cofactor is used is the formation of ethanol from acetaldehyde in the presence of the enzyme alcohol dehydrogenase (ADH) and the cofactor nicotinamide adenine dinucleotide (NAD): alcohol dehydrogenase acetaldehyde ( S , ) NADH ( S , ) ethanol (P ,) Derivation of the rate Iaws for this reaction sequence is given in PRS 7.4. ~eferenceShelf
The growth o f biotechnolopy S 16 billion
7.4
Biareactors
A bioreacror is a reactor that sustains and supports life for cells and tissue cultures. VirtuaEly all cellular reactions necessary 10maintain life are mediated by enzymes as they catalyze various aspects of cell metabolism such as the transformation nf chemical energy and the construction. breakdown. and digestion of cellular components. Because enzymatic reactions are involved in the growth of microorganisms, we now proceed to study microbial growth and bioreactors. Not surprisingly. the Monod equation. which describes the growth law for a number of bacteria, is similar to the MichaelisMenten equation. Consequently, even though bioreactors are nol truly homogeneous because of the presence of living cells. we include them in this chapter as a logical progression from enzymatic reactions. The use of living cells to produce marketable chemical products is becoming increasingly important. The number of chemicals, agricultural products and food products produced by biosynthesis has risen dramatically. In 2003. companier in this seclor raised over S16 biIlion of new financing.12 Both
" CC & E hltlrI'r.January
II . 2MU. p. 7 .
microorganisms and mammalian cells are being used to produce a variety of products, such as insulin, most antibiotics, and polymers. It is expected that jn the future a number of organic chemicals currently derived from petroleum will be produced by living cells. The advantages of bioconversions are mild reaction conditions; high yields k g . , 100% conversion of glucose ta pluconic acid with Aspergillus niger); the fact that organisms contain several enzymes that can catalyze successive steps in a reaction and, most important,act as stereospecific catalysts. A common example of specificity in bioconversion production of a single desired isomer that when produced chemically yields a mixture of isomers is the conversion of cisproenylphosphonic acid to the. antibiotic () cis1,2epoxypmpylphosphonicacid. Bacteria can also be modified and turned into living chemical factories. For example, using recombinant DNA, Biotechnic International engineered a bacteria ro produce fertilizer by turning nitrogen into niuates.I3 In biosynthesi~,the cells, also referred to as the biorna.~,consume nutrients to grow and produce more cells and important products. Internally, a cell uses its nutrients to produce energy and more cells. This transformation of nutrients to energy and bioproducts is accomplished through a cell's use of a number of different enzymes in a series of reactions to produce metabolic products. These products can either remain in the cell (jntracellular) or be secreted from the cells (extracellular). In the fonner case the cells must be lysed (ruptured) and the product purified from the whole broth Ireaction mixture). A schematic of a cell is shown in Figure 715.
Cell Membrane Cylo~h9rn Carl Nlrctasr reglon
_.. .. ..
. 
..
Figure 715 (a) Schematic of cell (h) Photo of cell ditiding L rolr. Counesy of D. L. Nelson a ~ i dM M. Cox. Lehninger Pn'nclples of B i ~ c h e r n i s i3rd ~ ~ d. (New York: Worth Publishers. 2000)
The cell consists of a ce!l wall and an outer membrane that encloses cyroplasm containing a nuclear region and ribosomes, The cell wall protects the cell from external influences. The cell membrane provides for selective transport of materials into and our of the cell. Other substances can attach to the cell membrane to carry out important cell functions. The cytoplasm contains the ribosomes that contain ribonucleic acid (RNA). which are irnportanr in the synthesis af proteins. The nuclear region contains deoxyri bonucleic acid
420
Reaction Mechanisms, Pathways. Bioreactions, and BioreacEors
CI
(DNA) which provides the genetic information for the production of prc and other cellular substances and structures. I' The reactions in the cell all take place simultaneously and are clas! as either class (1) nutrient degradation (fueling reactions), class (11) synt of small molecules (amino acids), or class (111) synthesis of large molel (polymerization. e.g., RNA, DNA). A rough overview with only a fractk the reactions and metabolic pathways i s shown in Figure 716. A more det model is given in Figures 5.1 and 6.14 of Shuler and Kargi.I5 In the C1 reactions. Adenosine triphosphate (ATP) participates in the degradation o nutrients to form products to be used in the biosynthesis reactions (Class 1 small molecules (eg.. amino acids), which ate then polymerized to form 1 and DNA (Class 111). ATP also transfers the energy it releases when it lor phosphonate group to form adenosine diphosphate (ADPI
ATP + HzO+ ADT + P + H 2 0+ Energy Nutrient (e.g., Glucose)
Waste (CO:,water, etc.)
Cell
Figure 716
Examples of reactions occumng i n the cell
Cell Growth and Division
The cell growth and division typical of mammalian cetls is shown schen cally in Figure 717. The four phases of cell division are called GI. S,GZ. M, and are also described in Figure 717.
New


+
01 Phase
S Phase:
G2 Phase:
MP k e :
Cells ~ncrease~n slre RNA ana
DNA dwbles. RNA and pmte~n 5ynlhests
RNA and protein
Mlrosrs. Nuclear reg& dmdss.
occurs.
No ONA
prweln synthesis MYXlrB No DNA synthmus
synlhss& murs
M Pha:
New
~ywklnssrs CeHs c h v t ~ o n mcum to g w ma OEW cd14
synmsas
Flgute 717 Phases OF cell division.
IJM.L. Shufer and F. Kargi, Bioprocess Engineering Basic Concepts, 2nd ed. (U] Saddle River, N.I.: Prentlce Hall. 2002). ISM.L. Shuler and F. Kargi. Biopmcess Engineering Basic Concepts, 2nd ed. (U1 Saddle River, N.J.: Prentice Hall. 2002). pp. 135, 185.
In general, the growth of an aerahic organism follows the equation
[ ~ ~ l+ l[Carbon ~ l source
Cell mult~pl~cation
] [ Nitrogm ] [ Oxygen ] [ Phospl~an ] source sutlrce source +
+
[CO?] + [H?O]+ [Products]
+
+
Culture media
[
MOR cells
1(pH:
conditions temperature,etc.1
+ ,,
.
1 (749)
A more abbreviated form oSEquntion (749) generally used is
Substrate
C'11S
More cells + Product
(750)
The products in Equation (750)include CO?, water. proteins. and other species specific to the particular reaction. An excellent discussion of rhe stoichiornetry (atom and mole balances) of Equation (749) can be found in Shuler and KargiIh and in Bailey and Ollis." The substrate culture medium contains all the nutrients (carbon, nitrogen, etc.) along with other chemicals necessary for growth. Because, as we will soon see, the rate of this reaction is proportiom1 to the cell concentration, the reaction is autocatalytic. A rough schematic of a simple batch biochemical reactor and the growth of two types of micmrganisrns, cocci (i.e., spherical) bacteria and yeast, is shown in Figure 718,
Qum Bacteria
Paddle Dlade
Sparger Oxygen
Batch B10reac:or
Figure 718 Batch bioreactor.
I6M. L. S h u h and F,K q i , Binprocess Engineering Basic Concepts, 2nd ed. (Upper Saddle River. N.J.: Prentice Hall. 2001,). ''1. E. Bailey and D. R Ollis, Biochemicd Engineering, 2nd ed. (New York: McGrawHill. L987).
Reaction Mechanisms, Pathways, Bioreactions, end Bloreactofs
Chap. 7
7.4.1 Cell Growth
Stages of cell growth in a batch reactor ate shown schematically in Figures 719 and 720. Initially, a small number of cells is inoculated into (i.e., added to) the batch reactor containing the nutrients and the growth process begins as shown in Figure 719. In Figure 720, the number of living cells is shown as a function of time.
Time
Figure 719
a

G W h (11) Phase
Lag (1) Pham
t =O
Increase in cell concentration.
Time
Figure 720 P h a e s of bacteria cell
Lap phase
Statlonary (ill) Phase
prowth
I n phase I. called the lag phase, there is little increase in ceIl concentration. During the lag phase the cells are adjus~ingto their new environment, synthesizing enzymes, and getling ready to begln reproducing. During this time the cells c a q out such functions as synthesizing transporl proteins for moving the substrate into the cell, synthesizing enzymes for utilizing rhe new substrate, and beginning the work for replicating the cells' genetic material. The duration of the lag phase depends upon the growth medium from which
the inoculum was taken relative to the reaction medium in which it is placed. If the inoculum is similar to the medium of the batch reactor, the lag phase will be almost nonexistent. If. however, the inmulum were placed in a medium with a different nutrient or other contents, or if the inoculum culture were in the stationary or death phase, the cells would have to readjust their metabolic path to aIlow them to consume the nutrients in their new environment.IY Exponential growth Phase I1 is called the exponential growth phase owing to the fact that the Phax cell's growth rate is proportional to the cell concentration. In this phase the cells are dividing at the maximum rate because all of the enzyme's parhways for metabolizing the substrate are in place (as a resuIt of the lag phase) and the cells are able to use the nutrients most efficiently. Phase 111 is the stationary phase, during which the cells reach a minimum biological space where the lack of one or more nutrients limits cell growth. During the stationary phase, the net grrnvth rate i s zero as a result of the depletion of nutrients and essential metabolites. Many important fermentation prodAnlibiotics ucts, including most antibiotics, are produced in the stationary phase. For produced during example, penicillin produced commercially using the fungus Penjcilliurn the starionary phase c*hql.ogenurnis formed only after cell growth has ceased. Cell growth is also slowed by the buildup of organic acids and toxic materials generated during the growth phase. Death phase The final phase, Phase IV, i s the death phase where a decrease in live cell concentration occurs. This decline is a result of the toxic byproducts, harsh environments. andlor depletion of nutrient supply. 7.4.2 Rate Laws

W i l e many laws exist for the cell growth rate of new cells. that is,
Cells
+ Substrate
More cells t Product
the most commonly used expression is the Mrv~odequation for exponential growth:
r, =
where
PC,
175 I )
P.,= cell growth rate. g/dm3.s C, = cell concentration. g/dmi p = specific growth rate. s  '
The specific cell growth rate can be expressed as

IXB. Wolf and H. S. Fogler, "'Alteration of the Growth Rate and Lag Time of hl,conostor. i~~rsenreroiCIrs NRRLB523." Ein~rchnologj and Binerlgir~cerirrg. 72 (6). 603 (2001). B. Wolf and H. S. Fogler. 'Growth of LPucorrosrnr r~lrsrr~rcrniJ~r NRRLB523. in Alkaline Med~u~n." Biozechnolo~rnttd Bin~ngir~eerirr~. 89 I 1. 96 (7OM).
424
Reaction Mechanisms, Pathways. Bioreacfions, and Bioreectorc:
Ch,
wherc F , , = i~ maximum specific growth reaction rate, sI K, = the Monod constant. _e/drnt C, = subsrrate lie.. nutrient) concentration. g/dm3 Representative valves of p, and K , are 1.3 h' and 2.2 x 1P5 rnollc respectively, which are the parameter values for the E. culj growth on g h s Combinins Equations (751) and (752). we arrive at the Monod equation bacterial cell growth rate Monod equation
For a number of different bacteria. the constant K , is small. in which case rate law reduces to
I
rg
l i
~rn;,Xc'
(7The growth rate, r e , often depends on more than one nutrient concentrati however, the nutrient that is limiting is usually the one used in Equar
C.
rR =
(753). In many systems the product inhibits the rate of growth. A classic ext ple of this inhibition is in winemaking, where the fermentation of glucosc produce ethanol is inhibited by the product ethanol. These are a number of I ferent equations to account for inhibition: one such rate law takes the empiri form
where Ernplrical form of Monud equation for product inhibition
with
C;
=
product concentration at which all metabolism ceases. gldm3
n = empirical constant
For the glucosetoethanol fermentation. typical inhibition parameters are n = 0.5
'
and
Cp* = 93 gldm3
In addition to the Monod equation, two other equations are also commor used to describe the cell growth rate; they are the Tessier equation.
r, = k,,,
[
I  exp
(31 
C,
Sec. 7.4
Bioteactors
and the Moser equation,
where X and k are empirical constanIs determined by a best fit of the data. The Moser and Tessier growth laws are often used because they have been found to better fit qxperirnental data at the beginning or end of Fermentation. Other growth equations can be found in Dean." The cell death rate is a result of harsh environments, mixing shear forces, local depletion of nutrients and the presence of toxic substances. The rate law is
Doubling tlrncs
where C, is the concentration o f a wbstance toxic to the cell. The specific death rate constants kd and kt refer to the natural death and death due to a toxic substance, respectively. Representative vatues of k, range from 0.1 h' to less than 0.0005 hI. The value of k, depends on the nature of the toxin. Microbial growth rates are measured in terms of doubling times. Doubting time is the time required for a mass of an organism to double. Typical doubling times for bacteria range from 45 minutes to I hour bur can be as h s t as 15 minutes. Doubling times for simple eukaryotes. such as yeast, range from 1.5 to 2 hours but may I x as fast as 45 minutes.
Effect of Temperature. As with enzymes (cf. Figure 791, there i s an optimum in growth rate with temperature owing to the competition of increased rates with increasing temperature and dennturizing the enzyme at high temperatures. An empirical Iaw that describes this functionality is given in Aiba et al." and is of the form
u T
r
where is the fraction of rhe maximum growth nte, T, is the temperature at which the maximum growth occurs. and p(T,j the growth at this temperature. For the rate of oxygen uptake of Rhicnbium rrifollic, the equation takes the form
The maximum growth occurs az 310K. I9A. R. C.Dean, Growth, Frmction, and Regtifation in Barrerial Cells (London: Oxford University Press, 1964). 20s.Aibn, A. E. Humphrey, and N. F. Millis. Biochemical Engineering (New York:Academic Press, 1973), p. 407.
426
R&on
Mechanisms, Pathways, Bioreactions, and Bioreactors
Chap. 7
The stoichiometry for celI growth is very complex and varies with microorganism/nuhent system and environmental conditions such as pH, temperature, and &ox potential. This complexity is especially true when more than one nutrient contributes to cell growth, as is usually the case. We shall focus our discussion on a simplified version for cell growth, one that is limited by only one nutrient in the medium. In general. we have
Cells + Substrate
+More cells + Product
In order to relate the substrate consumed, new celIs formed, and product generated, we introduce the yield coefficients. The yield coefficient for cells and substrate is YClB
Mass of new cells formed Mass of substrate consumed
=
ACc Acs
with
A representative value of Y , , might he 0.4 (gig). See Chapter 3, Problem
P314Bwhere the value of Y,, was calculated. Product fomatian can take place during different phases of the cell growth cycle. When product formation only occurs during the exponential growth phase, the rate of product formation is r, = y,.';
Growth arwciated
product formation
= yp,',,,pcc=
I.;,,, ~mxccc~ 4 4 C,
(763)
where
'' c
Mass of product formed =
 Mass o f new cells formed
5 AC,
v, that is, (qp= Y,, p) i s often called the specific rate of product formation. y,. (mass productJvoIume/tirne). When the product i s formed during the stationary phase where no cell growth occurs. we can relate the rate of product formation to ssbstrate conwnption by The product of Ypk and
Nongrowth
rf, = y,,,, (rr)
(765)
a r s m a t e d product
formation
The substrate in this case i s usually a secondary nutrient, which we discuss in more detail later.

The stoichiomemc yidd coefficient that relates the amount of product formed per mass of substrate consumed is
 Mass of product formed = S'f  Mass of substrate consumed
AC AC,
(766)
In addition to consuming substrate to produce new cells, part of the substrate must be used just to maintain a cell's daily activities. The corresponding maintenance utilization term is
m =
Cell maintenance
Mass of substrate consumed for maintenance Mass of cells Time
u
A typical value i s
m = 0.05 g != 0.05 hI g dry weight h
The rate of substrate consumption for maintenance whether ar not the cells are growing is
Neglecting cell maintenance
When maintenance can be neglected. we can relate the concentration of cells formed to the amount of substrate consumed by the equation
(768)
This equation can be used for both batch and continuous flow reactors. If it is possible to sort our the substrate (S)that is consumed in the presence of cells to form new cells (0from the substrate rhat is consumed to form product (P),that is,
the yield coefficients can be, written as y:j5 =
yp
Mass of substrate consumed to form new cells Mass of new cells formed
(7698)
Mass of substrate consumed to form product Mass of product formed
(769Bl
=
428
React~onMechanisms. Pathways. Bioreactions, and B~oreactors
Ch;
Substrate Utilization. We now come to the task of relating the rate of nl en1 consumption, r,. to the rates of cell growth, product generation, and maintenance. In general. we can write Substrate accounting
substrate
In a number of cases extra attention must be paid to the substrate balance product is produced during the growth phase, i t may not be possible to sepa out the amount of substrate consumed for cell growth from that consume( produce the product. Under these circumstances all the substrate consume( lumped into the stoichiornctric coefficient. Y,,, and the rare of substrate dis pearance is
The corresponding rate of product formation is Growthas~miated
(7
product h m a t ~ o n in the growth phafe
Because there is no growth during the stationaq phase. it is clear that Eq tion (770) cannot be used to account for substrate consumption, nor can the I of product fonnation be related to the growth rate [e.g.. Equation (763)l. M antibiotics, such ar penicillin, are produced in the stationan, phase. In this phi the nutrient required for growth becomes virtually exhausted. and a different nl ent, called the secondary nutrient. is used for cell maintenance and to produce desired product. Usually, the rate law for product formation during the station phase is similar in fom to the Monod equation, that is. Nongrowthassociated product
formation in the srationary
phaqe
where
k,, = specific rate constant with respect to product. (dm3/g s) C,, = concentration o f the secondary nutrient, gldrn3 C, = cell concentration, g/dm3 (g G gdw = gram dry weieht) K, = Nonod constant, g/dm3 rp = Y p t s n (  ~ r n ) (gldm3 s)
The net rate of secondary nutrient consumption during the stationary phase In the stationary phase, the: concentration of live cells
is constant.
Sec. 7.4
429
Bioreactors
Because the desired product can be produced when them is no cell growth, it i\ always best to relare the product concentration to the change in secondary nutrient concentration. For a batch system the concentration of product, C,,. formed after a time r in the stahnary phase can be related to the substrate concentration, C,. at that time. Wrgtcct~cell maintenance
We have considered two limiting situations for relating substrate consunlp tion to cell growth and product formation; product formation only during the growth phase and product formation only during the stationary phase. An example where neither of these situations applies is fermentation using Iactobacillus, where lactic acid is produced during both the logarithmic growth and stationary phase. The specific rate of product formation is often given in terms of the LuedekingPiret equation, which has two parameters cc (growth) and 0 (nonLuedekingPiret equation for the mte of product forrnatiun
growth) ql, = WB+P
(774)
with
The assumption here in using the Pparameter is that the secondary nutrient is In excess.
I
Example 75 Esrimate rke Yield Coeficicnts The following data was determined in a batch reactor for the yeast Sncchamm~ces cerevrsine Tmr E E75 1.
I
Glucose
Time.
0 t
(hrl 1
2 3
I
RAW DATA
C'11'. More cells + Ethanol
Cells.
Glucose.
Cc @!dm1) 1 i.33 1 87 2 55
Cs (gldm31 2Sfl
245 238.7 229.8
Ethanol, Cp (g/dm3) 0 2.14
5.03 8.96
.,
Determine YPIr. Y Y,,. Y,,, Y,,,,,b and Ks.Assume no lag and neglect rnaintenance at the stan of the growth where there are just a few cells,
(a)
Calculate the substrate and cell ~ i e Mcoeficients. Y,,,and Y,,, Between t = 0 and r = I h
430
Reaction Mechanlsms, Pathways, Bloreactlons, and Bioreactors
Chap. 7
Between t = 2 and t = 3 h
Taking an average
v,,
= 13.3 glg
We could alsa have used Polymath regression to obtain
@)
Similarly for the substrate and product yield rmfici~nfs
1 Up,,= 
Y#,
(c)
1 = 0.459 glg 2.12g!g
The producu'ceI1 ~'ieldcoqflcicnt is
y C'P
1 ==
YP,,
5.78 glg
 0.173 g/gl
We now need to deternine the rare law parameters h,and KT in the M o n d equation
For a batch system
How to regress the Monad equation for p,,, and K,
To find the rate law pararntkxs ha, and K,. we first apply the differenrial formulas in Chapter 5 to columns 1 and 2 o f Table E75.1 to find r,. Because C, >> K, initially. i t is best to repress the data using the HanesWoolf form of the Monod equation
Using Ptoymarh's nonIinear regression and more data points, we find p, hI and K, = 1.71dm1.
= 0.33
7.4.4 Mass Balances
There are two ways that we could account for the growth of microorganisms. One i s to account for the number of living cells, and the other is to account for the mass of ~e living ceIls. We shall use the latter. A mass balance on the microorganism in a CSTR (cbemostat) (shown in Figure 72 1) of constant volume is Cell Balance
Rate of
Rate of
c,
(775)
vdCc dt
Substrare Balance
=
u0C&

vC,
+
(rgrd)v
The corresponding substrate balance is Rate of
Rate of
Rate of
Rate of
(776)
In most systems the entering microorganism concentration C,, is zero.
Batch Operation For a batch system v = v, = 0 and the mass balances are as follows: Cell
vddtc ~= r,v
The mass halances
rdv
Dividing by the reactor volume V gives
Substrate The rate of disappearance of substrate, r,, results from substrate used for cell growth and substrate used for cell maintenance,
432
Reaction Mechanisms, Pathways. Bioreaciions, and Bioreacton
Ck
Dividing by Y yields the substrate balance for the growth phase
For cells in the stationary phase, where there i s no growth, cell rnainten and product formation are the only reactions to cansume the substrate. U these conditions the substrate balance, Equation (776), reduces to Stationay phase
Typically, r, will have the same form of the rate law as r, [e.g.. Quz (77 I)]. Of course, Equation (779) only applies for substrate concentrat greater than zero.
Product The rate of product formation. r,. can be related to the rate of substrate E sumption through the following balance: Batch stationary growth phase
During the growth phase we couId also relate the rate of formation of prod r,,, to the cell growth rate, r,. The coupled firstorder ordinary differer equations above can be solved by a variety of numerical techniques. Example 7 4
Bacteria Growth in a Batch Reactor
Glucosetoethanol fermentation is to be carried out in a batch reactor using organism such as Saccknrulrn~cescerertisrae. Plot the concentr~lionsof cells, : strate. and product and growth rates as functions of time. The initial cell concen tion is 1.0 g/dm3, and the substmte (glucose) concentration is 250 g/dm3. Additional doto [partial source: R. Miller and M. Melick, Chem. Eng., Feb. p. 1 13 (t983)l:
C,' = 93 gldrn3 n = 0.52
Y,,,= 0.08g l g Y,, = 0.45 g l g lest.)
I
Sulution
1. Mass balances:
Cells: The algorithm
Substrate: Pmduct:
Y
dC 2
= Y,,,(r,V
2. Rate laws:
'p
= Yp/c'g
4. Combining gives
Cells Substrate hoduc t
These equations were soIved on an ODE equation solver (see Table E76.1). The results an shown in Rgun E76.1 for the parameter valuer given in the pmb
lem statement.
POLYMATH Results Example76 Rsetrrln Emwb la s Bowb W
Living Ewample Problem
r #I&ZLW. Rn3,luz
434
I
Reaction Mechanisms, Pathways, Bioreactions, and Bioreactors
Chap. 7
Figure E76.1 Concentrations and rates as a func~ionof time.
The substrate concentration C, can never be less than zero. However, we note that when the subsrrate is completely consumed, the first term on the righthand side of Equation (E76.8) (and line 3 of the Polymath program) will be zero bklt the second term for maintenance, mC,. will not. Consequently, if the integration is carried further in time, the integration program will predict a negative value of C,! This inconsistency can be addressed in a number of ways such as including an ifstatement in the Polymath program (e.p.. if C, i s less than or equal to zero, then m = 0).
7.4.5 Chemostats
Chemostats are essentially CSTRs that contain microorganisms. A typical chemostat is shown in Figure 721. along with the associated monitoring equipment and pH controller. One of the most important features of the chemostat is !hat ir allows the operator to control the cell growth rate. This control of the growth rate iq achieved by adjusting the volumetric feed rate (dilution raw).
SZerlle Hedlurn Reservoir
..
Fermentor
Figvre f 21 Chemostat system.
7.4.6 Design Equations
CSTR
In this section we return to mass equations on the cells [Equation 1775)] and substrate [Equation 1776)] and consider the case where the volumetric Row rates in and out are the same and that no live (ie., viable) cells enter the chemostat. We next define a parameter common to bioreactors called the dilution rate, D. The dilution rate is
and i s simply the reciprocal of the space time T. Dividing Equations (775) and (7765by V and using the definition of the dilution rate, we have Accurnu1a;tion = In  Out t Generation CSTR mass balances
Cell:
Substrate:
dr
5 = DC,  DC, + r. dl
Using the Monod equation, the growth rate is determined to be
For steadystate operation we have
DC, = r,

r,
and
D(Cm  C,} = r,T
(782)
436
Reaction Mechanisms, Pathways, Bioreactions, and Bioreactors
Ch
We now neglect the death rate, Q , and combine Equations (751) (7835for steadystate operation to obtain the mass flow rate of cells out 01 system.
F,. F, = C,u, = r,V = @C,V
Afrer we divide by C,V, Dilution rate
An inspection of Equation (786) reveals that the specific growth rate of can be controlled by the operator by controlting the dilution rate Using Equation 1752] to substitute for in terms of the substrate concen tion and then solving for the steadystate substrate concentration yields
How tocontrol cells cell growth
Assuming that a single nutrient is limiting, cell growth is the only process c tributing to substrate utilization, and that cell maintenance can be negtec the stoichiometry is
r x = r8 Yslc
(7
Cr = Y c , s ( C d  Crl
(7 
Substituting for C, using Equation (787) and rearranging, we obrain
To learn the effect of increasing the dilution rate, we combine Equations (71 and (754) and set r,, = 0 to get
We see that if D > p, then dC,ldt will be negative, and the celE concentrati wil! continue to decrease until we reach a point where all cells will be washed o
The dilution rate at which washout wilI occur is obtained from Equati (789) by setting C, = 0. Row rate at which
washout mcun
Sec. 7.4
437
Bioreactors
We next want to determine the other extsenie for the dilution rate, which is the rate of maximum cell production. The cell production rate per unit volume of reactor is the mass flow rate of cells out of the reactor (i.e., h,. = C,v,) divided by the volume I/, or
Using Equation (789) 16 substitute for C, yields
Figure 722 shows production rate, cell concentration, and substrate concentration as functions of dilution rate. We observe a maximum in the production raze, and this maximum can be found by differentiating the production rate. Equation (792). with respect ta the dilution mre D:
Maximum rate of cell production CDC,)
Dmoxprod
0
Figure 722 Cell concentntion and production rate as a function of drlution rate.
Then Maximum rate of cell production
The organism Streptornyces aureofaciena was studied in a 10 dm3chemostat using sucrose as a substrate. The cell concentration, C, (rnglmlj, the substrate concentration, C, (mglml), and the production rate. L)C, (mglrnllh), were
438
Reaction Mechanisms, Pathways, Bioreactions, and Bioreacton
Chap. 7
measured at steady state for different diIutjon rates. The data are shown in Figure 723.21
1Symbol [
~ N D (D W va
Figure 723 Continuous culture of Strepromvces aureoficiens i n chemostats. (Note: X = C,) Coune~yof S. diba. A. E. Humphrey, and N. F. Millis. Biochemical Engineering. 2nd Ed. (New 'iork: Academic Press, 1973).
Note that the data follow the same trends as those discussed in Figure 722. 7.4.8 OxygenLimited Growth
Reference Shelf
Oxygen is necessary for all aerobic growth (by definition). Maintaining the appropriate concentration of dissolved oxygen in the bioseactor is impartant far efficient operation of a bioreactor. For oxygenlimited systems. it is neressary to design a bioreactor to maximize the o x y p n transfer between the injected air bubbles and the cells. ~ y p i c a l l ~a ,bioreactor contains a gas sparger. heat transfer surfaces, and an impeller. A chemmtat has a similar configuration with the addition of inlet and outlet streams similar to that shown in Figure 718. 7 h e
oxygen transfer rate (Om) is related to the cell concentration by OTR
=
Q,,C,
?'B. Sikyta, J. Slezak, and M. HernId, dppl. Microbial., 9, 233 (1961).
(795)
Sec. 7.5
Physiologically Based Pharmacohinetic {PBPK) Models
439
where Q , . is the microbial respiration rate or specific oxygen uptake rate and usually follow~s MichaelisMenten kinetics (Monod growth, e.g., QO2= YO?,, rg). (See Problem P713s.)The CDROM discusses the transport steps from the bulk liquid to and within the microorganism. A series of mass transfer correlations are also given.
Scaleup for the growth of microorganisms is usually based on maintaining a constant dissolved oxygen concentration in the liquid (broth), independent of reactor size. Guidelines for scaiing from a pilotplant biareactor to a commercial plant reactor are given on the CDROM. One key to a scaleup is to have the speed of the end (tip) of the impeller q u a 1 to the velocity in both the laboratory pilot reactor and rhe fullscak plant reactor. If the impeller speed is too rapid, it can lyse the bacteria; if the speed is too slow, the reactor contems will not be well mixed. Typical tip speeds range from 5 to 7 mts.
7.5 Physiologically Based Pharmacokinetic (PBPK)Models We now apply the material we have k e n discussing on enzyme kinetics to modeling reactions in living systems. Physiologically based pharmacokinetic models are used to predict the distribution and concentrationtime trajectories of medications, toxins, poisons, alcohol. and drugs in the body. The approach is to model the body components (e.g.. liver, muscle) as compartments consisting of PFRs and CSTRs connected to one another with inflow and outflow to each organ compartment as shown in Figure 724. Blood
F m Lung Richly Pertused
Slow4y Perfused
Liver
~jguie 72d
(a) camp an men^ model ul human hody. ( h ~ Generic srructurc of of Chew. E ~ y q Pnl,cr.i,.\r. . IM)(15)38 (June 2W1.
PRPK mode\\. Courtcqq
440
React~onMechanisms, Pathways, Bcoreactions, and Bioreac?ors
Ch
Associated with each organ is a certain tissue water volume, TWV, w we will designate as the organ compartment. The TWV for the different or4 along with the blood flow into and out of the different organ cornpartm (called the perfusion rate) can be found in the literature. In the models
cussed here, the organ compartments will be modeled as unsteady wellm C S T h with the exception of the liver, which will be modeled as an unstt PFR. We will apply the chemical reaction engineering algorithm (mole ance, rate law, stoichiomerry) to the unsteady operation of each compann Some compartments with similar fluid residence times are modeled to col of several body parts (skin, lungs, etc.) lumped into one compartment, suc the central compartment. The interchange of material between cornpartmen primarily through blood flow to the various components. The druglrnedica concentrations are based on the tissue water volume of a given compartrr Consequently, an important parameter in the systems approach i s the perfu rate for each organ. If we know the perfusion rate, we can determine exchange of material between the bloodscream and that organ. For exampl organs are connected in series on one in parallel by blood flow as show Figure 725,
Blood Ftaw
b0
vi
vl
Organ f
CAI Vo "l
Organ 3
t
v2
Organ 2
~2
z7o
4
Figure 725 Physiologically based model.
then the balance equations on species A in the TWVs OF the organs V,.V2, V, are
where r,,, and rA3are the metabolism rates of species A in organs I and 3, respectively, and CAj is the concentration of species A being men
lized in each of the organ compartments j = 1, 2, and 3. Pharmacokinetic models for drug delivery are given in Professional Referr Shetf R7.5.
Sec. 7.5
Physiologically Based Pharmacokinetic (PBPK) Models
I
Example 77 Alcohol Mef~bolismin the Bodsz2
/
E774.G ~ n a r a l We are going to model the metabol~smof ethanol in the human body using fundamental reaction kinetics along with five compartments to represent the human body. Alcohol (Ac) and acetaidchyde IDe) will flow between these compartments, but the aicohol and aldehyde wrll only be rneraboItzed in the liver compnrtmenf. Alcohol and acetaldehyde are meiabolized in the h e r by the following series reactions.
Example Problerr
The first reaction is catalyzed by the enzyme alcohol dehydrogenease (ADH) and the second reaction is catalyzed by aldehyde dehydrogenease (ADLH). The reversible enzyme ADH reaction is catalyzed reaction in the presence of a cofnctor, nicotinarnide adcnimc dinucleotide ( N A P )
C,H,OH t NAD'
(==?CH,CHO + H++ NADH ADA
The rate law for the disappearance of ethanol foIlows MichaelisMenton kinetics and i s
where V,, and KM are the MichaelisMenten parameters discussed in Section 7.2, and CAcand CD,are the concentrations of ethanol and acetadehyde, respectively. For the metabolism of acetaldehyde in the presence of acetaldehyde dehydrogcnast. and NAD*
NAD+ + CH,CHO + H,O lLDH t CHICOOH + NADH e H*
I
the enzymatic rate law is
The parameter values for the rate laws are Vm,,,D, = 2.2 mPYl/(min kg liver). = 32.6 mmoV(min * kg liver), KmADH= 1 rnM. V, = 2.7 rnmoY{min * kg liver), and KMALDH = 1.2 pM (see Summary Notes). The concentration time trajectories for alcohol concentration in the central compartment are shown in Figure E77.I . K a ~ H= 0.4 rnM.,,V
We are going to use as an example a fiveorgan compartment model for the metabolism of ethanol in humans. We will apply the CRE algorithm to the tissue warer volume in each organ. The TWVs are lumped according la their perfusion rates and
,* Summary Notes
M Urnulis, M. N. Curmen, P,Singh. and A. S. Fogler, Alcohol 35 (I), 2005, The complete paper is presented in the Summary Notes on the CDROM.
Q.
442
Reaction Mechanisms. Pathwsys. Bionactions, and Bioreactors
Chap. 7
residence times. That is, those compartments receiving only small amounts of blood Row will be lumped together (e.g., fat and muscle) as will those receiving large blood flows (e.g., lungs, kidneys, etc.). The following organs will be modeled as single unsteady CSTRs: stomach. gastrointestinal tract, central system. and muscle and fat. The metabolism of ethanol occurs primarily in the liver, which i s modeled as a PFR. A number of unsteady CSTRs in series approximate the PER. Figure E77.1 glves a d i a p m showing the connection blood flow (pfusion), and mean residence, T.
The physiologically based mudel
I
Stomach
I
&=
m' Muscle & Far 25.76 dm3
=
Central = 11.56 dm'
Liver = I . l dm3 G.I. = 2.4 dm?
LF=d
K~sidenueT t m ~
Muscle & Fat = 27 rnin
Muscle B Fat
Central = 0.9 min h e r = 2.4 rnin
Figure E77.1 Companmenr model of human body. The residence times for each organ were obtained from the individual perfusion rate? and are shown in the margin note next to Figure E7.71. We will now discuss the balance equation on the tissue water volume of each
of the organslcompartments. Stomach
As a fin1 approximation, we shall neglect the 10% of the total alcohoI ingested that is absorbed tn the sromach because the majority of the alcnhol (90R ) is ahsorbed at the entrance to the fasrmrntestinai (G I.) tract. The contents of the stomach are emptied into the G.1. tract at a rate proponional to the volume of fhe contents in the
stomach.
where i',,is the voluine of the contents of the stomach and Ir, is the rate constant
The flow of ethanol from the \lomach inlo the G.I. tract, where it is absorbed ally instantancov~ly.i\
Sm, 7.5
where CSAc is the ethanol concentration in the stomach, ,k
rAi ,
organ species
'
S = Stomach
G = G.1. Tract
443
Physiologically Based Pharmacokinetic (PBPK) Models
is the maximum enp
tying rate, D is the dose of ethanol in the stomach in (mmoI), and a is the emptying
parameter in (mm01)~.
Gastrointestinal (G.1,) Trsct Component Ethanol is absorbed virtually instantaneously in the duodenum ttt the entrance of the G.I. tract. In addition, the blood flow to the G.1, compartment from the centraI compartment to h e G.Z. tract is twothirds of the total blood flow with the other thjrd bypassing the G.I. h c t to the liver, as shown in Figure E77.1. A mole mass bslance on ethanol in the G.I. tract tissue water volume (W) VG,gives
C = Central
M = Muscle L = Liver
where CCAc is the concenmtion of alcohol in the G.I. mrnpamenent. B~causethe l%V remains conslant, the mass balance becomes
I
A similar balance on acetaldehyde gives
The centrat volume has the largest W V . Material enters the cenbal companment from the liver and the musclelfat compartments. A balance on ethanol in this corn
partment is
In
I
[Accumulation1
=
Ethanol: Vc
= uLCLAc + U CMAE 
d,
"
Out to
Out
uLCCA,  v , ~ OC ' ~(E77'8) ~ ~
Similarly the acetaldehyde balance i s Acetaldehyde:
Yc
dt
(E77.9)
444
Reaction Mschan~sms.Pathways. Broreactions, and Bioreactors
C
MuscleCFat Compartment
Very little rnztterial profuses in and out of the muscle and fat compartments pared to the other cornpartlnents. The muscle compartment mahs balances on el nnd acetaldehyde are
Ethanol:
d C, V , w d =uv {Cqc
Acetaldehyde:
Yt,
 C,tt,{)
dt
(E7.
 Uw (CcDr C,tfje)
(E7
Liver Compartment
The liver will be modeled as a number of CSTRs in series to approximate a aith a voturnr of I . t dm". Approxitnuting a PFR with a number of CSTRs in ! was discusqed in Chapter 2 . The total volume of the liver is divided into Death by alcohol poisoning can occur when the central cornpart
CSTRs.
ment concentration reaches 2 g/dmJ.
Figure E77.2Liver modeled as a number of CSTRs in series
Because the first CSTR receives inflow from the central compartment (113 u) from the G.I. compartment, it is treated separately. The balance on the first CS? Fiat reactor
Ethanol: AV,c
2
L =L(
dr
+)
C
) A L
(E77
Aceraldehyde:
*vL5fi
7
= u L ( ~ ~ c b ~ ~ cLm)rL,c(cL,,p7 ~ G D c 
+rLh(c~ne
(E77 where Ct4= is the concentntion of alcohol leaving the Arst CSTR. A balancc
reactor i gives
Later reactors
Ethanol:
dC
b v LdtL c= u~[C,,,,~, CIA=] + rEAc(CGC, C,D,)AVL
(E77
Sec. 7.5
Physiolog~callyBased Pharmacokinet~c(PBPK) Models
Acetaldehyde:
The concentrations exiting the tast CSTR are CLnAk and C,, . Equations (E77.1) through (E77.15)along with the parameter values are given on %e CDROM summary notes and the Polymath living example problem. The Polymath program can be loaded directly Fmm fhe CDROIM so that the reader can vary the model parametemt You can print or view the complete Polymath program and read the complete paper [AIcuhol35 ( 1 ), p. 10, 20051 in the Sumttlruy ,Votes on the CDROM. Summary dotes
In the near future, E. R. physicians will go to and interact with the computer to run
siniulations to help rreal pat~entsw ~ t h drug overdoses and dntg internction.
I
.!win& Example Problc
+
The Polymath program on the CDROMwas written for Polymath Version 5.1. If you use Version 6.0or higher, reduce the number of liver compartments from 10 TO 9 to avoid exceeding the maximum number of equations aIfowed in Version 6.0.
446
Reaclton Mechanisms, Pathways, Bioreactions, and Bioreactors
Chap. 7
Results figure E77.3 gives the predicted blood ethanal concentration trajectories and experimentally measured trajectories.The different curves are for different initial doses of ethanol. Mote that the highest initial dose of ethanol reaches a mwimum concentration of 16.5 m M of alcohol and that it takes between 5 and 6 hours to reach a level where it is safe to drive. A comparison of the mode1 and experimental data of Jones et al. for the acetaldehyde concentration is shown in Figure ET7.4. Because the acetaldehyde concentrations m three orders of magnitude smaller and more difficult to measure, there is a wide range of error bars. The model can predict h t h the alcohol and acetaldehyde concentration trajectories without adjusting any parameters. In s u r n m q , physiologically based pharamacoktnetic models can he used to predict concentrationtime trajectories tn the TWV of various organs in the body. These models find werincreasing application of drug delivery to targeted organs and regions. A thorough discuss~onof the following data and other trends is given in the paper (Ulrnuli~,Gurmen. Singh, and Fogler).
Comparison of model with experimental data
Tme {min) Fjrure EJ7.3. Blood alcoholtime trajectories from dam of W~lkinsonel al?"
0
20
PO
60
50
100 120 140 tBO 1BO
Time (min)
Figure E77.4.Blood nlcahollime rrajector~eqfrom data of Jones et a]."
'jP. K. WiIkinson, et a].. '.Pharmacokinetics of Ethanol After Oral Administration in the Fasting State." J'. P h n n ~ i ~ ~ o o kHiophonrr.. ei. 5(3):20724 ( 1977). "A. W.Jones, J. Neirnan. and 51. Hillhi>rn. "Concentrationtime Profiles of Ethanol and Acetaldehyde in Human Volua~eenTreated with the Alcohol~ensirizingDrug, Cslclurn Carhimide." BK J. CiEtt Phurrnarnl., 25. 2\321 (198%).
447
Summaty
Chap. 7
.  e. The theme running through most of this chapter is the steadystate hypothesis (PSSH) as it applies to gasphase reactions and enzymatic reactions. The reader should be able to apply the PSSH to reactions in such problems as W7 and P712 in ordw to develop rate laws. Reaction pathways were discussed in order to visualize the various interactions of the reacting species. After completing this chapter the reader should be able to describe and analyze enzymatic reactions and the different types af inhibition as displayed on a LineweaverBurk plot. The reader should be able to explain the use of microorganisms to produce chemical products along with the stages of cell growth and how the Monod equation for celI growth is coupled with mass balances on the substrate, cells, and product to obtain the concentrationtime trajectories in a batch reactor. The reader should be able to apply the growth laws and balance equations to a chemostar (CSTR)to predict the maximum product rate md the washout rate. Finally, the reader should be able to discuss the application the enzyme kinetics to a physiologicalIy based pharrnacokinetic (PBPK) model of the human body to describe ethanol metabolism.
SUMMARY 1. In the PSSH, we set the wte of formation of the active intermediates equal to zero. If the activc intermediate A * i q involved in rn different reactions, we set
ir
to
This approximation is justified when the active intermediate is highly reactive and present in low conceatrarions. 2. The azomethane (AZO) decomposition mechanism is
2AZO r(~r E~~c/o(r,r:l,ro.ri.~
Chaa. 7
Chap. 7
CDROM MATERIAL
Profdona1 Reference Shelf P7.1. Polymerization A. Srep Polymerization Mechanism ARB + ARB tAR,B + AB ARB + AR,B +AR3B + A% ARB + AR,B
+AR4B + AB
AR2B + ARIB+ AR,B + AB
Reference Shelf
Rate Law r, =
PIP,,
 2kqM
Concentration
Example R71 Determining the Concentration of Polymers for Slep Polymerization B. Chain Potymeriza~ions FreeRadical Polymerization
Iz 4 21
Initiation
Ri+M+R,,,
Propagalion
Termination Addition
R, + Rk + P,+,
Disproportionation
Ji, + R, +P,+ P,
Mole Fraction of Pnlymer of Chain Length
h a ? n p l ~R72
Parameters of M W Distribution
452
Reaction Mechanisms, Pathways, Bioreactmns, and Bioreactors
Cb
C. Anionic Polymerization Initiation by An Ion AB A+B
Initiation
A+M+R,
R,+ M + R ,
Propagation
R, +M+Rp1 Transfer to Monomer
R, + M t P, + R ,
Example 7PRS3 Calculating the Distribution Parameters from Ana
Expressions for Anionic Polymerization Example 7PRS4 Qetermination of Dead Polymer Disrribution \I Transfer to Monomer Is the Primary Termination R7.2. OxygenLimited Fetrnenmrion Scale Up
B Ye&
Reference Shelf
W ~ fM i a h 9 p ~ r ZIt, MiWQ0S
R7.3. Receptor Kinetics A. Kinetics of signaling
Chap. 7
GOROM MATERtAL
B. Endocytosis f
Adapted Fmm D.A. Lauffenburger and J . J. Lineman. Receptors (New York:Oxford University Press. 19931.
R7.4. Multiple Enzyme and Substrate System A. Enzyme Regeneration Example R7.4I Construct a LineweaverBurk Plot for Different Oxygen Concentration B. Enzyme Cofacton (1) Example 7.42 Derive a Rate Law for Alcohol Dehydrogenase (2) Example 7.43 Derive a Rate Law a Multiple Substrate System (31 Example 7.44 CaIculate the Initial Rate of Formation of Ethanol in the Presence of Propanediol
R7.5. Phanacokinetics in Drug Delively Pharmacokinetic models of drug delivery for medication administered either orally or intravenous1y are developed and analyzed. in.
Figurn A. %ocompanment mode?.
Figure B. Drug response c u m .
454
Reaction Mechanisms, Pathways, Bioreactions, and Bloreactors
Chap. 7
Q U E S T I O N S AND P R O B L E M S
In each of the following questions and problems, rather than just drawing a box around your answer, write a sentence or two describing how you solved the problem, ClomGworkPsobftmr the assumptions you made, the reasonableness of your answer, what you learned, and any other facts that you want to include, You may wish to refer to W. Strunk and E. B. White, The Elements qf Style, 4th ed. ((New York Macrnillan. 2000) and Joseph M. Williams, Style: Ten Lessons in Clariv & Grace* 6th ed. (Glenview, Ill.: Scott, Foresman, 1999) to enhance the quality of your sentences. See the Preface for additional generic p a t s (x), (y),
W1
Livlng Example Problem
(2)
to the home problems.
fa) Example 71. How would the results change if the concentration of CS, and M were increased? (b) Example 72. Over what range of time is the PSSH not valid? Load the Lrving . h ~ m p kProblem. Vary the temperature (800 < T c 1600).What temperature gives the greatest disparity with the PSSH results? Specifically compare the PSSH solution with the full numerical solution. (c) Example 73. ( 1 ) The following additional runs were carried out when an inhibitor was present.
(dl
(e)
(fl Liv~ngExample Problem
(g)
What type of inhibition is taking place? (2) Sketch the cutves for no inhibition, competitive, uncornpetitive, noncompetitive (mixed) inhibition, and substrate inhibition on a WoolfHanes plot and on an EadieHofstee plot. Exampie 74. (1) What wouId the conversion be after 10 minutes if the initial concenmtion of urea were decreased by a factor of 100? (2) What would be the conversion in a CSTR with the same residence trme, T, as the batch reactor? (3) A PFR? Example 75. What is the lotal mass of substrate consumed in grams per mass of cells plus what is consumed to form product? Is there disparity here? Example 76. Load the Living Example Problem. ( 1 ) Plot the concentration up to a time of 24 hours. Did you observe anything unusual? If so. what? 12) Modify the code ta carry out the fermentation in n fedbatch (semibatch reactor) in whrch the substrate is fed at a rare of 0.5 dm% and at concentration of 5 gfdmhto an initial liquid volume of 1.0 dm3 contain~nga cell mass with an initial concentra~ionof C, = 0.2 rng/dm3 and an initial substrate concentration of C,, = 0.5 rngldm3. Plot the concentration of cells, substrate, and product as a functron of time along with the mass of product up to 24 hours. Compare your result< with (1) above. (3) Repeat (2) when the growth uncornpetitively inhibited by the substrate with K, = 0.7 g/dm3. 14) Set C i = lU.000 gfdm3, and compare your results with the bare case. Example 73. This problem IS a gold mine for t h ~ n p slo he learned about the effect of alcohol on the human body Load the Polyn~arl~ LirrtJl: Exa~npleProgranr from the CDROM. ( 1 ) Start by varylng the initial doqes of nlcohol. 12) Next comider individuals who are ALDH enzyme
Chap. 7
Explorc the bloodalcohol simulation on the CDROM L i ~ i n gExample problem
Llulng E~ampfeProblem
Computer Modules Enzyme Man
455
Questions and Problems
deficient. which includes about 40% to 50% of Asians and Native Americans. Set V,,,,, for acetaldehydes between 10% and 50% of its normal value and compare the concentrationrime trajectories with the base cases. Hint: Read the journal anicle in the Summary Notes [Alcohol 35. p.1 (2005)l (h) Load the Ozone Poipiath Living Example Program from the CDROM. Vary the halogen concentrations and describe what you find. Where does PSSH break down? Vary the race constmts and other species concenuatians. (i) Load the GIowsh'cks Livinl: Example Problem from the CDROM. Vary the rate constants to learn how you can malie the luminescence last longer. Last shoner. CjJ Load the Russell's ICpm Polymath Living Examglc Program from the CDROM. Describe what wou!d happen if the victim rece~vedmore than one bite. In the cobra problem in Chapter 6 we saw that after 10 bites, no amount of antivenom would save the victim. What would happen if a victim received 10 bite5 from a Russell's riper? Replot the concentrationtime trajectories for venom, FOP, and other appropriate species. Next, inject different amounts of antivenom to learn if it 1s possible to negate 10 bites by the viper. What is the number of bites by which no amount of antivenom w ~ l lsave the victim? (k) Load the FerdeDance Polj~rnafhLivitrg Example Progmrn from the CDROM. Repeal 7 1 Cj) for the FerdeLance. (1) Load the Receptor Endocytosis Living Example Problem from the CDROM. Vary k,,, fR. and ,fL over the ranges in Table R7.31. Describe what you find. When will acute renal failure wcur? (m)List ways you can work this problem incorrectly. (n) How could yow make this problem more difftcult? ICM E;nzyme Man. Load the ICM on your computer and carry out the exercise. Performance number = (a) List ways you can work this probrern incorrectly. (hj How could you make this problem more difficult? (Fiml7e ~rmrrlnrrrs)Hydrogen radicals are important to sustaining combustion reactions. Consequently. if chemical compounds that can scnvange the hydrogen radicals are introduced, the flames can he extinpished. While many reactions uccirr during the cornbusttnn process. we shall choose CO flame< as a rnodcl system to illu~trate the procesq IS. Senkan er al.. Cinhrrrtir~rr n)lrl Flri~ne,69. I I3 (1987)J.In the ahrence of inhibitors
When HCI i< inttvThe last two reactions are rapid compared to the first TWO. duced to the flame, the fnllou rng additional reactions occur: H1 + C1.
+ HCI
H
C1.
t
HCI
456
Reaction Mechanisms, Pathways, B~areact~ons, and Bioceactars
Ch,
Assume that all reaction< are elzrncntary and that the PSSH holds for the OH .. and CI . radic;~l\. ( a ) Derive a rate law for the con$umptlon of CO when no retardant i s pre: (b) Derive an equation fur the concentratiun of !I as  a function or assuming constanr concentration of 0,. CO. and H,O for both unin ited cornbustion and cumbustion with HCI present. Sketch H. vc time for both cases. (c) Sketch a reaction pathway d~agramfor this reaction. (dl List ways you can work (hi5 problem incorrectly. ( e ) How could you make this problem more difficult? More elabornte forms uf t h i ~problem can be found in Chapter 6. where
PSSN is not invoked.
W4, The pyrolysis of acetaldehyde i x believed to take place according to the lowing sequence:
CHO
+ CH,CHQ
"
> CC,.
+ 2C0 +
(a) Derive the rate expression for the rate of disappearance of acetaldeh rAc.
Ih) Under what conditions does it reduce to Equation (73)? (c) Sketch a reaction pathway diagram for this reaction. (d) List ways you can work this problem incorrectly. (e) How could you make this problem mote difficult? WsR (a) The gasphase homogeneous oxidation of nitrogen monoxide (NO dioxide (NO:
Hall of Fame
),
is known to have a form of thirdorder kinetics. which suggests that reaction is elementary as written, at least for low partial pressures of nitrogen oxides. However, the rate constant k actually d f c r e f l ~ e7 ~ increasing absolute temperature. indicating an apparently negative acl tion energy. Because the activation energy of any elementztry reac must be positive, some explanation is in order.
Provide an explanation, starting from the fact that an active interned species. NO3,is a participant in some other known reactions that invr oxides of nrtrogen. Draw the reaction pathway. tb) The rate law for formation of phosgene. COCI2. from chlorine. Cl?, carbon monoxtde, CO, has the rate law
Suggest a mechanism for this reaction that is consistent with this rate and draw the reaction pathway. [Hint:CI formed from the dissociatioi C1, is one of the two active intmediate3.l (E) List ways you can work this problem incorrectly. (d) How could you make this problem more difficult?
Chap. 7
457
Questions and Problems
WtiR &(me is a reactive gaa that has been ac~ociatedwith resptralory illness an$ decreased lung tunct~on.The follow~ngreactions are involved in ozone formation [D AIfcn and D. Shunnard, Green En~irreesirrg (Upper Saddie River, N.J.: Prentice Hall, 2002)j. Green enfinrering
i q primarily generated by combustion in he automobilc engine. Show that the steadystate concentration of Ozone 1s directly proportional to NOI and inversely proportinnal to NO. (b) Drive an equation for the concentration of ozone in soleIy in rerms of the initial cor~centrationsChO,O, CVO,,, and 0,"and the rate law pamrneteE. (c) In the absence of NO and NOZ,'the rate law for ozone genemtion i s
NO1 (A)
.
Suggest a mechanism. (dE List ways you can work this problem incorrectly. (e) How could you make this problem more difficult? W7c ( T r i b n l o ~One ) ~f the major reasons for engine oiI degradation is the oxidation of the motor oil. To retard the degradation process, most oils contain an antioxidant [see blrl. Eng. Chrrn. 26, 902 (1987)l. Without an inhibitor to oxidation present. the suggested mechanism ar low temperatures is
Why you need to change the motor oil in your car?
ZRQ, % inactive whew I? is an initiator and RH is the hydrocarbon in the oil.
fzj Motor Oil
When an antioxidant is added to retard degradation at low temperatures, the following additional termination steps occur:
A+
R01
inactive
Reaction Mechanisms. Pathways. Bioreactions, and Bioreactors
W8,
Chap. 7
(a) Derive a rate law for the degradation of the motor oil in the absence of an antioxidant at low temperatures. (bl Derive a rate law for the rate of degradalion of the motor oil in the presence of an antioxidant for low temperatures. ( c ) Hou. would your answer to part (a) change if the radicals I  were produced at a constant rate in the engine and then found the~rway rnto the oil? (d) Sketch a reaction pathway diagram for both high and low ternvratures. with and \\ ithout anrioxida~it. (el See the openended probleins on the CDROM for rnclre on this problem. (fJ List ways lo11 can work h i s problem incorrectly. (g) Hnw could you make this problem more difficull? Consider the applicarion of the PSSH to epidemiology. We shall Ireat each nf the following steps as elementary in that the rate will he proportional to the number of people In a particular state of health. A healthy person. H, can become ill, I,
478
Steadystate Nonisothermal Reactor Design TABLE81.
Chap. El
ENERGY BALANCES OF COMMONREACTORS (COD)
7. For a variable codant temperature, T,
(781.K) These are the equations that we will use to sokve reaction engineering problems with heat effects.        +         +     
[Nomenclaturr: I: = overall heattransfercoefficient, (Jh? * s K):A = CSTR heatexchange area, Im2). a = PFR heatexchange area per volume of reactor. (m21m'l; CS = mean heat capacity of species i,(JlmolK): C p = the heat capacity of the coolant.
(JkJlkglK),ni, = coolant flour rate, tkgls); AH,, = hear of reactLon. Illmol):
AH;, =
kD
a
a
j in reaction i.(JEmoll;
e
b
+ f ~ ; H i
Q
1
Jimol.4: AH..,, = heat of reaction w n species
= heat added to the reactor, (Jls); and
C cPD+;cpc
Up=

illrno1.A * K ) AII other syrnhols are as dehned in
Chapter 3.1
Examples on How to Use Table 81. We now couple the energy balance equations in Table 81 with the appropriate reactor mole balance, rate law, smichiometry algorithm to solve reacdon engineering problems with heat effects. For example, recall rate law for a firstorder reaction, Equation (Eg1.5) in Example 8 1 .
If the reaction is carried out adiabatically. then we use Equation IT&I .B) for the reaction A d B in Example 81 to obtain Adiabatic
Consequently. we can now obtai~.r, as a function of X done by first choosing X. then calculating T from Equation (TE1 .B). then calculating k from Equation (E81.3). and then finally calculating (r,) from Equation (Eg1.5). Choose X 4 calculate T + calculate k
x Lel'enspiel plot
F~~ + caIcuIate rA + calculate 
 ?'*
We can use this sequence lo prepare a table of (FA,+r,) as a function of X . We can then proceed to size P F R ~and CSTRs. In the absolute worst case scenario, we could use (he techniques in Chapter 2 (e.g.. Levenspiel plots or the quadrature formulas in Appendix A ) .
Sec. 8.2
The Energy Balance
479
However, instead of using a Levenspiel plot, we will most likely use Polymath to solve our coupled differential energy and mole balance equations. If there is cooling along the Iength of a PFR. we could then apply Quation (T8I .€) to this reaction to arrive at two coupled differential equations. Nonadiabatic
PFR
which are easily solved using an ODE solver such as Polymath. Similarly, for the case of the reaction A + B carried out in a CSTR. we could use Polymath or MATLAB to solve two nonlinear equations in X and T. These two equations are combined moIe balance Nonadiabatic CSTR
and the application of Equation (T83 .C), which is rearranged in the form
why hother' Here is why'!
From these three cases, (1) adiabatic PFR and CSTR, ( 2 ) PFR and PBR with heat effects. and (3) CSTR with heat effects, ane can see how one couples the energy balances and mole balances. In principle, one could simply use Table 81 to apply to different reactors and reaction systems without further discussion. However, understanding the derivation of these equations will greatly facilitate their proper application and evaluation to various reactors and reacfion systems. ConsequentIy, the following Sections 8.2. 8.3. 8.4, 8.6. and 8.8 will derive the equations given in Table 81. Why bother to derive the equations in Table 81 ? Because I have found that students can a p l ~ these l ~ equations mucll more accurately to solve reaction engineering problems with heat effects if they have gone through the derivation to understand the assumptions and manipulations used in arriving at the equations in Table 8.1. 8.2.4 Dissecting the SteadyState Molar to Obtain the Heat of Reaction
FEow Rates
To begin our journey, we start with the energy balance equation (89) and then proceed to finally arrive at the equations given in Table 81 by first dissecting two terms.
480
SteadyState Noniscthermal Reactor Design
!. The molar Row sates. F,and FA, 2. The molar enthalpies, Hi. H,,,[H, = lnteractrvc
H,(n,and Ha
Ch
HAT,)]
An animated version of what foIlaws for the derivation of the energy ance can be found in the reaction engineering modules "Heat Effects 1" 'Weat Effects 2" on the CDROM.Here equations move around the sc
making substitutions and approximations to arrive at the equations show Table 81. Visual learners find these two ICMS a useful resource. We wiIl now consider Row systems that are operated at steady state. Cornpu:~: Modules steadystate energy balance is obtained by setting (dE',,,/dr)equaI to zel Equation (89)in order to yield
*~'v.h
Steady state energy balance
To carry out the manipulations to write Equation 1810) in terms of the he reaction, we shall use the generalized reaction
The inlet and outIet summation tems in Equation (8 10) are expanded. re: tively, to
In:
Z HIoF,o = HAoFAo+ ffBD FBo + Ha F*,, + H m Fm + H10 FIO
IF
and
We first express the molar Row rates in tems of conversion. In general. the molar Row rate of species i for the case of tion and a stoichiometric coefficient v, is
no accun
F, = FA,(Oi + v , X ) b SpecificalIy, for Reaction (221, A t  B a
FA= FAD(I X)
Steadystate operation
+aG C + ad D , we have
Sec. 8.2
The
fnergy Balance
481
We can substitute thege symbolc for the molar flow rates into Equations (81 1 ) and (813). then subtract Equation (812) From (81 1 ) to give
The term in parentheses that is multiplied by FAOXis called the heat of reaction at temperature T and is designated AHR,. Heat of reaction at temperature T
All d the enthalpies (e.g., H A , HB)are evaluated at the temperature at the outtet of the system volume, and, consequently. [AH,,(T)] is the heat of reaction at the specific temperature T.The heat of reaction is always given per mole of the species that is the basis of calculation [i.e., species A coules per mole of A reacted)].
Substituting Equation (814) into 1813] and reverting to summation notation for the species. Equation (8 13) becomes
Combining Equations (810) and (815), we can now write the steadysmte live.,(dESy/d! = O)] energy balance in a more usable form: One can use t h i ~ Form of the steadystate energy balance if the enthalpres m ava~lable.
If a phase change takes place during the course of a reaction, this form of the energy balance [i.e., Equation (81611 must be used (e.g., Problem 54,).
8.2.5 Dissecting the Enthalpies We are neglecting any enthalpy changes resulting from mixing so that the partial rnolal enthatpies are equaf to the mob1 enthalpies of the pure components. The molal enthalpy of species i at a particuIar temperature and pressure, Hi, is usually expressed in terms of an enthulpy offormarion of species i at some reference temperature T,. HI0(TR),plus the change in enthalpy AHQ,, that results when the temperature is raised from the reference temperature, TR.to some temperature T:
482
SteadyState Monisothermal Reactor Design
Chap. B
For example, if the enrhalpy of formation is given at a reference temperature where the species is a solid, then the enthalpy, H(?),of a gas a! tempemre T is


Enthalpy of
formation
Enthalpy of species Calculating the enthalpy when phase changes are involved
= i
in

at
T
1
at
+
I;P 
",:I
Heat of
AHQ in heating
intbtbid S:ptIp,"]
ofspecies
+
Here, in addition to the increase in the enthalpies of the solid, liquid. and gas from the temperature increase, one must include the heat of melting at the melting point, AH,, (T,,), and the heat of vaporizarion at the boiling point. AHvi (Tb). (See Problems P84c md P94B.) The reference temperature at which HP is given is usuaIly 25OC. For any substance i that is being heated from TI to T2 in the absence of phase change, No phase change
Qpical units of the heat capacity, C, , are ( C p l )=
(moi of i ) (K)
or
cal Btu or (mol of i ) (K) (Ib rnoi of i ) ( O R )
A large number of chemical reactions carried out in industry do no1 involve phase change. Consequently, we shall further refine our energy balance to apply to singlephase chemical reactions. Under these conditions, the enthalpy of species i at temperature T is related to the enthalpy of formation at the reference temperature T, by
H, = HP(T,)
+J'p, d~
(819)
If phase changes do take place in going from the temperature for which the enthalpy of formatron i s given and the reaction temperature T, Equation 1817] must be used insread of Equation (819). The heat capacity at temperature T i s frequently expressed as a quadratic function of temperature, that is.
S ~ C8.2 .
Reference Chef
The Energy Balance
483
However, while the text will consider only constant heat capacities, the PRS R8.3 on the CDROM has examples with variable heat capacities. To calculate the change in enthalpy (HI  H,v) when the reacting Ruid is heated without phase change from its entrance temperature, 4,, to a temperature T, we integrare Equation (819) for constant C, to write
Substituting for H,and
in Equation (816) yieIds
Result of dissecttng the enthalpies r=l
8.2.6 Relating A 4,( T ), A H",A TR),and A c,, The heat of reaction at temperature T is given in terms of the enthalpy of each species at temperature T, that is, Affk,(T) =
b
d
H~(T)+Ta,1 For endothermic reactions
{TIPTQTJ Halfpipe jacket
Figure 812 CSTR rank reactor with heal exchanger. [rh) Courtesy of Pfaudler. Inc.1
" Informat~onon lhe overall heattran\fer
coefficient may 'be found in C. J. Geankopli.
Transpon Pw)cesscr ur~dUni? Opemrinns, 3rd ed. Englewood Cliffs. N.J., PrentKe
Hall (2003). p. 268.
Sec. 8.6
CSTR wfth Heat Effects
523
The following derivations, based on a c ~ 1 m (exothermic t reaction) apply also to heating mediums (endothermic reaction). As a first approximation, we assume a quasisteady state fwthe coolant flow and neglect the accumulation term (i.e., dT'ldt = 0).An energy balance on the coolant fluid entering and leaving the exchanger is Rate of heat exchanger
to reactor
by flow
where Cpc is the heat capacity of the coolant fluid and TR is the reference temperature. Simplifying gives us
Solving Equation (846)for the exit temperature of the coolant fluid yields
T,, = T  ( T  T,,) txp

From Equation (846)
Substituting for T,, in Equation (848). we obtain Heat transfer to a
CSTR
For large values of the coolant flaw rate. the exponent wilI be small and can be expanded in a Taylor series (e" = 1  x . . ) where secondorder terms are neglected in order to give
+
Then Valid only for large Coolant flow raws!!
where To,2 Taz = TO.
I
524
SteadyState Nonisothetrnal Reactor Design
Chi
With the exception of processes invoIving highly viscous materials s as Problem P84c, the California P.E exam problem, the work done by the : rer can usualiy be neglected. Setting W,in (827) to zero. neglecting ACp, ! stituting for Q and rearranging, we have the following relationship betw conversion and temperature in a CSTR.
Solving for X IT
To)+ SO,C, (T To)
X=
[A%AT,?)l Equation (852) i s coupled with the mole balance equation
to
size CSTRs. We now will further m a n g e Equation (85 1) after letting CO,Cpr= C
Let
KT, + T,
and T, = Then
XbGK= Cpo(l+ K ) ( T  T,)
(3
The parameters K and T, are used to simplify the equations for nonadiab; operation. Solving Equation (854) for conversion
Forms of the energy
Solving Equation (854) for the reactor temperature
balance for a CSTR with heat exchange
Figure 813 and Table 84 show three ways to specify the sizing o CSTR. This procedure for nonisothermal CSTR design can be illustrated considering a firstorder irreversible liquidphase reaction. The aIgorithm working through either case A (Xspecified), B (T specified), m C (Y specified shown i n Table 84. Its application is ilfustrated in the following example.
Sec. 8.6
CSTR with Heat Effects
Algorithm
Example: Etementary ~rreversibleliquidphase reaction
A+B Given F A OCM, ~ ko, E,CpA,AH~r, ACp=O, q = O CSTR I
Oedgn equation
v = !d
Rate law
+A
Combining
v=
+A
= kcA
FAOX ~ C A(1O X I
+ I I
X specified: Calculate V and T
Need k(T)
JI
Calculate T
Two quatians and h o unknowns
1 I
I
Calculate k
1
S
E:: Plot X vs. T
I

4 Calculate V
I
X
T Flgure $13
Algorithm for adiabatic CSTR design
526
SteadyState Nonlsotherrnal Reactor Design TABLE84.
WAYSTO SPECIFY THE SIUNG
A
CSTR
A
B
C
Specify X
Find V and T
Spe~ifyT Find X and V
W m d T
Calculate T From Eqn. (856)
Calculate X From Eqn. 1855]
1
1
Use Eqn (8553 to plat XElgVS. T
Calculate k k = ApCtRT
Calculate k k = kCEIRT
Solve Eqn. (853)
L
Spify
L
v
1 J.
for
xus = p n
to find XMnvs.
ce.e.,  r ~= kcAO() XI)
fe.g.. ra =
kcAo(]  x))
T
le.g., XhlB = TA
expp[E/(RT)]
1 + 0.exp[E/(RT)l
J.
S
Calculate r,(X,TE
Calculate r,{X,X.T)
1
1
Calculate V
CalcuIate V
y = Fad: r~
v= FAJ r,
Chap. 8
1
1 Plot XE8 and X,
as a function of T
xp L
T
XMR=conversion calculated from the mole balance XER= conversson calculated from the energy balance
Example
g8
Prnduction of Proplena Glycol in on Adiabah'c CSTR
Propylene glycol is produced by the hydrolysis o f propylene oxide:
Production, usesa and economics
Over 800 million pounds of propylene glycol were produced in 2004 and the selling price was approximak!y 50.68 per pound. Propylene glycol makes up about 25% of the major derivatives of propylene oxide. The reaction takes place readily at room temperature when catalyzed by sulfuric acid. You are the engineer in charge of ap adiabatic CSTR producing propylene glycot by this method. Unfortunately, the reactor i s beginning to leak, and you must replace it. (You told your boss seveml times that sulfuric acid was corrosive and that mild steel was a poor material for construction.) There is a nicelooking overflow CSTR of 300gal capacity standing idle: it is glasslined. and you would l ~ k eto use it. You are feed~ng2500 lblh (43.04 IIb morlh) of propylene oxide (P.O.) to the reactor. The feed stream consists of (1) an equivolumetric mixture af propylene o x ~ d e(46.62 ft'/h) and methanol (46.62 ft'lh). and (2) water contaming 0.1 wt % H2S0,.The volurneuic RON rate of water is 233.1 ftqlh, uhich is 2.5 times the methanolPO. Row rate. The cornspanding moIar feed rates of methanol and water are 71.87 and 802.8 Ib mol/h, respectively. The waterpropylene oxidemethanol mixture undergoes a slight decrease in volume upon mixing
S e .8.6
527
CSTR with Heat Effects
(approximately 38). but you neglect this decrease in your calculations. The temperature of both feed streams is 58°F prior to mixing, but there is an j m d i i r t e 1 7 O F temperature rise upon mixing of the two feed smams caused by the heat of mixing. The entering temperature of all feed streams is thus taken to be 75°F (Figure E88.1). pmwlene @xidm
T~ = w
Methanol
;y F~~ Water
To = 75" F
Furusawa et state that under conditions similar to those at which you are operating, the reaction is firstorder in prepylene oxide concentration and apparen! zeroorder in excess of water with the specific reaction rate
The units of E are Btullb mol. There is an important constraint on your operation. Propylene oxide i s a rather lowboiling substance. With the mixture you are using, you feel that you cannot exceed an operating temperature of 125°F. or you will lose too much oxide by vaporization through the vent system. Can you use the idle CSTR as a replacement for the leaking one if if will be operated adiabatically? If so, what will be the conversion of oxide to glycol?
(All data used in this problem we= obtained from the Handbook uf Clremfsr~ a11d Physics unless otherwise noted.) Let the reaction be represented by
A is propylene oxide (CpA= 35 Btullb mol OF) 8
B is water (CPB= 18 BtuJlbmol."F) ?
T. Furusawa. H. Nishimura, and T. Miyauchi, 3. Che~n.Eng. Jprr.. 2. 95 119691. .CpA and Cpc are estimated from the observation that the great major~ty of lowmolecularweight oxygencontaining organic liquids have a mas$ hear capacity of 0.6 callg% 2 1 5 % .
528
SteadyState Nonisothermal Reactor Design
Cha
C i s propylene glycol (Crc= 46 B t u lb ~ mol  "F) M i s methanol ICp = 19.5 Btullb rnol . "F)
In this problem neither the exit conversion nor the temperature of the a1 hatic reactor IS given. By application of the material and energy baIances we solve two equations with two unknowns (X and T). SoIving these coupled equatir we determine the exit conversion and temperature for the glasslined reactor to if it can be used to replace the present reactor.
I
I. Mole Balance and design equation: FA"F*+r*V= 0
The design equation in terns of X is
2. Rate Law:
1
3. Stoicbiometry (liquid phase, u = u,): C A = C A , , ( l q
4. Combining yields Following the Algorithm
1
Solving for X as a Function o f T and recalling that r = V l v , gives
This equation relates temperature and conversion through the mole bnlanl 5. The energy balance for this adiabatic reaction in which there IS neglig energy input provided by the stimr is
Two equations. two unknown
This equation relates X and T through the energy balance. We see that I equations [Equations (EB8.5) and (ES8.6)] must be solved with X,, = . for the two unknowns, X and T 6. Calculations: a. Heat of reaction at temperature T : ~
II
AH,, (T)= AH;,(T,)+ AC, (T TR)
H i (68°F) : 66,600
Stullb moi
Hland H t are calcurated from heatofcombustion data.
Sec.8.6
CSTR with Heat Elfects
Calculating the parameter values
Hg (68°F):  123,000 Btuilb mol H,9(68°F) : 226.000 Btu/lb mol
AH", ((68°F)= 226,000  (  123,000)  (66,600)
(E88.7)
= 36,400 BtuIIb mol propytene oxide ACp = CpcCPB CPA
 =46I&35=7Btullbmol~"F A W l , ( T ) = 36,400 ( 7 ) ( T  528)
b.
Stoichiometry
T is in "R
(C,,, @, , z ): The total liquid volumetric flow rate entering
the reactor is
For methanol:
(3, = F~~  7E.87 lb m0l/h = FA, 43.0 lb mollh
For water:
lb mollh = 18,65 aB FFBO A = 802.8 43.0 Ib moI1h
c. Evaluate mok balance terms: The conversion calculated from the mole balance, XM,,is found from Equation (EX8.5).
(16.96 X 10rshW1)(0. 1229 h) exp ( 32,4001 1.987T)
Plot XM, as a function of
= 1
+ (11.96X 10il hL1)(0.1229h) exp(32,40011.987T) (E88.10)
temperature. ~ M
(2.084 X 1012) exp (  16,3061T) , is = B I + (2.084 X 1012)exp( 16,3061 T)
d. Evaluate energy balunct terms:
530
SteadyState Nonisoihermal Reactor Design
Chap. B
The conversion cal~uIatedfrom the energy balance, Xm,for an adiabatic reaction is given by Equation (829):
Substituting all the known quantities into !he energy balances gives us
Btullb mol  OF)(T  533°F xm= [(403.3 36,400  7 ( T  528)l Btullb mol
I. Solving. There are a number of
different ways to solve these two simultaneous equations. The easiest way is ro use the PoIymath nonlinear equation solver. However. to give insight into the functional relationship between X and T for the mole and energy balances, we shall obtain a graphical solution. Here X is plotted as a function of T for the mole and energy balances, and the inersection of the two curves gives the solntion where both the mole and energy balance solutions are satisfied. In addition, by plotting these two curves we can l e a n if there is more than one intersection (i.e., multipIe steady states) for which both the energy balance and mole balance are satisfied. If numerical motfinding techniques were used to solve for X and T. ir would be quite possible to obtain only one root when there are actualry mare than one. I f Polymath were used. you couId learn if multiple roots exist by changing your initial guesses in the nonlinear equation solver. We shall discuss multiple steady states further in Section 8.7. We choose T and then calculate X (Table E88. I ). The calculations are plotted in Figure EX8.2. The virtually straight line c m s p o n d s to the energy balance [Equation {E88.12)] and the curved line corresponds to the mole balance [Equation (E8R. lo)]. We: observe from this plot that the only intersection point 1s at 85% conversion and 61 3"R. At this point, both the energy balance and moIe balance are satisfied. Because the temperature must remaln below 125OF (585"R). we cannot use rhe 300gal reactor as tt 1s now.
Don't give upv Head back ro the storage shed lo check out thc heat exchanFe equipment!
1
Sec. 8.6
CSTR with Heat Effects
'Zhe reactor cannot be used becai~se~t uoiHexceed the specified maximum temperature
of 585%.
T ('RI
Figure E88.2
I
Tbe conversions XEa and Xw, as a function of temperature.
EsornpIe 89 CSTR wilh o Cooling Coil A cooling coil has been located in equipment storage for use in the hydration of propylene oxide discussed in Example 84. The cooling coil has 40 ft2 of cooling surface and the cooling water flow rate inside the coil is sufficiently l q e that a constant coolant temperature of 85°F can be maintained. A typical overall heattransfer coefficient for such a coil is 100 Btu/hft2."E W111 the seactor satisfy the previous constraint of 125°F maximum temperature if the cooling coil is used?
SnEutidn If u7eassume that the conling coil rakes up negligible reactor volume, the conversion calculated as a function of temperature from the mole balance is the same as that in Example 88 [Equation (EB8.10)]. 1. Combining the mole baiance, stoichiornetry, and rate law, we hav,=. from
Example 88.
XWB= ~k  (2 084 X 10") n p ( 16,3061 T) I + t k 1 + (2.084 X 1 OJ2) exp ( 16.306) T )

(E88,
T i s i n "R. 2. Energy halance. Neglecting the work hy the stirrer, we combine Quations (527) and (850) to write LTA(To  T ) FA"
I((AHORr(TR)+ ACptT TH))= SO,Cp,[T T,)
(Eg9.1)
I
532
SteadyState Nonisothermal Reactor Design
Cht
SoEcin~Ae enerxy balance for X,, yields Energy Balance
The cooling coil term in Equation (E89.2) is
Ud  (IW F~~
h) Btu ft?."F +
(40 ft2) (43.03 lb moIl'h)
 92.9 Btu
(E8
lb mold "F
Recall that the cooling temperahre is
T, = 8S°F
=
545"R
The numerical values of all other terns of Equation (E89.2) are identic: those given in Equation (E88.12) but with the addition of the heat excha
tern.
We can now use the glass lined reactor
We now have two equations [(ES8.80) and (E89,411and two unknowns. X and The POLYMA~Iprogram and solution to these two Equations (EB8.10), X and (E89.4). XEa.are given in Tables ES9.1 and E89.2. The exiting tempera and conversion are 103.7"F (563.7"R)and 36.46, respectively. i.e..
IT
= 5 6 4 " ~and x = 0.361
Equations: Nonlmear quationr [ II
rrX) = X(M3.3'(T*5.5uH92.9'flMS))I(~7'(T528)1 =0
I21 IrTI 5 X~au*W:
R(p =
as the heatremoved
term
(860)
Heatremoved term
To study the multiplicity of steady states. we shall pIot both R(T)and G(T) as a function of temperarue on the same graph and analyze the circurnstance.w under which we will obtain multiple intersections of R(T) and G ( 0 .
534
SteadyState Nonisothermal Reactor Design
Chap. 8
8.7.1 HeatRemoved Term, f l T ) Vary Entering Temperature. From Equation (860) we see that RIT) increases linearly with temperature, with slope C,(1 + K) As the entering temperature To Is increased. the line retains the same slope but shifts to the right as shown in Figure 814.
.
Heatremoved curve R(T)
Figure 814 Variation of heat removai line with inlet temperature.
Vary Nonadiabatic Parameter K. If one increases K by either decreasing the molar flow rate FA, or increasing the heatexchange area. the slope increases and the ordinate intercept moves to the left as shown in Figure 815, for conditions of T,< To:
If T, > To, the intercept will move to the right as K increases.
Figure 815 Varialion of hear removal line with
K (K
= Url!C,F,,).
0.7.2 Heat of Generation, G( f ) The heatgenerated term, Equation 1859], can be written in terms of conversion. (Recall: X =  r,VI FA,.)
G ( T )= {AH",,
)X
(861 )
Sec. 8.7
535
Multiple Steady States
To obtain a plot of heat generated, CCT), as a function of temperaturr, we must solve for X as a function of T using the f3TFt mole balance, the sate law, and stoichiometry. For example, for a firstorder liquidphase reaction, the CSTR mole balance becomes
v= FA&
VOCAOX
kcA,(] X )
kc*
Solving for X yields I storder reaction
rk x= I
+ zk
Substituting for X in Quation (861), we obtain
Finally, substituting for k in terms of the Amhenius equation, we obtain
Note that equations analogous to Equation (863) for G ( T ) c3n be derived for other reaction orders and for reversible reactions simply by solving the CSTR mole balance for X . For example, for the secondorder liquidphase reaction
the corresponding heat generated is
At very low temperatures, the second term in the denominator of Equation (863) for the firstorder reaction can be neglected so that G(T)varies as LOW
T
G ( T ) = AHi,zAeE'Rr (Recall that AH,", means the heat of reaction is evaluated at T,.) At very high temperatures. the second term in the denominator dominates. and G(T)is reduced to
G(T) is shown as a function of Tfor two different activation energies, E, in Figure 816. If Ehe flow rate 1s decrea\ed or the reactor volume increased so as to increase 7, the heat of generation term, G(T),c h a n ~ e sas shown in Figure 817.
536
SteadyState Nonisothermal Reactor Design
Chal
High E

s
S
Figure 816
Heat generation curve.
Figure 817 Variation of heat generatton curve with spacetime
Heatgenerated curves. G(Tj
8.7.3 IgnitionExtinction Curve The points of intersection of R(T) and GtT) give us the temperaturn which the reactor can operate at steady state. Suppose that we begin to feed reactor at some relatively low temperature. To,. If we construct our G ( T ) i R ( T ) curves, illustrated by curves y and a, respectiveIy. in Figure 818, we that there will be only one point of intersection, point 1. From this point of in section. one can find the steadystate temperature in the reactor, T,,, by follc ing a vertical line down to the Taxis and reding off the temperature as show!
Figure 8 18. If one were now to increase the entering temperature to T,. the G curve, y, would remain unchanged, but the R(Tf curve would move to the ri) as shown by Iine b in Figure 8 t 8, and will now intersect the G(T)at point 2 : be tangent at point 3. Consequently, we see from Figure 818 that there are 1 steadystate temperatures. T,, and T,,, that can be reaIized in the CSTR for
entering temperature TO:.If the entering temperature is increased to ir;,, R(T) curve, line c (Figure 819). intersects the G(T) three times and there three steadystate temperatures. As we continue to increase To, we finally re line e, in which there are only two steadystate temperatures. By further incrt ing T, we reach line f, corresponding no T, , in which we have only one tern1 ature that will satisfy both the mole and energy balances. For the six enter temperatures, we can Form Table 85, relating the entering temperature to possible reactor operating temperatures. By plotting T, as a function of T,, , obtain the wellknown ignitionextinction cune shown in Figure 820. FE this figure we see that as the entering temperature is increased. the steadys temperature increases along the bottom line until To,is reached. Any f n c t i o ~ a degree increase in temperature beyond Tm and the steadystate reactor tern1 ature will jump up to T,,, , as shown in Figure 820. The temperature at wk
Sec. 8.7
537
Multrpfe Steady States
+
+.
5 Both the mole and energy bnlnnccs are mtidicd at the points of intsncction or tangency.
A
Figure I(I8 F~ndingmultiple steady ~tatekwith T,, varred
Figure 819 Finding ~nulriplesteady ~ t a t ew~~ t hT, vaned.
We must exceed
a certain feed temprilture to operate at the upper ~ t e a d ystate where the
temperature and conversion are higher.
this jump occurs is called the ignition temperature. If a reactor were operating at T,,, and we began to cool the entering temperature down from To,, the sredystate reactor temperature T,, wouId eventually be reached. corresponding to an entering temperature To,.Any slight decrease below To?would drop the steadystate reactor temperature to TF3.Consequently, To? is called the e.~tincfiantemperarure. The middle points 5 and 8 in Figures 819 and 820 represent unstabIe steadystate temperatures. Consider the heat removal line d in Figure 5 19 along with the heatgenerated curve which is replotted in Figure 821.
SteadyState Nonisotherrnal Reactor Desigl
Chap. 8
Figure 820 Temperature ignitionextinction curve.
T ~ 7
Figure 821
T&
T~
T
Stab~litjon multiple state temperatures
If we were operating at TsB! for example, and a pulse increase in reactor temperature occurred. we would find ourselves at the temperature shown by vertical Iine CZ) be~weenpoints 8 and 9. We see along this vertical line @ the heatgenerated curve. G. is greater than the heatremoved line R (G> R).Consequently. the temperature in the reactor would continue to increase until p n t 9 i s reached at the upper cteady state. On the other hand. if we had a pulse decrease helween in temperature from point 8. we would find ourselves a vertical line ~3
Sec. 8.7
539
Multiple Steady States
points 7 and 8. Here we see the heatremoved curve is greater than the heatgenerated curve so the temperature will continue to decrease until the lower aeady state is reached. That js a srnajl change in temperature either above or below the middle steadystate temperature, T,, will cause the reactor temperature to move away from this middle steady state. Steady states that behave in the manner are said to be unstable. In contrast to these unstable operating points, there are stable operating points. Consider what would happen if a reactor operating at T, were subjected to a pulse increase in reactor temperature indicated by Iine O in Figure 82 1. We see that the heatremoved line (d) is greater than the heatgenerated curve (y), so that the reactor temperature will decrease and return to T* On the other hand, if there is a sudden drop in temperature below T* as indicated by line we see the heatgenerated curre {y) is greater than the, heatremoved line (d) and the reactor temperatue will increase and return to the upper steady slate at T&. Next let's look at what happens when the lower steadystate temperature at T,, is subjected to pulse increase to the ternperature shown as line 3 in Figure 821. Here we again see that the heat removed, R, is greater than the heat generated, G. so that the reactor temperature will drop and return to Ts7. If there is a sudden decrease in temperature below T,, to the temperature indicated by Iine @, we see that the heat genesated is greater than the heat removed, G > R, and that the reactor temperature will increase until i t returns to T,,.A similar analysis could be carried out for ternperature TS1, T,?. T,,, Ts6, T,,,, and T,,, and one would find that reactor temperatures would always return to local sleadvsrare values, when subjected to both positive and negative fluctuations. While these points are locally stable, they are not necessarily globally stable. That is, a perturbation in temperature or concentration. while srnaIl, may be sufficient to cause the reactor to fall from the upper steady state (corresponding to high conversion and temperature such as point 9 in Figure 821) to the lower steady state (corresponding to low temperature and conversion, point 7 ) . We will examine this case in detail in Section 9.4 and in Problem P916B. An excellent experimental investigation that demonsrrates the muItiplicity of steady states was camed out by Vejtasa and Schmitz (Figure 822).They studied the reaction between sodium thiosulfate and hydrogen peroxide:
a,
in a CSTR oprated adiabatically. The multiple steadystate temperatures were examined by varying the Row rate over a range of space times, r, as shown in Figure 823. One observes from this figure that at a spacetime of 12 s, steadystate reaction temperatures of 4, 33, and 80°C are possible. If one were operating on the higher steadystate temperature line and rhe volwmetric flow rates were steadily increased (i.e., the spacetime decreased). one notes that if the space velocity dropped below about 7 s. the reaction temperature would drop from 70°C to Z°C. The flow rate at which this drop occurs is referred to as the ~ I O M ' O velocin'. U~
540
Ch
SteadyState Nonisothermal Reactor Design
100
0" Y,

o
f h@oreticul Expenmentalstabk SMtes
P+
Expenmental~nterrwdare states

I
0
X)
40
60
80
0
[I
rl'cl
Figure 822 Heat generation and removal functions for feed mixture of 0.8 M NalSIO, and 1.2 M H,Ot at PC.
1
4
I
I B
I I I I 1 2 1
# 6
t I M :
f (5)
Figure 823 Multiple steady
states.
By S.A. Vejtasa and R. A. Schmitz. AlChE J., 16 (3). JIS (1970). (Reproduced by permission uf the American Inst~tuteof Chemical Engineers. Copyright Q 1970 AIChE. All right reserved.) See Journal Critique Problem PSC4.
8.7.4 Runaway Reactions in a CSTR
In many reacting systems, the temperature of the upper steady state ma! sufficiently high that it is undesirable or even dangerous to operate at this I dition. For example, at the higher temperatures, secondary reactions can place, or as in the case of propylene glycol in Examples 88 and 89, evap tion of the reacting materials can mcur. We saw in Figure 820, that we operated at the upper steady state i we exceeded the ignition temperature. For a CSTR,we shall consider tuna (ignition) to wcur when we move from the lower steady sbte to the ul steady state. The ignition temperature occurs at the point of tangency of heal removed curve to the heatgenerated curve. If we move slightly off point of tangency as shown in Figure 824, then runaway is said to t occurred. At this point of tangency, T,we have not only
Sec. 8.7
Multiple Szeady States
Tc
T*
T
Flgure 824 Runaway in a CSTR.
but also the slopes of the R(7)and G(T) curves are also equal. For the heatremoved curve. the slope is
and for the heatgenerated curve, the slope is
Assuming that the reaction is irreversibIe and follows a power law model and that the concentrations of the reacting s p i e s are weak functions of temperature.  r ~ = (Aem',
fn(Ci)
then
Substituting for the derivative of (r,) wrt Tin Equation (867)
1868]
542
SteadyStele Nonisotherrnal Reactor D ~ i g n Chap. 8
where
Equating Equations (866) and (869) yields
Next, we divide Equation (865) by Equation (870) so obtain the fellowing AT value for a CSTR operating at T = T:
I f t h i . ~diflerence herween rhe reacfor temperazure and T,,AT,, is exceeded, transifion m the rrpper sready stare will occur. For many industrial reactions, U F T is typically between 16 and 24, and the reaction temperatures may be between 300 to 500 K. Consequently, this critical temperature difference AT, wiIl be somewhere around 15 to 30°C.
Stability Diagram. We now want to develop a stability diagram that will show regions of stable operation and unstable operation. One such diagram would be a pIot of S* as a function of T,. To construct this plot, we first solve Equation (871) for T.the reactor temperature at the point of tangency,
and recalling
T [Equation ( 8  J Z ) ] , calculate k" at T from rate law. calculate G ( T ) [Equation (85411, and then finally calculate S* to make a plot of S* as a function of T, as shown in Figure 825. We see that any deviation to the right or below the intersection of Cw ( 1 + K ) and S' will result in runaway. We can now vary 7,. then caIcutate
(k*= AeE/RT*),calculate
r*,
Sec, 8.8
Nonisothermal Mdliple Chrnical Reactions
EYpre 825 CSTR stability diagram.
For example, for a firstorder reaction, the equation for S is
Shelf
We simply combine Equation (872) and the equation for T, and then substitute the resdt into Equation (873) and plot S* as a function of Tw From Figure 825, we see that for a given value of [C,(I+u)I, if we were to increase the entering temperature To from some Iowvalue To',, (T,,) to a higher entering temperature value Tm IT,,), we would reach a point at which runaway would occur. Further discussions are given on the CDROM professional reference shelf R8.2. Referring to Equation (8701, we can infer
we will not move to the upper steady state, and runaway will not occur. How
ever, if
runaway will occur.
8.8 Nonisothermal Multiple Chemical Reactions Most reacting systems involve more than one reaction and do not operate isothermally. This section is one of the most important sections of the book. It ties together all the previous chapters to analyze multiple reactions rhar do not take place isothermally.
544
Steady+StateNonisolherrnaT Reactor Design
CV
8.8.1 Energy Balance for Multiple Reactions in PlugFlow Reacts
In this section we give the energy balance for multiple reactions. We begi recalling the energy balance for a single reaction taking place in a PFR H is given by Equation (8351,
When q multiple reactions are taking place in the PFX and there are m cies, ir is easily shown that Equation (835) can be generalized to
i = Reaction number j = Species
The heat of reaction for reaction i must be referenced to the same species i rate, r , , by which AHRxbis multiplied, that is,
[rqll[AH~,l
=
s= eI,,,[
Joules "released"'in reactio Moles of j reacted in reactic
Moles o f j reacted in reaction i Volume  time
I
~'releasd" in reaction i Volume time
where the subscript j refers 10 the species, the subscript d refers to the parti reaction, q is the number o f independent reactions, and m is the numb species. Consider the following reaction sequence cartied out in a PFR:
) B
Reaction 1:
A
k'
Reaction 2:
B
k l > C
The PIX energy balance becomes
where AHRxlA = FJlmol of A reacted in reaction 11 and AH,,, = [kllrnol of B reacted in reaction 21.
(:
(:
Sec. 8.8
Nonisothermal Multiple Chern~calReactions
Example 818
1
ParalIeL Reoc~bnsin a PFR with Heat
EflecIs
The foilowing gasphase reaciions occur i n r PFR:
Reaction 1: Reaction 2:
A 2A
&B k,
C
 r , , = k,,C,,
IE8 10.1)
rlA = kZ4?*
(E8 10.2)
Pure A i s fed at a rate of 100 molls. a ternprature of 150PC, and a concentration of
0.I rnolldm3 . Determine4the temperature and Row rate profiles down the reactor.
AHR,,, = 20.000 J/(rnol of A reacted in reaction 1)
AHR,?, = 60,000 J/(mol of A reacted in reaction 2) Living Example Jroblem
Solslrion The PFR energy balance becomes [cf. Equation (876)]
Mole balances:
One of the major goais of this text is
I
hat the reader will
be able ta solve multiple reactions with heat effects.
Rate laws, relative rates, and net rates: Rate Eaws
546
Reaction 1:
SteadyState Nonisothermal Fleactor Oestgn
'IA = 5 . 1 I , ~
Chap. 8
=~ ~ A C A
I B
Reaction 2: Ner mtes:
Stoichiometry (gas phase AP = 0):
The algorithm for multiple reactions w t h heat effects
[
I):
k , , = 10 exp 3000    sL1
[
k2* = 0.09 exp 9000
(3;
 h)]
(3;
(Tin K)
2
Energy balance:
The Polymath program and its graphical outputs are shown in Table E810.1 and Figures E810.1and E810.2.
Sec. 8.8
Nonisotherrnal Multipls Chemical Reactions
547
TABLEE810.1. POLYMAWPROGRAM Equarim:
I
Lfving Eram~icProblem
POLYMATH Results Examplt &111F*tnllcl R n a b
In 4 PFR aILb Hm En=@m1Mwc.
Ra3.1.u~
Splculrtd values or !C DEO mrlablra initial value
Variablm
v Fd
m FC

T kla CTO PC TO Cb.
1
9.738206
I00
o
0 D 423 482 .a247 551.05566
35.04326 w.41a369 812.19122
2 .?>BE06 55.04326 22.478369 722.08816
4.4BPE+01 l.4BE*07 0.1
?.4266+04
100 PO3
77.521631
0.1 0.0415941 0.0169B6
2.069E09 0.04i5941 0.016986
373 39077
5.019E05
5.019E05
840.11153
1.591E11
1.591E11
0.1
0.1
100 433
77.521631
123
0.1 0
2.069E09 0
Cc
0
0
48.28947 1.5305566
ria
M ~ value
1
Cb
r21
fL
0
La3 402.8247 553.05566
k21
minim1 valua msximl vaf u t
0 100 0
3.716I+O6 0.1
423
lRKF451
rt
I)IVWMIMB ~ U * I D M aa meted by me uaer
1:I d(FaYd(W = r l u+Ra i 2 l dffbYdtW = rla
f xpliil w & t M a as anratedby !he uner I I I k ~ =a ~ O ' ~ ~ ( * O W ' ( I / ~ [ K ~ . I ~ Q ) I 2 l k2e O.OB'erp(QMXl'(l~9C&1rr)J

1 7 1 Cto = 0.1

14 I FI Fa+Fb+Fe 151 T o = 4 2 3 1 6 1 Cs Cto'(FaFl)'(Tdr) I?I Cb = Clo'{FWF!).(TdQ
l a ] Cc Cto'{FMt)'(TPrT) 1 9 1 r l a  hla'Cn I1 0 l Ra P k2a'Cm
850
Why does the
550
temperalure go through a maximum va rue:'
4M
00
02
04
06
08
V
Figure E810.1 Temperature profile.
10 DO
a2
04
v
06
06
10
Figure ES10.2 Profile of molar flow rates F,.F,,and F,.
548
SteadyState Nonisothermal Reactor Design
Cht
8.8.2 Energy Balance for Multiple Reactions in CSTR RecalI that F,,,X = r,V fur a CSTR and that AH,,(Ir) = AH;, + AC,(T so that Equation (827) for the steadystate energy balance for a single reac may be written as
For g multiple reactions and m species, the CSTR energy balance becomes
1
Energy balance
for multiple reactions in a CITR
1
Substituting Equation (850) for Q, negIecting the work term. and assum constant heat capacities, Equation (880) becomes
For the two parallel reactions described in Example 810. the CSTR ene balance is
Mdor goal of CRE
(8
One of the major goals of this text is to have the reader solve problems invc ing multiple reactions with heat effects (cf. Problem P&26c). Example 811 Multiple Reactions in a CSTR
The elementary liquidphase reactions
take place in a 10dm3 CSTR. What are the effluent concentrations for a volume feed rate of 1000 drn31rnin at a concentration of A of 0.3 molldm3?
The inlet z e r n p t u r e is 283 K.
Additional information:
k, = 3.3 minI at 300 K. with E, = 9900 cat/rnol k2 = 4.58 minI
at 500 K,with
El = 27,000 callmol
Sec. 8.8
I
I
hfi',,~, = 55.00fi Jlmol A
The &ons
UA = 40,000 JJrninK with
follow elcmentay rate laws
I. Mole Balance on Every Species A: Comb~nedmole balance and rate law for A:
Solving for C, g'~ v e sUS
I
I 1
I
549
Nonisolherrnal Multiple Chemical Reactions
B: Combined mote balance and rate INI f a B:
Solving for C, yields
2. Rate Laws:
3. Energy Balances: Applying Equation (882)to this system gives Substituting for r l , and r13 and rearransing, we have
r, = 57'C
550
SteadyState Nonisothermal Reactor Design
Chap. 8
We are now going to write a Polymath program to increment temperature 10
obtain
G(T) and R(r). The Polymath program to plot R ( T ) and G ( T ) vs. T i s shown in Table E811.1, and the resulting graph is shown in Figure E811.1.
POLYMATH Results Example 811 Multiple Reactions in a CS7"R 0813zW. Rev5.1.232
Incrementing temperature in this manner is an easy way to generale RU)and G(T) plots
Living Example Problen
Differentialequations as entered by the user [ 1 1 d(T)ld(t) = 2 Expl'iclt equations as entered by the user t l l Cp=200 [ a ] Cao=0.3 [ 3 J Toe283 [ 4 1 tau=.Ol [ 5 ] DHt =55000 16 1 DH2 = 71500 171 v0=1000 [ 8]
€2 = 27000 =9900
191 El
[10] VA=400M) t l l l Ta=330 [ 12 1 k2 = 4.58'exp((E2/1.987)*(1/5007 1 13 3 kl = 3.3*exp{(El/l.907)*(113001TT)) [ 14 1 Ca = Caol(1+lau'kf ) [ 15 I kappa = UN(vo'Cao)lCp 11 6 1 G = tau'kl/{l +kl * t a u ~ ~ ~ 1  k l ' t a u + k 2 " t a u * D ~ 2 ~ ~ ( i + t ~ u ~ k l ) * ( l + ( 17 I TC= (To+kappa'Ta)l( 1+kappa) [ 18 I Cb = tau'kl'Cd(1 + k 2 r ~ u ) [ 19I R = Cp'f 1+kappa)'(TTc) 120 I Cc = CaoCaCb [Ill F=GR
m)
When F = O
G(T) = R(T) and the steady states can be found.
We see that five steady states ( S S ) exist. The exit concenrrarions and tempemtures listed in Table E8 1 1.2 were interpre~edfrom the labular output of the ~ o l ~ ~ n a t h program.
Sec. 8.9
Radlal and Axial Variations in a Tubular Reactor
Wow! Five (5) multiple steady states!
71K)
Figure E81I.I Heatremoved and heatgenerated curvec
We note there are five steady states (SS) whose values are given in Table ES11.2. What do you think of the value of tau? Is it a realistic number? TABLE ESI 1.2.
EFFLC'EVI C O ~ ' ~ A T I AND O YTEMPERANRES S
SS
T
cA
CE
cc
1
330 363 449
0.285
2 3
0.189
0015 0.111
4
558
0.033 0.N
0.265 0.163
5
677
0.001
0.005
0 0.0 0.002 0.132 0.294
8.9 Radial and Axial Variations in a Tubular Reactor EMLAB application
In the previous sections we have assumed that there were no radial variations in velocity. concentration, temperature ar reaction rate in the tubular and packed bed reactors. As a result the axial profiles couId be determined using an ordinary differential equation (ODE) solver. In this section we will consider the case where we have horh axial and radial variations in the system variables in which case will require a panial differenrial IPDE) solver. A PDE solver such as FEMLAB. will allow us to solve tubular reactor problems for horh the axla1 and radial profites, as shown on the web module. CVe are going to carry out differential mole and energy balances on the differential cyllndricai annulus shown in Figure 326.
SteadyState Nonisothermal Reactor Design
Cha
Front View
w1,
Figure 826 Cylindr~calshell of thickness Ar, length h, and volume 2 n r S r k
Molar Flux
In order to derive the governing equations we need to define a couple of ten The first is the moIar flux of species i, Wi {mollm2 s). The molar flux has t cornpsnents. the radial component W,,, and the axial component, W,,.1 molar flow rates are just the product of the molar fluxes and the crosssectio areas normal to their direction of flow A,. For exampte, for species i flow in the axial (i.e., z) direction
where W,: is the molar flux in the z direction (rnoVm2/s), and A, (m2) is the crosssectional area of the tubular reactor. In Chapter 11 we discuss the molar fluxes in some detail, but for now us just say they consist of a diffusional component, D,(aC,/az) , and a ct vective flow component, U:C,
ac.t.upci
wir=  D ~ 
a2
13r
where D, is the effective diffusivity (or dispersion coefftcienl) (m2/s). and is the axial molar average velocity (mls).Similarly, the flux in the radial din tion i s
ac.+ UrCi
Wir =  D,
Radial Direction
ar
(85
U,( m l s ) is the average velocity in the radiaI direction. For now we n neglect the veiocity in the radial direction, i.e., U, = 0. A mole balance or cylindrical system volume of length Az and thickness Ar as shown in Fig1 826 gives where
Rad~aland Rxlal Variations In a Tubular Reactor
Sec 8.9
Mole Balances on Species A
(
Moles of A = i n ) .
(
(
1
Crasssectionai area = WAr.2~rAr normal t o axial flux
)
*
( M ~ E ~ A ) Moks of* +(M;:~:*)( out at ( r + At)
out at ( z + Az)
Dividing by 2xrAr4 and faking the limit as Ar and Az
hz ,O
hfOlesof
)
Similarly, for any species i and steadystate conditions.
Using Equation (8831 and (884) to substitute for W!:and W,,in Equation (885)and then setting the radial velocity to zero, U,= 0, we obtain
This =quation For steadystate conditions and assuming 157, does not vary also be d~scussed funher in
Chapter 14.
in the axial direction,
554
SteadyState Nonisothermal Reactor Design
Chap. 8
Energy Flux
Wen we applied the first law of thermodynamics to a reactor to relate either temperature and convtrsian or molar flow rates and concentrati~n.we arrived at Equation (89). Negleaing the work term we have for steadystate conditions Conduction
Convection
In terms of the molar fluxes and the crosssectional area and (q [email protected],)
The q term is the heat added to the system and almost always includes a conduction component of some fom. We now define an energy flux vector. e. (J/rn2 s), to include both the conduction and convection of energy. +
e =e
n e r ~flux J/s.rnZ
e = Conduction
+ Convection
where the conduction term q (kJlm2 . s) is given by Fourier's law. For axial and radial conduction Fourier's laws are 4; =
k,aT 8,
and
q,=kca T
ar
where k, is the thermal conductivity (JJm.s.K). The energy transfer (flour) is the vector flux times the crosssectional area, A,, normal to the energy flux
Energy flow = e  A, Energy Balance Udng the energy Bux. e, to carry out an energy balance on our annulus [Figure 826) with system volume 7nrhr&, we have
(Energy flow in at I.) = e, A,, = e; 2 n r h (Energy flow in at z ) = e,A, = e; 2arAr Accumulat~on o f Energy In VoIumc ( 2 r r ~ r . k )
See. 8.9
Radia! and Axial Vanetlons in a Tubular Reactor
Dividing by 2nrArbz and taking the Ijmit as & and Az
+ 0,
The radial and axial energy fluxes are e, = g,
+ E' W,,HE
e. = g,+CW,, H,
Substituting for the energy fluxes, e, and e;,
and expanding the convective energy fluxes, Z WiHi,
Axial: Substituting Equations (893) and (894) into Equation (892), we obtain upon rearrangement 'I
Recognizing that the term in brackets is related to Equation (885) for steadystate conditions and is just the rate of formation of species i. rimwe have 1 a % Z H j f j 3 ?yzmio (rqr) rar
Recalling
and
ilz
az
(W5)
556
SteadyStale Nonisothermal Reactor Design
Ch:
we have the energy in the form
Some Initial Approximations
I . Neglect the diffusive term wrt the convective term in the exp sion involving heat capacities
Assumption
With this assumption Equation (896) becomes
Energy
with radial and axial grddientq
Assumption 2. Assume that the sum CPm= ZCp,C,= C,,Z8,Cg
is cons1
The energy balance now becomes
Equation (898) is the form we will use in our FEMLAB problem. In rr instances, the term C P is~just the product of the soIution density and the capacity of the solution (kJkg K).
Coolant Balance
We also recall that a balance on the coolant gives the variation of coolant t peralure with axial distance where Uhris the overall heat transfer coeffic and R i s the reactor wall radius
For laminasfiow, the veIociv profile is
where Uois the average velocity inside the reactor.
S x . 8.S
Radial and Axial Varia!ions In a Tubular Reacior
Boundary and initial conditions A. Initial conditions ifother than steady state t = O , C,=O. T=T,? f o r z > O a l l s B. Boundary condition 1) Radial
(a) At r = 0, we h a ~ esymmetry a T l d r = 0 and aC,/ar = 0 . (bj At :he tube wall r = R, the temperature flux to the wall on the reaction side equals the convective flux out of the reactor into the shell side of the heat exchanger.
( c ) There is no mass flow through the tube walls aC,/ar = 0 at
Id)
r = R.
At the entrance to the reactor .z = 0,
T = Toand Ci = Cn (e) At the exit of the reactor z = L,
The following examples wiIl solve the preceding equations using FEMLAB. For the exothermic reaction with cooling, the expected profiIes are
Example 812 Radial Effects in Tubuhr Reactor This example will highiight the radial effects in a tubular reactor, which up until now have been neglected to simplify the calculations. Now, the effects of panmeters such as inlet Temperature and Row rate will be studied using the software program FEMLAB. Follow the stepbystep procedure in the Web Module an the CDROM. We continue Example 88, which discussed the reaction of propylene oxide (A) with water ( 8 ) to form propylene glycol (C). The hydrolysis of propySene oxide takes place readify st room temperature when catalyzed by sutfuric acid.
558
SteadySlate Nonisothermal Reactor Design
Chap. B
This exothermic reaction is approximated as a firstorder reaction given that the reaction rakes place in an excess of water. The CSTR from Example .88 has been repIaced by a tubular reactor 1.0 m in length and 0.2 m in diameter. The feed to rhe reactor consists of two streams. One stream is an equivolumetric rnlxture of propylene oxide and methanol, and the orher stream is water containing 0.1 wt Yr sulfuric acid. The water is fed at a volumetric rate 2.5 times larger than the propylene oxidemethanol feed. The molar flow rate of propylene oxide fed to the tubular reactor IS 0.1 moUs. There is an immediate temperature rise upon mining the two feed streams caused by the heat of mixing. In these calculations, this temperature rise is already accounted for, and the inlet temperature of both streams is set to 312 K. The reaction rare law i s
with where E = 75362 Jlrnol and A = 16.96 x 10" hI, which can also be put in the form
With k, = 1.28 h' ar 300 K. The thermal conductivity kt of the reaction mixture and the diffusivity D, are 0.599W l m K and 1 p rn2/s, respectively. and are assumed to be constant throughout the reactor. In the case where there is a heat exchange between the reactor and its surroundings, the overall heattransfer coefficient is 1300 Wlm2/K and the temperature of the cooling jacket is assumed to be constant and is set to 273 K. The other propeny data are shown in Table E8 12.1.
See. 8.9
Radlal and Axial Variations In a Tubular Reactor
Solution
Mole Balances: RecdIing Equation (886) and applying it to species A
 !)]c, RT, TJ
rA = 4 T,) cxp[f(l
(
(E8 12.2)
Stoichiometry: The conversion along a streamline (r) at a distance z
The overall conversion is
I
The mean concentration at any distance z
I
For plug flow she velocity prohle is
v: = u, The Laminar flow velacily profile is
I
Recalling the Energy Balance
Assumptions .iviog Example Problcm
1. U, is zero. 2. NegIect axial diffusion/dispersionflux wn convective flux when summing the heat capacity times their fluxes.
3. Steady state.
560
1
SteadyState Nonisothermal Aeactor Destgn
Cha
Cooling jacket
 2aRUh,(T,(z) nl'C, AT0 1
a=
 T,)
Roundary conditions
At r = R. then ~t
2
= 0,then
aca  0 and  k ,  =ST dr
ar
Lrh,(T,(z)T,}
C,= C , andT= TO
(E812.9
(E812. I0
These equations were solved using FEMLAB for a number of cases inctud~ngas batic and nonadiabatic plug flow and laminar flow: they were also solved with without axial and radial dispersion. A detailed accounting on how to change parameter values in the FEMLAB program can be found in the FEMLAB Inst1 t~onssection on the web in screen shots simllar to Figure EX12. I . Figure E8I gives the data set in SI units used for the FEMLAB example on the CDROM.
Nore There is a ~tepbystep FEMLAB tutorial uslng screen fhots for th~sexample on
the CDROM.
Define expressian
Figure FA12.1 FEMLAR screen shot of Data Set.
Color surfaces are used to show the concentration and temperature profiles. sirr to the black and white figures shown in Figure ES12.2. Uqe the FEMLAB pmg on the CDROM to develop temperature concentration profiles similar to the r shown here. Read through the FEMLAB web module entitled "Radial and A Temperature Gradient." before running the program. One notes in Figure E&that the conversion is lower near the wall because of the coofer fluid temperat These same profiles can be found in color on the web and CDROM in the mdutes. One notes the maximum and minimum in these profiIes. Near the wall. ternperamre of the mixture is lower because of rhe cold wall temperature. Co quently. the rate will be lower, and thus the conversion will be lower. However, r next to the waH, the velocity through the reactor IS almost zero so the react spend a long dme In the reactor; therefore. a greater conversion is achieved as nl by the upturn right next to the walt.
Sec 810
I Result5 of the FEMLAR si~nulation
The Pnctrcat Sree
fa) Tarnwrstura S u k e
Radial Temperature Profllss
Aadrar Cocation rm)
RMaisl Lmdlh (m)
Conversion S m e
Radial Clxrversron Profllm
Radial Locattm (m]
(dl Figure E812.2 (a) Temperature surface, (b) ternperamre surface profiles. (c) convenion surface. and Id) radial conversion profile.
8.1 0 The Practical Side Scaling up exothermic chemical reactions can be very tricky. Tables 8.6 and 8.7 give reactions that have resulted in accidents and their causes, respectively.IO
'OCourtesy of J. Singh, Chemical Engineen'ng, 92 (1997) Engineering, 54 (2002).
and B. Venugopal, Chemical
562
SteadyState Nonisothermal Reactor Design
Chap. 8
Conttiburion, Cause
Lack of knowledge of reaction chemistsy
AlkyIation (FriedelCrafts)
9t 20
Problems with malerial quality
9
Temperaturecontrol problems
19
Agitation problems
10
MISchargingof reactants or catalyst
21
Poor ma~ntenance
15
Op~ratorerror
5
More information is given in the Summary Notes and Professional Reference Shelf on the web. The use of the ARSST to detect potential problems will be discussed in Chapter 9. Summary Motes
Closure. Vinually all reactions that are carried out in industry involve heat effects. This chapter provides the basis to design reactors that operate at steady state and involve heat effects. To model these reactors, we simply add another step to our algorithm; this step is the energy balance. Here it is important to understand how the energy bdance was applied to each reaction type so that you will be able to describe what would happen if you changed some of the operating conditions (e.g., To).The living example problems (especially 8T83) and the ICM moduIe will help you achieve a high level of understanding. Another major goal after studying this chapter is to be able to design reactors that have multiple reactions taking place under nonisothemia~conditions. Try working Problem 826 to be sure you have achieved his goal. An industrial example that provides a number of practical details is incIuded as an appendix to this chapter. The last example of the chapter considers a tubular reactor that has both axial and radial gradients. As with the other living example problems, one should vary a number of the operating parameters to get a feel of how the reactor behaves and the sensitivity of the parameters for safe operation.
Ghap. 8
Summary
SUMMARY For the reaction
A + bB + $ C +  Dd a
1.
a
a
T h e heat of reaction at temperature T. per mole oFA, is
2. The mean heat capacity difference. ACp. per mole of A is
where Cp, is the mean heat capacity of species i between temperatures T, and T . 3. When there are no phase changes, the heat of reaction at temperature T is related to the heat of reaction at the standard reference temperature T, by
4.
Neglecting changes in potential energy, kinetic energy, and viscous dissipation, and for rhe case of no work done on or by the system and all species entering at the same temperature, the steady state CSTR eneqy balance is
5.
For adiabatic operation of a PFR,PBR, CSTR, or batch reactor. the temperature conversion relationship is
Solving for the remperature, T.
6.
The energy balance on a PFR/PBR
564
SteadyState Nonisothermal Reactor Design
Ch:
In terms of conversion,
7 . The temperature dependence of the specific reaction rate is given in the fc
8. The temperature dependence of the equilibrium constant is given by v HoFs equation for AC, = 0,
9.
Multiple steady states:
10. The criteria for Runaway Readions accurs when (T, T,)> R> 7 ~Alrio . see the Fauske web site: wltwfauske.com.
9.3 Semibatch Reactors with a Heat Exchanger
'
t i c ci:\tril~ly lutrll wac
In our past discussions of reactors with heat exchanges, we assumed that the ambient temperature T, was spatially uniform throughout the exchanger. This assumption is true if the system is a tubular reactor with the external pipe surface exposed to the atmosphere or if the system is a CSTR or batch where the coolant flow rate through exchanger is so rapid that the coolant temperatures entering and leaving the exchanger are virtually the same. We now consider the case where the cmlant temperature varies along the length of the exchanger while the temperature in the reactor is spadally unifonnThe cooEant enters he exchanger at a mass flow rate m, at a temperature Tel and leaves at a temperature To? (iee Figure 93). As a first approximation. we ;issurne a quasisteady state for the coolant flow and neglect the accumulatian term (i.e.. dT,/dt = O 1. As a result, Equation (849) will give the rate of heat transferfrom the exchanger ro the reactor:
Using Equation 1849) to substitute for
Q in Equation (99), we ohtain
Sec. 9.3
Semibatch Reactors with e Heat Emhanger
615
Heat Exchanger Coiled Tubing
F i ~ r 93 e Tank reactor with heat exchanger.
At steady state ( l d t = 0 ) Equation (921) can be solved for the conversion X as a function of reaction temperature by recalling that FAOX = rAV and
and neglecting ACp and then rearranging Equation (921) to obtain Steadystate energy balance
X=
We are assuming that there is vinually no accumulation of energy in the coolant fluid, that is,
61 6
UnsteadyState Nonisothermal Reactor Desrgn
Cha
The secondorder ~aponificationof ethyl acetate is to be carried out in a selnib; reactor shown schematrcally i n Figure E94.1.
Living Example Problem
Aqueous sodium hydroxide is to be fed at a coocenaation OF t h l l m B . a tempentun 300 K,and a rate of 0.004 m3/s to an initial volume of 0.2 m" of water and ethyl ace1 The initial concentsat~ons of ethyl acetate and water are 5 kmol/m3 and 3 kmollm" respectively. The reactlon is exothermic, and it is necessary to add a 1. exchanger to keep its temperature below 315 K. A heat exchanger with UA = 331 U s . K is available for use. The coolant enters at a rate of tOO kgls and a temp ture of 285 K. Is the heat exchanger and coolant flow rate adequnte to keep the reactor t r perature below 315 K? Plot temperature, C,4bCB.and C, as a function of time. Additionul informatiun: 3
KC ]03885.44/T 
AH",
=
79+076kJ/ kmol
CpA= 170.7 J/rnol/K
Cwo=55kmollrn3
Feed: Initially:
C,
= 30.7 kmolJm3
CBn=f.0km01Jm3
CA,= 5 krnol/m3
C,, = O
Figure E94.1 Semibatch reactor with heat exchange.
k from J. M. Smith, Chemical Engineering Kinetics, 3rd ed. (New York: McGw Hill. 1981). p. 205. AHR, and Kc calculated from vatues given in Perry's Chemk Engineem' Hrurdbook, 6th ed. (NewYork: McGrawHill. t984), pp. 3147.
Sec. 9.3
Semrbatch Reactors with a Heat Exchanger
Snl~rtiut~
Mole Balances: (See Section 4.10.2. )
I
Initially, lVwi
= V,Cwr= (0.2)(30.7) = 6.14 kmol
Rate Law:
Stoichiornetry:
(
1 =
However. CpB= CPw: n
1
1
F,Cp, in Equation (99).Because only B . P. and water continually Row into the reactor Energy Balance: Next we replace
where
618
UnsteadyState Nonisofherml Reactor Design
Chap. 9
d ~ ~, C F $ T , , T ) I 1  exp( UAlk,Cp)l F,,C,{I + B,)(T To)+ (r,V) AH, C P / N B + N ~%+NW1f + CPANA (E94.10) RecaIIing Equation (847) for the outlet temperature of the Auid in the heat exchanger
r m:;pl
T,, = T  ( T  T,,)exp 
The Polymath program is given in Table E94.1.The solution results are shown in Figures E94.2 and E94.3.
Living Example Problem
Set. 9.4
I
Unsteady Operation of a CSTR
Figure E94.2 Temperaturetime trajectory in a sernibarch reactor
Figure E943 Concentrationtime trajectories in a semihatch reactor.
9.4 Unsteady Operation of a CSTR 9.4.1 Startup Startup of a CSTR
Rcfercnce Sheff
In reactor startup it is often very important faow remperature and concenrrations approach their steadystate values. For example, a significant overshoot in temperature may cause a reactant or product to degrade, or the overshoot may be unacceptable for safe operation. If either case were to occur. we would say that the system exceeded its prtlcrical stabiliv limit. Although we can solve the unsteady temperaturetime and concentrationtime equations numerically to see if such a Iimit is exceeded, it is often more insightful to study the approach to steady state by using the temperatureconcenrra~io~? phase plane. To illustrate these concepts we shall confine our analysis to a liquidphase reaction carried out in a CSTR. A qualitative discussion of how a CSTR approaches steady state is given jn PRS R9.4. This analysis, summarized in Figure SI in the Summary for this chapter, is developed to show the four different regions into which the phase plane is divided and how they allow one to sketch the approach to the steady state.
(
Example 95
Startup of o CSTR
Again we consider the production of propylene gIycnl IC) in a CSTR with a heat exchanger in Example 88.Initially there i s only water at 75'F and 0.1 wt % H2S04 in the 500gallon reactor. The feed stream consists o f 80 Ib mallh of prnpylene oxxde (A), 1000 Ib molth of water (3)containing 0.1 wt % H2S04, and 100 rb mollh of methanol (M}. Plot the temperature and concentration of pmpylene oxide as a functron of time, and a concentration vs. temperature graph for different enrering temperatures and initial concentrations of A in the reactor.
620
UnsteadyState Nontsothermal Reactor Design
Chal
The water coolant flows through the heat exchanger at a rate of 5 lbls (1( Ib mollh). The molar densities of pure propyiene oxide (A). water (B), and met no1 (M) are p, = 0.932 Ib mollft3. = 3.45 lb mollft3, and p, = 1.54 rnollft3, respectively.
Solution
Mole Balances: Initial Conditions A:
d%=r,+
Rate Law: Stoichiometry:
dr
(c~, CA)~, V
 r A = kcA rA =  r B = rc
Enew Balance:
with
and
Ta2= T  (T T,,)exp
0
(E95
Sec. 9.4
Unsteady Operation of a CSTA
Evaluation of parameters:
Neglecting ACp because it changes the heat of reaction insignificantly over the temperature range o f the reaction, the heat of ofaction i s assumed constant at

The Polymath program is shown in Table E95.1. TABLE E95.1.
POLYMATH~ o G I I A MFOR C S m START~TP
622
Unacceptable sranup
UnsteadyStale Noniaothsrmal Reactor Design
Chap. 9
Figures (E95.1)and (E95.2) show the reactor concentration and temperature of propylene oxide as a function of time, respectively, for an initial temperature of 75°F and only water in the tank (LC., C,, = 0). One observes, both the temperatute and concentration oscillate mound their steadystate values (T= 138"E Cn = O.Q.79 Ib rnol/ft'). Figure (E95.3) shows the phase plane of tempemture and propylene oxide concentration for three different sets of initial conditions (T, = 75"F, C,, = 0; T, = lSO°F, CAi= 0;and = 160%. CAr= 0.14 Ib mollft3), keeping Toconstant. An upper limit of 180°F should not be exceeded in the tank. This temperature is the practical stahilip limit. The practical stability limit represent a temperature above which it is undesirable to operate because of unwanted side reactions. safety considerations, or damage to equipment. Cansequently, we see if we started at an initial temperature of 160°F and an initial concentration of 0.14 molldrn'. the practical stability limit of 180°F would be exceeded as the reactor approached its steadystate temperature of 1 3 8 T See the concentrationtempemture trajectory in Figure E95.4.
Figure E95.1 Propylene oxide conceneation as a function of time.
Figure E95.2 Ternperamretime trajectory for CSTR startup.
Figure E95.3 Concentrationtemperature
Figure E95.4 Concentrationtemperature phase plane.
Oops! The practical s~abilitylimit was exceeded.
phaseplane ~rajeclon:
.
After about 1 .h h the reactor isoperating at steady state with the foHowing vaIues: 

Sac.9.4
Unsteady Opratlan of a CSTR
9.4.2 Falling Off the Steady State
We now consider what can happen to a CSTR operating at an upper steady state when an upset occurs in either the ambient temperature, the entering temperature, the flow rate, the reactor temperature, or some other variable. To ilIusaate, Iet's reconsider the production of propyIene glycol in a CSTR,which we just discussed.
I
Example 96 F d h g Oflthe Upper Steady S w e
In Example 95 we saw how a 500gal CSTR used for the production of propylene glycol approached steady state. For the Row rates and conditions (e.g., To = 75"F, T,,= 60°F). the steadystate temperature was 138"E and the corresponding mnversion was 75.5%. Determine the steadystate temperature and conversion that would result if the entering temperature were ro drop from V0F to 7PF,assuming that all other conditions remain the same. First. sketch the steadystate conversions calculated from [he mole and energy balances as a function of temperature before and afrw the drop in entering temperature mcurred. Next, plot the "conversion," concentration of A, and the temperature in the reactor as a function of time after the entering temperature drops from 75°F to 70°F.
)
The steadystate convenions can be calculated from the mote balance.
(
mi h r n tbe energy balance.
before (To = 75°F) and after IT, = 70T) the upset occurred. We shall use the parameter values given in Example 95 (e.g., FA, = 80 Ib rnol/h. UA = 16,000 Btulh . "E)lo obtain a sketch of these conversions as a function of temperature, as shown in Figure E96. I . We see that for To =7OoF the reactor has dropped below the extinction ternPrature and can no longer operare at the upper steady state. In Problem P416. we will see it i s not always necessary for the temperature to drop below the extinction temperature In order to fall to the lower steady state. The equations describing the dynamic drop from the upper steady state to the lower steady smte arc identical ro those given in Example 95: only the in~tialcondit~onsand entering temperature are different. Consequently, the qame Polymath and MATLAB programs can be used wtth these modifications. (See Liilirtg ErampIc 96 on the CDROM.)
624
UnsteadyState Nonisotherrnal Reaclor Design
Figure E96.1 Convetxion from mole and energy balances as
Ch,
a Function
of temperature.
Initial conditions are taken from the final steadystate values given in Ex ple 95.
C,, = 2.12 Ib mollft~ C, = 0.226 ib molJft' T,= I38.5"F Change T, to 70°F Because the system is not at steady state, we cannot rigornusly define a convers in terms of the number of moles reacted because of the accumulation within the re tor. However, we can approximate the conversion by the equation X = ( 1  C,IC, This equation is valid after the steady state is reached. Plots of the temperature i the conversion as a function of time are shown in Figures E96.2and E96.3,r a p tively. The new steadystate temperature and conversion are T = 83.6"Fand X = 0.
Figure E96.2 Temperature versus time.
Figure T5962 Conversion versus time.
We could now see how we can make adjustments for upsets in the reac operating conditions (such as we just saw in the drop in the entering temperatu so that we do not fall to the Iower steadystate values. We can prevent this drop conversion by adding a controller to the reactor. The addition of a controller is d cussed in the Pmfessional Reference Shew R9.2 on the CDROM.
Sec. 9.5
625
Nonrsothermal MulttpFe Reactions
9.5 Nonisothermal Multiple Reactions For multiple reactions occurring in either a semibatch or batch reactor, Equation (921) can be generalized in the same manner as the steadystate energy balance, to give 4
~ C C P ~ ( T , I  Texp(UA/mcCp,)l+ )[~
dT 
1
r,,
I=
c N, CP,
dt
V A H R q ( T )  Z FjoCp,(TTo)
I
For large coolant rates Equation (923) becomes
I
Example 97 MuItipk Reaclions in a Semibotch Rebctor The series reactions
Livlng Example Proble
are catalyzed by H2S04.All reactions are first order in the reactant concentration. The reaction ts to be carried our in a semibatch reactor that has a heat exchanger inside with UA = 35.000 callh K and an exchanger temperature, T,, of 298 K. Pure A enters at a concentnition of 4 moI/dm3, a volume&ic flow sate of 240 dm3/h, and a temperature of 305 K. Initially there is a total of I00 dm3in the reactor, which contains 1.0 rnoI/drn3 of A and 1.0 mol/dm3 of the catalyst W2S04.The reaction rate is independent of the catalyst concentration. The initial temperature of the reactor i s 290 K. Plot the species concentrations and temperaturn as a function of time.
626
UnsteadyStnTe NonisoFmrmal Reactor Design
Chap. 9
AaYiriomI information:

klA = 1.25 h' at 320 K with EIA= 9500 caIIrno1
CpA = 30 call mol K
bB = 0.08 hI
CpB = 60 cal/mol K
at
300 K with
= 7000 cd/mol
AHRxl, = 6500 cal/mol A &IR =, +8000 ,, callmol
Cpc = 20 cat/ mol
B
c%,SO,
SoIuiion
Mole Balances:
Rate Laws:  r , , = RIACA
rza = k2RCB
Stoichiomern (liquid phase): Use C,, C,, Cr Relative rates: flB=
=
I r 2
I*
3 rZ8
Net rates:
~ ~ , = @ m o ~dm3 x ~  = gmot mdr n 3
h
h
K
= 35 cal/mol . K
Sac. 9.5
1 ,
Nonisothermal Multl~leReactions
Enecgy Balance:
Equations (E97.1through ) (E97.3) and (E97.8)through (E97.12) can be solved simultaneously with Equation (E97.14) using an ODE solver. The Polymath propram is shown in Table E97.1 and the Matlab program is on the CDROM. The time graphs are shown in Figures E97.1 and E97.2.
kving Example Problem
1
Figure E97.1 Cancentrationtime.
Figure E97.2 Temperature (K)time (h).
628
UnsteadyState Nonisothermal Reactor Design
Ch
9.6 Unsteady Operation of PlugFlow Reactors In the CDROM. the unsteady energy balance is derived for a PFR.Neglec changes in tota! pressure and shaft work, the following equation is derive[ Reference Shelf Transient energy balance on a PFR
This equation must be coupled wih the mole balances: Numerical solution required for these three coupled equations
and the rate law,
~ e f e r e n cShelf l
and solved numerically. A variety of numerical techniques for solving eq tions of this type can be found in the book Applied Numerical method.^.^ One can use FEMLAB to solve PFR and laminar flow reactors for timedependent temperature and concentration profiles. See the FEML. problems and web module in Chapter 8 and on the FEMCAB CDRE
enclosed with this book. A simpler approach would be to model the PFR a number of CSTRs in series and then apply Equation (99) to each CSTR.
Closure. After completing this chapte~ o appl r, the teatIer shoddI be able 1t nd batch Iwctot; the unsteadystate energy balance to CSTRs, selmibatch a~ The reader should be able to discuss R ~ L L U L oalcW usine t w ~*,.~+..m l r t r l e on s: e ARSST to he1 a case study of an explosion and the other the ihouId be how F prevent explosions, Included in the redader's dL sfart up a reactor so as not to exceed thle practicaJ stability limit. Aft:erread .  . . .ing these examples, the reader should be able to descnbe Row to opera1 reactors in a safe manner for both single and multiple reactions. ,.A
* &. Camahan, H.A. Luther, and I. 0.Wilkes, Applied Nurncrical Methods (New Yo1 Wiley, 1969).
Chap. 9
Summary
SUMMARY I . Unsteady operation of CSTRs and semibatch reactors
For large heatexchanger coolant rates (To\= To?)
For moderate to low coolant rates
[ ( sf11
Q = m,Cpc(T T,,) 1  exp  2. Batch reactors a.
Nonadiabatic
Where Q is given by either Equation (S92) or ( 9  3 ) . b. Adiabatic
3. Startup of a CSTR (Figure SI) and the approach to the steady state (CDROM). By mapping out regions of the concentrationtemperature phase plane, one can view the approach to steady state and learn if the practical stability limit is exceeded.
UnsteadyState Nonisothermal Reactor Design
Figure S1
Chap. 9
Startup of a CSTR.
4. Multiple reactions (q reactions and n species)
CDROM MATERIAL Learning Resources Summary Motes
Sofved Problems
Living E~ampleProblem
I . Summay Notes 2. Weh links: SACHE Safety web sire www.soche.org. You will need to get the user name and password from your department chair. The kinetics line..CRE) text. examples, and problems are marked K in the product sections: Safety. Health, and the Envimnment (S.H,& E). 3. Solved Problems Example CD91 Startup of a CSTR Example CD92 Falling Off the Steady State Example CD93 ProportionalIntegral (PI)Control Living Example Problems I . Example 91 Adiuhric Barch Reactor 2. Example 92 Safcv ~n CkernicaE PInnrs with Exothermic Reactions 3. h m p S e 93 U3e of the A,?SST 4. Example 94 Heat Effects in a Semibafch Rcacror 5. Example 95 Srartup of a CSTR 6. Exomple 915Falling of the Upper Sfeady Stare 7. Example 97 Mul~ipleReaclrons in a Semiharch Reactor 8. Example RE9I Ifiregral Con~ml# f a CSTR 9. Exumplt REP2 Proportionit~lexrolConrml of a CSTR 10. Example RE93 Lineariz~dSrabiiiy
Chap. 9
CDROM Material
Pmfessional Reference Shelf R9.1 The Complete ARSST In this section further details are given to size safety valves
to
prevent run
away reactions m.
.*.. .
TI.
Refcrcncc Shcff
r*
1M
I#.
E
tm.
Figure W3.1 Temperaturetime trajectory for hydrolysis of acetic anhydride. Controt of a
CSTR
In this section we discuss the use of proponiond (P) and integral (I) control of a CSTR.Examples include I and PI control of an exothemic reaction,
Reactor with control system
Prnprtional integral action
R9.3. Lineoriied Stabili~Theop Zn this section we learn i f a perturbation will decay in an exponential manner ((a\ below) in an oscillatory manner (h). grown exponentialIy (c), grown exponentially with oscillations (d), or just oscillate (e3.
632
Uns!eabyS!a!e
T ~ B L9C1 E A.
Tr < 0
Non~so!hermaFReactor Design
Chal
EIGENVALCES OF COUPLED ODES &.&
Det > 0
[Tr2iM) 1DenM)I > 0
h
(a)
I
~cfercnckShelf
8.
Tr r 0
C.
Tr
Det z 0
(Mj=0
Uns&
Det (MI > 0
R9.4. Approach
10 the Stend!State PhasePlane Pbts and Tmjectnrie.~ of Conct rrrlrion versus Temperafur< Were we team if the practical stability is exceeded during startup. Example RE94.1 Start Up of a CSTR Example RE94.2 Fafling Off the Steady State Example RE94.3 Revisit Example RE92.
Figure CD95 Approach to the steady state.
R9.S. Adiabaric Operution of a Batch Reacfor R9.6. Onrrendy Opemtfon of PlugFlow Reactors
Chap. !3
Questions and Problems
QUESTIONS AND PROBLEMS WI,+ Read over the problems at the end of this chapter. Refer to the guidelines given in Problem 41. and make up an originat problem that uses the concepts presented in this chapter. To obtain a soIution: (a) Make up your data and reaction. (b) Use a real reaction and real data. Creative Prsblcmr Also, (c) Prepare a list bF safety considerations for designing and operating chemical reactors. See R. M. Felder. Chem. Eng. Educ.. I9 (41, 176 (1985). The August 1985 issue of Chemrcal Engineering Pmgress may be useful for part (c). P92, Review the example problems i n this chapter, choose one. and use a software package such as Polymath or MATLAB to carry out a parameter sensitivity analysis, What if.,. (a) Example 91. How much time would it take to achieve 90% conversion if the reaction were staned on a very cold day where the initial temperature was 20'F? (Methanol won't freeze at this temperature.) (h) Example 92. Explore the ONCB explosion described in Example 92. Show that no exploszon would have mcurred if the cooling waq not shut off for the 9.DGkmol charge of ONCB or if the cooling war; shut off for 10 min after 45 rnin of operation for the 3.17kmol ONCB charge. Show that if the cooling had been shut off for EO min after 12 h of operation, no explosion would have occurred for the 9.04kmol charge. Develop a set of guidelines as to when the reaction should be quenched should the cooling Bit. Perhaps safe opention could be discussed using a plot of the time after the reaction began at which the cooling failed, t,, versus the length of the cmling failure period, tr, for the different chilrges of ONC8. Parameter d u e s used in this example predict that the reactor will explode at midnight. What parameter values would predict the time the Hall cf Famc reactor wouId explode at the acnral time of 18 rnin after midnight9 Find a set of parameter vaIues that would cause the explosion to occur at exactly 12: 18 A.M. For example. include heat capacities of metal rencror and/or make a new estimate of CIA. Finally, what if a 112in. rupture disk rated at 800 psi had been installed and did indeed rupture at 800 psi (27OoC)? Would the explosion still have occurred? (Note: The mass flow rate ni varies with the crosssectional area of the disk. Consequently, for the conditions of the reaction the maximum mass flow rate out of the 112in. d ~ s kcan be found by comparing it with the mass flow rate of 830 kglrnin of The 2in. d ~ s k . (c) Example 93. What would be the conversion at the o n x t temperature if the hearing rate were reduced by a factor of ID? Increased by a factor of lo?
(d) Example 94. What would the X versus t and T versus t trajectories look like if the coolant rate is decreased by a factor of lo? Increased by a factor of SO?
634
UnsteadyState Nonisathermel Reactor Osslgn
(el Example 95. Load the Living Exampie Problem for
P93,
Chap. 9
Stanup of a CSTR, for an entering temperature of 7VF, an initial reactor temperature of lm,and an initial concentration of propylene oxide oF0.J M. Try other combinations of To, T,,and C,,, and report your results in terms of temperaturetime trajectories and temperatureconcentration phase planes. (0 Example 96. Load the Living Example Problem for FalIing Offthe Upper Steadv Ssate. Try varying the entering temperature, To, to between 80 and 68°F and plot the steadystate conversion as a function of T,,, Vary the coolant rate between 10,000 and 400 mollh. Plot conversion and reactor temperature as a function of coolant rate. (g) Example 97.What happens d you increase the hear transfer coefficient by a factor of 10 and decrease T, to 280 K? Which trajectories change the most? (h) Example RE91. Load the Living Example Problem. Vary the gain, kc, between 0.1 and 500 for the integral controller of the CSTR. Is there a lower value of kc that will cause the reactor to fall to the lower steady state or an upper value to cause it to become unstable? What would happen if Towere to fall to 65°F or 6 0 W (i) Example RE92. Load the Living Example Problem. Learn the effects of the parameters kc and 7,. Which combination of parameter values generates the leas1 and greatest oscilIations in temperature? Which values of k,and T, return the reaction to steady state the quickest? Cj) Reactor Safety. Enter the SACHE web site, wfiw.sache.org. [Note you will need to obtain the user name and password for your school from your department chair or SACHE represenrative.] After entering hit the current year (e.g., 2004). Go to product: Safety, Health and the Environment (S,H, & E).The problems are for KINETICS [i.e., CRE). There are some example problems marked K and explanations in each of the above SOH,& E selections. Solutions to the probEems are in a different section of the site. Specifically look at: Loss of Cooling Water ( K  J ) Runowa!. Reocrions (HT1). Design of Relief Values (D2). Temperature Cuntrol and Runaway {K4)and (K5).and R u n a w u ~and rhe Critical Temperarure Region (K7). The following is an excerpt from 711e Morning News. Wilmington, Delaware (August 3, 1973): "Investigators sift through the debris from blast in guest for the: cause {that destroyed the new nitrous oxide plant]. A company spokesman said it appears more likely that the [fatal] blast was caused by another gasammonium nltsateused to produce nitrous oxide." An 83% (wt) ammonium nitrate and 1 7 8 water solution is fed at 200T to the CSTR operated at a temperature of about 5 2 0 T Molten ammon~umnitrate decomposes directly to produce gaseous nitrous oxide and steam. It is beliwed that pressure fluctuations were observed in tht system and as a result the m i t e n ammonium nitrate feed to the reactor may have heen shut off approximately 4 min prior to the explosion. Can you explain the cause of the blast7 If the feed rate to the reactorjust before shutoff war 310 Ib of solution per hour, what was the exact temperature in the reactor ju5t prior to shutdown? UGng the following data. calculate the time it took to explode after the feed was shut off for the reactor. How would you star1 up or shut down and control such a react~on?
Chap. 9
PI4,
635
Questions and Problems
Assume that at the time the feed to the CSTR stopped, there ms 50Q Ib of ammonivm nitrate in the reactor at a temperature o f 520"E me conversion in the reactor is virtually complete at about 99.998. Additional data for this problem are given in hoblem 83. How would your answer change if 100 Ib of solution were in the reactor? 310 Ib? 800 Ib? What if To= 100°F? 5wQp The firstorder irreversible reaction is carried out adiabatically in a CSTR into which 100 rnollrnin of pure liquid A i s fed at 400 K. The reacljon goes virtually to completion (i.e., the feed raw into the reactor equals the product of reacrion mtc inside the reactor and &e reactor volume).
CSTR
How
many moles of liquid A are in the CSTR under steadystate condirionsq Plot \he temperature and moles of k in the reactor as n function of time after
the feed to the reactor has been shut off. Additinnall ilefnrmarion:
reaction in Prnhlem P85 is to he carried out in a semibatch P95, The liq~id~phase reactor. There is 503 mol of A initially in the reactor at 25°C. Species B is fed to the reactor at 50°C and a rate of 10 mollmin. The feed to the reactor is stopped after 500 mol of B has been fed. (a) Plot the temperature and conversion as a function of time when the reactian is carsied out adiaba~ically.Calculare to t = 2 h. (b) Plot the conversion as s function o f time when a heat exchanger (UA = 100 callminK) is placed in the reactor and the arnbtent temperature is constant at 50°C.Calculate to I = 3 h. {c) Repeat part (b) for the case where he reverse reaction cannot be neglected.
k = 0.01 (dmVmol min) at 300 R with E = 10 kcatlmoI Vo = SO dm3, u, = 1 dm3Jrnin, CAn= CB0 = I0 molldm3 Far the reverw reaction: k, = 10 s  I at 300 K with E, = 15 kcallmol
636 P96,
UnsteadyState Nonisothermal Reactor Design
Chz
You are operating a batch reactor and the reaction is tirstorder. liquidph and exothermic. An inert coolant is added to the reaction mixture to cor the temperature. The temperature is kept constant by varying the flow rat the coolant (see Figure P46).
Mhture OF A. 8, and C
(a) Calculate the flow rate of the coolant 2 h after the start of the react] (Anr.: Fc = 3.157 lbls.) (b) It is proposed that rather than feeding a coolant to the reactor, a sol\ be added that can be easily boiled off, even at moderate ternperant The solvent has n heat of vaporization of 1000 Btullb and initially th are 25 Ib mol of A placed in the tank. The initial volume of solvent i reactant is 300 ft3. Determine the solvent evaporation rate as a funct of dme. What is the rate at the end of 2 h? Additional information:
Temperature of reaction: 100°F Value of k at 100°F: 1.2 X sI Temperature of coolant: 80°F Meat capacity of all components: 0.5 Btu/lb.T Density of at1 components: 50 lblft3 P H L : 25,000 Btutlb mol Initially: Vessel contains only A (no B or C present) CAo:0.5 Ib mollft3 Initial volume: 50 ft3
P97, The reaction i s carried out adiabatically in a constantvolume batch reactor, The rate law
Plot the conversion, temperature, and concenmtions of the reacting species a function of time.
Chap. 9
Questions and Pmblems
Additional inJormatinn:
Entering Temperature = 100°C k , (373 K) = 2 x lo' 5  1 kl (373 K) = 3 X lQTF SI Cka= 0.1 mol/dm3 CBO= 0.125 rnol/drn3 A H L ( 2 9 8 K) = 40,OM) Jlmol A
P98,
El = 100 kJlrnol E2 = 150 kJlm01 CpA= 25 Jf m01 .K Cpg = 25 Jfm01.K Cpc = 40 JEmo1.K
The biomass reac%ion Substrate
C"1".
More cells + Product
is carried out in a 25 dm3 batch chernostat with a heat exchanger.
The initial concentration of cells md substrate are 0.1 and 300 g/drn3. respectively. The temperature dependence o f the growth rate follows that given by Aiba et al.. Equation (761)5
(76 I ) For adiabatic operation and initial temperature of 278 K, ptot 'l; I', r,, &, and Csas a function of time up to 300 hours. Discuss the trends. (b) Repear (a) and increase the initial temperature in 10°C increments up to 330 K and describe what you find. Plot the concentration OF ceIls at 24 hours as a function of inlet temperature. (c) What heat exchange area should be added to maximize the total number of cells at the end of 24 hours? For an initial temperature of 3 10 K and a constant coolan1 temperature of 290 K, what wnuld be the ceI1 concentration after 24 hours? {Ans. Cc = eldrnJ.) (a)
rs,
AddjIiorral Information:
Yus = 0.8 g celVg substrate
K, = 5.0 ddm3 pIm, = 0.5 hI (note p = & at 3 10 K and Cs 4 m) Cps = Heat capacity of substrate solution induding all cells
S.Abia. A. E. Humphrey, and N. E Mills. Biochemical Engineering (New York: Academic Press. 1973).
UnsteadyState Nonisotherrnal Reactor Design
Chap. 9
p = density of soIution including ce1Is = 1OOO AHk = 20,000 Jlg e l l s Cpc = Heat capacity of cooling water 74 JIgK
U = 50,000 J k d K h z
P9%
The fttsr order exothermic liquidphase reaction is carried out at 85OC in a jacketed 0.2m3 CSTR.The coolant temperature in the reactor i s 32°F. The heattransfer coeficient i s 120 WlmZ.K. Determine the critical value of the heattransfer area below which the reaction will sun away and the reactor wit1 explode [Chem. Eng., 91 (lo), 54 (1984)j. Additional informarion:
Specific reaction rate: k = 1 . 1 min' at 41PC k = 3.4 minI at 5VC The heat capacity of the solution is 20 J1g.K. The solution density is 0.90 kg/dm3. The heat of reaction is 2500 Jlg. The feed temperature is 40°C and the feed rate is 90 kglmin. M W of A = 80 glrnol. CAo= 2 M. P9loc The ARSST adiabatic bomb calorimeter reactor can also be used to determine the reaction orders. The hydrolysis of acetic anhydnde to form acetlc acid was carried out adiabatically
The rate law is postulated to be of the form
The following temperature time data were obtained for
two different critical concentrations of acetic anhydride under adiabatic operation. The hearing rate
was Z0Clmin.
Chap. 9
639
Questions and Problems
Figure P910.1 Daa from Undergraduate Laboratory University of Michigan.
la) Assume ACp = 0 and show for complete conversion, X = I, the differenoe between the find temperature, Tf.and the initial temperature, Tb
(b) Show that the concentration of A can be written as
and Cg as
and  r ~as
(c) Show the unsteady energy balance can be written as
(d) Assume first order in A and in B and
that OB= 3
then show
640
UnsteadyState Nonisotharmal Reactor Desrgn
Ct
te) Rearrange Equat~on(P410.6) in the form
(f) Ptot the data to obtain the activation energy and the specific reactior
k,. (g) Find the heat of reaction.
Additioml information: Hear cupaciry Chemical
Densit?! ( g h l )
(J/gm0C)
MW
1.0800
1.860
I02
Water
1.0000
4.187
I8
Glass cell (bomb)
0.1474
0.837
Acetic anhydride
Tor4 volume
Water
Hear cupacit (J/molmaC 189.7 75.4 0.84 J/$C
10 ml with
3.638 g
Acetic anhydride
6.871 g (MsCPs = 28.01 2 JPC and $I = 1.CQ4 and msCp,= $ iUs Cps )
P9llB The elementary irrweraible liquidphase reaction
is to k carried out in a semibatch reactor in which B i s fed to A. The vo3 of A in the reactor is 10 dm3. the initial concentration of A in the react( 5 rnol/dm3, and the initial temperature in the reactor is 27°C. Species B is at a temperature of 52°C and a concentration of 4 M.It is desired to obtai least 80% conversion of A in as short a time as possible, but at the same I the temperature of the reactor must not rise abDve 130°C. You should tr make approximately 120 mol of C in a 24hour day allowing for 30 rnin to empty and fill the reactor between each batch. The coolant flow
through the reactor is 2000 motimin. There is a heat exchanger in the reac (a) What volumetric feed rate (drn3/rnin)do you recommend? (b) How would your answer or strategy change if the maximum coolant dropped to 200 moIlmin? To 20 molimin? Additional infontsarion:
AHO, = 55,000 cal/mol A CpA=35cal/mol.K, Cpg =20cal/molK,
k = 0.0005
Cpc = 7 S c a l l m o l ~ K
dm6 at 2Y°C with B = 8000 cal/rnol mol2. rnin
caI UA = 2500 with T, = 17°C min.K
Gp(coolant) = 18 callrnoI. K
[Old ex:
Chap. 9
64 1
Questions and Problems
P91& Read Section R9.2 on the CDROM and then rework Example RE9I using (al Only a propurtiunal controller.
t ~ntagol~ Contralhr
(b) Only an integral controller. (c) A combined proponional and Integral controller. PP13, Apply the different types OF controllers to the reactions in Problem P9I I . P914, (a) Rework Example R9I for the case of a 5°F decrease in rhe outlet temperature when the controlled input variable is the reactant feed rare. (bh Consider a SaF drop in the ambient temperature, T,,, when the controlled variahle is the. inlet temperature, T,. (c) Use each of the controllers (P with kc = 10. I with t, = 1.0 h. D with To = 0.1 h) to keep the reactor temperature at the unstable !V~ in F i p r c 1010. S~artingfrom the upper left, u'c xee that singlecr>st;il hilrcnn inpol\ are grotvn in a C7c>chr;il\ki cryrtallizer. sliced into wafers. and chemically al~dph\.\icall> poli5hed. T1iec;e porished wafer\ cerve ac \tarting material5
Sec. 10.5
Reaction Engineering in Microelectronic Fabrication
699
for a variety of rnicmelectronic devices. A typical fabrication seguence is shown for processing the wafer beginning with the formation of an SiO, Iayer on top of [the silicon. The Si02 layer may be formed either by oxidizing a silicon layer or by laying down a SiOz layer by chemical vapor deposition (CVD). Next, the wafer is masked with a polymer photoresist, a template with the pattern to be etched onto the SiO, layer is placed over the photoresist. and the wafer is exposed to ultraviolet irradiatjon. If the mask is a positive photoresist, the light will cause the exposed areas of the polymer to dissolve when the wafer is placed in the developer. On the other hand, when a negative photoresist mask is exposed to ulrraviolet irradiation, crosslinking of the polymer chains occurs, and the unexposed areas dissolve in the developer. The undeveloped portion of the photoresist [in either case) will protect the covered areas from etching.
CVD {two Rlmrl
Mm&, Etzh, then Strip Mslk
E i l h %#Id4
Wiw
CW.Muk. Etch. S ~ m pMh.L
D
m bu Phmphmr m w h
CYO of Rnal Layer
Figure 1020 Micrmltctronic fabrication steps.
700
Catalysis and Catalytic Reactors
Chap.
Afier the exposed areas o f SiO? are etched to form trenches [eirher wet etching (see Problem P512) or by plasma etching), the remaining pha resist i h removed. Next. the wafer i s placed in a furnace containing gas mc cules of the desired dopant. which then diffuse into the exposed silicon. Al diffusion of dopant to the desired depth in the wafer, the wafer is removed : covered with S i 0 2 by CVD. The sequence o f masking, etching, CVD, and rr allization continues until the desired device is formed. A schematic of a fi chip is shown in the lower righthand corner of Figure 1020. In Section 10. we discuss one of the key processing steps, CVD. 10.5.2
Etching
We have seen in Figure 1020 that etching (i.e,. the dissolution or physical chemical removal of material} is also an important step in the fabrication F crss. Etching takes on a priority role in rnicrmlectronics manufactur because of the need to create welldefined structures from an essentialiy hor geneaus material. I n integrated circuits, etching is necess~uy to remi unwanted material that could provide alternative pathways For electrons :
Reference Shelf P1U.3
thus hinder opersltion of the circuit. Etching is also of vital importance in fabrication of micr~rnechanical and optoelectronic devices. By selectit etching semic~nducrossurfaces. it is possible to fabricate motors and v a l ~ ultrasmall diaphragms that can sense differences in pressure, or cantile beams that can sense acceleration. In each of these applications, proper etch is crucial to remove material that would either short out a circuit or hin movement of the rnicromechanicrtl device. There are two basic types of etching: wet etching and dry etching. ' wet etching process, as described in Problem P51ZB, uses liquids such as or KOH to dissolve the layered material that is unprotected by the photore mask. Wet etching is used primarily in the manufacture, of rnicrornechan devices. Dry etching involves gasphase reactions, which form highly reac species, usually in plasmas. that impinge on the surface either to react with surface, erode the surface, or both. Dry etching is used almost exclusively the fabrication of optoelectronic devices. OptoeZectronic devices differ f~ microelectronic devices in that they use light and electrons to c a q out tl particular function. That function may be detecting light. transmitting light emitting light. Etching is used to create the pathways or regions where I can travel and interact to produce the desired effects. Appliances using s devices include remote controls for I T sets, LED displays on clocks microwave ovens, laser printers, and compact disc players. The material on CDROM gives examples of both dry etching and wet etching. For dty etch the reactive ion etching (RIE) of InP is described. Here the PSSH is usec arrive at a rate 3aw for [he sate of etching. which is compared with experime observation. In discussing wet etching, the idea of dissolution catalysis is in duced and rate laws are derived and compared with experimental obsewatic In the formation of microcircuits. electrically interconnected films laid down by chemical reactions (see Section 12.10). One method by wl these films are made is chemical vapor deposition.
Sec. 30.5
Reaction Engineering in Mic~oelectronicFabricallan
10.5.3 Chemical Vapor Deposition The mechanisms by which CVD occurs are very similar to those of heterogeneous catalysis discussed earlier in this chapter. The reactant(s1 adsorbs an the surface and then reacts on the surface to fom a new surface. This process may be followed by a desorption step. depending on the particutat reaction. The growth of a germanium epitaxial film as an interlayer besetween a gal
lium arsenide layer and a silicon layer and as a contact layer is receiving Ge used in %Iar
increasing arteation in h e microelectronics industry.20Epitaxial germanium is also an important materia1 in the fabrication of tandem solar cetls. The growth
of germanium films can be accomplished by CVD.A proposed mechanism is
Gasphase dissociation: GeCt4(g)
GeCl,(g) k*
+ Cl, (g)
GeCI, .S
Adsorption:
GeC12(gj + S
Adsorption:
Hz+ 2 S lo uo!lsunj r! rr! x pur? .i ,old ,SyC ~ =O m piln do.lp u n c w d ~u!l~;ncn.,t:(p) md ,cadla 11) t'7.a uo!lsag aagaJd aaS :IYHJ.?tl!llurql I~:JI!I.I~ sa.l!nba~uo!~sanh m0h' hym urydxa uaql pup S U ~ ~ U (ns!l!Ja ! ~ I .;al!nllal 1l:yl unysanb r? ~ Y U M (a) 'ay 
dx
~ W A+: r A
a:
 ac,
ar
The corresponding balance in cylindrical coordinates wirh no variation in he rotation about the zaxis is FEMLAB
We will now evaluate the flux terms W,. We have taken the time to derive the molar fiux equations in this form because they are now in a form that is consistent with the partial differential equation (PDE) solver FEMLAR. which is included on the CD with this textbook. l f . 1 . 2 Molar Flux
The molar flux of A. W,, is the result o f two contributions: J,. the molecular diffusion flux relative to the bulk motion of the fluid produced by a concenlration gradient. and B,. the flux resulting from the bulk motion of the fluid: k ~ n Rux l diffusion
=
hulk mo~ion +
The bulk flow term for species A i s the total flux of a l l molecules relative io a fixed coordinate times the mole fraction of A, g,: i.e.. B, = y, 2 W,. The bulk f OW term B, can also be exprex~edin terms of ~ h cconcentration of A and the molar average veloci~yV:
External Diffusion Ef!ecls cn Heterogeneous Reactions
R , = CAV
mol  m2.s
Chap.
(I I:
moi . rn
m3 s
where the molar average velocity is Molar average
velocity
Here V, is the particle velocity of species i, and yi is the mole fraction of spc cies i. 3 y particle wlocities, we mean the vectoraverage velocities of rnilIior of A molecules at a point. For a binary mixture of species A and B, we let h and Y, be the pnnicle velocities of species A and B. respectively. The flux ( A with respect to a f xed coordinate system le.g.. the lab bench), 1%. Is ju the product of the concentration of A and the particle velocity of A:
The molar average velocity for a b i n a ~ ysystem is (1 I; V = vAVA+ ynVs The total molar ffux of A is given by Equation ( 1 14). BA can b expressed either in terms of the concentration of A, in which case
or in terms of the mole fraction of A: Binary system ot' AanJ B
We now need to evaluate the molar flux of A,
JA,
that is superimposed on th
molar average velocity V.
11.1.3 Fick's First Law Our discussion on diffusion will be restricted primarily to binary systems con taining onty species A and B. We now wish to determine how the molar diffu Expcriment~tion sive Rux of a species (i.e.. J,) is related to its concentration gradient. As an ail with frog "ps led lu in the discussion of the transport law that is ordinarily used to describe diffu F~ck's% r ~law t sion, recall similar laws from other transport processes. For example. in con ductive heat transfer the constitutive equation relating the heat flux q and thl
temperature gradient is Fourier's law:
where k, is the thermal conductivity. In rectangular coordinates, the gradient is in the form
Sec. 11.2
milks
rrmclur
762
Binary Diffus~on
The onedimensional form of Equation ( 1 1  10) i 5
Heat Transfer
4: = k,
dT dz

( 1 112]
In momentum transfer, the constitutive relationship between shear stress, t,and shear rare for simple planar shear Raw is given by Newton's Inw of viscosity:
r=
Momentum Transfer
,J

du d:
The mass transfer flux law is analogous to the laws for heat and rnornentuln transport. i.e., for constant total concentration J,=
Mass Transfer
D A B
c/C, dz
The general 3dimensional constitutive equation for JA, the diffusional flux of A resulting from a concentration difference. is related to the mote fraction gradient by Fick's first law:
JA =  c D ~ ~ V ~ ~ (1 113) where c is the total concentration (rnol/dm3), D,, is the diffusivity of A in B (drn2/s), and y,, is the mole fraction of A. Combining Equations (I 19) and I1 113]. we obtain an expression for the molar flux of A:
[WA = 4DABv~,l + .vA(WA+ wB)/
Molar flux equation
(1 114)
In terms of concentration for constant total concentration Molar Hux equation
112 Binary Diffusion Although many syrtems involve more than fwo components, the diffusion of each species can be treated as if it were diffusing thmugh another sin@ species rather than through a mixture by defining an effective diffusivity. Methods and examples for calculating this effective diffusivity can be Found in Hill.'
I
11.2.1 Evaluating the Molar Flux The task is ta now the bulk flow term.
We now consider four typical conditions that arise in mass transfer problems and show how the molar flux is evaluated in each instance. The first two
' C. G . Hi!l.
Chemical Engineering Kirletics nrtd Reocror Design (New York: Wiley.
19771. p. 480.
762
External Diffusion Effects on Heterogeneous Reactions
Chap. 11
conditions, equal molar counter diffusion (EMCD) and dilute concentration give the same equation for WA,that is. Summery tJotes
WA= DABVCA
The third condition. diffusion through a stagnant film,does not occur as often and is discussed in the summary notes and the solved problems on the CD. The fourth condition is the one we have been discussing up to now for plug flow and the PFR, that is, FA= vc*
We will first consider equimolar counter diffusion (EMCD). colvcd problems
112.1A Fduimolar Counter Diffusion. In quimolar counter diffusion (EMCD), for everv molt of A that diffuses in a given direction, one mole of B diffuses in the opposite direction. For exarnplc,consider a species A that is diffusing at steady stare from the bulk fluid to a catalyst surface, where it isomerizes to form B. Species B [hen diffuses back into the bulk (see Figure 1 I  I). For every mole of A that diffuses to the surface, 1 rnol of the isomer B diffuses away from the surface. The fluxes of A and B are equal in magnitude and flow counter to each other. Stated mathematically,
Figure 111 EMCD in isomerization reaction.
An expression for 1% in (ems of the concentration of A. CA.for the case of EMCD can be found by first subaituting Equation (1 116) into Equation ( 1 191:
For constant total concentrarion EMCD flux equation
11.2.1R Dilute Concentrations. When the mole fraction of the diffusing solute and the bulk motion In the direction of the diffusion are small, the second term on the rrghthand side of Equation ( 1 114) [i.e.. y,(W, + Wn)] can usually he neglected compared with the f rst term, J A . Under these conditiondirection (W,,= 0). and Qe are at steady state so that Equation (1 121 seduces to
which i s the wrne as Equation (E I 1  l .?I. Similarly one could apply Equation ( I 121 1 to this example realizing we are at steady state, no reaction, and there is no variation in concentration in either the xdirection or the direction
so that Equation ( 112t ) reduces to
770
External Diffusion E m s on Heterogenws Reactions
Chap. 11
After dividing both sides by the diffusivity, we realize this equation is the same as Equation (El 115), This problem is reworked for diffusion through a stagnant film in the solved example problems on the CDROMlweb solved problems. Colved Problems
11.2.4 Temperature and Pressure Dependence of DAB
Before closing this brief discussion on mass transfer fuodamentaIs, further mention should be made of the diffusion coefficienr."quations for predicting gas diffusivities are given by Fuller' and are aEso given in Perry's I/andbnnk." The orders of magnitude of the diffusivities for gases. liquids: and solids and the manner in which they vary with temperature and pressure are given in Table 112. We note that the Knudsen, liquid, and solid diffusivities are independent of total pressure.
lr is importan1 to know the magnitude and the T and P dependence of the diffusivity
Oidcr of Mugr~itude
Phare
cm21s
m p Ik. . liquid viscosit~esat
m71r
T~mperatrrreand Pressacre Drpnde~rcesP
temperarures T,and T2, sespcctively: ED. d~ffusionactivat~onenerg).
' For funher discussion of mass transfer fundamentals,
see R. B. Bird. \V, E. Stewan. {New York: Wiley. 1960). '' E. N.FuIIer. P.D.Schettler, and J. C. Giddings, Ind Eng. Chem.. 5815). 19 (f966). Several other equat~onsfor predicting diffusion coefficientscan he found in B. E. Polling, J. M . Prausnitz, and I. P. O'ConneII. Tile Properires ofcases unn' Liqrrids, 5rh ed. ( New York: McGrawHill. 21K)I ). R. H. Perry and D.N'. Green, Cl~~nriral Ensrneerb tirindbook. 7rh ed. (New Y d and E.N. L~ghtfoot,Tr~nsporrPhinorn~na,2nd ed.
McGrdwHiIE. 1999). To es~imate llqufd drffusivittes fur binary syslerns, see Doraiswamy, h d En#. Cl~enl.F Z I I Ia,~ 77 , , (1967).
K. A. Reddy iind L. K.
Sec. 11.3
External Resistance to Mass Transfer
ni
f 1.2.5 Modeling Diffusion with Chemical Reaction
The method used in solving diffusion problems similar to Example 111 is shown in Table 1 13. Also see Cussler? TABLE 1 13.
Expanding the previous s,x
mdeling steps jusr a bit
STEPSIN MODELFNG CHEMICAL SYSTEMSuwi D I ~ S S O AMIN REnm03
I. Define the problem and state the assumptions. (See Problem Solving on the CD.) 2. &fine the system on which the balances are to be madc. 3. Perfom a differential mole bakance on a panicular species. 4. Obtain a differentinl equatioli In W, by rearranping your balance equation properly and ukIng thelimit as the volume of the elernen1 goes lo zero 5. Substirure the appropriate expression involving the concentration gradient for W, fmm Section E 1 2 lo abrain a secondorder differential equation for the conccnwation of A,& 6 Express the reaction rate r, ( ~ any) f In terns of concentration and substitute into h e differ1. R 9. 10 I 1.
ential equation. Stale the appropriate boundary and initial condi~ions. PUI the differential equations and boundary condrtions. in dimensionless Tom. Solve the resultmg differential equation for the concenrraiion profile. Differentiate thrs concentration profile to obtain an expression for the molar flux of A. Suhstllute numerical values for symbols.
"In some instances it may be ea~ierto ~ntegratethe resulting differential equation in Sfep 4 before substituting for WA.
The purpose of presenting algorithms (e.g., Table 1131 to solve reaction engineering probIems is to give the readers a starting point or framework with In ;3 of the algorithm which to work if they were to get stuck. It is expected that oncereaders are 6, familiar and comforrable using the algorithdframework, they will be able to to generate creati~esnlutions~move in and out of the framework as they develop creative solutions to nonstandard chemicaI reaction engineering problems. Move
+
11.3 External Resistance to Mass Transfer
'
11.3.1 The Mass Transfer Coefficient
To hegin our discussion on the diffusion of reactants frum the bulk fluid to the external surface of a catalyst. we shalI focus attention on the flow past a single catalyst pellet. Reaction takes place only on the external catalyst surface and not in the fuid surrounding it. The fluid velocity in the vicinity of the spherical pellet will vary with position around the sphere. The hydrodynamic boundary layer is usually defined as the dismnce from a solid object to where the fluid velocity is 99% of the bulk velocity U U .Similarly. the mass transfer boundary layer thickness, 6, is defined as the distance from a solid object to where the concentration of the diffusing species reaches 99% of' the bulk concentration. A reasonable representation of the concentration profile for a reactant A diffusing to the external surface is shown in Figure 1 13. As illustrated, the
E.L. Cussler, Drffusion Mu1.7
Tmnsfer in Flltid S~rte~ns, 2nd ed. (New
bridge University Press. 1997).
York: Carn
772
External Drftus~onEffects on Heterogeneous React~ons
Chap.
Side Note: Transdermal Drug Deliver)r The principles of steady state diffusion have been used in a number of drug delivery systems. Specifically, medicated patches are commonly used tc attach to the skin to deliver dmgs for nicotine withdrawal, birth control, and motion sickness, to name a few. The U.S. transdermal drug deIivq markel was $1.2 billion in 2001. Equations similar to Equation f 126 have been used to model the release, diffusion, and absorption of the drug from the patch into the body. Figure SN11.1 shows a drug delivery vehicle (patch) along with the concentration gradient in the epidermis and dermis skin layers. Skin Layers
Figure SN 11.1 Tmnsderrnal dryg delivery schematic.
As a first approximation, the delivery sate can be written as
where
R=R
6 + ' + A DAB,
8
DAB^
Where A, is the area of the patch; CAP, the concentration of drug in the patch; R, the overall resistance; and R,, the resistance to reIease from the patch. There are a number of situations one can consider, such zts the patch resistance limits the delivery, diffusion through the epidermis limits deIivery, or the concentra€ion of the drug is kept constant in the patch by using solid hydrogels. When difision through the epidermis layer limits, the nte of drug delivery rate is
Other mwers include the use of a quasisteaay anruysls t~ coupIe the diffusio!n equation dance on the drug in the patcl3 or the zrm order (iissoluti.on of the n the patch.t Roblem 11210 explores these situalions.
Fw t u r n infomutm see:Y. H.K d a and R Guy, Advmced Unrg Uellvery Km'ews 48,159 (2001); B.MuHer. M.Kasper, C.Surber. and G.Jinanidis, Ei4mpem J o u d of PhnmtaceuticaI Science 20. 18 t (2003); www.dmgdeLiverytech.codcgibid articIes,cgi?irlim'cLe=143: w w w . p h a m q ,umoiyland,ediu$%culry/rdalb?r//dul~/ Teaching%ZOWeb%2OPuge/Teaching.khn
Sac. 1 t .3
773
E~fernalR ~ 9 i s t a m eto Mass Transfer
chanze in concentration o f A from CAhto CA, takes place in a very narrow fluid Layer next to the surfslce of the sphere. Nearly all of the resistance to mass trancfer i s found in this I:iyer.
Ph
Figt~re113 Boundary layer around the surface of a cat;ilyst pellet.
11.3.2 Mass Transfer Coefficient
The concept or a b~~thrticnl stagnanr tilm within which the wsistance D I
external mas$ transfer exlsts
A useful way of modeling diffusive transport i s to treat the: fluid layer next to a solid boundary as a stagnant film of thickness 8. We say that all the resistance to mass transfer is found [i.e.. lumped) within this hypothetical stagnant film of thickness 6. and the properties (i.e., concentration, temperature) of the fluid at the outer edge of the film m identical to those of the bulk fluid. This model can readily be used so solve the differential equation for diffusion through a stagnant film, The dashed line in Figure l I3b represents the concentration profile predicted by the hypothetical stagnant film model. while the solid line gives the actual profile. If the film thickness is much smalIer than the radius of the pellet, curvature effects can be neglected. As a resutt. only the onedimensional diffusion equation must be solved. as was shown in Section 1 1. L (see also Figure L 14).
c A, Figure 114 Concentration protilr for E.VCD in stagnant film model.
For either EMCD or dilute concentrations, the solution was shown in Example El 11 to be in the form
While the boundary layer thickness will vary around the sphere, we will take it to have a mean film thickness 6. The ratio of the diffusivity DAB to the film thickness F is the mass transfer coefficient, kc, that is,
External Diffusion Effects on Heterogeneous Reactions
Chap. 71
The mass transfer cwfhcient
Combining Equations (1 126) and (1 127), we obtain the average molar flux from the bulk fluid to the surface Molar flux of A to the surface
In this stagnant film model. we consider all the resistance to mass transfer to be lumped into the thickness 6. The reciprocal of the mass transfer coefficient can be thought of as this resistance W, = Flux = Driving force  Cnh C A ~
Resistance


(w)
113.3 Correlations for the Mass Transfer Coeflicient
The mass transfer coefficient kc is analogous to the heat transfer coeficient h. The heat flux q from the bulk fluid at a temperature Toto a solid surface at T,is
For forced convection. the heat transfer coefficient is normally cotrelated in terms of three dimensionless groups: the Nusselt number, Nu; the Reynolds number, Re; and the Prandtl number, Pr. For the single spherical pellets discussed here, Nu and Re take the following forms:
The Prandtl number is not dependent on the geometry of rhe system. The Nusselt. handtl. and Reynolds numbers are used in forced convection heat
transfer
cornlalions
\
' *
'
where or, = k,/pC, = thermal diffusivity, m*/s v =
= kinemarfc visco'sity (momentum diffusivity). m2/s P
d, = diameter o f pellet. m U = freestrearn velocity, mls k, = thermal conductivity. J I K . m a s p = Fluid density. kglrn3 h = beat transfer coefficient, Jlrn2.sK or Watts/m2 K The other symboIs are as defined previously.
Sec. 1 I.3
775
External Resistance to Mass Transfer
The heat transfer correlation relating the Nusselt number to the Prandtl and Reynolds numbers for flow around a sphere is8 Nu = 2 +
( I 133) Although this correlation can be used over a wide range of Reynolds numbers, it can be shown theoretically that if a sphere is immersed in a sragnant fluid (Re = 0), then
and that at higher Reynolds numbers in which the boundary layer remains laminar, the Nusselt number becomes Although further discussion of heat transfer correlations is no doubt worthwhile, i t will not help us to determine the mass transfer coefficient and the mass flux from the bulk fluid to the external pellet surface. However, the preceding discussion on heat transfer was not entirely futile, because, for similar geometries, rhe hear and mass transfer cor~lationsare analogous. If a heat transfer correlation for the Nusselt number exists, the mass transfer coefficient can be estimated by replacing the Nusselt and Prandtl numbers in this correlation by the Shemood and Schmidt numbers, respecrively: Convening n heat transfer correlation to a maw transfer
correlation
Sh SC

 NU fi
The heat and mass transfer coefficients are analogous. The corresponding fluxes are 4: = h(T
 T,)
( I 136)
Wk = k,(CA  )C, The onedimensiona1 differential forms of the mass flux for EMCD and the heat flux are. respectively. For EMCD the hear and molar flux equations are analvpous.
If we replace h by k, and k, by DABin Equation (1 130), i.e..
a \hl. E. Ran? and W. R. Marshall. J r , C l ~ ~ rEng. n . Pmg.. 48, 141146, 173180 (19521.
776
External Diffusion Effects on Heterogeneous Reactions
Chao.
we obtain the mass transfer Nussel t number (j.e., the S herwood number):
The Prandtl number is the ratio of the kinematic viscosity (i.e., the momentu dihsivity) to [he thermal diffusivity. Because the Schmidt number is anal gous to the Prandtl number. one would expect that Sc is the ratio of tl momentum diffusivity (i.e.. the kinematic viscosity), v, to the mass diffusivi DAB.Indeed, this is true:
a, + D A B The Schmidt number is
s C = Y  m2/s
Schmidt number
The Sherwood. Reynnlds. and Sehmldt number< are used in forced convection
maw tmnGer correlation\.
D,, ..
m2/s
dimensionless
( 1 131
Consequently, the correlation for mass transfer for flow around a spherical pe let is analogous to that given for heat transfer [Equation ( 1 133)]. that is. Sh = 2 6 0.6ReT'2Sc"3 11141 This relationship is often referred to as the Friisslirrg correl~~rion.~ 11.3.4 Mass Transfer to a Single Particle
In this section we consider two limiting cases of diffusion and reaction on catalyst particle.") In the first case the reaction is so rapid that the rate of di fusion of the reactant to the surface limits the reaction rate. In the second casl the reaction is so slow that virtually no concentration gradient exists in the g~ phase ( i t . , rapid diffusion with respect to surface reaction). Example 112 IF rapid reaction, then diffusion timits the overall rate.
Rapid Reactinn on a Catalyst Surface
Calculate the mass flux of reactant A to a single catalyst pellet I cm in diameter su! pended rn n Imge body of liquid. The reactant is present in dilute concentration: and the reaction is considered to take place instantrtoeousIy at the external peHt surface (i.e.. C,, = 0). The bulk concentration of the reactant is I.0 M. and th freesystem liquid velocity is 0. I mls. The kinematic viscosity is 0.5 centistoke (cI I centistoke = 10AhmZls), and the liquid diffusivity of A is to" rn2/b. SoI~rtion For dilute concentrations of the solute the radial flux is W4r =
k,(C~h Ckr)
( 1 128
N. Frossfing, Gerlnndx Beit!: Geophy., 52, 170 (1938). "A comprehensive list of correlations for mass transfer to panicles is given by G. A Hughrnark, Ind Eng. Chem. fund.. 19I2), 198 (1980).
I
Set. 11.3
777
Exiernal Rasistance to Mass Transfer
Becauqe reaction i\ dcsumed to wcur in\tantanenuqly un the extem:~l c ~ ~ r f i l cnt' e the pellet. C4, = 0. A l w . C',,,, i\ given sc 1 molldrn7.The n1nh.c tr;~ri\trr ct~tficienlfor siugfe yheres i q calculated froom the Frridinp corrzlat~on:
Substituting these value!, into Equation I 1140) gives us Sh = 2
+ 0.611Wo)0~5(500R)1'~ = 360.7
(El lZbl)
Substituting for k, and CAhin Equation ( 1 128), the molar Rux to the surface is LVn, = (4.61 X
.
m/s (1.0.'
 0) mul/rn3 = 4.61 X lo' mol(m2.s
Because W,,=  r,", thisr;lteisalsethente of~eactionper unit surface areaofcatalyst.
In Example 1 12, the surface reaction was extremely rapid and the rate of mass transfer to the surface dictated the overall rate of reaction. We now consider a more general case. The isornerizatian is taking place on the surface of a solid sphere (Figure 1 151. The surface reaction folIows a LangmuirHinsheLwood singlesite mechanism forwhich the rate law is
Figure 115 Diffusion to, and reaction on. external surface of pellet.
External Diffusion E M S on Heterogeneous Reactions
Chap, 1 7
The temperature is sufficiently high that we need only consider the case of very weak adsorption (i.e., low surface coverage) of A and B: thus (KBCBT+ K~Ckrl 1
Therefore,
Udng boundary conditions 2b and 2c in Table I I  I , we obtain
The concentration CAI is not as easily measured as the bulk concentration. Consequent?y. we need to eliminate CA,from the equalion for the flux and rate of reaction. Solving Equation (1 144) for C, yields ~,CA c*, = kr+k,
and the rate of reaction on the surface becomes Molar flux of A to lhe surface is equal to the rate o f consumption of A on the surface
One will often find the flux to or from the surface as written in termt of an efecrive transport coefficient kt,:
where
Rapid Reaction. We first consider how the o~erallrate of reaction ma!, he increased when the rate of mars tiansfer to the surface limits the overall rate of reaction. Under these circumstances the specific reaction rate constant is much greater than the mass transfer coefficient k, % k,
and k,. 41
k,
Sec. 11.3
M e r n a l Res~stancsto Mass Transfer
779
To increase the rate of reaction per unit surface area of solid sphere, one must increase C, andlor kc. In this gasphase catalytic reaction example, and for most Iiquids, the Schmidt number is sufficiently large that the number 2 jn Equaljon (1 140) is negligible with respect to the second term when the ReynoIds number is greater than 25. As a result, Equation (1 140) gives It i s important to know how the mass transfer cwffic~ent vanes with fluid velocity, pangcle siw, and physrcal proprtles.
kc = 0.6 X (Term I ) X (Term2) Mass Transfer
;/ r
Llrnlled
U
Reaction Rate Limited
Term I is a function of the physical properties DAB and v, which depend on temperature and pressure only. The diffusivity always increases with increasing temperature for both gas and liquid systems. However, the kinematic viscosity v increases with temperature (v w T p 2 ) for gases and decreases exponentialiy with temperature for liquids. Term 2 is a function of flow conditions and particle size. Consequently, to increase k, and thus the overall rate of reaction per unit surface area, one may either decrease the particle size or increase the velocity of the fluid flowing past the particle. For this particular case of flow past a single sphere. we see that if the velocity is doubled, the mass transfer coefficient and consequently the rate of reaction is increased by a factor of
Slow Reaction. Here rhe specific reaction rate constant is small with respect to the mass transfer coefficient:
1 u
The specific reaction rate i s independent of the velocity of fluid and for considered here. independent of particle size. Howet8er. for porous catalyst pellets, k, may depend on particle size for certain situations, as shown in Chapter 11. Figure 116 shows the variation in reaction rate with Term 2 in Equation ( 1 149), the ratio of velocity to particle size. At low velocities the mass
Mars tran~fer the solid sphere
effectr are no1
important when rhe reaction rate 1 5 Iimi~ing.
780
External Diffusion Effects on Heterogeneous Reaclions
Chap :
transfer boundary layer thickness is large and diffusion limits the reaction, A the veltxity past the rphere i s increased, the boundary layer thickne? decreases. and the mass transfer across the boundary layer no Ionger limits th rate of reaction. One also notes that for a given velocity, reactionlimiting cot ditions can be achieved by using very small particies. However, the smaller ~h particle size, the greater the pressure drop in a packed bed. When one I obtaining reaction rate data in the laboratory, one must operate at sufficient1 high velocities or sufficiently small particle sizes to ensure that the reaction I not mass transferlimited.
Limited
When collecting rate law data, operate in the reactionlimited region.
Figre 11.6
Regions of mass transferlimited and reactionlimited reactions.
1 1.3.5 Mass TransferLimited Reactions in Packed Beds
A number of industrial reactions are potentially mass transferlimited becaus they may be carried out at high temperatures without the occurrence of unde sirable side reactions. In mass transferdominated reactions, the surface reac tion is so rapid that the rate of transfer of reactant from the bulk gas or liqui phase to the surface limits the overall rate of reaction. Consequentty. mas transferlimited reactions respond quite differently to changes in ternpentur and flow conditions than do the ratelimited reactions discussed in previou chapters. In this section the basic equations describing the variation of conver sion with the various reactor design parameters (catalyst weight, Row condi [ions) will be developed. To achieve this goal, we begin by carrying out a mol balance on the following mass transferlimited reaction:
carried out in a packedbed reactor (Figure 117). A steadystate mole balanc on reactant A in the reactor segment between z and r + ht is
Sec. 1 1.3
781
External Resistance to Mass Transfer
rate out FA=!:  F
A
+
I;cI,(A~Az)=
0
( 1 151]
Z+AZ Figure 117 Packedbed reactor.
where r i = rate of generation o f A per unit catalytic surface area. mol/s.m? a, = external surface area of catalyst per volume of catalytic bed. m2/m" = 6( 1  +)Id, for packed beds, m21m' 4 = porosity of the bed (i,e.. void fraction)" d,, = particle diameter. m A, = crosssectional area of tube containing the catalyst, rn2 Dividing Equation ( I I51) by A,& and taking the limit as k ,0,we have
We now deed to express F, and r(; in terms of concentration. The molar Aow rate of A in the axial direction is
FA.= A, Wk = (Jk
+ B,)A,
( l t 53)
I n aImost all situations involving flow in packedbed reactors. the amount of material transported by diffusion or dispersion in the axial direction is negligineelwted. ble compared with that transported by convection h e . , bulk flow):
Axial diffusion is,
Jk < Bk (In Chapter 14 we consider the case when dispersive effects must be taken into account.) Neglecting dispersion. Equation ( 1 120) becomes Fk = A, W, = R,Bk = UC, A, f 1 154] where U is the superficial rnoIar average velocity through the bed Imls). Substituting for Fk in Equadon (I I52) gives us
For the case of constant superficial velocity U .
"In the nomenclature for Chapter 4, for Ergun Equation for pressure dmp.
782
External Dlffusron Effects nn Heterogeneous Reactions
Chap. I t
Differential equation describing Row and reaction in a packed bed
For reactions at steady state, the molar flux of A to the particle surface, WA, (mol/rn2~s)(see Figure 118), is equal to the rate of disappearance of A on the surface ri (mol /m2s); that is,
Boundary
I;+'
Layer
Figure 118 Diffusion across stagnant film s u m n d i n g catalyst pellet.
?A"
From Table 111, the boundary condition at the external surface is
ri = Whr = kr(CA CArl where
kc = mass transfer coefficient = 1DAB/8(s1) CA = bulk concentration (moUm"> C,, = concentration of A at the catalytic surface (molJm")
Substituting for
I n mact~onsthat c o m ~ l e t c 1mass ~ transferltmired, it IF not necewary to knoa the rate law.
ri in Quation
( 1 156), we have
In most mass transferlimited reactions. the surface concentration is begligible with respect to the bulk concentration (i.e.. CAs CAT): ( 1 160)
integrating with the limit, at z = 0. C, = CAo:
Sec. 1 1 3
783
External Resistance lo Mass Transfer
me corresponding variation of reaction rate along the length of the reactor is
The concentration and conversion profiles down a reactor of length L are shown in Figure 1t9. Reactor concentration profile for a mass
fransferlimited reaction
1 .o
C~
X
,c
+
0 0
rlL
1.0
slL
0
1.0
Figure 119 Axial concentration (a) and conversion (b) profiles i n a packed bed.
To determine the reacto; length L necessary
to
achieve a conversion X,
we combine the definition of conversion,
with the evaluation of Equation (1 161) at z
=
L to obtain
I t .3.6 Robert the Worrier Robert is an engineer who is always worried. He thinks something bad will happen if we change an operating condition such as flow rate or temperature or an equipment parameter such as particle size. Robert's motto is "If it ain't broke, don't fix it." We can help Robert be a little more adventuresome by analyzing how the important parameters vary as we change operating conditions in order to predict the outcome of such a change. We first look at Equation I1 164} and see that conversion depends upon kc, a, U. and L. We now examine how each of these parameters will change as we change operating conditions. We first consider the effects of temperature and flow rate on conversion. To l e m the effect of flow rate and temperature on conversion, we need to know how these parameters affect the mass transfer coefficient. That is. we must determine the c o r n Iation for the mass transfer coefficient for the particular geometry and flow field. For flow through a packed bed. the correlation
784
E ~ t e ~ n D~%lsion al Effects on Weterogewaus Reac!io~s
given by Thoenes and KrarnersI2 For 0.25 I < Sc < 4000 is
Cbap 1
< + < 0.5. 40 < Re' < 4000, an
Sh' = I.O(Rc')"?Sc"?
( 1 165
Thurne+Kramer?;
corrrlatio~~ for Ruw through packed bedr
where Re' =
Re ( I b)w
GI,, = particle diameter (equivaient diameter of sphere o f the sanl
voIume). m (volume of peElet)l1/
: u
=/ \,A
Chap. 12
Questions and Problems
A. Chemical Reaction Engineering in Microelectronic b e s s i n g B. Fundamentals of CVD C, Effectiveness Factors for Boar Reactors
Example R129 Diffusion Between Wafers Example R 1210 CVD Boat Reactor
~efcrcnceShelf
QUESTIONS AND PROBLEMS The subscript to each of the problem numbers indicates the level of difficulty: A. least dificult: 1). most difficult.
Mall of Fame
Fl2lc Make up an original problem using the concepts presented in Section (your instructor will specify the section). Extra credit will be gtven if you obtain and use real data from the literature. (See Prohrem P41 for thc guidelines.) P122R (a) Example 121. Efferril?~Drffusivin: Make a sketch of a diffus~onpath for n h ~ c hthe tortuosity is 5. How ~ ~ o u your l d effective gasphase diffuslv~ty change if the ahsolute pressure were tnpled and the temperature were increased by SO%? How would your answers change if (b) Example 122. 7issr1eEtlginrerir~~. the reaction kinetics were ( I ) first order In O? concentration with k , = lW2 hI? ( 2 ) Monod kinetics with p,,,, = 1.33 x hI and K, = 0.3 rnolldm'. (3) zeroorder kinet~cscarry out a quasi steady state analysis uslnf Equations (E122.19) along with the overall balance
to predict the 0?Aux and collagen buildup as a function of time. ,V01r: V = A,L. Assume o: = 10 and the ~toichiorne~ric coefficient for oxygen In collagen. v,. is 0.05 mass liaction of cclllmol 0:. (c) ExampFe 153. ( I ) What is the percent of the total reri? (3I ~f the Fac vclnc~tywere
D~Rus~on ano Rsact~on
Chap.
tripled? (4) if the particle v i ~ ewere decreased by a factor of 27 would the reactor length change in each case? (5)What length would required to achieve 99.99% conversion of the pollutant NO? What it.. (e) you app11cdthe Mean and WeiszPrater criteria to Examples E 14 a 124'? What would you find? What would you learn if AHR, = kcal/moI. h = 100 Btulhft:."F and E = 20 k callmol? (f) we let y = 30, = 0.4, and d~ = 0.4 in Figure 127? What wot~ldcau you to go from the upper steady state to the lower steady state and vi versa? (g) your internal surface area decrrnsed with time because of sintering. Hr would your effectiveness factor change and the rate of reaction chan with time if k, = 0.01 h1 and q = 0.01 at r = O? Explain. [h) someone had used the False kinetics (it..wrong E. wrong n)'? Wou their catalyst weight be overdesigned or underdesigned? What are 0th positive or negative etficts that occur? (i) you were asked to compare the conditions (e.g.. catalyst charge, convt sions) and s~zesof the reactors in CDROM Example R l 2 . l . What diffc ences would you find? Are there any fundamental discrepancies betwe1 the two? If so, what are they, and what are some reasons for them? (j) you were to assume the ressstance to gas abwrption in CDROM Examp Rl2.1 were the same as in Example RZZ.3 and that the liquid phase rea tor volume in Example R12.3 was 50% o f the total, could you estima the controlling resistance? If so, what is it? What other things could yc calcuIat~in Example R12.1 (e.g., selectivity. conversion, molar flow rat, in and out)? Hint: Some of the other reactions that occur include
a
Green engineering Web site zrrvw row~lrrr.ed~r/
.qrtenengineerirrg
(k) the temperature in CUROM Example R12.2 were increased? How woul the relative resistances in the slurry reactor change? (I) you were asked for alI the things that could go wrong in the operation ( a slurry reactor, what would you say? P123R The catalytic reaction
takes place within a fixed bed containing spherical porous catalyst X22. Fil ure P123 shows the overalf rates of reaction at a p i n t in the reactor as function of temperature For various entering total molar flow rates. Fm. (a) Is the reaction limited by external diffusion? (b) If your answer to part (a3 was "yes." under what conditions [nf thos shown (i.e.. T, F,)] is the reaction limited by external diffusion? (c) IS the reaction "reactionratelimited"? (d) If your answer to part (c) was "yes," under what conditions [of those show (ie., T, Fm)] is the reaction limited by the rate of the surface reactions' (e) Is the reaction limited by internal diffusion? (fl If your answer to part (e) was "yes;' under what conditions [of thos shown (i.e., 7; FTu)]is the reaction limited by the rate of internal diffusion (g) For a flow ratate of ID g molth. deterrmne (if possible) the overall effec tiveness factor, R, at 360 K. Ih) Estimate (if possible) the internal effectiveness factor, r(. at 367 K.
Chap. 12
Quest~onsand Problems
I
0 350
1
I
I
I
I
360
370
380
390
400
T(k) Figure P123 Reaction ntes in a catalyst bed.
i
If the concentration at the external catalyst surface is 0.01 mol/dm3, calculate (if possible) the concentration at r = R12 inside the porous catalyst at 337 K. (Assume a firstorder reaction.)
Addirionnl information:
Bed properties:
Gas properties:
Biffusivity: 0.I cmzls Density: 0.001 g/cm3 Viscosity: 0.0001 g l c m  s P12.la The reaction
Tortuosity of pellet: 1.415 Bed permeability: 1 millidarcy Porosity = 0.3
A  B
is carried out in a differential packedbed reactor at different temperatuses, flow rates, and particle sizes, The results shown in Figure P124 were obtained. Mall c f Fame
t
Figure P I 2 4 Reaction rates in a camlyst hzd.
8 58
P125,
Diffusion and Readion
Chap. 12
ta) What regions (i.e.. conditions d,, 7, F,) are external mass transferlimited? (b) What regions are reactionratelimited? lc) What region is internaldiffusioncontrolled? (d) What i s the internal effectiveness factor at T = 400 and d, = 0.8 cm? Curves A, B. and C in Figure P125show the variations in reaction rate for three different reactions catalyzed by solid catalyst pellets. What can you say ahout each reactlon?
! . I '
Figure P125 Temperature dependence of three reactions
PI26,
A firstorder heterogeneous irreversible reaction is ttking place within a spherical catalyst pellet which is plated with platlnum throughout the pellet (see Figure 123). The reactant concentration halfway between the external surface and the center of the pellet f i x . . r = RI?) is equal to onetenth the
concentratlon of the pellet's external surface. The concentratlon at the external surface is 0 001 g molfdrn3.the diameter ( 2 R ) is 2 X 30hcm. and the dtffuslon c r ~ f f i c i c nis~ 0.1 cm=ls.
crn in (a) What is the concentration of reactant at a distance of 3 X from the external pellet surface? {Ans.: C, = 2.36 X 10%ol/dm4.) (bk To wwh atdiameter should the pellet be reduced if the effectiveness factor . = 6.8 X I F 4 cm. Critique this answer!) is lo be 0.8? I A I I ~ .dp (c) Ifthe catalyst suppon were not yet plated with platinum. how would you suggest that the catalyst support be plated afrer it had been reduced by Applioaflon Pending rut Pmblem Hall of
grinding? PIZ7,, The SMimming rate of a small organism fJ. Theorpi. Biol., 26, 1 1 (I970)] is related to the energy released by the hydrolysis of adenosine triphosphate (ATPI to adenosine diphosphate (ADP). The rate of hydroly~isis equal to the rate of diffuvon of ATP from the rnidp~eceto the tail (see Figure P 1271. The diffusion coefficient of ATP in the midpiece and tail is 3.6 X cmZls.AQP is converted to ATP ~n the midsection. where its concentrafion is 4.36 X loc' mollcm'. The crossrcctional area of the tail is 3 X IOIU cm2.
Figure P127 Swirnming of an organicrn. (a1 Derive an equation for diffusion and reactlon in the tail.
Ihl Derirc an equation Tor the effectiveness Factor in the tail
Chap, t 2
Questions and Problems
859
(c) Taking the reaction in rfie tail lo be of zero order. calcufate the lmgth of the tail. The rate of reaction in the tail is 23 X 10 I R molts. (d) Compare your answer with the average tail length of 41 pm. What
possible sources of error? P12.BB A firstorder, heterogeneous, irreversible reaction is taking place within a catalyst pore which i s plated with platinum entirely along the length of the pore (Figure P128). The reactant concentration at the plane of symmetry (i.e., equal distance from the pore mouths) of the pore is equal to onetenth the concentmtion of the pore mouth. n e concenmlion at the pore mouth is 0.001 g moIldm3, the pore length (2L) i s 2 x cm. and the diffusion coefficient is 0.1 cmYs.
Figum P128 Single catalyst pore.
(a) Derive an equation for the effectiveness factor. (b) What is the concentration of reactant at LIZ? (c) To what length should the pore length be reduced if the effectiveness factor is to be 0.8? (d) If the catafyst suppon were not yet plated with platinum, bow would you suggest the catalyst support be plated after the pore length, L, had been
reduced by grinding? P129, A firstorder reaction is taking place inside a porous catalyst. Assume dilute concentrations and neglect any variations in the axial (x) direction. (a) Derive an equation for both the internal and overall effectiveness factors for the rectangular porous slab shown in Figure PI 29, (b) Repear part (a) for a cylindrical catalyst pellet where the reactants diffuse inward in the radial direct~on.
Figure P129 Flow over porous catalyst slab.
Pf210, The irreversible reaction
is taking place in the porous catalyst disk shown in Figure P129. The reaction is zero order in A. (a) Show that the concentration profile using the symmetry B.C.is
Hall of Fame
Diffusion and Reaction
Char
where
(b) For a Thiele modulus of 1.0, at what point in the disk is the ioncenml zero? For = 4? (c) What i s the concentration you calculate at z = 0.1 L and = 10 us Equation (P1210.I)'? What do you conclude about using this equatio, (d) Plot the dimensionless concentration profile y = CAiCAIas a function h = z/L for Q,, = 0.5. 1, 5 , and 10. H i ~ r there : are regions where the c~ centration i s zero. Show that h, = 1  I/@, is the start of this reg where the gradient and coocedtntion are both zero. [L. K. Jang, R. York. J. Ctln. and L. R. Hile, Inst. Chem. Engr., 34, 319 (2003).]Sh that v=O; X'2@,,($,) 1 ) A+(&I ) = f o r & 5 h c 1. (e) The effectiveness factor can be written ns
@,,
where z , (&)is the point where both the concentration gradients and flux , zero and A, is the crorssectional area of the disk. Show for a remord
to
reaction that for ib,, 5 1.0
q = { i  x c = ~ torhzl $0
(f) Make a sketch for versus dosimiEar to the one shown in Figure 125 (g) Repeat parts (a) to ( f ) for a spherical catalyst pelIet. (h) Repeat parts {a) to (0 for a cylindrical catalyst petlet. ti) What do you believe to be the point of thls probiem? P12.11c The secondorder decomposition reaction
is carried out in a tubular reactor packed with catalyst pellets 0.4 cm in diarnt ter. The reaction is inkmaldifft~sionlimited. Pure A enters the reactor at superficial velocity of 3 mls, n temperature of 25O"C, and a pressure of 500 kPi Experiments carried out on smaller gellets where surfilca reaction is lirmtin, yielded a specific reaction rate of 0.05 mblrnoE g cat s. Calculate the length o bed necessary to achieve 80% conversion. Critique the numerical answer.

Effective diffusivity: 2.66 X LO8 m2ls lneffectlve diffusivity: 0.00 m2/s Bed porosity: 0.4 Pellet density: 2 X 1P g/m3 Internal surface area: 400 m2/g
Chap. 72
861
Questions and Promems
P12lZc Derive the concentration profile and effectiveness factor for cylindrical pellets 0,2 crn in diameter and 1.5 crn in length. Neglect diffusion through the ends of the pellet. (a) Assume that the reaction is a firstorder isomerization. (Hint: Look for a Bessel function.) (b? Rework Problem 812 I1 for these pAlets. P1213c Reconsider diffusion and reaction in a sphericaIcatalyst pellet forthe case where the reaction i s dot isothermal. Show rhat the energy balance can be written as
where k, is the effective thermal conductivity, calls  cm  K of the pellet with dTldr = 0 at r = 0 and T = T, at s = R. (a) Evaluate Equation (1211) for a firstorder reaction and combine with Equation (P1215.1) to arrive at an equation giving the maximum temperature in the peIIet.
w e . ' At Tm,, c, = 0. (b) Choose representative values of the parameters and use a software package to solve Equations (1211) and (P1213.1) simultaneously for T(r) and CA(r)when the reaction is carried out adiabatically. Show that the resuiting solution agrees qualitatively with Figure 127. P1214cDetermine the effecrrveness factor for a nonisothermal spherical catalyst pellet in which a tirstorder isomerisation is taking place.
Additional infomtion:
A, = 100 m2/m3
hH, = 800,000 Jlmol D, = 8.0 X m2/s CAs= 0.01 kmol/m3 External surface temperature of pellet, T, = 400 K
E = 120.000 Jlrnol ThermaE conductivity of pellet = 0.004 JErn .s
.K
d, = 0.005 m Specific reaction rate = 10I m/s at 400 K Density of calf's liver = 1.1 gJdm3
How would your answer change if the pelIets were I t 2 , 1 P , and l P 5 m in diameter? What are typical temperature gradients in catalyst pellets? P1215, Extension of Problem P128. The elementary isomerization reaction
is taking place on the walls of a cylindrical catalyst pore. (See Figure P128.) In one run a catalyst poison P entered the reactor together with the reactant A. To estimate the effect of poisoning, we assume that the poison renders the catdyst pore walls near the pore mouth ineffective up to a distance z,, so that no reaction takes place on the walls in this entry region.
862
DifIuslon and Reaction
C h a ~ 12 .
(a) Show that before poisoning of the pore occurred, the effectiveness factor was given by
where
with
k = reaction rate constant (lengthltirne) r = pore radius (length) I), = effective molecular diffusivity larealtime)
(b) Derive an expression for the concentration profile and also for the molar flux of A in the ~neffecriveregion 0 < x < 2,. in iems of:,. DAD,,. C,,, and CAs.Without solving any funher differential equations. obtain the new effectiveness factor T ' far the poisoned pare. P1216B Fals$ed Kit~erics.The irreversible gasphase dirnerization
Is carried out at 8.2 atm in a stirred containedsolids reactor to which only pure A is fed. There is 40 g of catalyst In each of the four spinning baskets. The following runs were carried out at 227°C: Toral Molar Feed Rote, F , (g rnollmin)
1
2
4
6
11
20
The following experiment was carried out at 237°C:
(a) What are the apparenl reaction order and the apparent activation energy? (b) Determine the true reaction order, specific reaction rate. and activation energy. (c) Calculate the Thiele modulus and effectiveness factor.
(d) What diameter of pellets should be used to make the catalyst more effective? (el Calculate the rate of reaction on a rotating disk made of the catalytic matenal when the gasphase reactant concentration 1s 0.01 g mallL and the temperature is 527°C. The disk i s flat, nonporous, and 5 crn in diameter.
Effective diffusivity: 0.23 cm7s Surface area of porous catalyst: 49 m31g cat Density of caraly\t pellets: 1.3 g/crn3 Radius of catalyst pellet$: 1 cm Color of pellets: blushrng peach
Chap. 12
863
Journal Article Problems
P1217, Derive Equation (1235). Hint: Multiply both sides of Quation (1225) for nth order reaction, that is.
by 2&I&
rearrange to get
and solve using h e boundary conditions d p l d = O at h = 0.
JOURNAL ARTICLE
PROBLEMS
P12JI The article in Tmns. ini. Chcrn. Eng.. 60, 131 (1982) may be advantageous in answering the fotlowing questions. (a) Describe the various types of gasliquidsolid reactors. (b) Sketch the concentration profiles for gas absorption with: (3) An instantaneous reaction (2) A very slow reaction (3) An intermediate reaction rate P12J2 After reading the journal review by Y.T. Shah et al. [AIClrE J., 28. 353 (1982)j. design the following bubbie column reactor. One percent carbon dioxidd in aiT is to be removed by bubbling through a solution of sodlum hydroxide. The reaction i s massuansferlimited. Calculate the reactor size (length and diameter) necessary to remove 99.9% of the COz. Also specify a type of sparger. The reactor is to operate in the bubbly flow regime and still process 0.5 m"ls of gas. The liquid flow rate through the column is lo' rn3/s.
JOURNAL CRITIQUE PROBLEMS P12C1 Use the WeiszPraler criterion to determine if the reaction discussed in AIChE J., 10. 568 (1964) is diffusiunratelimited. P12C2 Use the references given in Ind. Eng. Chem. Pmd. Res. Dei:, 14. 226 (1975) to define the iodine value, saponification number. acid number, and experimental setup. Use the slurry reactor analysis to evaluate the effects of mass uansfer and determine If there are any mass transfer I~rnitations. Additional Homework Pmblems CDPlZAB Determine the catalyst size that gives the highest conversion in a packed bed reactor. CDP12BB Determine importance of concentration and temperature gradients in a packed bed reactor. CDP12C, Determine concentration profile and effecttveness factor for the first order gas phax reaction
864
Diffusion and Reaction
Cha~.
Slurry Reactors
CDPIZD,
Hydrogenation of methyl linoleatecomparing cataiyst. 13rd 1 PI2IYJ CDPlZE, Hydrogenation of methyl linolente. Find the ratelimiting step. 13rd Ed. PI2201 CDP12FR Hydrogenation of 2butyne1.4diol to butenediol. Calculate perc, resistance of total of each for each step and the conversion. [3rd E PI?211 %
CVD Boat Reactors CDPIZGI, Derennine the temperature profile to ach~eve\>nuniform thickne \>
CDPXtH, CDPl2IR CDP12.Ic CDPIZKc
P Mcrcbcr
Mall of Fame
Green Engineering
, [Znd Ed. P1 118) Explain how varying a number of the parameters in the CVD bc rerrctor will affect the wafer shape. [Znd Ed. P11141j Determine the wafer shape in a CVD boat reactor for a series of opt ating condttions. [Znd Ed. P 1 3 201 Model the buildup of a silicon wafer on parallel sheets. [2nd E PI I21J Rework CVD boat reactor accounting for the reaction
SiH4 SiH2 + HI [Znd ed. Pll221 Trickle Bed Reactors CDP12LB Hydrogenation of an unsaturated organic is carried out in a trick.klebt reactor. [2nd Ed. PI271 CDPlZM, The oxidation of ethanol is camed out in a trickle bed recrcror. [2r Ed. P1291 CDPltNc Hydrogenation of aromatics in a srickle brd wauror ['nd Ed. P12R: Fluidized Red Reactors CDPIZ0, Openended Ruidizrttion problem that requires critical thinking 1 compare the twophase fluid models with the threephase bubblin bed model. CDP12P, Calculate reaction ntes at the top and the bottom of the bed fc Example R12.33. CDP12Q, Calculate the conversion for A + I3 in a bubbling fluidized bed. CDP12Rk Calculate the effect of operating parameters on conversion for th reaction limited and transport limited operation. CDP12Se Excellent Problem Calculate all the parameters in Example R 123. for a different reaction and different bed. CDPf2Ta Plot conversion and concentration as a function of bed height in bubbling fluidized bed. CDP12UB Use RTD studies to compare bubblingkd with a fluidized bed. CDP12VB New problems on web and CDROM. CDPIZWB Green Engineering, www.rowan.edu/greenengineering.
Ghap. 12
Supptemefltary Reading
865
SUPPLEMENTARY READING I. There are a number of hooks that d ~ ~ internat c u ~ diffusion In catalyst peliets: however. one of the first b ~ k that s should be consulted on this and other topics on heterogeneous catnlyhis is
L ~ P m u s L.. , and N.R. AMU~DSOY. Chernicnl ReachtorTheor?.:A Review Upper Saddle River, N.J.:Prentice Hall. I977. In addition, see ARIT.R.. E f e m e t t ~ nChemicnl Rrrrcror Ann(v3i.r. Upper Saddle River. N.1.: Prentice Hall. 1969. Chap. 6. One should f n d the references listed at the end of this reading particularly useful. Lvss. D., "DiffusionReaction Interactions in Catalyst Pellets," p. 239 in ChernicnI Rerrcrion and Rerrcror Err,yineeritrg. New York: Marcel Dekker. 1987. The effects o f mass transfer on reactor performance are also discussed in DENBIGH, K.. and J. C. R. TURNER, Chemical Reactor Theov. 3rd ed. Cambridge: Cambridge Univer~ityPress. 1984. Chap. 7. SATERFIELD, C . N . , Heremgcneous Catnivsis in Industrinl Pmcrice, 2nd ed. New York: McGrawHill, 199 1 .
2. Diffusion with homogeneous reaction is discussed in ASTARITA, G.,and R. OCONE,Speci~rlTopics in Transport Phenomena. New York: Elsevier, 2002.
DANCKWERTS. P.V.. EnsLiqrtid Rencrions. New York: McGrawHill. 1970. Gasliquid reactor design is also discussed in CHARPG~E J. RC., , review article. Trans. lnst. Chem. Eng,60, 131 (L982). S H A HY . .T., GasLiquidSolid Rencror Design. New York: McGmwHill. 1979.
3. Modeling of CVD reactors is discussed in HESS, D.W.. K. F. JENSEA,and T. I. ANDERSON,"Chemical Vapor Deposition: A Chemicat Engineering Perspective." Rev. Chrm. Eng.. 3, 97, 1985. JENSEN, K. E, "Modeling of Chemical Vapor Deposition Reactors for the Fabrication of Microelectronic Devices," Clternicul and Cntal,vric Rencror Modeling. ACS Syrnp. Ser. 237. M. F. Dudokavic, P.L. Mills. eds., Washington, D.C.: American Chemical Society. 1984. p. 197. LEE, Id. H.. fundamm~nis of ,Wicroelectronics Pmcessing . New York: McGnwHill. 1990. 4. Multiphase reactors are discussed in
R~UACHANDRAN, P.A.. and R. V. CHAUDHARI.ThreePlrase Caraiyfic Reactors. New York: Gordon and Breach, 1983. RODRIG~~ES, A. E..J. M. COLO,and N. H. SWEED. eds.. Mulriphnse R~actorx, Vol. I : F~ndDme??~~ljs. Alphen aan den Rijn, The Netherlands: Sitjhoff and Noordhoff, 198 1. RODRIGUES. A. E., J. M. COLO.and N. H . SWEED, eds.. Muftiphase Renctors, Val. 1: Design Metitoris. Alphen aan den Rijn. The Netherlands: Sitjhoff and Nmrdhoff. 198 1.
866
Diffusion and Reaction
Chap. 12
SHAH. Y. T.. 8 . G. KELKAR, S. P. GODBOLE,and W. D. DECKWEK"Design Parameters Estimations for Bubble Column Reactors" (journal review), AIChE J.. 28, 353 (1982). TARHAN. M.O.,Catalytic Reactor Design. New York: McGrawHill, 1983. YATES,J. G.,Fundamenrals of FluidizedBed Chemical Processes, 3rd ed. London: Butternorth, 1983. The following Advances in Chemistry Series volume discusses a number of mul
tiphase reactors:
F'OGLERH, S., ed., Chernical Reactors. ACS Syrnp. Ser. 168. Washington. D.C.: Arnencan Chemical Society, 198 1, pp. 3255. 5. Fluidization
In addition to Kunii and Levenspiel's book, many comlations can be found in DAV~DSON, J. F., R. CLIFF,and D. HARRISON, Fluidization, 2nd ed. Orlando: Academic Press, 1985. A discussion of the different models can be found in
YATES.I. G.. Fluidizad Bed Chemical Processes. London: ButterworthHeinemann, 1983. Also qee GELDART, D. ed. Gas Fluidi:ation Technolop. Chichester: WileyInterscience, 1986.
Distributions of Residence Times for Chemical Reactors
13
Nothing in life is to be feared. It is only to be understood. Marie Curie
In this chapter we 'learn about nonideal reactors, that is, reactors that do not follow the models we have developed for ideal CSTRs, PFRs, and PBRs. I n Parl 1 we describe how to characterize these nonideal reactors using the residence time distribution function E(t), the mean residence time t,, the cumulative distribution function F(t), and the variance a2.Next we evaluate Elf), E(r), t,, and aZfor ideal reactors, so that we have a reference point as to how far our real (ie., nonideal) reactor is off the norm from an ideal reactor. The functions E(t) and F ( t ) will be developed for ideal PPRs, CSTRs and lamjnar flow reactors. Examples are given for diagnosing problems with reaI reactors by comparing t, and E ( f ) with ideal reactors. We will then use these ideal curves to help diagnose and troubleshoot bypassing and dead volume in reaI reactors. In Part 2 we will leam how to use the residence time data and functions to make predictions of conversion and exit concenmtions. Because the sesidence time distribution is not unique for a given reaction system, we must use new mcdels if we want to predict the conversion in om nonjdeal reactor. We present the five most common models to predict conversion and then close the chapter by applying two of these models, the segregation model and the maximum mixedness model, to single and to multiple reactions. After studying this chapter the reader will be able to describe the cumulative F(t) and external age i 3 t ) and residencetime distrihtion functions, and to recognize these functions for PFR, CSTR,and laminar flow reactors. The reader will also be able to apply these functions to calculate the conversion and concentrations exiting a reactor using the segregation model and the maximum mixedness model for both single and multiple reactions. Overview
Distributions of Residenc$Times !or Chemical Reactor7
868
Chzp. 1
13.1 General Characteristics
We want to analyze and characterize nonrdeal reactor behavior.
The reactors treated in the book thus farthe perfectly mixed batch. th plugflow tubular, the packed bed, and the perfect1y mixed continuous tan reactorshave been modeled as fdeal reactors. Unfortunately, in the real worl we often observe behavior very diflerent from that expected from the exem plar: this behavior is true of students. engineers. college professors. and chem ical reactors. Jusl as we mug[ learn to work with people who are not perfec~ so the reactor analyst must learn to diagnose and handle chemical reactor whose performance deviates from the ideal. Nonideal reactors and the princi ptes behind their analysis form the subject of this chapter and the next. J
Part 1 The basic ideas s
characterization and Diagnostics
d
t ;ue used in the distribution of residence times to charac terize and moddnonideal reactions are really few in number. The two majo,
uses of the residence time distribution to characterize nonideal reactors are
I. To diagnose problems of reactors in operation 2. To predict conversion or effluent concentrations in existinglavaiIab1~ reactors when a new reaction is used in the reactor
In a gasliquid continuousstirred tank reactor (Figure 13I), the gaseous reactant was bubbled into the reactor while the Iiquid reactant was fed through an inlet tube in the reactor's side. The reaction took place at the gasliquid interface of the bubbles, and the product was a liquid. The continuous liquid phase couId be regarded as perfectly mixed, and the reaction rate was proportional to the total bubble surface area. The surface area of a particular bubble depended on the time it had spent in the reactor. Because of their different sizes, some gas bubbles escaped from the reactor almost immediately, while others spent so much time in the reactor that they were almost comSystem 1
I Gas AFigure 131 Gasliquid reacror.
Sec. 13.1
~ o at1t molecutes vending the same ttme in the
,,
General Characteristics
869
pletely consumed. The time the bubble spends in the reactor is termed the bltbble reside~zcetitne. What was important in the analysis of this reactor was not the average residence time of the bubbles but rather the residence time of each bubble (i.e., the residence time distribution). The total reaction rate was found by summing over all the bubbles in the reactor. For this sum. the distribution of residence times of the bubbles leaving the reactor was required. An understanding of residencetime distributions (RTDsf aad their effects on chemical reactor performance is tlws one of the necessities o f the technically competent reactor analyst,
System 2 A packedbed reactor is shown in Figure 132. When a reactor is packed with catalyst, the reacting fluid usually does not flow through the reactor uniformly. Rather, there may be sections in the packed bed that offer little resistance to flow, and as a result a major portion of the fluid may channel through thrs pathway. Consequenrly, the molecuIes following this pathway do not spend as much time in the reactor as those flowing through the regions of high resistance to flow. We see that there is a distribution of times that molecules spend in the reactor in contact with the catalyst.
Figure 132 Packedtred reactor.
System 3 In many continuousstirred tank reactors, the inIet and outlet pipes are close together (Figure 133). In one operation it was desired to scale up pitar plant results to a much larger system. It was realized that some short circuiting wcurred, so the tanks were modeled as perfectly mixed CSTRs with a bypass stream. In addition to short circuiting, stagnant regions (dead zones) are often encountered. In these regions there is little or no exchange of material with the wellmixed regions. and. consequently, virtually no reaction occurs
We want to find ways of
determining the dead volume and amount of bypassing.
Dead zone
Figure 133 CSTR.
870
Tfie three concepts * RTD * Mixing
Model
D~str~but~ons of R e s l d e m Tomes tor Chernml Reaclors
Chap 13
there. Experiments were carried out to determine the amount of the material effectively bypassed and the volume of the dead zone. A simple modification of an ideal reactor successfully modeled the essential physical characteristics of the system and the equations were readily solvable. Three concepts were used to describe nonideal reactors in these examples: the disrribution uf residence time$ in the sysrem, rhe quolip of mixing, and the model used to describe the system. All three of these concepts are considered when describing deviations from the mixing patterns assumed in ideal reactors. The three concepts can be regarded as characteristics of the mixing in nonideal reactors. One way to order our thinking on nonideal reactors is to consider modeling the flow patterns in our reactors as either CSTRs or PFRs as afirsr approximation. In real reactors, however, nonideal flow patterns exist. resulting in ineffective contacting and lower conversions than in the case of ideal reactors. WE must have a method of accounting for this nonideality. and ro achieve this goal we use the nexthigher level of approximation, which involves the use of macromixing information (RTD) (Sections 13.1 to 13.4). The next level uses microscale (micromixi~tg)information to make: predictions about the conversion in nonideal reaclors. We address this third level of approximation in Sections 13.6 to 13.9 and in Chapter 14. 13.1.1 ResidenceTime Distribution (RTD) Function
The idea of using the distribution of residence times in the analysis of chemical reactor performance was apparently first proposed in a pioneering paper by MacMulIia and Weber.' However, the concept did not appear to be used extensively until the early 1950s. when Prof. P. V. Danckwerts2 gave organizational structure to the subject of RTD by defining most of the distributions of interest. The everincreasing amount of literature on this topic since then has generally foIlowed the nomenclature of Danckwerts, and this will be done here as well. In an ideal plugflow reactor. all the atoms of material leaving the reactor have been inside i t for exactly the same amount of time. Similarly. in an ideal batch reactor, all the atoms of materials within the reactor have becn inside it for an identical lenph o f time. The time the atoms have spent in the reactor is called the residence tirne of the atoms in the reactor. The idealized plugflow and batch reactors are the only two classes of reactors in which all the atoms in the reactors have the same residence time. In all other reactor types, the various arorns in the feed spend different times inside the reactor; that is, there is a distribution of residence times of the marerial within the reactor. For example. consider the CSTR; the feed introduced into a CSTR at any given lime becomes completely mixed with the material already in the reactor. En other words. some of the atoms entering the CSTR
I
R. &. MacMullin and M. Wehcr. Jr.. Trullr. Am. Insr.
?
P.y. Danck~ens,Cl~rtli.En8. Sri.. 2. 1 4 1953).
Cltrrr~ D t g . . 31. 309 ( 1935).
Sec. 13.2
Measurement of itre RTD
871
leave it almost immediately because material is being continuousIy withdrawn from the reactor; other atoms remain in the reactor almost forever because all the material is never removed from the reactor at one time. Many of the atoms. of course, leave the reaclor after spending a period of time somewhere in the welcome. vicinity of the mean residence time. In any reactor, the distribution of e s i dence times can significantly affect its performance. The residencetinre distribution (RTD) of a reactor is a characteristic of the mixing that occurs in the chemical reactor. There is no axial mixing in a We will use the plugRow reactor. and this omission is reflected in the RTD. The CSTR is thorRTD to oughly mixed and possesses a far different kind of RTD than the plugflow characteri~e ~ ares unique to a particular reacnonideal reacton. reactor. As will be illustrated Inter. not all R tor type: markedly different reactors can display identical RTDs. Nevertheless. the RTD exhihited by a given reactor yields distinctive clues to the type of mixing occurring within il wd is one of the most informative characterizations of the reactor.
The "RTD": Some molecules Fwve quickly, orhcrs oversmy their
13.2 Measurement of the RTD The RTD is determined experimentally by injecting an inert chemical, molecule. or atom. calIed a rracer. into the reactor a1 some time f = 0 and then ineasuting rhe rracer concentration. C, in the effluent stream as a function of time. In addition to being a nonreactive species that is easily detectable, the tracer should have physical properties similar 10 those of the reacting mixture and be completely soluble in the mixture. Et also should not adsorb on the walls or other surfaces in the reactor. The latter requirements are needed so that the tracer's behavior will honestly reflect that of the material flowing through the reactor. Colored and radioactive materials along with inen gases Use of tracers 10 determine the R m are the rnost common types of tracers. The two most used methods of injection are prrlsc ilrpu? and .TIP]?i ~ ~ p u f . 13.2.1 Pulse Input Experiment
~h~
c cllr\e
In a pulse input, an amount of tracer A', is suddenly injected in one shor into the feedstream entering the reactor In as short a time as possible. The outlet concentration is then measured as a funct~enof time. Typical concentmtiontime curves at the inlet and outlet of an arbitrary reactor are shown in Figure 134. The effluent concentrationtime cun7e is referred to as the C curve In RTD analysis. We shall analyze the injection of a tracer pulse for a singleinput and singleoutput system in which only f70w (i.e., no dispersion) carries the tracer material acrosq system boundaries. First, we choose an increment of time Ar sufficiently small that the concentration of tracer. C(r). exiting between time 1 and ! + A1 is ersentially the rame. The amount of trnccr matei i u l . Ah'. leaving the reactor between time r and r + Ar is then
872
Distributions of Res~denceTimesfor Ghernicai Reactors
Chap 1:
where u is the effluent volumetric flow rate. In other words. AN is the anoun of material exiting the reactor that has spent an amount of time between t an[ r At in the reactor. If we now divide by the total amount of material that was
+
injected into the reactor. No.we obtain
which represents the fraction of material that has a residence time in the reactor between time t and r + At.
For pulse injection we define
so that
The quantity E(t) is called the residencetime distri'brifioonfunction. It is the function that describes in a quantitative manner how much time different fluid elements have spent in the reactor. The quantity E{t)(it is the fraction of fluid exiting rhe reactor that has spent between time f and r + db inside the reactor.
+@
Reactor
Pulse injection
Pulse response I
I
Step i!jection
/
I
Step response
Figure 134 RTD measurements.
Sec. 13.2
The C curve
If No is not known directly, it can be obtained from the outlet concentration measurements by summing up all the amounts of materials, AN.between time equal to zero and infinity. Writing Equation ( 1 3 1 ) in differential form yieIds
t
djV = uC(t)dr
/135)
lx
(1 361
CUI
=lo
h
C(f)dt
~rea
COI
873
Measuremenl of the RTD
t
We find the RTD
and then integrating.
we
obtain N, =
u ~ ( id)
0
The volumetric flow rate v is usuaIly constant, so we can define E ( t ) as
function. E f t ) . from bhe tracer concentration
( 1 37)
C(1)
The E curve
t
The iniegral in the denorinrtor is the area under the C curve, An alternative way of interpreting the residencetime function is in its integral form:
Fraction of material leaving the that has resided in the reactor for times between r , and t2 We know that the fraction of a13 the material that ha.. resided for a time r in ithe reactor between t = 0 and t = TJ is 1: therefore, Eventually aIl must leave
The following example will show haw we can calculate and interpret E(t) from the efRuent concentrations from the response to a pulse tMCer input to a reaI (i.e., nonideaI) reactor. Example 131
Conshucting the C(t) and E(t) CUNCS
A sample of the tracer hytane at 320 K was injected as a pulse to a reactor, and the efRuent concentration was measured as a function of time. ~ s u l t i n gin the data shown in Table E13t 1.
Pulse Input
The measurements represent the exact concentrations at the times listed and not average values between the various sampling tests. ( a } Construct figures showing C(r) and E(t) as functions o f time. (b) Determine both the fraction of rnatenal leaving
874
Distriberiions of ResidenceTimes lor Chemical Reactors
Chap. 13
the reactor that has spent between 3 and 6 min in the reactor and the fraction of material leav~ngthat bas spent between 7.75 and 8.25 min In the reactor, and (c) determine the fraction of material leaving the reactor that has spent 3 min or less in zhc reactor. Solurion
(a) By plotting C as a function of time. using the daza in Table El 31 .l. the curve shown in Figure El>1.1 is obtained. To obtain the E ( r ) curve from the C(r) curve. we just divide
C ( r )by rhe integral
0
C ( i ) dt .
The C curve
Figum E131.1 The C curve.
which is just the area under the C curve. Becaufe one quadrature (integration) Formula will not suffice over the entire range of data in Table E 13 I . I . we break the data inlo two regions. 010 minutes and 10 ta 14 minutes. The area under the C curve can now be found using the numerical integratron forn~ulas(A31 ) and (A25) in Appendix A.4:
+2(10>+41R)+2(6) +4(4)+2(3O)+412.2lr = 47.4 g  min !m7
l(1.513
Sec. 13.2
Measurement of the RTD
I We now calculate
with the folFowing results: TAHLE E131.2 C(t) AND E(t) I
(min)
1
2
1
3
4
5
6
7
8
9
10
1 2 1 4
(b) These data are plotted in Figure E131.2. The shaded area represents the fraction of material leaving the reactor that has resided in the reactor between 3 and 6 min.
Tne E curve
Figure E131.2 Analyzing the
E curve.
Using Equation (A22) in Appendix A.4:
E (f) dr
= shaded area
3
=
]AfV,+ 3 , 1 ; + 3 S , + f , j
Evaluating this area. we find that 51% of the material leaving the reactor spends between 3 and 6 rnin in the reactor. Because the time between 7.75 and 8.25 rnin is very smalf relative to a time scale of 14 min. we shall use an alternative technique to determine this fraction to reinforce the inrerprelation of the quantity Elr)dt. The averaye value of E ( r ) between these times 1s 0.06 min :
The tail
Consequently. 3.0%of the fluid leaving the reactor has been in the reactor between 7.75 and 8.25 min. The longtime portion of the Elt) curve is called the toil. In this example the tail IS that ponion of the curve between sny 10 and 14 min.
876
DIstrrbutians of Resrdence T!res tor Chemical aeaflors
Chap 1:
(c) Finally, we shall consider the fraction of mater~aithat has been in the rent tor for a time !or less. hat is. the fraction that has y e n t between O and I minute> rn the reactor. Thiq fraction i s just the rhaded area under the curve up to I = I min. uter. This area is qhomn in Figure E131.3 for r = 3 mln. Calculating the area undel the cune, we see that 20%of the material has spent 3 min or 1e.r.~in the reactor.
0
1 2 3 4 5 6 7 8 91011121314 ! (min)
I
Drawbacks ur the Puix injection '0 obtain the RTD
Figure E131.3 Analyzing the E curve.
Tbe principal difficulties with the pulse technique lie in the problem: connected with obtaining a reasonable pulse at a reactor's entrance. The injec tion must take place over a period which is very short compared with residenc~ times in various segments of the reactor or reactor system, and there must be : negligible amount of dispersion between the point of injection and the entranc~ to the reactor system. If these conditions can be fulfilled. this technique repre sents a simple and direct way of obtaining the RTD. There are probiems when the concentrationtime curve has a long tai because the analysis can be subject to larse inaccuracies. This probIem princi pally affects the denominator of the righthand side of Equation (137) 1i.e the integration of the C(t) curve]. It is desirable to extrapolate the tail and ana lytically continue the calculation. The tail of the curve may sometimes b approximated as an exponential decay. The inaccuracies introduced by thi assumption are very likely to be much less than those resulting from eithe truncation or numerical imprecision in this region. Methods of fitting the tal are described in the Professional Reference Shetf 13R. 1. 13.2.2 Step Tracer Experiment Now that we have an understanding of the meaning of the RTD curve from pulse input, we will formulate a more general relationship between timevarying tracer injection and the corresponding concentration in the e H t ent. We shall state without development that the output concentration from vessel is related to the input concentration by the convolution integral:3 A development can be found in 0,Levenspiel, Chemical Reaction Engineering, 2n ed. (New York: Wiley, 1972). p. 263.
Sec. '3.2
Step input =I"
Ir
The inlet concentration most oAen takes the form of either a perfect pulse irtpitl (Dirac detta function). impetjfectp~ilwitjecrion (see Figure 1341. or a step inplit. Just as rhe RTD function E(t1 can be determined directly fmm a pulse input, the cumulative distribution F(t) can be determined directly from a step input. We will now anaryze a srcp itlplit in the tracer concentration for a system with a constant volumetric flow rate. Consider u constant race of tracer addition to a feed that is initiated at time r = 0. Before this time no tracer was added to the feed. Stated symbolically, we have CoIO =
 
[IJ
t
877
Measurement of the RTD
tX,,
Sec. 13.8
then
Important point
923
Using Software Peckages
and
We note that in some cases X, is not too different from Xm. However, when one is considering the destruction of toxic waste where X > 0.99 is desired, then even a small difference is significant!! In this section we have addressed the case where all we have is the RTD and no other knowledge about the flow pattern exists. Perhaps the flow pattern cannot be assumed because of a lack of information or other possible causes. Perhapswe wish to know h e extent of possible ersor from assuming an incorrect flow pattern. We have shown how to obrain the conversion, using only the RTD. for two limiting mixing situations: the earliest possible mixing consistent with the RTD,or maximum mixedness, and mixing only at the reactor exit, or complete segregation. Calculating conversions for these two cases gives bounds on the conversions that might be expected for different flow paths consistent with she observed RTD.
13.8 Using Software Packages Example 137 could have been solved with an ODE solver after fitting E(t) to a potynornial.
Fitting the E(t) Curve to s Polynomial Some forms of the equation for the conversion as a function of time multiplied by E(t) wilE not be easily integrated analytically. Consequently, it may be easiest to use ODE software packages. The procedure is straightforward. We recall Equation (1 353
where is the mean conversion and X(t) is the batch reactor conversion a t time I. The mean conversion is found by integrating between r = 0 and t = w or a very large time. Next we obtain the mole balance on X ( I ) from a batch reactor
and would write the rate law in terms of conversion, e.g., 5 = k d O ( 1q2 The ODE solver will combine these equations to obtain X ( I ) which will be used in Equation (1356). Finally we have to specify E(r). This equation can be an analytical funcrion such as those for an idea! CSTR,
924
Distributions of Residence Times for Chemical Reactors
or it can be polynomial
Chao. 1
or a combination of polynomials that have been use
to fit the experimental RTD data
We now simply combine Equations ( 1352), (1369), and (1370) and use a, ODE sotves. There are three cautions one must be aware of when fitting E(t) ttl
a poIynomia1, First. you use one poIynomia1 E l ( { ) as E(rj increases with tim to the top of the curve shown in Figure 1327. A second polynomial E2(1)i used from the top as E(r) decreases with time. One: needs to match the tWr curves at the top.
Match
E
h
t
Figure 1327 Matching E,(r) and E,(r).
Summary Motes polymath~~~~~~l
Second. one should be certain that the polynomial used for E2(t) does no become negative when extrapolated to Iong times. If it does, then constraint: must be placed on the fit using IF statements in the fitting program. Finally one should check that the area under the E(t) curve is virtually one and that thr cumulative distribution F(t) at long times is never greater than 1. A tutorial or how to fit the C(r1 and EIt) data to a polynomial is given in the Summary Note: for Chapter 5 on the CDROM and on the web. Segregation Model
Here we simply use the coupled set af differential equations for the mean exit conversion, 2 ,and the conversion X(t) insidea globule at any time, t.
The rate of reaction is expressed as a function of conversion: for example,
01
Sec. 13.8
Using S o h a r e Packages
and the equations are then solved numericaIIy with an ODE solver.
Maximum Mixedness Model Because most software packages won't integrate backwards. we need to change the variable suchthat the integration proceeds forward as h decreases from some large value to zero. We do this by forming a new variable, s, which is the difference between the longest time measured in the E ( t ) curve. T , and 1.In the case of Example 137, the longest time at which the tracer concentradon was measured was 200 minutes (Table E137.13. Therefore we will set T = zoo.

X = T z=200z Then,
One now integrates between the limit z = O and 2 = 200 to find the exit conversion at z = 200 which corresponds to h = 0. In fitting E ( t ) to a polynomial. one has to make sure that the polynomial does not become negative at large times. Another concern in the maximum mixedoess calculations is that the tern 1  F ( X ) does not go to zero. Setting the maximum value of F ( t ) at 0.999 rather than 1.0 will eliminate this problem. It can also be circumvented by integrating the polynomial for E ( I )to get F ( t ) and then setting the maximum value of F ( I ) at 0.999. If F(r) is ever greater than one when fitting a polynomial. the solution will blow up when integrating Equation 11372) numericaIIy. Example 138
Using Software to Make Maximum Miredrpess Model Calc1dat3ons
Use an ODE solver to determine the conversion predicted by the maximum mixedm e s s model for the E(r) curve given in Example E137.
Because of the nature of the E ( r ) curve, it i s necessary to use two polynomjals, a third order and a fourth order, each for a different part of the curve to express the RTD, E ( t ) , as a function of time. The resulting E t t ) curve is shown in Figure
E138.1. To use Polymath to cany out the integration, we change our variable from X to z using the largest time measurements that were taken fmm E(r) in Table E137.1,which i s 200 min:
926
Distributions of ResidenceTmes for Chernlcal Reactors
Chap 13
First, we fit E ( I ) .
Figure E138.1 Polynomiel fit of E ( f ) .
z = 200  A The equations to be solved are A = 200  2 Maximum rnixedness modti
For values of h less than 70, we use [he polynomial
EI(~)=4.M7c10h41.I80e7h3+1.353eL5h28.657e4k+0.028{E138.3) For values of A greater than 70,we use the polynomial
E2()c)= 2.640eqh3
+ 1.361S C  ~ A~ 2.407ejh + 0.U15
E138.4)
(El 38.5) with z = 0
[I at
[
F(h)]
= 200). X = 0, F = 1 [i.e., F(A) = 0,9993.Caution: Because tends za infinity at+ = 1, (z = 0).we set the maximum vatue of F
0.999 at z = 0. The Polymath equations are shown in Table E138.1. The solution is
The conversion predicted by the maximum rnixedness model i s 56.3%.
Sec. 13.9
RfC)
snd Munipte Reactions
ExplH equalbs as mmwd
927
uw usor
[11
LfvingExample Problem
121 k t . 0 1 131 lam=2#z I 4 l cs r uto'(lx) 1 S l E l = 4 ME€.bl[nswP+f . 1 8 ( t 2 e ~ ~ 1 3 6 3 ~ h ~ * . ~ 5 5 5 2 ' I ~ ~ 1 . 0 2 B D W I 6 1 E2 m 2.We91em*a+1,381~7emh2.0M12*08~gI~m+.o~ 5011 171 F1 = 4 ~ 5 8 a l [ Y 5 ? a ~ 1 . 1 8 0 2 b ? / 4 % m F d + 1 . 3 5 3 ~ R:'FZ +9 3 0 7 6 ~  B ' b ~ t i . 0 2 ~ 5 5 1 a ~ 2  ."tarnc.618nl2) 0W1 191 r a =  P W 2 110 1 E il (lam*~70)Fo) ((El)e h (E2) [ll l F = tf (lamc=7Q)ihen (Fl) s$s (F2) I121 EF = B(1Q
I

Polynomiafs used to fir Err) and F(r)  I
uanaMe v a n W name : t lnn~alvalue O f~nalvalue : 200
.
13.8.1 Heat Effects If traces tests are carried out isothermally and then used to predict nonisotherma1 conditions, one must couple the segregation and maximum mixedness models with the energy balance to account for variations in the specific reaction rate. This approach will only be valid for liquid phase reactions because the vo1umetrjc flow rate remains constant. For adiabatic operation and
A?~=o,
As before, the specific reaction rate is

,
I."[;IkJj r
k=k,exp 
I

Assuming that E(r) is unaffected by temperature variations in the reactor, one simply soIves the segregation and maximum mixedness models, accounting for the variation of k w ~ t htemperature [i.e., conversion; see Problem P132(j)].
13.9 RTD and Multiple Reactions As discussed in Chapter 6.when multiple reactions occur in reacting systems, it is best to work in concentrations,moles, or molar flow rates rather than conversion. 13.9.1 Segregation Model
In the segregation model we consider each of the globules in the reactor to have different concentrations of reactants, CA,and products, Cp.These globules
928
Distnbutions of Residence Ttmes for Chemical Reactors
Chap.
are mixed together immediately upon exiting to yield the exit concentration I , which is the average of all the globules exiting: A,
The concentrations of the individual species, CA(t)and C B ( t ) ,in the difTere globules are determined from batch reactor calculations. For a constantvolun batch reactor. where q reactions are taking place, the coupled mole balan equations are
These equations are solved simultantousIy with
to give the exit concentration. The RTDs. E(t), in Equations (1377) and (137 are determined from experimental measurements and then fit to a polynomial.
13.9.2 Maximum Mixedness For the maximum rnixedness model, we write Equation (1362) for each sp cies and replace r, by the net rate of formation
After substitutionfor the rate laws for each reaction (e.g., r i =~k,CA),these equ tions are solved numericalIy by starting at a very large value of X, say T = 201 and integrating backwards to A = 0 to yield the exit concentrations CA,CB,.. We will now show how different RTDs with the same mean residenl time can produce different product distributions for multiple reactions.
Sec. $3.9
RTD and Multiple Reactions
Example 139
RTD and Complex Reactions
Consider the following set of liquidphare reactions: Living Exarcple Problc
A+B
k', C
which are occurring in two different reacton with the same mean residence time r, = 1.26 min. However, the RTD is very different for each of the reactors, as can k seen in Figures E139.1 and E l 39.2.
t
Figure EI39.1
El (t): asymmetric distrihation.
t
Figure E139.2 E2(t): bimdal distribut~on
930
Distributions af ResidenmTlmes far Chernbl Reamts
Chap. 13
(a) Fit a polynomial to the m s .
(b) Determine the prduct distribution (e.g., S,. 1. The segregation model 2. The maximum mixedness model
)S,
for
Additional lnforrnat~on
k , = k2 = k3 = 1 in appropriate units at 350K. Solution
Segregation Model
Combining the mole balance and rate jaws for a constantvolume batch reactor he., globules), we have
and the concentration for each species exiting the reactor is found by integrating the equation
over the life of the E(r) curve. For this example the life of the E, I r ) is 2.42 mlnutes (F~gureE139.1f. and the life o f is 6 minutes (Figure El 39.2). The initial conditions are r = 0, C, = C, = 1, and C, = C, = C, = Q. The Pnlyrnath program used to solve these equations is shown in Table El 39.1 for the asymmetric RTD,J, ( r ) . With the exception of the polynomial for E&), an idenlical program to that in Table E139.1 for the bimodal distribution is given on the CDROM.A comparison of the exit concentration and selectlvities of the two RTD curves is shown in Table E 139.2.
See. 53.9
931
RTO and Multiple Reactions
Ltvlng Example Problem
Aqmmctric Disriibwtion
The solution
for E , ( t ) is:

C, = 0.E51
= 0.178
X
Bimodal Disrrihrian The solution for E2(r) is:
C, 
=0245
C, = 0.357
Sm = 1.18

C,
SD, = 1.70
= 0.265
= 0.454


0.303
=
84.94
Maximum Mixwines Model The equations for each s p i e s are
C, =0.5!0
C, =0.162
X=75.5%
0.321 ,3
Sm
= 1.21 = 1.63
932
Solved .Prohlcn?r
Oistribut~onsof Resrdence Trrnes for Chemical Reactors
C h a ~1:
The Polymath program for the bimodal distribution, Eltt), i s shown in Tab11 E139.3.The Polymath program for the asymmetric distribution is identical with thl exception of the polynomial fit for E, ( I ) and is given on the CDROM. A compari son of the exit concentration and select~vit~es of the two RTD distributions is show1 in Table E139.4.
FOR
TABLE E139 3. FOLYMATH PR~RAY MAXIMUM MIXEDYESSMODEL WITH BIMODAI.DISTRIBCTION (MULTIPLE RBA~ONS)
As!mmernc Disrribrrrion
BSmotIuI Dirtriburion
The solution hr E l ( 0 ( I 1 is:
The wlulion for & ( t ) (21 is:
Catcutations similar to those in Example 139 are given in an example on
*
Solved Problemr
kt
B
k?
>
c
I n addition, the effect of the: variance of the RTD on the parallel reactions in Example 139 and on the series reaction in the CDROM is shown on the
CDROM.
Chap. :3
Summary
Closure After completing h i s chapter the reader will use the tracer concentration time data to calculate the external age distribution function E(r), the cumuIative distribution function F(r), the mean residence time,r, and the variance, rsZ. The reader will be able to sketch E(t) for ideal reactors, and by comparing E(t) ftom experiment with E(t) for ideal reactors (PFR,PBR, CSTR, laminar flow reactor) the reader will be able to diagnose problems in red reactm. The reader will also be able to couple RTD data with reaction kinetics ta predict the conversion and exit concentrations using the segregation and the maximum rnixedness models without using any adjustable parameters. By analyzing the second derivative of the reaction rate with respect to concentration, the reader wiIl be able to deternine whether the segregation model or maimurn rnixedness model wiIl give the greater conversion. SUMMARY I. The quantity Err) dt is the fraction of material exiting the reactor that has spent between time t and r * dr in the reactor. 2. The mean residence time
:1
=
d t r ) dr = r
(~131)
is equal to the space time T for constant volumetric flow, v = vU. 3. The variance about the mean residence time is
4. The cumulative distribution function Fit) gives the fraction of emuent material that has been In the reactor a time r or less:
I
 F ( t ) = fraction of effluent material that has been in
( 5 133)
the reactor a time t or longer
5 . The RTD functions for an ideal reactor are
CSTR: Laminar flow:
E(t) = 0
!< f 2
(51361
934
Dishibrrfim of Residence T i m e br Chemical Reactors
Chap. 13
6. The dimensionless residence time is
7. The internalage distribution, [I(&) da],gives the fraction of material inside the reactor that has been inside between a time ol and a time (u + da). 8. Segregation model: The convwsion is
and for multiple reactions
9. Maximum fixedness: Conversion can be calculated by soiving the following equations:
and for multiple reactions
from h,,, to h = 0. To use an ODE solver let z = h,,,
k
CDROM MATERIAL
Cummary tJotcs
& Links
Learning Resources I . Summaw Now5 2. Web ~ a i e r i a lLinks A, The Attainable Region Analysis www,engin.umich.edu/cdChapters/ARpgcs/lntm/intm, h rm and w~~:wi~~.u~.:a/fac/engine~ring/pmmar/ar~gion 4. Solved Problems A. Example CD 131 Calculate the exit concentrations series reaction AElC
8 . Example CD 132 Deteminalion of the effect of variance on rhe exit concentrations for the series reaction Sobed Problems
A+BC
Chap. 13
Livrng Example Problem
CDROM Materiat
Living Example Problems I . Exansple I 3 4 Laminar Flow Reactor 2. Example 138 Using Sofware 10 Make Maximum Mixrdness Model Cnlculations 3. Example 139 RTD and Complex Reactions 4, ~ x m p l eCD13J A + B + C Effect of RTD 5. Example CD132 A +JB + C Effect of Variance Professional Reference Shelf 13R.1. Frrrirrg the Toil Whenever there are dead zones into which the material diffuses in and out,  the C and E curves may exhibit long tails. This section shows how to analytjcaliy describe fitring these tails to the curves. E(I) =
b = slope of In E vs. r MI
a = be [ l  F ( t , ) ]
Reference Shelf
13R.2.h1temalAgeDistribution The internalage distribution currently in the reactor is given by the distribution of ages with respect to how long the molecules have been in the reactor. The equation for the internalage distribution is derived along and an example is given showing how it i s applied to catalyst deactivation in a "fluidized CSTR."
Example 13R2.1 Mean Catalyst Activity In a Fluidized Bed Reactor. 1 3R.3.Con~paringX,,g with X,, The derivation of equations using the second derivative criteria
i s carried out.
@jstriSutio?sof Reside~ceTirnec
$7
Chemical Reactors
Chap. 13
QUESTIONS AND PROBLEMS The subscript to each of the problem numbers indicates the level of difficulty: A, least

A=.
B=.
C = *
D=**
Clomehmork Problems
P131, Read over the problems of this chapter. Make up
an original problem that
uses the concepts presented in this chapter The guidelines are given in hoblem P4I. RTDs from real reactors can be Found i n Ind. Eng. Chem., 49, LW ( 1957); Ind. Enx. Chern. Pmce.rr Des. Den, 3, 38 1 (1964). Can. J. Chsm. Eng., 37, l OT 11 959 ): It~dEn p. Chcm.. 41, 2 18 ( 1 952); Cliem. Et1.p. Sci.. 3. 26 (1954): and Inri. Eng. Ckertz., 53. 38 I I196 I ) . P132, What if ... (a) Example 131. What fractlon of the Ruid spends nine minutes or longer in the reactor? (hj The combinations of ideal reactors modeled the following real reactors, given E(B),F(Q), or I  F(Q).
(c) Example 133. How would the E(r) change if ,T, as reduced by 50% and q , w a increased by 50%? (d) Example 134. For 75% conversion. what are the relative sizes of the CSTR. PFR. and LFR? (e) Example 135. How does the mean conversion compare with the conversion calculated with the same, I, applied to an ideal PFR and CSTR? Can you give examples of E(t) where this calculation would and would not be n good estimate of X? (fi Example 136. Load the Living Erample Problem. How would your results change if T = 40'C? How would your answer change i f the reaction was pseudo first order with kcAa = 4 x lP3ks? What if the reaction were carried out adiabatically where C,, = C,, = 20 cnVmoUK. 6 = 10 kcaVmal k = 0.01 dm3/ml/min at 25'C with E = 8 kcaUml
Chap. 13
937
Ouestrons and Problems
(g) Example 137. Load the Living ExampIe Pmblem. How does the X,,, and X,$, compare with the conversion calculated for a BFR and n CSTR at the mean residence time'' (h) Example 138. t o a d the Lirina Eramplc Probletrt. Ilow would your results change if the reaction was pseudo first order with k, = CA&= 0.08 mind? If the reaction was third order with kdAo = 0.08 minI? The reaction was half order with = 0.08 minI. Describe any trends. (i) Example 139. Load the Living Example Pmbletn. If the activation energtes in callm~rland E , = 5.000.E: = 1,OIK). and E3 = 9,000. how would the selectives and conversion of A change as the temperature was raised or lowered around 350 KQ G) Heat Effects. Redo Lit'E~rgEmmple PmbSems 137 and 138 for the case when the reaction is carried out adlnbatically with (1) Exothermic reaction with
with k given at 320 K and E = 10.000 catlrnol. (2) Endothermic reaction with
and E = 45 Id/mol. How will your answers change? (k) you were asked to compare the results from Example 139 for the asymmetric and bimodal distributions in Tables E139.2 and E1394. What similarities and differences do you observe? What generalizations can you make? (I) Repeat 132(h) above using the RTD in Polymath program E E 38 to predict and compare conversions predicted by the segregation m d e l . (m) the reaction in Example 135 was half order with = 0.08 minI? How would your answers change? Hint: Modify the Living Example 138 program. (n) you were asked to vary the specific reaction rates k, and in the series
k,
Heat effects
reaction A B bC given on the Solved ProbIems CDROM? What would you find? ( 0 ) you were asked to vary the isothermal temperature in Example 139 from 3 0 K, at which the rate constants are given, up to a temperature of 509 KT The activation energies in callmol are El = 5000. E, = 7000,and E, = 9000. How would the selectivity change for each RTD curve? (p) the reaction in Example 137 were carried out adiabatically with the same panmeters as those in Equation [PIS2(j).ljqHow would your answers change? (q) If the reaction in Examples 138 and 135 were endothermic and carried out adiabatically with
TCK) = 320
 IOOX
and
E = 45 Hlmol
[?132u). I ]
how would your answers change? What generalizations can you make about the effect of temperature on the results (e.g., conversion) predicted from the RTD? (r) If the reaction in Example 812 were carried our in the reactor described by the RTD in Example 139 with the exception that RTD is in seconds rather than minutes (i.e., r, = 1.26 5). how would your answers change'?
938
Dtstr~b~rt~ons of Res~denceT~mesfor Chem~calReactors
Chap. 13
P133c Show that for a firstorder reaction AI3 the exit concentration maximum rnixedness equation
dC1 d)l
E("
1 F ( h )
(C*C*,)
is the same as the exit concentration given by the segregation mode!
[Hinc Verify
is a solution to Equation (P133.I).]
P134, The firstorder reaction with 1: = 0.8 minI is carried out in a real reactor with the following RTD function
0
r
22
t, min
For then E(t) = circle) For t > 22 then E(r) = 0 (a) What 1s the mean residence rime? (b) What is the variance? (c) What is the conversion predicted by the segregation model? (d) What is the conversion predicted by the maximum mixedness model? P135, A step tracer input was used on a real reactor with the following results: For I S 10 min. then C, = 0 For 10 5 t 5 30 rnin. then C, = 10 g/drn3 For 1 2 3Q min, then c;= 40 gldrn) The secondorder react~onA + B with k = 0.1 drn'lmol . min is ro be carried out in the real reactor with an entering concentration of A of 1.25 molldm' at a volumetric flow rate of 10 dm3/mln. Were k is given at 325 K. (a) What is the mean residence lime r,? (b) What is the variance a?'? (t) What conversions do you e x v c t from an ideal PFR and an ideal CSTR in a real reactor w ~ f hr,?
Chap. 13
Question&and ProMems
(d) What i s the conversion predicted by (1) the segregation model? (2) the maximum mixedness model? (e) What conversion is predicted by an ideal laminar flow reactor? (0 Calculate the conversion using the segregation model assuming T(K) = 325 500X. P136# The following E(f) curve was obtained from a Vawr test on a tubular reactor in which dispersion is believed to occur.

t
(min)
F i g n ~P135 RTD.
A secondorder reaction A
4 B with
kc,,,= 0.2 minI
is to be carried out in this reactor. There is no dispersion wcurring eather upstream or dowosaeam of the reactor, but there is dispersion inside the reactor. Find the quantities asked for in parts (a) through (e) in pmblem PI 3SB? P137, T h e irreversible $quid phase reaction is half order in A. The reaction is carried out in a nonideal CSTR,which can be modeled using the segregation model. RTD measurements on the reactor gave values of t = 5 min and u = 3 rnin. For an entering concentmtion of pure A of 1.0 molldm" the mean exit conversion was 108. Estimate the specific reaction rate constant, k , . Hinc Assume a Gaussian distribution. P13SB The thirdorder liquidphase reaction
a reactor that has the following RTD E(r) = 0 for 1
Consequently, we are going to solve Equation ( 1430) for the solution concentration as a function of r and then substitute the solution c (r, z, t) into Equation (1431) to find ,;( 0. All the intermediate steps are given on the CDROM R14.1. and the partial differential equation describing the variation of the average axial concentration with time and distance is
Reference Shelf
ac "C+uz=,*a*
a=*
f 1432)
ar*l
where D* is the ArisTaylor dispersion coefficient: ArisTaylor disprsion
coefficient
That is, for laminar flow in a pipe
Figure 1410 shows the dispersion coefficient. D' in terms of the ratio D' lU(2R) = D*lUdf as a function of the product of the Reynolds and Schmidt numbers. 14.4.5 Correlations for
D,
14.4.5A Dispersion for Laminar and Turbulent Flow in Pipes
An estimate of the dispersion coefticient, D,, can be determined from Figure 1411. Here d, is the tube diameter and Sc is the Schmidt number discussed in Chapter 11. The Bow is laminar (streamline) below 2,100. md we see the ration {DoU/d,) increases with increasing Schmidt and ReynoIds numbers. Between Reynolds numbers of 2,100 and 30,000, one can put bounds on D, by calculating the maximum and minimum values at the top and bottom of the shaded region.
Sec
Id4
965
Flow, React!on, and Dispersion
Figure 1410 Correlation for dispersion for streamline flow in pipes. (From 0.Levenspiel, Chemical Reaction Engineering, 2nd ed. Copyright O 1972 John Wiley & Sons. Inc. Repnntd by permission of John Wlley Sr Sons, Inc. All nghtr reserved.) [Note: D = D, and D =DAB]
Once the Reynolds number is calm
Iated. D, can be found.
ail lol
I
! ~rlltrl
103
I ! ]
111rr8l tQ4
I
I 1111111
106
I
I I I I I ~
10"
Figure 1411 Comefation for dispersion of fluids flowing in pipes. (From 0. Levenspiel. Chemical Reucrion En~irl~ering, 2nd pd. Copyright Q 1972 John Wiley & Sons, Inc. Reprinted by p e m i s ~ l o nof John Wiley & Sons. rnc. AH rights reserved.) [Nnie: D = D m ]
966
Models h r Nonideat Reac!ors
Chap. f 4
14.4.50 Dispersion in Packed Beds
For the case of gassolid catalytic reactions that take place in packedbed reactors, the dispersion coefficient, D,, can be estimated by using Figure 1412. Here d, is the particle diameter and E is the porosity.
Figure 1412 Experimental findings on dispersion of fluids flowing wirh mean axial velmity u m packed beds. (From 0.Levenspiel, Chemical Rcacrion Engtneerins, 2nd ed. Copyright Q 1972 John Wiley & Sons, Inc. Reprinted by perm~ssionof John Wiley & Sons, Inc. All rights reserved.) [Note: D = Do]
14.4,6 Experimental Determination of DD,
The dispersion coefficient can be determined from a pulse tracer experiment. Here, we will use I, and u2to solve for the dispersion coefficient D, and then the Peclet number, Pe, Here the effluent concentration of the reactor is measured as a function of time. From the effluent concentration data, the mean resi d e time, ~ ~ r,, and variance, u2,are calculated, and these values are then used to determine Do. To show how this is accompljshed, we wiIl write
in dimensionless form, discuss the different types of boundary conditions at the reactor entrance and exit, solve for the exit concentration as a function of dimensionless time (O= r l z ) , and then relare Do, u2. and T. 14.4.6A The UnsteadyState Tracer Balance
The first step is to put Equation (1413) in dimensionless form to arrive dimensionless group(s) that characterize the process. Let
at the
Sec. 14.4
967
Flow, Reaction, and Dispersion
For a pulse input, Cm is defined as the mass of tracer injected, M,divided by the vessel volume. K Then
The initial condition is At t = 0, 2 > 0, CdO",O) = 0,
Initial condition
${a+>= 0
(1435)
The mass of tracer injected, M is
14.4.68 Solution for a ClosedClosed System
In dimensionless form, the Danckwerrs b o u n d q conditions a+e /
Equation (1434) has been solved numericaIly for a pulse injection, and the resulting dimensionless effluent tracer concentration. JJ,,;,, is shown as a function of the dimensionless time O in Figure 1413 for various Peclet numbers. Although analytical solutions for $ can be found, the result i s an infinite series. The corresponding equations for the mean residence time, 1,. and the variance. . ' u areg
0 t,, = T
and
which can be used with the solution to Equation (1434) to obtain
See K. Bischoff and 0.Levenspiel. Ad\: Chent. Eng . 4, 95 ( 1963).
'
Models for Nonideat Reactors
Cham 14
Intermediate amount d dispemon
Effects of dispersion on the efffuent tracer
concentration rga amounl ol d~spersmn.
Figure 1413 C curves in closed vessels for various extents of backmixing as predicted by the dispersion model. (From O Levenspiel. Cltemicflf Reactton Eng~necring,2nd ed. Copyright 0 1972 John Wiley & Sons. lnc. Reprinted by p e m i s ~ i o nof John Wley & Sons, Inc. All rights m e w e d ) [Note: D D,]I0
Calculating Pe, using t, and rr? determined from RTD data for a closedclosed system
Consequently, we see that the Peclet number, Pe, (and hence D,).can be found experimentally by determining t, and uZ from the RTD data and then solving Equation ( 1439) for Pe,. 14.4.6C OpenOpen Vessel Boundary Conditions
When a tracer is injected into a packed bed at a location more than two or three particle diameters downstrenm from the: entrance and measured some distance upstream from the exit, the openopen vessel boundaty conditions apply. For an openopen system, an analytical solution to Equation ( 14 13) can be obtained for a pulse tracer input. For an openopen system, the boundsry conditions at the entrance are
'"0. Levenspiel, Chemical Reaction Engineering, 2nd ed. (New York: Wlley. 1972).
pp. 282284.
Sec. 14.4
Flow. Readion, and D~spers~on
969
Then for the case when the dispersion coefficient is the same in the entrance and reaction sections:
Eecause there are no discontinuities across the boundary at z = 0
As the exit Open at the exit
There are a number of perturbations of these boundary conditions that can be applied. The dispersion coefficient can take on different values in each of the three regions (z < 0. 0 2 2 5 L,and z > L), and the tracer can also be injected at some point z, rather than at the boundary, z = 0. These cases and others can be found in the supplementary readings cited at the end of the chapter. We shall consider the case when there is no variation in the dispersion coefficient for aIl ;and an impulse of tracer is injected at z = 0 at t = 0. For tong tubes (Pe > 100) in which the concentration gradient at k .s will be zero, and the solution to Equation (1434) at the exit is1' Vnlid For Pe, > 100
The mean residence time for an openopen system is Calculate r for an openopcn system
where T is based on the volume between z = 0 and z = L (i.e., reactor voIume measured with a yardstick). We note that the mean residence time for an open system is greater than that for a closed system. The mason i s that the molecules can diffuse back into she reactor after they exit. The variance for an
openopen system is Calculate Per for en
openpen
system.
"W. Jost. Dgusion in Solids, Liquids gad Gases (New York: Academic Press. 1960). pp. 17, 47.
970
Mcdels for Nonideal Reactors
Chap. 14
We n o w consider two cases for which we can use Equations (1439) and (1446) to determjne the system parameters: Case J . The space time s is known. That is, V and vo are measured independently, Here we can determine the Peclet number by determining t,,, and u2 from the concentrationtime data and then using Equation (1446) to calculate Pe, We can also caiculate r, and then use Equation (1445) as a check, but this is usually less accurate. Case 2. The space time T is unknown. This situation arises when there are dead ar stagnant pockets that exist in the reactor along with the dispersion effects. To analyze this situation we firsf calculate r,, and cr' from the data as in case 1. Then use Equation (1445) to eliminate .c2 from Equation (1446) to arrive at
Finding the effective reactor voume
We now can solve for the Peclet number in terms of our experimentally determined variables cr? and t i . Knowing Pe,, we can solve Equation (1445) for z. and hence V. The dead volume is the difference between the measured volume (ie., with
a yardstick) and the effective volume calculated from the RTD. 14.4.7 Sloppy Tracer Inputs It is not always possible to inject a tracer pulse cleanly as an input to a system because it takes a finite time to inject the tracer. When the injection does not approach a perfect pulse input (Figure 1414), the differences in the variances between the input and output tracer measurements are used to calculate the Peclet number:
where rr?" is the variance of the tracer measured at some point upstream (near the entrance) and o:, is the variance measured at some point downstream (near the exit).
z =L Measure Figure 1414 Imperfect tracer input.
971
Flow, Reaction, and Dispersion
Sec. 14.4
For an opeaspen system,it has been shownll h a t the Peclet number can be calculated from the equation
Now let's put all the material in Section 14.4 together to determine the conversion in tubular reactor for a firstorder reactor. Exampic 1 4 2 Conversion Using Disgersion and TanksinSeries Models The firstorder reaction
is carried out in a EOcmdiameter tubular reactor 6.36 m in Ienpth. The s p i f i c reaction rate is 0.25 minI. The results of a tracer test carried out on this reactor are shown in Table E142.1.
Calculate conversion using (a) the closed vessel dispersion model, (b) PFR, (c) the tanksinseries model, and (d) a single CSTR. Solution (a) We will use Equation (1427) to calculate the conversion
X=l
4q exp (Per/2) (I + q I 2 exp{Pe,q/Z) ( I  q ) 2 exp(Pe,q/2)
( I 427)
JwT Do = ~ k and , Pe, = ULID,. We can calculate Per from
where g = Equation (1439):
2 o'=tZ
First cafculate I, and u2 Fmm RTD data.
2 (,ePt') Pe, Pe:
However, we must find T~ and 02 fmm the tracer concentration data first.
'2R.
Aris, Chem. Eng. Sci.,
9,266 (1959).
972
Models tor Nonrdeal Reactors
Chap. 14
Conrider the data listed in Table E142.2. TO DETERUIWF I , AND TABLE E142.2. CALCCLATIOYS
U'
Here again spreadsheets can be used to calculate r2 and rr:.
To find E l f ) and then
t,
w e first find the area under the
1
C curve, which is
C ( r ) di = 50 g m i n
0
Then
Calculating the first term on the righthand side of Equation (E142.2). we find
Substituting these values to Equation (E142.2). we obtain the variance, ul.
Most people, including the author. would use Polymath or Excet to form Table E142.2 and to calculate t,,, and rr2. Dispersion in a closed vessel is represented by
Calcuinte Pe, from r, and d.
Solving for Pe, either by trial and error or using Polymath, we obtain Next. calculate Do, q. and X.
Next we catcvlate Do to be
Sec. 14.4
I
TwoParameter Models
Using the equations for y and X give\
Then
Substitution into Equation (1440) yields X= l 
Dispersion M d e l
4( 1.30)e''S1?'
(5.3)' exp (3 87)  (0.3)' exp (4.87) 68% conversion for the dispersion model
When dispersion effects am present in this tubular reactor, 68% conversion is achieved. (bl If the reactor were operating ideally as a plugflow reactor. the conversion would be X = 1  eTk = 1  pDa = 1
PFR Tanksinseries model

29
= 0.725
That is, 72.5% conversion wouId be achieved In an ideal plugflow reactor. (c) Conversion using the tanksinseries model: We recall Equation (1412) to calculate the number of tanks in series:
To calculate the conversion. we recaIE Equation (411). For a firstorder reaction for n tanks in series. the conversion is 1  1I = 1[ I + ( t / n )kIn (1 ( I + t,kp X = 67.7% for the tanksinwries model
x= I
1
+ 1 .29/4.35)435
(d) For a single CSTR,
CSTR
So 5 6 2 % conversion would be achieved in a singte ideal tank. Summary:
Summary
Dispersion: X = 68.0% Tanks in series:X = 67.7%
In this example. correction for finite dispersion, whether by a dispersion m d e t ar a tanksinseries model, is significant when c o m p a ~ dwith a PFR.
974
Moclels for Nonideal Reactors
Chap. 14
14.5 TanksinSeries Model Versus Dispersion Model We have seen that we can apply both of these oneparameter models to tubular reactors using the variance of the RTD.For firstorder reactions, the two models can be applied with equal ease. However, the tanksinseries model is mathematically easier to use to ob~brainthe effluent concentration and conversion for reaction orders other than one and for multiple reactions. However, we need ta ask what would be the accuracy of using the tanksinseries model over the dispersion model. These two models are equivalent when the PecletBodenstein number is related to the number of tanks in series, n, by the equation1"
Equivalency
or
between models of
tan ksinsenes and dispersion
where Bo = U r n " , where U is the superficial velocity, L the reactor length. and D, the dispersion coefficient. For the conditions in Example 142, we see that the number of tanks calculated from the Badenstein number, 330 (i.e., Pe,), Equation (14SO), is 4.75, which is very close to the value of 4.35 caIculated from Equation (1412). Consequently, for reactions other than first order, one would solve successively for the exit concentration and conversion from each tank in series for both a battery of four tanks in series and of five tanks in series in order to bound the expected values. In addition to the oneparameter models of tanksinseries and dispersion, many other oneparameter models exist when a combination of ideal: reactors is to model the reai reactor as shown in Section 13.5 for reactors with bypassing and dead volume. Another example of a oneparameter model would be to model the real reaclor as a PER and a CSTR in series with the one parameter being the fraction of the total volume that behaves as a CSTR. We can dream up many other situations that would alter the behavior of ideal reactors in a way that adequately describes a real reactor. However. it may be that one parameter is not sufficient to yield an adequate comparison between theory and practice. We explore these situations with combinations of ideal reactors in the section on twoparameter models. The reaction rate parameters are usually known (i.e., Da), but the Peclet number is usually not known because it depends on the Raw and the vessel. Consequently. we need to find Pe, using one of the three techniques discussed earlier in the chapter.
I3K. Elgcti, Chm. Eng. Sci.,51, 5077 (1996).
Sec. 14.6
Numerical Solutions to Flows with Dispersion and Reaetion
975
14.6 Numerical Solutions to Flows with Dispersion and Reaction We now consider dispersion and reaction. We first write our mole balance on species A by recalling Equation (1428) and including the rate of formation of A, r,, At steady state we obtain
AnaIytical solutions to dispersion with reaction can only be obtained for isothermal zero and firstorder reactions. We are now going to use FEMLAB to solve the flow with reaction and dispersion with reaction. A FEMLAB CDROMis included with the text. We are going to compare two soIurions: one which uses the ArisTaylor approach and one in which we numerically solve for both the axial and radial concentration using FEMLAB. Case A. ArisT~ylsrAnalysis for hrninar
Flow
For the case of an nthorder reaction, Equation (1415) is
If we use the ArisTaylor analysis, we can use Equation (1415) with a caveat that $ = CA/cAO where ?A is the average concentration from r = 0 to r = R as given by
where
P+=
and Da
=r
kqi
Do
For the closedclosed boundary conditions we have
Danckwens boundary conditions
For the openopen boundary condition we have
Models !or Nonideal Reac!ors
Chap. 14
Equation (14531 is a nonlinear second order ODE that is solved on the FEMLAB CDROM. Case
B. Full
Numerical Solution
To obtain axial and radial profile we now solve Equation (145 1)
First we wiIl put the equations in dimensionless form by letting $ = C,/C,\, , k = rtL, and 4 = r/R. Following our earlier transformation of variables. Equation ( l 45 l ) becomes
Example 1 4 3 Dispersion wifh Reactinn
(a)
(b) (c)
First, use E M L A B to solve the dispersion part of Example 142 again. How does the FEMLAB result compare with the solution to Example 142? Repeat (a) for a secondorder reaction with k = O,5 dm3/mol mln. Repeat (a) but assume laminar flow and consider radial gndients in concentration. Use D,, for both the radial and axial diffusion coefhcients. PIor the axial and radial profile%.Compare your results with p;ln (a).
Additional information:
U, = UT = 1.24 mlmin, D, = UorJPe, = 1.05 ml/min. DAB= 7.6E5rn2/rnin. Nore: For part (a), the twodimensional model with no r;ldial gndients (plug flow) becomes a onedimensional model. The inlet boundary condition for part (a) and part (b) is a closedclosed vessel (flux[; = 01 = flux[i = 0'1 or U;CA0 = flux) at the inlet boundary in EEMLab format is: Nim = UORCAO.The boundary condition for laminar flow in FEMLAB format for part (c) is: N;n = 2*UO*(I(r/Ra)?}*CAO. CAn= 0.5 moVdm3.
The different types of E M L A B Boundary Conditions are given in Problem P 14 2 9,
Solution
(a)
Equation (1452) was used in the FEMLAB program along with the nte law
977
Nurnerlcal Solutions to Flows wlih 51saem1onand React~on
Sec. 14.6
We see that we get the bame result\ nh the: analytical solution in Example 141. W ~ t hthe Ari+Taylor n n a l y ~ i sthe twodimensional profile becomes onedimensional plug Row t e l r m ty profile. Figure E l43. I(a) shows a uniform concenrration surface and shows the plug flow k h a t lor OF the reactor. Figure E 143. I(b) shows the corresponding croshsection plots at the inlet, half axial location, nnd outlet. The average outlet conversion ij 67.99.
The average outlet concentration at an axial distance z is foond by integrating across the radtus as hhown below
From the avenge concentrations at the inlet and outlet we can calculate the average conversion as
See FEMLAB 'rutorinls with w e e n shots
Concentration Surface
Radial Concentration Profiles
Load enclosed FEMLAB CD
Llulng Example Prcblem 015r
,
0
Radial Location (m)
3
001
OD1
DO3
OM
OD6
Radial Location (m)
FEgure E143.1 FEULAB results for a plug Row reactor w ~ t hfirstorder rcactlon. (Concen~mtionsin moVdm3.)
(b)
Now we expand our resu!rs to cpnsider the case when the reaction is second order (7, = kc,=kCA,$) with k = O . S dm3/moImin and CAo=0.5 moI/drn3. Let's assume the radial dispersion coefficient is equaI to the molecu3ar diffusivity. Keeping everything else constant, the average outlet conversion is 52.3%. However. because the flow inside the reactor i s modeled as plug flow the concentration profiles are still flat, as shown in Figure Ell3.2.
P78
(
Models for Nonideal Reactors
Concentration Surface
Chap. 14
Radial Concentration Profiles
Radial Locat~on{m)
Radial Location (m)
(h)
Figure E143.2 E h 1 L A B result$ for a plug flow reactor with secondorder reaction. (Concentmilons in moWdm3.)
{c)
Now. we will change the flow assumption from plug flow ta laminar flow and solve Equation (145 1 ) for a firstorder reaction.
Radial Concentration Profiles
Concentration Surface 7
I
D5
0 6 ~
rmrt
1
oe. Awragc Outlel Converssn = 68 8%
401
.Dl
en!
c

3 5
C i
Radial Lucat~on(m)
C1
0
OPi
002
003
ow
DO5
Radial Location {m)
Figure EI433 FEMLAB ou~putfor lan~inarflow in the reacror. (Concentratton< in mol/dmf 1
(dl
The average outlet converrion becomer 68 8 9 , not much d~fferentfrom the one i n part ( a ) In agreement w ~ t hthe ArisTaylor analyris. Hnuever. due to the laminar iron7 assumption i n the reactor. the radial concenlrat~onprofiler are very d~fferentthroufhoui the reactor, As a homexnrk exercise, repeat pan (c) for the secondorder reaction given in pad (b).
Sec. 14.7
TwoParameter Models
14.7 TwoParameter ModelsModeling Real Reactors with Combinations of Ideal Reactors We now will see how a real reactor might be modeled by one of two different combinations of ideal reactors. These are but two of an almost unlimited number of combinations that could be made. However, if we limit the number of adjustabIe parameters to two (e.g., bypass flow rate, uh, and dead volume, VD). the situation becomes much more tractable. After reviewing the steps in Table 141, choose a model and determine if it is reasonable by qualitatively comparA tracer ing it with the RTD. and if it is, determine the model parameters. Usual13 the expenmenl is used simplest means of obtaining the necessary data is some form of tracer test. 10 evaluate the model parameters. These tests have been described in Chapter 13. together with their uses in determining the RTD of a reactor system. Tracer tests can be used to defermine the RTD, which can then be used in a similar manner to determine the suitability of h e model and the value of its parameters. In determining the suitability of a particular reactor model and the parameter values from tracer tests, it may not be necessan, to calculate the RTD function Eli). The model parameters (e.g., V,) may be acquired directly from measurements of efffuent concen~ationin a tracer test. The theoretical prediction of the particular tracer test in the chosen model system is compared with the tracer measurements from the rea3 reactor. The parameters in the model are chosen so as to obtain the closest possible agreement between the model and experiment. If the agreement is then sufficiently close, the model is deemed reasonable. If not, another model must be chosen. The quality of the agreement necessary to fulfill the criterion "sufficiently close" again depends on creatively in developing the m d e l and on engineering judgment.The most extreme demands are that the maximum error in the prediction not exceed the estimated error in the tracer test and that there be ne observable trends with time in the difference between prediction (the model) and observation (the seal reactor). To illustrate bow the modeling is carried out, we will now consider two different models for a CSTR. Creativity and
engineering judgment am necessary for model formulation
f4.7.2 Real CSTR Modeled Using Bypassing and Dead Space
A real CSTR is believed to be modeled as a combination of an ideal CSTR of volume V,, a dead zone of volume V,, and a bypass with a voFurnetric flow rate v, (Figure 34 15). U% have used a tracer experiment to evaluate the parameters of the model Vs and u , . Because the total volume and volumetric Row rate are known, once V , and v , are found, u, and Vd can readily be calculated. 14.7.jP, Solving the Model System for C, and X
We $hall calculate the conversion for this model for the firstorder reacfion
980
Models for Nonideal Reactors
Chap. 74
The model system
Deod zone
(01
Figure 1415 (a) Real system: Ib) model system. A
Bnlnnce at junction
The bypass stream and effluent stream from the reaction volume are mixed at point 2. From a balance on species A around this point, [It11 = [Out]
We can solve for the concentration of A leaving the reactor,
For a firstorder reaction, a mole balance on V, gives Mole balance on CSTR
U, CAD V ,
or, in terms of ol and
, C  kcA, V, = 0
(1459)
p.
Substituting Equation (1460) into (1453) gives the effluent concentration of species A: Conversion as a function of model
panmeters
We have used the ideal reactor system shown in Figure 1415 to predict the conversion in the real reactor. The model has two parameters. a and P. If these parameten are known,we can readily predict the conversion. In the following section, we shall see how we can use tracer experiments and RTD data to evaIuate the model parameters.
Sec. 14.7
981
TwoParameter Models
14.7.1B Using a Tracer to Determine the Model Parameters En CSTRwithDeadSpaceandBypassModel
In Section 14.7.1A, we ured the system shown in Figure 141 6, with bypass flowrate ub and dead volume y,. to rnodei our real reactor system. We shall inject our tracer, T, as a positivestep input. The unsteadystate balance on the noareacting tracer T i n the reactor volume V, is I n  out = accumulation Tracer balance for step input
Mdel system
Figure 1416 Model system: CSTR with dead volume and bypaqsing.
The conditions for the positivestep input are
mejunction A balance around junctiofl balance
AttCO
C,=O
Atf2O
C,= C ,
point
2 gives
As before, = aV Vh
= Puo
t=
v
uo
Integrating Equation (1462) and substituting in terms of a and P gives
982
Models for Nonideal Reactos
Chap. 14
Combining Equations (1463) and (14641, the effluent tracer concentration is CT CiT =
i l  p l e X P [
I).(
aI
(1465)
We now need to remange this equation to extiact the model parameters, a and
p, either by regression (Polyrnath/MATLAB/Excel)or from the proper plot of the effluent uacer concentrations as a function of time. Rearranging yields Evaluating the m d e l parameters
Consequently, we plot In[Cm/(Cm  CT)]as a function of t. If our model is correct, a straight line should result with a slope of (1 P ) / z a and art intercept of In[l/(l P)].



Example 1 4 4 CSTR with h a d Space a d Bypass
The elementary reaction A+B
C+D
is to be carried out io the CSTR shown schematically in Figure 1415. There is both bypassing and a stagnant region i n this reactor. The tracer output lor this reactor is shown in Table E144.1. The measured reactor volume is 1.Q m3 and the flow rate to the reactor is 0.1 m3/min. The reaction rate constant i s 0.28 m31kmol.min. The feed is equjmolar in A and B with an entering concentration of A equal to 2.0 kmollm'. Calculate the conversion that can be expected in this reactor (Figure E144.1).
Tkroparameter mdel
Fipm E144.1 Schematic of real rcactor modeled with dead space ( V d ) and bypass ( v , ) .
Sec. 14.7
TwoParameter Models
Solution Recalling Equation ( 1466)
Equation (1466) suggests that we construct Table E144.2from Table E144.1 and plot Cml(Cm  C), as a function of time on semilog paper. Using this table we get Figure E144.2.

TABLE E144.2.
PROCESSEu DATA
We can find a and 0 from either a semilog plot as shown in Figure E144.2or by regression using Polymath, MATLAB. or Excel.
Evaluating the parameters a and p
Figure E144.2 Response ta a step input.
The volumeuic flow rate to the wellmixed ponion of the reactor. u s , can be determined from the intercept, I:
The volume of the wellmixed region, V,, can be calculated from the slope:
Ia= s = , . 1 l 5 m i ,  ~ 017
UT
=
1  0.2 = 7 min 0.115
Mwrers for Nonrdeal Aaaclors
984
Chap. fq
We now proceed to determine the conversion corresponding to these model parameters I. Balance on reactor volume V , :
[In] Out + Generation = Accumulation v,CAOU,C*, +r,,l/,
=Q
2. Rate law: r*s = RCJ,CB, Equimolar feed
:. C,
= C, r,,
=
3. Combining Equnt~ons(E144. I) and
(E144.7) gives
Rearranging. we have
T~RC:, + CAT CAb= O The Duck Tupc Cormcil would like to point out the new wri~~kle: The Junction Balance.
Solving for C,,yields
4. Balance around junction point 2:
[In] = [Outl
Rearranging Equation (E1446] g'lves us
Sec. 14.7
(
TwoParameter Modeis
5. Parameter evaluation:
r, =
5 = 8.7 min v,r
(
Substituting into Equatian (ELJ4.7) yields
Finding the
conversion
I f the reat reactor were acting as an ideal CSTR. the conversion would be 
I
(E131.9)
Other ModeIs. I n Section 14.7.1 it was shown bow we formulated a model consisting of ideal reactors to represent a real reactor. First, we soived for the exit concentration and conversion for our model system in terms of two parameters a and p. We next evaluated these parameters from data of tracer concentration as a function of time. Finally. we substituted these parameter values into the mole baIance, rate law, and stoichiomewic equations to predict the cortvecsian in our real reactor. To reinforce this concept, we will use one more example. 14.7.2 Real CSTR Modeled as Two CSTRs with Interchange I n this particular model there is a highly agitated region in the vicinity of the
impeller: outside this region, there is a region with less agitation (Figure 1417). There is considerable material transfer between the two regions. Both inlet and outlet flow channels connect Pa the highly agitated region. We shall
986
Models for Nonideal Reactors
Chap. 14
model the highly agitated region as one CSTR, the quieter region as another CSTR. with material transfer between the two.
The model system
Figure 1417 (a) Real reaction sysrem; (b) model reaction system.
14.7.214 Solving the Model System
Let
for C, and X
p represent that fraction of the total flow that is exchanged between reac
tors 1 and 2, that is,
ol represent that fraction of the total volume V occupied by the highly agitated region:
and let
Two parameters: a and
p
Then
The space time i s VQ
As shown on the CDROM 1 4 ~ ~for 2 , a firstorder reaction, the exit concentration and conversion are ~eferenccShelf
988
Models
14.8 Use of Software Packages to
!or Nonideal Reactors
Chap. ? 4
Determine
the Model Parameters If analytical solutions to the model equations are not available to obtain the parameters from RTD data, one could use ODE solvers. Here, the RTD data would first be fit to a polynomial to the effluent concentrationtime data and then compared with the model predictions for different parameter values.
I
Example 145
CSTR with Bypass and Dead Volume
(a) Determine parameters a and p that can be used to model rwo CSTRs with interchange using the tracer concentrntion data Iisted in Table E 145.1.
TABLE E145.1. RTD DATA 
r (min)
0.0
Cr, (plm') 2000
I

20
40
60
80
120
160
200
240
1050
52fl
280
160
61
29
16.4
10.0
(b)Determine theconversion offirstorder reaction with k = 0.03 minI and r = 40rnin
First we will use Polymath to At the RTD to a polynomial. Because of the steepnes! of the curve, we shall use two polynomials. Far f 5 80 rnin,
Trial and error using sortware packages
where CT, is the exit concentration of tracer determined experimentally. Next ws would enter the tracer mole (mass) balances Equations (147 1) and (1472) into a ODE solver. The Polymath program is shown in Table E145.2.Finally. we vary th parameten a aod j 3 and then compare the calculated effluent concentration Cn wit the experimental effluent tracer concentration CT,.Affer a few trials we converge o the values a = 0.8 and P = 0. I.We see from Figure E145.1and Table E145.3 thz the agreement between the RTD data and the calculated data are quite good, indical ing the validity of our values of a and P. The graphical solution to this problem i given on the CDROM and in the 2nd Edition. We now substitute these values i Equation (1468). and as shown in the CDROM, the comsponding conversion j SIR for the model system of two CSTRs with interchange:
Sec 14 8
Use of Software Packages ?a Determine the Mode! Parameters
Comparing models, we find
Livtng Example Problem
Differentla1 aquation8 as entered by the user I II d(CTl)ld(t) m fbeta'CT2(l+beZa)*CTl)lalphdtau 12 J d(CT2)ld(t) = (betafCT1beta*CT2)/(1alpha)/tau
Expllc# equations as entered by the user 111 beta0.1 [ 2 1 alpha = 0.8 [ 3 1 tau=40 [dl Cfel = 2MH)69.B~+0.g4*t"2O000146"P3f .047'IP(5)W [ 5 1 CTe2 = 921 17.3't+O. 129'P20.000438T2"3+5.&'10"(7)'th4 L S I tl =ta0 I7 1 CTe = if(tc8Q)thsn(CTel)e1se~CTe2)

2.0
Scale:
Y: 10m3 KEY:
 m1
 CTe
(a/m3)
f .2
0.8
0.4
0.0 0.0
40.0
80.0 120.0 160.0mo.0
Figure E145.1 Cornparison of model and elrperimenrnl exit tracer co~centrations.
Models for Nunideal Reactors
Chap. 14
14.9 Other Models of Nonideal Reactors Using CSTRs and PFRs
A Ease hislor?, for terephthalic acid
Several reactor models have been discussed in the preceding pages. A11 are based on the physical observation that in almost all agitated tank reactors, there is a wellmixed zone in the vicinity of the agitator. This zone is usually represented by a CSTR. The region outside this wellmixed zone may then be modeled in various fashions. We have already considered the simplest models, which have the main CSTR combined with a deadspace volume: if some shortcircuiting of the feed to the outlet is suspected, a bypass stream can be added. The next step is to look at all possible combinations that we can use to model a nonjdeal reactor using only CSTRs, PFRs, dead volume, and bypassing. The rate of transfer between the two reactors is one of the model parameters. The positions of the inlet and outlet to the model reactor system depend on the physical layout of the real reactor. Figure 1418(a) describes a real PFR or PBR with channeling that is modeled as two PERsJPBRs in parallel. The: two parameters are the fraction of flow to the reactors li.e.. p and (1  P)] and the fractional voIume [i.e., ol and (1  a ) ] of each reactor. Figure 14I 8(b) describes a real PFRlPBR that has a backmix region and is modeled as a PFRJPBR in parallel with a CSTR.Figures 14191a) and (b) show a reaI CSTR modeled as two CSTRs with interchange. I n one case, the fluid exits from the top CSTR (a) and in the other case the fluid exits from the bottom CSTR. The parameter P represents the interchange volumetric flow rate and'cr the fractional volume of the top reactor, where the fluid exits the reaction system. We note that the reactor in model 14 19(b) was found to describe extremely well a real reactor used En the production of terephrhalic acid." A number of other combinations of ideal Wactions can be found i n Levenspiel.
'"
'Trot. Indian Insi. Chem. Eng. Golden Jubilee, a Congress, Delhi, 1997, p. 323 ''Lcvenspiel. 0 Chrmicnl Reacriotl Ertgit~rrring,3rd ed. (New York: Wilcy. 19991, pp. 284292.
Sec. 14.10
Appllcafions to Pharmacokinetic Modeling
Real System
Model S y s t m
Real System
Model System
Ih)
Figure 1418 Combinations of ideal reactors used to m d e l real tubular reactors. (a) two ideal PFRs in pmllel (b) ideal PFR and Ideal CSTR in parallel
14.10 Applications to Pharmacokinetic Modeling The use combinations of idea1 reactors to model metabolism and dmg distribution in the human body is becoming commonplace. For example. one of the simplest models for drug adsorption and elimination is similar ro that shown in Figure 149(a). The drug is injected intravenously into a central companment containing the blood (the top reactor). The blood distributes the drug back and forth to the tissue compartment (the bottom reactor) before being eliminated (top reactor). This model will give the farnitjar linear semilog plot found in pharmacokinerics textbooks. As can be seen in the figure for P~~fe.rsionul Refererice Shelf R7.5 on pharmacokinetics on page 453 there are two different slopes. one for the drug distribution phase and one for the elmination phase. More elaborate models using combinations of ideal reactors to model a real system are descrikd in section 7.5 where alcohol metabolism is discussed.
998
Models Icr Nonldeal Reacto~s
Chap. 14
P I 4 4 The elementary liquidphase reaction
is carried out in a packed bed reactor in which dispersion is present. What is the conversion? Additional informution
Porosity = 50% Particle size = 0.1 cm Kinematic viscosity = 0.01 cm2/s
Reactor length = 0.1 rn Mean velocity = 1 cmls
P145, A gasphase reaction is being carried out in a 5cmdiameter tubular reactor that is 2 m in length. The velocity inside the pipe is 2 c d s . As a very first appmximation, the gas properties can be taken as those of air (kinematic viscosity = 0.01 cm21s), and the diffusivities of the reacting species are approximately O.QO5 cmr/s. ( a ) How many tanks in series would you suggest to model this reactor? (b) If the secondorder reaction A + B d C + D is carried out for the case of equal molar feed and with CAO= 0.01moWdm3, what conversion can be expected at a temperature for which k = 25 drn3/molns? (c) How would your answers to parts (a) and (b) change if the fluid velwity were reduced to 0.1 c d s ? ~n&easedto 1 m/s? Id) How would your answers to parts (a) and @) change if the superficial velocity was 4 cm/s thmugh a packed bed of 0.2cmdiameter spheres? (e) Wow would your answers to parts (a) to (d) change if the fluid were a liquid with properties similar to water instead of a gas, and the diffusivity was 5 X cm2/s? Plrdd, Use the data in Example 132 to make the following determinations. {The volumetric feed rate to this reactor was M) drn3/snin.) (a) Calculate the Peclet numbers for both open and closed systems. (b) For an open system, determine the spacetime T and then calculate the 8 dead volume in a reactor for which the manufacturer's specifications give a volume of 420 dm3. (c) Using the dispersion and tanksinseries models, calculate the conversion for a closed vessel for the firstorder isorncrizatian
A  B with k = 0.18 minI. Id) Compare your results in part (c) with the conversion calculated from the tanksinseries model, a PFR, arid a CSTR. P147, A tubular reactor has beensized to okain 98%conversion and to prMess 0.03 m3/s. The reaction is a firstorder irreversible isomerization. The reactor is 3 m long, with a crosssectionaI area of 25 cm2. After being built. a pulse tracer test on the reactor gave the following data: t, = 10 s and u2 = 65 s2. What conversion can be expected in the real reactor? P14BB The following E(t) curve was obtained from a tracer test on a reactor.
E in minutes
Chap. 14
999
Ouesllons and Problems
The conversion predicted by the tanksinseries model for the isothermal elementary reaction
was 50% at 300 K. (a) If the temperature is to be mised 10°C (E = 25.000 callrnol} wd the reaction carr~edout isothermally, what will be the conversion predicted by the maximum mixedness model? The TIS model?
(b) The elementary reactions
k, = k, = k, = 0.1 min
I
at
300 K, ,C , = 1 rnolldm3
were carried out isothermally at 300 K in the same reactor. What is the concentration of B in the exrl stream predicted by the maxlmum rnixedness model? (c) For the multiple reactions giveen in part (b). what is the conversion of A predicted by the dispersion model in an isothermal closedclosed system? P149B Revisit Problem P134, where the RTD function is a hemicircle. What is the conversion predicted by (a) the tanksinseries model? (b) the dispersion model? P1410, Revisit Problem P135B. (a) What combination of ideal reacton would you use to model the RTD? (b) What are the model parameters? (c) What is the conversion predicted for your model? P14llR Revisit Problem PI 36,. (a) What conversion is predicted by the tanksinseries model? (b) What is the Peclet number? (c) What conversion i s predicted by the dispersion model? P1412, Consider a real tubular reactor In which dispersion is occurring. (a) For small deviations from plug flow. show that the conversion for a firstorder reaction is ~ i v e napproximately as
(b) Show that to achieve the same conversion. the relationship between the volume of a plugflow reactor. I'p and volume of a real reactor, K in which dispersion occurs is
(c) For a Peclet number of 0,1 based on the Pm length, how much bigger than a PFR must the real reactor be to achieve the 999 conversion pred~cledby the PFR?
.
996
Models br Nanideal Reactors
Chan. 14
Q U E S T I O N S AND P R O B L E M S The subscript to each of the problem numbers indicates the level OF difficulty: A. least difficult; D. most difficult.
\I
\'Q Crea tivc TninXlng
P141, Make up and soIve an original problem. The guidelines arr given in Pmblem P41. However, make up a problem in reverse by first choosing a model system such as a CSTR in parallel with a CSTR and PFR [with the PFR modeled as four small CSTRs in series; Figure P14.lla)l or a CSTR with recycle and bypass [Figure P14 lo)].Write tracer m a s balances and use an ODE solver to predict the effluent concentrations. l o fact, you could build up an arsenal of tracer curves for different model systems to compare itgqinst real reactor RTD data. In t h s way you could deduce which model best describes the real reactor.
(b)
Figure PIQ1.1 Model systems.
? Lultrnne,
Hall of R m e
P14*2f, (a) Example 141. How lsrge would the ermr term be in Equation E141.4 i F t k = 0.1? ~ k l? = .tk= 10? (b) Example 142. Vary Do, k, U,and L. To what parameters or groups of parameters (e.g.. kL2/D,) would the conversion be most sensitive? What i f the firstorder reaction were carried out in tubular reactors of different diameters, but with the space time, t, remaining constant? The diameters would range from a diameter of 0.1 dm to a diameter of f rn for v = ClJp = 0.01 cm2/s. U = 0.1 c d s . and DAB = 1W5cm2/s. How would your conversion change? Is there a diameter that would maximize or minimize conversion in this range?
Chap 14
997
Ouest~onsand Problems
(c) Example 143. ( 1 ) Load the reaction and dispersion program from the FEMLAB CDROM. V ~ r ythe Darnkbhler number for a secondorder reaction using the AnrTaylor approximation (part (b) in Example 1431. 12) Vary the Pecler and Darnk6hler numbers for a secondorder reaction in laminar flow. What values of the Peclet number affect the conversion significantly? (d) Example 144. How would your answers change if the slope was 4 minI and the inrercepr was 2 in Figure E144.2? (el Example 145. Load the Livzng Example Pol~rrmthPmgmm. Vary a and p and describe what you find. What would be the conversion if a = 0.75 and p = 0.15? (i? Whnt if you were asked to design a tubular vessel that would minimize dispersion? What would be your guidehnes? How would you maximize the dispersion? How would your design change for a packed bed? (g) What if someone suggested you could use the solution to the flowdispcrsionreactor equation, Equation (1427). for a secondorder equation by s (kCAd2)CA= A'C, ? linearizing the rate law by lettering r, = Under what circumstances might this be a good approximation? Would you divide CAoby something ofher than 2? What do you think of linearizing other nonfirstorder reactions and using Equation (1427)? How could you test your results to learn if the approximation is justified? (h) What if you were asked to explain why physicaIly the shapes of the curves in Figure 143 Look the way they do. what would you say? What if the first pulse in Figure 14.l(b) broke through at @ = 0.5 and the second pulse broke through at @ = 1.5 in n tubular reactor in which a secondorder liquidphase reaction
k'e
P143,
was occurring. What would the conversion be if T = 5 min, CAo= mol/drn3. and k = 0.1 drn31mol*min? The secondorder liquidphase reaction
2
is to be &ed out isothermally. The entwing concentmuon of A is 1.0 mol/dm3. The spec~ficreaction rate is 1.0 drn3lrnolmin. A number of used reactom (shown klow) are available, each of which has been characterized by an RTD. There are hvo crimson and white reactors and three maize and blue reactors available.
Maize and blue Green and white Scarlet and gray Onnge nnd blue Purple and white Silver and black Crimson and white
2
2
4 3.05 2.3 1 5.17
4 4
2.5 25
f 25,000 50,000
4
50,000 50.003
4
50.000
4
50,W
2
25.N
(a) You have $50,000 available to spend. What is the greatest conversion you can achieve with the available money and reactors? (b) How would your answer to (a) change if you had $75,000available to spend? (c) From which cities do you think the various used reactors came from?
Modets for Nonideal Reactors
Chap. 14
For an openopen system. use
5. If a real reactor is modeled as a combination of ideal reactors, the m d e l should have at most two parameters.
CSJR w i t h bypass ond deod volume
Two CSf R s with inlerchonqe
6. The RTD is used to exmct model parameters. 7. Comparison of conversions for a PFR and CSTR with the zeroparameter and twoparameter models. X, symbolizes the conversion obtained from the segregation model and X,, that from the maximum mixedoess model for reaction orders greater than one.
Cautions: For rate Iaws with unusual concentration functionaljties or for nonisothemal operation, these bounds may not be accurate for certain types of rate laws.
CDROM MATERIAL
Summary Notes
RJ
A
PLE
tlvtng Example Probltm
Learning Resources I. Slrrnrnay Notes 2. Web Material FEMLAB CDROM Living Example Problems 1. Example 143 Dispersion wirh ReactionFEMUB
2. Example 145 CSTH wirh Rjpass and Dead Volume
Chap. 14
CDROM Materiel
995
FEMLAB results
Radial Location (m)
Professional Reference Shelf R14.1 Derivation of Equation for ToylorAris Dispersion
ta 
r?U
Refcrcncc S ~ C B
C+$2=DddiC d*
z
R 14.2 Real Reuczor Modeled in on Ideal CSTR with Exchange Voluma Example R141 l b o CSTRs with interchange.
a ~ * ~
Models for Non~UealReactors
(a)
Chap. 14
(b)
Figure 1419 Combinations of ideal reactorsform. 8 1982? effective dtffunivity in. RI&Xlh for tissue engineering. 823857 CDROM material, 801802. 852855
Chemical Vapor Deposition in. 849852 convectton In. 763765 definitions. 758159 dtffvs~onand reactionlimited regime estimation in. 838R42
model. 955 externa! reqiszance to mtlrs transfer in example. 783788 mass ~ransfercoefficient In. 771776 mass transferlimited reactions, 780783 mass tranrfer to single panicles. 776780 falsified kineucs In. 833815 Fick's first law in. 7AC7hl through him. 7 6 6 7 7 0 f o r ~ r dconvec~innin. 763764 fundamentals. 7.58 il~trrnal.660461 I nrernal effectiuenc~sfac~clor. 827433 journal urticlc problems. 863 journal cril~queproblemr. 8098 10. 863864 limiting siruiltionc for. 848849 Inaw tranller in packed beds. 111 dispersion
842K48
Mean' crrterlon fur. 84 184! modeling with. 77 1 mnlilr flux in. 75X760. 763765 nlultiphase reactors ill. 849850
operating condition changes in.
783788 overall effectiveness factor in,
835838 in pharrnacokinetics, 798799 questions and problems. 8028 10, 855863 shrinking core model. 792799 through stagnant film, 762, 774 summary. 800801, 851852 supplcmentaty reading, 810811. 865866 l c m ~ r a l u r eand pressure dependence in, 770 for transdermal drug deli\wy. 772 DigttaIage problems. multiple reactions for. 356557 Dilute concentrations iri diffusiul~, 762763
Dilution rate in hioreactors. 4 3 5 4 3 6 In chernostat5. 434 in washout. 437 Dimens~onlesscumulative distributlons. 890891 D~mensionlessgroups in ma\s transfer cwfficient, 774 Dirnerize propylene, 6 M 1 Dimethyl ether (DME) decomposition of. 297 from methanol. 742 Diphenyl in reversible reactions. 8990
Dmc delta function in PFR RTD.886 in step tracer experiment, 877 Di~appearanceof ~uhwrate.100. 431432 Dl\appe:rrance rate. 56 Dlqguised kinetu. 873835 Dtsk rupture In nrtmanalinc production. 605 Dispers~nn of catalysts. 631 FEMLAB lor. 975;978. 1032 uneparameter model\. 947 in packed bed