CD_ROM_Elements of Chemical Reaction Engineering_3ed

Elements of Chemical Reaction Engineering


Third Edition By H. Scott Fogler University of Michigan Ann Arbor, Michigan

Welcome! Welcome to the CD-ROM that accompanies the Third Edition of Elements of Chemical Reaction Engineering by H. Scott Fogler. Follow the links (below) to learn how to get the most out of this CDROM. Introduction Begin: Chapter 1

© 1999 Prentice-Hall PTR Prentice Hall, Inc. ISBN 0-13-531708-8 Legal Statement

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APPENDICES

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Introduction

Welcome Navigation Components Usage

Welcome This CD-ROM is intended to be used as a learning resource; the material on this CD-ROM supports the chemical reaction engineering concepts covered in the text. You are encouraged to use the CD-ROM to supplement and expand upon your own studies. We are certain that you will find the extra knowledge you gain will be worthy of the time you invest to obtain it.

Warnings! Macintosh Users The majority of the files on the CD-ROM are HTML files. Great care was taken in trying to insure that these files would work on both PCs and Macs. However, many of the non-HTML files on the CD-ROM (i.e., Polymath, all of the Interactive Computer Modules, and most of the plug-ins included on the CD) are Windows/DOS-based programs, for which there are (unfortunately) no Macintosh equivalents. You may still use these files, if you have a PC emulator program on your Mac, such as Virtual PC. ICMs Some users have experienced problems, trying to run the Interactive Computer Modules directly from the ICM directory on the CD-ROM. If you have trouble with being able to run the Interactive Computer Modules from the CD-ROM, then try installing them on your hard drive. Hidden Files To clear up some of the confusion about which files to use in certain directories (e.g., Polymat4 and ICMs), some files and folders were hidden. You may find it easier to navigate the CD-ROM, if you make certain that you are NOT viewing hidden files.

Recommended Software Before you begin, we advise you to download or install the following software programs and plug-ins on your computer, if they are not already present on your system:

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Introduction

Adobe Acrobat Reader and Plug-in The Appendices (and certain other files on the CD-ROM) are in PDF format. You must have Adobe Acrobat Reader installed to access these files. You will also need the Adobe Acrobat Reader Plug-in to read these files from your web browser. Note: The PC version of Adobe Acrobat Reader 3.02 has been included on the CD-ROM in a directory called Software\Adobe. If you run the ar302.exe program, it will install the reader software and the browser plug-in on your computer. Apple QuickTime Plug-in There are a few QuickTime movies on the CD-ROM, which will require the QuickTime Plug-in to view them. Note: QuickTime 3.0 (for PCs) has been included on the CD-ROM in a directory called Software\Quick. If you run the quick3.exe program, it will install the movie viewer and the plug-in on your computer and in your browser, respectively. WinZip (Optional) Some of the Polymath files for the Living Example Problems are archived as Zip files. You may need an unzip utility, such as WinZip to access them. See the Polymath section of the CD-ROM for more information. MATLAB (Optional) In addition to Polymath, you may use MATLAB to access the Living Example Problems. See the MATLAB section of the CD-ROM for more information. IMPORTANT! Different browsers and font sizes may affect the alignment and general appearance of the HTML content of the CD-ROM. To ensure that items are aligning properly, you may need to adjust your browser's font size. The HTML content of this CD-ROM is also available at the University of Michigan's Chemical Reaction Engineering Website: file:///H:/htmlmain/intro.htm[05/12/2011 16:54:01]

Introduction

http://www.engin.umich.edu/~cre

Next Step Once you have downloaded and/or installed this software, you should proceed to the section on Navigation, to learn how to get around this CD. The Components section will give you information on the various "modules" that are available on this CD. The Usage section will give you information on the best way to integrate the information on this CD with the information in your book.

Welcome Navigation Components Usage Begin: Chapter 1


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Elements of Chemical Reaction Engineering, Credits

The following people have conspired to bring you this CD-ROM:

The University of Michigan Team H. Scott Fogler, Author Dieter Andrew Schweiss, Media Project Manager Ellyne Buckingham, Artist, ToPS Scott Conaway, Wetlands Susan Fugett, MATLAB Examples Anuj Hasija, HTML Designer Lisa Ingall, ToPS Brad Lintner, ToPS Timothy Mashue, Reactive Distillation James Piana, ICMs Susan Stagg, Cobra Problem Author Gavin Sy, Cobra Problem (ToPS = Thoughts on Problem Solving) Special thanks to: Nicholas Abu-Absi John Bell Michael Cutlip, Polymath Sean Connors Anurag P. Mairal Professor Susan Montgomery Mordechai Shacham, Polymath Probjot Singh Ibrahim "Abe" Sendajarevic Mayur Valanju

The Team at Prentice-Hall PTR: Bernard Goodwin, Executive Editor Diane Spina, Assistant to the Executive Editor Sophie Papanikolaou, Director of Production Lisa Iarkowski, Manager, Production Yvette Raven, Media Project Manager Talisman Desktop Productions, Developer Scholar's Net Academic Multimedia, Design/Programming file:///H:/htmlmain/credits.htm[05/12/2011 16:54:02]

Elements of Chemical Reaction Engineering, Credits

Additional Credits Membrane Reactors Parts of this site was originally presented as an Open-Ended Problem in the Winter 1997 Chemical Reaction Engineering Class at the University of Michigan. The students who developed this module were Kim Dillon, Namrita Kumar, Amy Miles, and Lynn Zwica. The module was further expanded and improved by Ellyne Buckingham, Dieter Andrew Schweiss, Anurag Mairal, and H. Scott Fogler for use with the Chemical Reaction Engineering Web Site and CD-ROM.


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Easter Egg

Easter Egg This is the only easter egg that I know of on the CD-ROM (but there may be more). My name is Dieter Andrew Schweiss, and after devoting a year of my life to this project, I had to put something hidden on this thing! I've really enjoyed working with Scott, the University of Michigan Team, and the people from Prentice-Hall on this CD-ROM project. It has helped me fulfill a long-time dream of contributing to the accumulated knowledge that is the field of Chemical Engineering. (That, and the fact that nothing quite like this CD had ever been done before in ChE!) Anyways, good luck with your classes. Be sure to use the resources available to you on this CD, especially the Lecture Notes and the worked example problems. They'll really come in handy. (Trust me, I know!) I like what this symbol represents: life is a balance between order and chaos, yet pure order still retains some element of chaos and pure chaos retains some element of order (kind of like my desk...)

Special Recognition Professor H. Scott Fogler would like to extend a special thanks to Dieter Andrew Schweiss, without whom the Elements of Chemical Reaction Engineering CD-ROM would never have been possible. Dieter worked countless days and nights to bring this project to completion, coordinating his efforts with both the University of Michigan Team and the Prentice-Hall Team. Thank you, Dieter!


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Chapter One

CHAPTER 1

1

Mole Balances After completing Chapter 1 of the text and associated CD-ROM material the reader will be able to: Define the rate of chemical reaction. Apply the mole balance equations to a batch reactor, CSTR, PFR, and PBR. Describe two industrial reaction engineering systems. Describe photos of real reactors. Describe how to surf the CD-ROM attached with this text.

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Legal Information

About Prentice Hall PRENTICE HALL CD-ROM/WEB SITE: LEGAL DISCLAIMER Unless otherwise indicated, this CD-ROM/Web Site and its contents are the property of Prentice-Hall, Inc. ("Prentice Hall") and are protected, without limitation, pursuant to U.S. and foreign copyright and trademark laws. PRENTICE HALL MAKES NO REPRESENTATIONS OR WARRANTIES WITH RESPECT TO THIS CD-ROM/WEB SITE OR ITS CONTENTS, WHICH ARE PROVIDED FOR USE "AS IS." PRENTICE HALL DISCLAIMS ALL WARRANTIES, EXPRESS OR IMPLIED, INCLUDING WITHOUT LIMITATION THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE, WITH RESPECT TO THE CD-ROM/WEB SITE AND ANY CD-ROM/WEB SITE WITH WHICH IT IS LINKED. PRENTICE HALL ALSO MAKES NO REPRESENTATIONS OR WARRANTIES AS TO WHETHER THE INFORMATION ACCESSIBLE VIA THIS CD-ROM/WEB SITE, OR ANY CD-ROM/WEB SITE WITH WHICH IT IS LINKED, IS ACCURATE, COMPLETE, OR CURRENT. Price information is subject to change without notice. In no event shall Prentice Hall or its employees, agents, suppliers, or contractors be liable for any damages of any kind or character, including without limitation any compensatory, incidental, direct, indirect, special, punitive, or consequential damages, loss of use, loss of data, loss of income or profit, loss of or damage to property, claims of third parties, or other losses of any kind or character, even if Prentice Hall has been advised of the possibility of such damages or losses, arising out of or in connection with the use of this CD-ROM/WEB Site or any CD-ROM/WEB Site with which it is linked. Portions of this Site (such as "chat rooms" or "bulletin boards") may provide users an opportunity to post and exchange information, ideas and opinions (the "Postings"). BE ADVISED THAT PRENTICE HALL DOES NOT SCREEN, EDIT, OR REVIEW POSTINGS PRIOR TO THEIR APPEARANCE ON THIS SITE, and Postings do not necessarily reflect the views of Prentice Hall. In no event shall Prentice Hall assume or have any responsibility or liability for the Postings or for any claims, damages or losses resulting from their use and/or appearance on this Site. You hereby represent and warrant that you have all necessary rights in and to all Postings you provide and all information they contain and that such Postings shall not infringe any proprietary or other rights of third parties or contain any libelous, tortious, or otherwise unlawful information. You hereby authorize Prentice Hall to use and/or authorize others to use your Postings in any manner, format or medium that Prentice Hall sees fit.

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Legal Information

Removing pieces of our Companion CD-ROM/Websites and incorporating them into class homepages always requires formal permission in writing. We are very cooperative in arranging permission for use of our components by faculty who have adopted our book. Note that the creation of a class homepage linked to Companion CD-ROM/Websites using Syllabus Builder does not require permission, if you are using our textbook for your course. Please read our legal statement for details. To contact our permissions department, please fax a request to Attention: CD-ROM/Website Permissions Dept fax 201-236-3290 Note that your fax must include a printed copy of the CD-ROM/Webpages that contain the material you'd like to re-use, with a clear indication of the pieces you desire. You must also include a short written description of your intended use for these components. This MUST include: the intended audience, an estimate of how many persons will view the content you've requested, and the dates that you expect the material to be available to your audience. a description of access control methods (is a password required to access the site) a description of the cost to access your site if possible, an example of how you would display our content and copyright notice © Prentice-Hall, Inc. A Simon & Schuster Company Upper Saddle River, New Jersey 07458

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Chapter Two

CHAPTER 2

2

Conversion and Reactor Sizing After completing Chapter 2 of the text and associated CD-ROM material the reader will be able to: Define conversion and space time. Write the mole balances in terms of conversion for a batch reactor, CSTR, PFR, and PBR. Size reactors either alone or in series once given the rate of reaction, -r A, as a function of conversion, X. Write relationship between the relative rates of reaction.

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Chapter Three

CHAPTER 3

3

Rate Law and Stoichiometry After completing Chapter 3 of the text and associated CD-ROM material the reader will be able to: Write a rate law and define reaction order and activation energy. Set up a stoichiometric table for both batch and flow systems and express concentration as a function or conversion. Calculate the equilibrium conversion for both gas and liquid phase reactions. Write the combined mole balance and rate law in measures other than conversion. Set up a stoichiometric table for reactions with phase change.

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Chapter Four

CHAPTER 4

4

Isothermal Reactor Design After completing Chapter 4 of the text and associated CD ROM material the reader will be able to: Describe the algorithm that allows the reader to solve chemical reaction engineering problems through logic rather than memorization. Size batch reactors, semibatch reactors, CSTRs, PFRs, and PBRs for isothermal operation given the rate law and feed conditions. Discuss solutions to problems taken from the California Professional Engineers Registration Examination. Account for the effects of pressure drop on conversion in packed bed tubular reactors and in packed bed spherical reactors.

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Chapter Four

APPENDICES


Chapter Five

CHAPTER 5

5

Collection and Analysis of Rate Data After completing Chapter 5 of the text and associated CD-ROM material the reader will be able to: Determine the reaction order and specific reaction rate from experimental data obtained from either batch or flow reactors. Describe how to use equalarea differentiation, polynomial fitting, numerical difference formulas and regression to analyze experimental data to determine the rate law. Describe how the methods of half lives, and of initial rate, are used to analyze rate data. Describe two or more types of laboratory reactors used to obtain rate law data along with their advantages and disadvantages. Describe how to plan an experiment.

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Chapter Five

APPENDICES


Chapter Six

CHAPTER 6

6

Multiple Reactions After completing Chapter 6 of the text and associated CD-ROM material the reader will be able to: Define different types of selectively and yield. Choose a reaction system that would maximize the selectivity of the desired product given the rate laws for all the reactions occurring in the system. Describe the algorithm used to design reactors with multiple reactions. Size reactors to maximize the selectivity and to determine the species concentrations in a batch reactor, semibatch reactor, CSTR, PFR, and PBR, systems.

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Chapter Seven

CHAPTER 7

7

Nonelementary Reaction Kinetics After completing Chapter 7 of the text and associated CD-ROM material the reader will be able to: Discuss the pseudo-steadystate-hypothesis and explain how it can be used to solve reaction engineering problems. Discuss different types of polymerization reactions and rate laws. Describe Michealis-Menton enzyme kinetics and enzyme inhibition. Write material balances on cells, substrates, and products in bioreactors.

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Chapter Eight

CHAPTER 8

8

Steady-State Nonisothermal Reactor Design After completing Chapter 8 of the text and associated CD-ROM material the reader will be able to: Describe the algorithm for CSTRs, PFRs, and PBRs that are not operated isothermally. Size adiabatic and nonadiabatic CSTRs, PFRs, and PBRs. Use reactor staging to obtain high conversions for highly exothermic reversible reactions. Carry out an analysis to determine the Multiple Steady States (MSS) in a CSTR along with the ignition and extinction temperatures. Analyze multiple reactions carried out in CSTRs, PFRs, and PBRs which are not operated isothermally in order to determine the concentrations and temperature as a function of position (PFR/PBR) and operating variables.

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Chapter Eight Legal Statement

APPENDICES


Chapter Nine

CHAPTER 9

9

Unsteady-State Nonisothermal Reactor Design After completing Chapter 9 of the text and associated CD-ROM material the reader will be able to: Analyze batch reactors and semibatch not operated isothermally. Analyze the start up of nonisothermal CSTRs. Analyze perturbations in temperature and presence for CSTRs being operated at steady state and describe under what conditions the reactors can be unsafe (safety). Describe the effects of adding a controller to a CSTR. Analyze multiple reactions in batch and semibatch reactors not operated isothermally.

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Chapter Nine

APPENDICES


Chapter Ten

CHAPTER 10

10

Catalysis and Catalytic Reactors After completing Chapter 10 of the text and associated CD-ROM material the reader will be able to: Define a catalyst, a catalytic mechanism and a rate limit step. Describe the steps in a catalytic mechanism and how one goes about deriving a rate law and a mechanism and rate limiting step consistent with the experimental data. Size isothermal reactors for reactions with LangmuirHinschelwood kinetics. Discuss the different types of catalyst deactivation and the reactor types and describe schemes that can help offset the deactivation. Analyze catalyst decay and conversion for CSTRs and PFRs with temperaturetime trajectories, moving bed reactors, and straight through transport reactors. Describe the steps in Chemical Vapor Deposition(CVD). Analyze moving bed reactors that are not operated isothermally.

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Chapter Ten


APPENDICES


Chapter Eleven

CHAPTER 11

11

External Diffusion Effects on Heterogeneous Reactions After completing Chapter 11 of the text and associated CD-ROM material the reader will be able to: Define the mass transfer coefficient, explain what it is function of and how it is measured or calculated. Analyze PBRs in which mass transfer limits the rate of reaction. Discuss how one goes form a region mass transfer limitation to reaction limitation. Describe how catalyst monoliths and wire gauze reactors are analyzed. Apply the shrinking core model to analyze catalyst regeneration.

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Chapter Eleven

APPENDICES


Chapter Twelve

CHAPTER 12

12

Diffusion and Reaction in Porous Catalysts After completing Chapter 12 of the text and associated CD-ROM material the reader will be able to: Define the Thiele modules and the effectiveness factor. Describe the regions of reaction limitations and internal diffusion limitations and the conditions that affect them. Determine which resistance is controlling in a slurry reactor. Analyze trickle bed reactors. Analyze fluidized bed reactors. Describe the operation of a CVD Boat Reactor.

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Chapter Twelve

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Chapter Thirteen

CHAPTER 13

13

Distributions of Residence Times for Chemical Reactors After completing Chapter 13 of the text and associated CD-ROM material the reader will be able to: Define a residence time distribution RTD [E(t), F(t)] and the mean residence time. Determine E(t) form tracer data. Write the RTD functions (E(t), F(t), I(t)) for ideal CSTRs, PFRs, and laminar flow reactors. Predict conversions from RTD data using the segregation and maximum mixedness models. Predict effluent concentrations for multiple reactions using the segregation and maximum mixedness models.

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Chapter Thirteen

APPENDICES


Chapter Fourteen

CHAPTER 14

14

Models for Nonideal Reactors After completing Chapter 14 of the text and associated CD-ROM material the reader will be able to: Describe the tanks-in-series and dispersion one parameter models. Describe how to obtain the mean residence time and variance to calculate the number of tanks-in-series and the Peclet number. Calculate the conversion for a first order reaction taking place in a tubular reactor with dispersion Describe how to use combinations of ideal rectors to model a real reactor and how to use tracer data to determine the model parameters.

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Appendices

APPENDICES

The Appendices are in PDF format. You must have Adobe Acrobat Reader installed to access these files. You will also need the Adobe Acrobat Reader Plug-in to read these files from your browser. (See the CDROM Introduction for more information.) Appendix D: Measurement of Slopes on Semilog Paper Appendix E: Software Packages Appendix H: Open-Ended Problems Appendix J: Use of Computational Chemistry Software Packages


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Navigation


Navigation The Objectives

At the opening of every chapter is the Objectives. Clicking on BEGIN takes you to the Chapter Outline -- the contents for that chapter.

Chapter Outline - The Main Interface

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Navigation

The Main Interface lists the contents for that chapter. There are links to the entire chapter's contents from this page. Return to this page by using the left-hand navigation bar and click on Chapter Outline. In the Professional Reference Shelf and Learning Resources sections, there are Examples that link from within these sections. When you click on an Example, a new browser window will open. Some Examples can also be accessed from the Chapter Outline, where a new browser window will not open.

An Example in a New Browser Window

This new browser window is not the main interface. To return to the previous page, close this window by clicking on the top-left button of the new browser window (for Macs), or on the top-right button of the new browser window (for PCs). file:///H:/htmlmain/intro2.htm[05/12/2011 16:54:33]

Navigation

Throughout this material there will be Footnotes. By clicking on a footnote, you are going to open a new browser window.

Footnotes in a New Browser Window

Again, this new browser window is not the main interface. To return to the previous page, close this window by clicking on the top-left button of the new browser window (for Macs), or on the top-right button of the new browser window (for PCs).

Left Navigation Bar

Chapter Number takes you to the Objectives page for that chapter, while Chapter Outline takes you to Contents page for that chapter. Software Toolbox takes you to the Software Toolbox, e.g. Polymath. Interactive Computer Modules takes you to the ICMs main menu. Thoughts on Problem Solving takes you to the main problem solving menu.


Navigation

Updates & FAQs takes you to updates and corrections for the book. Representative Syllabi takes you to the sample 3- and 4-credit course syllabi. Help takes you to the help menu.

Except for Chapter Number and Chapter Outline, information you access from this navigation bar is not specific to any chapter. You can go directly to the Thoughts on Problem Solving section to see examples of the 10 types of home problems or visit the Interactive Computer Modules, without leaving the chapter that you are in. All the material is ordered by chapter. The lower navigation bar will take you to a specific Chapter or the Appendices. The HOME button will take you to the welcome screen for this CD.

Lower Navigation Bar

To find out more, go to the Components section of this CD.

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Navigation

Begin: Chapter 1



Components


Components Components of the CD-ROM Each chapter is broken into the following bulleted components: Learning Resources Summary Notes Web Modules Interactive Computer Modules Solved Problems Living Example Problems Professional Reference Shelf Additional Homework Problems

Learning Resources These resources give an overview of the material in each chapter and provide extra explanations, examples, and applications to reinforce the basic concepts of chemical reaction engineering. The learning resources on the CDROM include: 1. Summary Notes The Summary Notes of the lectures given at the University of Michigan will serve as an overview of each chapter. They contain a logical flow of the equations being derived, along with additional examples and material that can be viewed either before or after reading the text. The first 26 lectures are covered in a four-credit hour undergraduate course. The last 11 (27-37) are taken from the graduate course at the University of Michigan. 2. Web Modules These modules show how key concepts of chemical reaction engineering can be applied to nonstandard problems (e.g. the use of Wetlands to degrade toxic chemicals). Current modules focus on Chapters 4 and 6 and include Wetlands, Cobra Bites, Membrane Reactors and Reactive Distillation modules. Additional web modules (http://www.engin.umich. edu/~cre) are expected to be added over the next several years. 3. Interactive Computer Modules Most chapters have one or more interactive computer modules (ICMs) to accompany them as a learning resource. For these chapters, students can use the corresponding ICM(s) to review the important material and then apply it to real problems in a unique and entertaining fashion. Each module contains: Menu Review of concepts Interactive problem


Components

Solution to the problem For example, in the Murder Mystery module students take on the role of assistant sleuth as they use basic chemical engineering principles to solve the strange disappearance of several of the Nutmega Spice Company's employees. This particular module has long been a favorite with students across the nation. 4. Solved Problems A number of solved problems are presented along with problem solving heuristics. Problem solving strategies and additional worked example problems and are available in the Thoughts on Problem Solving section of the CD-ROM. The Ten Types of Homework Problems section contains two worked examples for each of the ten homework problem types. These examples are based on the material from Chapter 4, and they provide useful information on how one can attack homework problems. The section on Getting Unstuck is especially helpful. Living Example Problems The example problems that use an ODE solver (e.g., POLYMATH) are referred to as "living example problems," because the students can load the program directly on their own computers in order to study it. Students are then encouraged to "play" with the example's key variables and assumptions. Students can change parameter values, such as the reaction rate constants to learn to deduce trends or predict the behavior of a given reaction system, and gain a better understanding of the concepts being studied. Using the living example problems provides students with the opportunity to practice critical and creative thinking skills as they explore the problem and ask "what if...?" questions. Professional Reference Shelf This section of the CD-ROM contains: (1) material that is important to the practicing engineer, although it is typically not included in the majority of chemical reaction engineering courses. A short synopsis of each of the following topics is given at the appropriate point in the text. These sections are: i. ii. iii. iv. v. vi. vii. viii. ix. x. xi. xii. xiii.

Photographs of real reactors Recycle reactors Weighted least squares Experimental planning Laboratory reactors Inhibition and cofactors in enzymatic reactors Bifurcation analysis Wet and dry etching of semiconductors Catalytic monoliths Wire gauze reactors Trickle bed reactors Fluidized bed reactors CVD boat reactors (2) material that gives a more detailed explanation of derivations that were abbreviated in the text. The intermediate steps to these derivations are given on the CD-ROM: a) First order reaction in a semibatch reactor b) Temperature-conversion relationship for an adiabatic reactor c) Aris-Taylor dispersion Additional Homework Problems New problems were developed for this edition that provide a greater opportunity to use today's


Components

computing power to solve realistic problems. In addition, parts of problems were designed to promote and develop critical thinking skills. Many instructors alternate the homework problems they assign from year to year, ususally taken from a limited number of problems at the end of each chapter. Consequently, some of the more traditional, yet excellent problems of previous editions were placed on the CD-ROM and they can serve as practice problems along with those unassigned problems in the text. Table CDI-1 gives the resources available in each chapter. TABLE CDI-1 Chapter: Learning Resources Summary Notes

1

Web Modules

2

3

4

7

8

9

10 11 12 13 14

Solved Problems Professional Reference Shelf Additional Homework Problems

6

Interactive Computer Modules Living Example Problems

5

In addition to the components listed at the end of each chapter the following components are also included on the CD-ROM Software Toolbox Instructions on how to use the different software packages to solve examples are described for: POLYMATH MatLab ASPEN PLUS All living example problems on the CD are in both POLYMATH and MatLab Format. The POLYMATH program can either be loaded to a computer and executed directly from the CD-ROM. The POLYMATH examples may also be loaded on a computer or run directly from the CD-ROM. In order to execute MatLab examples, MatLab must be available on a server with a site license or the student version of MatLab must be purchased. Similarly, in regard to ASPEN, the CD shows an example of how to use ASPEN to solve chemical reaction engineering problems, however, a site license must be available to actually use ASPEN to solve the homework problems. Representative Syllabi for both 3 and 4 Credit Courses The syllabi give a sample pace at which the courses could be taught as well as suggested homework problems. Virtual Reality Module This module provides an opportunity to move inside a catalyst pellet to observe surface reactions and coking. This module also allows students to navigate through a catalyst pore and see the catalytic steps of diffusion, adsorption, surface reaction, and coking occurring on a catalyst pellet.


Components

Credits Special recognition to the students who contributed so much to the CD-ROM. In particular, Dieter Schweiss, Anuj Hasija, and Susan Fugett. In addition, Gavin Sy, Scott Conaway, Tim Mashue, and Ellyne Buckingham also worked on the CD-ROM. Clicking on the topic you wish to view will bring up the following information: Learning Resources A. Summary Notes 1. Click on this hot button and a listing of all the lectures in pairs of two (e.g. Lectures 1 and 2, Lectures 3 and 4, etc.) will appear on the screen. Under each lecture pair will be a short listing of the topics covered in those two lectures at the University of Michigan along with the chapters that the lectures are based upon. 2. Click on particular lecture pairs of interest to view the Summary Notes.

B. Additional Homework Problems 1. Click on the topic you choose under Web Modules (e.g. Ch 6 - Pharmacokinetics of Cobra Bites) and the module will appear on your screen. 2. Click on the Interactive Computer Modules title (e.g. Ch 4 - Mystery Theater) and a description of that modules specific will appear. Next click on (2) the instructions that describe how to install the module on to your computer. Finally, load and run the interactive computer module. C. Living Example Problems If you wish to run the POLYMATH examples you can run them directly. If you wish to run the examples on MatLab you will have to purchase the student edition of MatLab or have MatLab available on the server and use an interface to load and run them on your own computer. All the examples are in the POLYMATH directory "POLYMATH/EXAMPLES." To access them, you can run POLYMATH dorectly or install POLYMATH on your computer. If you want to study the examples which use the ODE solver for example enter 1 when the blue POLYMATH screen appears. Type F9, and F9 again and the list of examples should appear. D. Web Modules Click on the web module of interest (e.g. Wetlands (Ch 4)) to pull up the module. These modules provide supplementary Examples on how the principles of chemical reaction engineering can be applied to nontraditional situations. E. Software Toolbox 1. Click on this hot button and the following menu will appear POLYMATH MatLab ASPEN 2. Click on the hot button of your choice (e.g. POLYMATH) and instructions will appear on the screen describing how to use the software to solve the homework problems. F. Syllabi 1. Click on Syllabi and the menu


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a. 3 credit hour course b. 4 credit hour course will appear on your screen. 2. Click on the Syllabus of your choice and a week by week (lecture by lecture) listing of the topics and chapter pages covered along with assigned homework problems will appear. G. Thoughts on Problem Solving (1) Click on this hot button and the following menu appears Closed-Ended Problems (CEP) Open-Ended Problems (OEP) Ten Types of Home Problems (10 types) Strategies for Problem Solving H. Credits Click on this hot button (on the HOME Screen) to learn about the people who helped develop the web page and the CD for this text. To find out about the ways to use this CD, go to the Usage section of this CD.




Navigation


Usage How to Use the CD-ROM The primary purpose of the CD-ROM is to serve as an enrichment resource. The objectives are fourfold: (1) To provide the option/opportunity for further study or clarification on a particular concept or topic through Lecture Notes, additional examples, interactive computing modules and web modules, (2) To provide the opportunity to practice critical thinking skills, creative thinking skills, and problem solving skills through the use of "What if" questions and "living example problems," (3) To provide additional technical material for the professional bookshelf, (4) To provide other tutorial information, such as additional homework problems, thoughts on problem solving, how to use computational software in chemical reaction engineering, and a representative course structure. There are a number of ways one can use the CD in conjunction with the text. The general guideline is that the CD provides enrichment resources for the reader. Pathways on how to use the materials to learn chemical reaction engineering are shown in Figure I-1 and I-2.

The keys to the CRE learning flow sheet are: Squares = Primary Resources Circles/Ovals = Enrichment Resources I. University Student

Figure CDI-1 "A" Student Pathway to Integral Class, Text, CD.


Navigation

II. Practicing Engineer

Figure CDI-2 "A" Problem Solving Pathway to Integrate text, CD.

III. Instructor

Figure CDI-3 Resources for Instructors

The author recommends that instructors use the living example problems before assigning home problems, but they may be by-passed if time is not available. This is, of course, true for all of the enrichment resources. Please note however, that class testing had shown that the enrichment resources not only aid students in learning the material, but they also motivate students by the novel use of CRE principles. Possible Implementation Strategies I. Learning Resources A. Lecture Notes: This material could be reviewed before reading the chapter get an overview of the material. B. Interactive Computer Modules - (ICM): Each module requires approximately 30 minutes to complete. If a module is not assigned or required, the student could quickly go on through the Review of Fundamentals Section to get an overview or to review (ca. 10 min.). The complete modules could be used by the student as a self test to check their level of understanding. A number of schools assign one either every week or every other week. C. Web Modules: This material can be used to motivate students by showing them the wide range of CRE applications of or as a basis for special projects or open-ended problems. The general problem solving algorithm could be one of the first modules to be reviewed.


Navigation

D. Solved Examples: After reading the material in the chapters and studying the example problems, students have the opportunity to see one or two more solved problems before embarking on solutions of the homework problems. II. Living Example Problems These examples are meant to be used in conjunction with the second problem of every chapter, beginning with Chapter 4 (i.e. P4-2). Typically one might assign a living example problem (e.g. P6-2) as one of the first problems assigned in a chapter to get students familiar and comfortable with the material. III. Professional Reference Shelf This material is important to the practicing engineer, but is not included in the majority of undergraduate or graduate courses in chemical reaction engineering. Consequently, instructors my pick and choose from this material along the lines of special topics. Material from Chapters 5, 8, 9 and 12 are used in the graduate course at the University of Michigan (i.e. Experimental Planning, Bifurcation Analysis, Control of Chemical Reactors, and the K. L. Model of Fluidized Beds, respectively). IV. The Web (http://www.engin.umich.edu/~cre) The web will be used to update the CD-ROM and text material, provide new examples and more solved problems, and correct of typographical errors from the first printing of the 3rd Edition.





The following Interactive Computer Modules (ICMs) are contained on the Elements of Chemical Reaction Engineering CD-ROM: Kinetics Challenge 1 -- Quiz Show Introduction to Kinetics Learning Resource for Chapter One Staging -- Reactor Sequencing Optimization Game Learning Resource for Chapter Two Kinetics Challenge 2 -- Quiz Show Stoichiometry and Rate Laws Learning Resource for Chapter Three Murder Mystery CSTR Volume Algorithm Learning Resource for Chapter Four Tic Tac Isothermal Reactor Design: Ergun, Arrhenius, and Van't Hoff Equations Learning Resource for Chapter Four Ecology A Wetlands Problem Collection and Analysis of Rate Date: Ecological Engineering Learning Resource for Chapter Five Heat Effects 1 Basketball Challenge Mole and Energy Balances in a CSTR Learning Resource for Chapter Eight Heat Effects 2 Effect of Parameter Variation on a PFR Mole and Energy Balances in a PFR Learning Resource for Chapter Eight Heterogeneous Catalysis Learning Resource for Chapter Ten Some users have experienced problems, trying to run the Interactive Computer Modules directly from the ICM directory on the CD-ROM. If you have trouble with being able to run the Interactive Computer Modules from the CD-ROM, then try installing them on file:///H:/htmlmain/interac.htm[05/12/2011 16:54:35]


your hard drive. Instructions for installing the ICMs and for using the ICMs are available.


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CRE -- Appendices

The Appendices are in PDF format. You must have Adobe Acrobat Reader installed to access these files. You will also need the Adobe Acrobat Reader Plug-in to read these files from your browser. (See the CDROM Introduction for more information.) Appendix D: Measurement of Slopes on Semilog Paper Appendix E: Software Packages Appendix H: Open-Ended Problems Appendix J: Use of Computational Chemistry Software Packages


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Polymath

Polymath Polymath is a DOS-based program that can help you solve differential equations, analyze rate data (with non-linear regression, etc.), and more. The program is fairly straight-forward, but you will want to read through the Polymath Manual before you start. Then we recommend that you take the Polymath Short Course to start you on your way.

Polymath-Related Files Polymath Program Files Polymath has been included on your CD-ROM in a directory called Polymat4. You can run Polymath directly from your CD-ROM, or you can install Polymath on your hard drive.

Polymath Examples (aka, Living Example Problems) You may have noticed that certain chapters have links to Living Example Problems. The example problems are actually in the text for the 3rd edition of Elements of Chemical Reaction Engineering, not on the CD-ROM. The chapter links for these examples direct you to the Polymath code for these problems. The Polymath code for the Living Example Problems from Chapters 2-10 and Chapters 13-14 is in the Html\Toolbox\Polymath\Examples directory on your CD-ROM. Each chapter is represented by a folder named Ch#, which is short for chapter number, of course. The Polymath code for these examples has been included on the CD-ROM for your convenience, so you won't have to waste time duplicating the examples from the text. (See the section on accessing the example problems or the Polymath Short Course for more information.) Once you load up an example, you are encouraged to "play around with it" by modifying the values of constants, varying key parameters, etc. In this way, you can get a feel for how modifying different variables will affect your results.

Using Polymath Polymath is easy to use -- once you know how! See the section on Using Polymath for more information, and don't forget to take a look at the Polymath Manual for instructions on how to run and generally use Polymath:

On-Line Information Polymath 4.0.2 Manual in PDF format. Polymath 4.1 Manual in PDF format.

file:///H:/html/toolbox/polymath/index.htm[05/12/2011 16:54:36]

Polymath

NOTE: You will need the Adobe Acrobat Reader Plug-in to read this file. (See the CD-ROM Introduction for more information.)

Why Two Versions of Polymath? This is the second printing of the text and the CD. Polymath 4.1 has better printing features than Polymath 4.0.2, but it was not available for the first printing of the CD. We could have replaced Polymath 4.0.2 with Polymath 4.1, but we decided to make both versions available instead.

Polymath Short Course You can also take a look at the Polymath Short Course for a quick-and-dirty introduction to Polymath. You won't learn everything about Polymath, but you will learn enough to get you started, so you can "play around" with the Living Example Problems. Polymath Main | Using Polymath | Installing Polymath | Short Course

References Polymath was created by Mordechai Shacham and Michael B. Cutlip. They make use of Polymath in their own text, Problem Solving in Chemical Engineering with Numerical Methods, also from Prentice-Hall. These pages on Polymath were created by Dieter Andrew Schweiss. Many thanks to Tim Hubbard and Jessica Hamman for proof-reading and error-testing them.

file:///H:/html/toolbox/polymath/index.htm[05/12/2011 16:54:36]

index.htm

MATLAB Information by Susan Fugett, D.A. Schweiss, and Mayur Valanju

Ordering MATLAB In order to use the MATLAB programs included on this CD-ROM, you must have your own copy of MATLAB. The latest version, MATLAB 5, is available from: The MathWorks, Inc. University Sales Department 24 Prime Park Way Natick, MA 01760-1500. Phone: (508) 653-1415 Fax: (508) 653-2997 Email: [email protected] Web: http://www.mathworks.com A Student Edition is also available.

On the CD-ROM Appendix E Appendix E contains detailed instructions for using MATLAB to solve the problems from the text. It is included in the Appendices section of the CD-ROM as an Adobe Acrobat Reader file (PDF format, see the CD-ROM Introduction for more information). It is also available in Word 6 format (for the PC) in the Html\Toolbox\Matlab\Word directory. Even if you are an experienced MATLAB user, we encourage you to read Appendix E to learn how to use the m-files in MATLAB. Otherwise, you may have difficulty using them.

M-Files

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index.htm

Included on this CD-ROM are the m-files for all the example problems that were solved in the text using POLYMATH. For PC users, these files can be found in the directory Html\Toolbox\Matlab\m-files. These files may be copied onto the hard drive of your computer or used directly from the CD-ROM. Mac users will be able to open these files, but they may need to edit the contents slightly.

MATLAB Notebook, M-book The Microsoft Word 6.0 Notebook files for each example problem have also been included on the CD-ROM for users who wish to use this utility. These files may be found in directory Html\Toolbox\Matlab\Word on the CD-ROM. The Notebook option provides an interface with the Microsoft Word 6.0 program, the M-book. Please note, however, that the M-book files are not included on the CD-ROM, but are included with the MATLAB software. The Notebook interface allows you to run MATLAB within Word, enabling you to fully explain and document your MATLAB operations. By typing Control + Enter at the end of a line of text, the user instructs MATLAB to perform the commands written on that line. Input into MATLAB is then changed to a different font from the text and appears green on the screen. The output from MATLAB is in another font which is blue. Plots generated in MATLAB are also added to the Word file using this interface. The M-book function is a convenient tool to prepare a detailed presentation of your MATLAB work.

Using the M-Files See the page on Using and Modifying the M-Files for more information.

Remember! We included Appendix E on the CD-ROM, because it contains detailed instructions for using MATLAB with our m-files to solve the problems from the text. It is in the Appendices section of the CD-ROM. If you plan to use MATLAB to solve these problems, then read Appendix E first!

file:///H:/html/toolbox/matlab/index.htm[05/12/2011 16:54:37]

CRE -- Chapter One-Objectives

1

Mole Balances After completing Chapter 1 of the text and associated CD-ROM material the reader will be able to: Define the rate of chemical reaction. Apply the mole balance equations to a batch reactor, CSTR, PFR, and PBR. Describe two industrial reaction engineering systems. Describe photos of real reactors. Describe how to surf the CD-ROM attached with this text.

BEGIN


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CRE -- Chapter One

1

Learning Resources 1. Summary Notes for Lectures 1 and 2 2. Web Modules A. Problem Solving Algorithm for Closed-Ended Problems B. Hints for Getting Unstuck on a Problem 3. Interactive Computer Modules A. Quiz Show I 4. Solved Problems A. CDP1-AB Batch Reactor Calculations: A Hint of Things to Come Professional Reference Shelf 1. Photographs of Real Reactors Additional Homework Problems CDP1-A Calculate the time to consume 99% and 80% of species A in a constant-volume batch reactor for a first order and for a second order reaction, respectively. Solution Included CDP1-B Derive the differential mole balance equation for a foam reactor.


file:///H:/html/01chap/html/one.htm[05/12/2011 16:54:38]


How to Create and Simulate Chemical Reaction Models with: Polymath Two versions of Polymath have been included on your CD-ROM, 4.0.2 and 4.1. You can run version 4.0.2 directly from the CD, so that you examine and modify the Living Example Problems specific to each Chapter. You can also install it on your computer. To use version 4.1, you will have to install it on your computer. Polymath 4.1 has better printing capabilities than Polymath 4.0.2. MATLAB MATLAB m-files have been included on your CD-ROM, but you will have to purchase your own copy of MATLAB, or MATLAB must be available on your school's computers, to be able to use them. Aspen Plus Read these pages to learn how to use Aspen Plus to design chemical engineering reaction systems. You will have to purchase (or acquire a site license for) your own copy of the Aspen Plus software, since it is not included on this CD.


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The Thoughts on Problem Solving area of this CD-ROM offers students step-by-step instruction for the purpose of further developing problem solving skills.

Closed-Ended Problems - these single answer homework problems include two example problems along with techniques for getting unstuck when stopped along the solution path.

Open-Ended Problems - solution heuristic to problems that may be ill-posed, have no solution as posed, or allow for the possibility of multiple solutions.

Ten Types of Home Problems - describes how different types of home problems can be used to improve critical and creative thinking skills.

Strategies for Creative Problem Solving - the award winning book on developing creative problem solving skills


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Elements of Chemical Reaction Engineering, Updates

3rd Edition Updates & FAQs This area contains updates to the 3rd Edition of Elements of Chemical Reaction Engineering as things like typographical errors, etc. are found. This information is current, as of just prior to the 2nd printing of the text and CD-ROM. You are encouraged to visit the Chemical Reaction Engineering Web Site every few months for new updates as more typos are found, or as new problems, activities, etc. are added to the web site. This is also the location our reaction engineering Frequently Asked Questions (FAQs) page.

Links Typos in the First Printing Which Printing Do I Have? Frequently Asked Questions (FAQs)

On the CD Updates to the 3rd Edition can be viewed in PDF format. NOTE:You must have Adobe Acrobat Reader installed to access PDF files. You will also need the Adobe Acrobat Reader Plug-in to read these files from your browser. (See the CDROM Introduction for more information.)

On the CRE Web Site CD-ROM users are encouraged to check the Updates Section of the Chemical Reaction Engineering Web Site every few months for new material.

© 1999 Prentice-Hall PTR Prentice Hall, Inc.

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Elements of Chemical Reaction Engineering, Updates

ISBN 0-13-531708-8 Legal Statement

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Representative Syllabus for a 3 Credit Hour Course from the University of Illinois, ChE 381, Fall 1998 Professor Richard Braatz

Representative Syllabus for a 4 Credit Hour Course from the University of Michigan, ChE 344, Winter 1998 Professor H. Scott Fogler

Representative Syllabus for a 4 Credit Hour Course from the University of Michigan, ChE 344, Winter 1999 Professor H. Scott Fogler


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1. Welcome 2. Navigation 3. Components 4. Usage 5. Downloading Software Interactive Computer Modules Polymath, MATLAB, Aspen 6. About this CD


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CRE -- Chapter Two-Objectives

2

Conversion and Reactor Sizing After completing Chapter 2 of the text and associated CD-ROM material the reader will be able to: Define conversion and space time. Write the mole balances in terms of conversion for a batch reactor, CSTR, PFR, and PBR. Size reactors either alone or in series once given the rate of reaction, -r A, as a function of conversion, X. Write relationship between the relative rates of reaction.

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CRE -- Chapter Two

2

Learning Resources 1. Summary Notes for Lectures 1 and 2 3. Interactive Computer Modules A. Reactor Staging 4. Solved Problems A. CD P2-AB More CSTR and PFR Calculations -- No Memorization Additional Homework Problems CDP2-AB Use Levenspiel plots to calculate PFR and CSTR reactor volumes given -r A = f(X). Solution Included CDP2-BA An ethical dilemma as to how to determine the reactor size in a competitor's chemical plant. CDP2-CA Use Levenspiel plots to calculate PFR and CSTR volumes. CDP2-DA Use Levenspiel plots to calculate CSTR and PFR volumes for the reaction A+B

C


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CRE -- Chapter Three-Objectives

3

Rate Law and Stoichiometry After completing Chapter 3 of the text and associated CD-ROM material the reader will be able to: Write a rate law and define reaction order and activation energy. Set up a stoichiometric table for both batch and flow systems and express concentration as a function or conversion. Calculate the equilibrium conversion for both gas and liquid phase reactions. Write the combined mole balance and rate law in measures other than conversion. Set up a stoichiometric table for reactions with phase change.

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CRE -- Chapter Three

3

Learning Resources 1. Summary Notes for Lectures 3 and 4 Summary Notes for Lectures 5 and 6 3. Interactive Computer Module A. Quiz Show II 4. Solved Problems A. CD P3-AB Activation Energy for a Beetle Pushing a Ball of Dung B. CD P3-BB Microelectronics Industry and the Stoichiometric Table Additional Homework Problems CDP3-AB Data on the tenebrionid beetle whose body mass is 3.3g shows it can push a 35g ball of dung at 6.5 cm/s at 27 C, 13 cm/s at 37 C and 18 cm/s at 40 C. How fast can it push at 41.5 C (Heinrich, B., The Hot-Blooded Insects. Harvard Press, Cambridge, 1993). Solution Included CDP3-BB Silicon is used in the manufacture of microelectronic devices. Set up a stoichiometric table for the reaction: SiHCl 3 (g) + H2 (g)

Si(s) + HCl(g) + Si x Hg Cl z(g)

Solution Included CDP3-CB The elementary reaction A(g) + B(g) C(g) takes place in a square duct containing liquid B, which evaporates into the gas phase to react with A. CDP3-DB Condensation occurs in the gas phase reaction: C2 H4 (g) + 2Cl 2 (g) CDP3-EB Set up a stoichiometric table for the reaction:

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CH 2 Cl 2 (g,l) + 2HCl(g)

CRE -- Chapter Three

C6 H5 COCH + 2NH 5

C6 H5 ONH2 + NH 2 Cl


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CRE -- Chapter Four-Objectives

4

Isothermal Reactor Design After completing Chapter 4 of the text and associated CD ROM material the reader will be able to: Describe the algorithm that allows the reader to solve chemical reaction engineering problems through logic rather than memorization. Size batch reactors, semibatch reactors, CSTRs, PFRs, and PBRs for isothermal operation given the rate law and feed conditions. Discuss solutions to problems taken from the California Professional Engineers Registration Examination. Account for the effects of pressure drop on conversion in packed bed tubular reactors and in packed bed spherical reactors.

BEGIN



CRE -- Chapter Four

4

Learning Resources 1. Summary Notes for Lectures 3 and 4 Summary Notes for Lectures 5 and 6 Summary Notes for Lectures 7 and 8 Summary Notes for Lectures 9 and 10 2. Web Modules A. Wetlands B. Membrane Reactors C. Reactive Distillation 3. Interactive Computer Modules A. Murder Mystery B. Tic Tac -- A Game of Reaction Engineering Tic-Tac-Toe 4. Solved Problems A. CD P4-AB A Sinister Gentleman Messing with a Batch Reactor B. Solution to a California Registration Exam Problem C. Ten Types of Home Problems: 20 Solved Problems 5. Analogy of CRE Algorithms to a Menu in a Fine French Restaurant 6. Algorithm for Gas Phase Reaction Living Example Problems The following examples can be accessed through the Software Toolbox. 1. Example 4-7 Pressure Drop with Reaction -- Numerical Solution 2. Example 4-8 Dehydrogenation in a Spherical Reactor 3. Example 4-9 Working in Terms of Molar Flow Rate in a PFR 4. Example 4-10 Membrane Reactor 5. Example 4-11 Isothermal Semibatch Reactor with a Second-Order Reaction Professional Reference Shelf

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CRE -- Chapter Four

1. Time to Reach Steady State for a First Order Reaction in a CSTR 2. Recycle Reactors 3. Critiquing Journal Articles Additional Homework Problems CDP4-AB A sinister looking gentleman is interested in producing methyl perchlorate in a batch reactor. The reactor has a strange and unsettling rate law. [2nd Ed. P4-28] Solution Included CDP4-BC Ecological Engineering. A much more complicated version of problem 4-17 uses actual pond (CSTR) sizes and flow rates in modeling the site with CSTRs for the Des Plaines river experimental wetlands site (EW3) in order to degrade atrazine. CDP4-CB The rate of binding ligands to receptors is studied in this application of reaction kinetics to bioengineering. The time to bind 50% of the ligands to the receptros is required. [2nd Ed. P4-34] CDP4-DB A batch reactor is used for the bromination of p-chlorophenyl isopropyl ether. Calculate the batch reaction time. [2nd Ed. P4-29] CDP4-EB California Professional Engineers Exam Problem, in which the reaction B + H2

A

is carried out in a batch reactor. [2nd Ed. P4-15] CD P4-FA The gas-phase reaction A + 2B

2D

has the rate law -r A = 2.5 CA 0.5 CB. Reactor volumes of PFRs and CSTRs are required in this mulitpart problem. [2nd Ed. P4-21] CD P4-GB What type and arrangement of flow reactors should you use for a decomposition reaction with the rate law -r A = k1 CA 0.5 / (1 + k2 CA )? [2nd Ed. P4-14] CD P4-HA Verify that the liquid-phase reaction of 5, 6-benzoquinoline with hydrogen is psuedo-first-order. [2nd Ed. P4-7]


CRE -- Chapter Four

CD P4-I B The liquid-phase reaction 2A + B

C+D

is carried out in a semibatch reactor. Plot the conversion, volume, and species concentrations as a function of time. Reactive distillation is also considered in part (e). [2nd Ed. P4-27] CD P4-J B The reaction A B is catalyzed by H2 SO 4 . The reaction is carried out in a semibatch reactor, in which A is fed continuously to H 2 SO 4 . Here plots of concentrations as a function of time are required. [2nd Ed. P4-27] CD P4-KB Calculate the overall conversion for a PFR with recycle. [2nd Ed. P4-28] CD P4-LB The overall conversion is required in a packed-bed reactor with recycle. [2nd Ed. P4-22] CD P4-M B A recycle reactor is used for the reaction A+B

C

in which species C is partially condensed. The PFR reactor volume is required for a 50% conversion. [2nd Ed. P4-32] CD P4-NB Radical flow reactors can be used to good advantage for exothermic reactions with large heats of reaction. The radical velocity varies as:

Vary the parameters and plot X as a function of r. [2nd Ed. P4-31] CD P4-OB The growth of a bacterium is to be carried out in excess nutrient. nutrient + cells

more cells + product

The growth rate law is:

CD P4-PB California Registration Examination Problem. Second-order reaction in different CSTR and PFR arrangements. CD P4-QB


CRE -- Chapter Four

An unremarkable semibatch reactor problem, but it does require assessing which equation to use.


CRE -- Chapter Five-Objectives

5

Collection and Analysis of Rate Data After completing Chapter 5 of the text and associated CD-ROM material the reader will be able to: Determine the reaction order and specific reaction rate from experimental data obtained from either batch or flow reactors. Describe how to use equalarea differentiation, polynomial fitting, numerical difference formulas and regression to analyze experimental data to determine the rate law. Describe how the methods of half lives, and of initial rate, are used to analyze rate data. Describe two or more types of laboratory reactors used to obtain rate law data along with their advantages and disadvantages. Describe how to plan an experiment.

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CRE -- Chapter Five

5

Learning Resources 1. Summary Notes for Lectures 9 and 10 2. Interactive Computer Module A. Ecology -- A Wetlands Problem 3. Solved Problems A. CD P5-B B Oxygenating Blood B. Example CD 5-1 Integral Method of Analysis of Pressure-Time Data Living Example Problems The following examples can be accessed through the Software Toolbox. 1. Example 5-6 Hydrogenation of Ethylene to Ethane Professional Reference Shelf 1. Weighted Least-Squares Analysis 2. Experimental Planning 3. Laboratory Reactors Additional Homework Problems CDP5-AB The reaction of penicillin G with NH2OH is carried out in a batch reactor. A colorimeter was used to measure the absorbency as a function of time. [1st Ed. P5-10] CDP5-B B

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CRE -- Chapter Five

The kinetics of the deoxygenation of hemoglobin in the blood were studied with the aid of a tubular reactor. [1st Ed. P5-3] Solution Included CDP5-C C The kinetics of the formulation of an important propellant ingredient, triaminoguandine, were studied in a batch reactor where the ammonia concentration was measured as a function of time. [1st Ed. P5-6] CDP5-DB The half-life of one of the pollutants, NO, in automotive exhaust is required. [1st Ed. P5-11] CDP5-EB The kinetics of a gas phase reaction A2 2A were studied in a constant-pressure batch reactor, in which the volume was measured as a function of time. [1st Ed. P5-6]

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CRE -- Chapter Six-Objectives

6

Multiple Reactions After completing Chapter 6 of the text and associated CD-ROM material the reader will be able to: Define different types of selectively and yield. Choose a reaction system that would maximize the selectivity of the desired product given the rate laws for all the reactions occurring in the system. Describe the algorithm used to design reactors with multiple reactions. Size reactors to maximize the selectivity and to determine the species concentrations in a batch reactor, semibatch reactor, CSTR, PFR, and PBR, systems.

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CRE -- Chapter Six

6

Learning Resources 1. Summary Notes for Lectures 11 and 12 Summary Notes for Lectures 13 and 14 2. Web Module A. Cobra Bites 3. Solved Problems A. CDP6-B B All You Wanted to Know About Making Maleic Anhydride and More 4. Clarification: PFR with feed streams along the length of the reactor Living Example Problems The following examples can be accessed through the Software Toolbox. 1. Example 6-6 Hydrodealkylation of Mesitylene in a PFR 2. Example 6-7 Hydrodealkylation of Mesitylene in a CSTR 3. Example 6-8 Calculating Concentrations as a Function of Position for NH 3 Oxidation in a PFR Additional Homework Problems CDP6-AB Suggest a reaction system and conditions to minimize X and Y for the parallel reactions A X, A B, and A Y. [2nd Ed. P9-5] CDP6-B B Rework the maleic anhydride problem, P6-14, for the case when reaction 1 is second order. [2nd Ed. P9-8]

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CRE -- Chapter Six

Solution Included CDP6-C B The reaction sequence A CSTR. [2nd Ed. P9-12]

B, B

C, and B

D is carried out in a batch reactor and in a

CDP6-DB Isobutylene is oxidized to methacrolum, CO, and CO 2. [1st Ed. P9-16] CDP6-EB Given a batch reactor with A Ed. P9-11]

B

D

, calculate the composition after 6.5 hours. [1st

CDP6-F B Chlorination of benzene, CO, monochlorobenzene, and dichlorobenzene in a CSTR. [1st Ed. P9-14] CDP6-GC Determine the number of independent reactions in the oxidation of ammonia. [1st Ed. P9-17] CDP6-HB Oxidation of formaldehyde: HCOOH HCHO + 1/2 O2 HCOOCH 3 2HCHO CDP6-IB Continuation of CDP6-H: HCOOH HCOOH

CO 2 + H2 CO + H2O

CDP6-J B Continuation of CDP6-H and -I: HCOOCH 3

CH 3OH + HCOOH

CDP6-KC Design a reactor for the alkylation of benzene with propylene to maximize the selectivity of isopropylbenzene. [Proc. 2nd Joint China/USA Chem. Eng. Conf. III, 51, (1997).] CDP6-LD Reactions between paraffins and olefins to form highly branched paraffins are carried out in a slurry reactor to increase the octane number in gasoline. [Chem. Eng. Sci. 51, 10, 2053 (1996).]


CRE -- Chapter Six

CDP6-M A Design a reaction system to maximize the production of alkyl choride. [1st Ed. P9-19] CDP6-NC Design a reaction system to maximize the selectivity of p-xylene from methanol and toluene over a HZSM-8 zeolite catalyst. [2nd Ed. P9-17] CDP6-OB Rework the maleic anhydride problem (P6-14) for the case when reaction 1 is second order. [2nd Ed. P9-8] CDP6-P C The oxidation of propylene to acrolein [Chem. Eng. Sci., 51, 2189 (1996)].



CRE -- Chapter Seven-Objectives

7

Nonelementary Reaction Kinetics After completing Chapter 7 of the text and associated CD-ROM material the reader will be able to: Discuss the pseudo-steadystate-hypothesis and explain how it can be used to solve reaction engineering problems. Discuss different types of polymerization reactions and rate laws. Describe Michealis-Menton enzyme kinetics and enzyme inhibition. Write material balances on cells, substrates, and products in bioreactors.

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CRE -- Chapter Seven

7

Learning Resources 1. Summary Notes for Lectures 25 and 26 Summary Notes for Lectures 36 and 37 Summary Notes for Lectures 38 and 39 4. Solved Problems A. Hydrogen Bromide Example CD7-1 Deducing the Rate Law Example CD7-2 Deriving the Rate Law from the Reaction Mechanism Living Example Problems The following example can be accessed through the Software Toolbox. 1. Example 7-2 PSSH Applied to Thermal Cracking of Ethane Professional Reference Shelf 1. Enzyme Inhibition A. Competitive Example CD7-3 Derive a Rate Law for Competitive Inhibition B. Uncompetitive C. Non-Competitive Example CD7-4 Derive a Rate Law for Non-Competitive Inhibition Example CD7-5 Match Eadie Plots to the Different Types of Inhibition 2. Multiple Enzymes and Substrate Systems A. Enzyme Regeneration Example CD7-6 Construct a Lineweaver-Burke Plot for Different Oxygen Concentrations

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B. Enzyme Co-factors Example CD7-7 Derive an Initial Rate Law for Alcohol Dehydrogenates C. Multiple Substrate Systems Example CD7-8 Derive a Rate Law for a Multiple Substrate System Example CD7-9 Calculate the Initial Rate of Formation of Ethanol in the Presence of Propanediol D. Multiple Enzyme Systems 3. Oxidation-Limited Fermentation 4. Fermentation Scale-up Additional Homework Problems CD P7-AB Determine the rate law and mechanism for the reaction: 2GCH 3

2G + CH 2 + H2

[2nd Ed. P7-6A] CD P7-B B Suggest a mechanism for the reaction: I- + OCl -

OI- + Cl -

[2nd Ed. P7-8B ] CDP7-C A Develop a rate law for substrate inhibition of an enzymatic reaction. [2nd Ed. P7-16A] CDP7-DB Use Polymath to analyze an enzymatic reaction. [2nd Ed. P7-19B ] CDP7-EB Redo Problem P7-17 to include chain transfer. [2nd Ed. P7-23B ] CDP7-F B Determine the rate of diffusion of oxygen to cells. [2nd Ed. P12-12B ] CDP7-GB Determine the growth rate of amoeba predatory on a bacteria. [2nd Ed. P12-15C ] file:///H:/html/07chap/html/seven.htm[05/12/2011 16:54:47]


CDP7-HC Plan the scale-up of an oxygen fermentor. [2nd Ed. P12-16B ] CDP7-IB Assess the effectiveness of bacteria used for denitrification in a batch reactor. [2nd Ed. P1218B ] CDP7-J A Determine rate law parameters for the Monod equation. [2nd Ed. P12-19A]


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CRE -- Chapter Eight-Objectives

8

Steady-State Nonisothermal Reactor Design After completing Chapter 8 of the text and associated CD-ROM material the reader will be able to: Describe the algorithm for CSTRs, PFRs, and PBRs that are not operated isothermally. Size adiabatic and nonadiabatic CSTRs, PFRs, and PBRs. Use reactor staging to obtain high conversions for highly exothermic reversible reactions. Carry out an analysis to determine the Multiple Steady States (MSS) in a CSTR along with the ignition and extinction temperatures. Analyze multiple reactions carried out in CSTRs, PFRs, and PBRs which are not operated isothermally in order to determine the concentrations and temperature as a function of position (PFR/PBR) and operating variables.

BEGIN



CRE -- Chapter Eight-Objectives Legal Statement


CRE -- Chapter Eight

8

Learning Resources 1. Summary Notes for Lectures 13 and 14 Summary Notes for Lectures 15 and 16 Summary Notes for Lectures 17 and 18 Summary Notes for Lecture 35A 3. Interactive Computer Modules A. Heat Effects I B. Heat Effects II 4. Solved Problems for Heat Capacities Expressed as Quadratic Functions of A. Example CD 8-1 Temperature B. Example CD 8-2 Second Order Reaction Carried Out Adiabatically in a CSTR 5. PFR/PBR Solution Procedure for a Reversible Gas-Phase Reaction Living Example Problems The following examples can be accessed through the Software Toolbox. 1. Example 8-5 CSTR with a Cooling Coil 2. Example 8-6 Liquid Phase Isomerization of Normal Butene 3. Example 8-7 Production of Acetic Anhydride 4. Example 8-10 Oxidation of SO2 5. Example 8-11 Parallel Reaction in a PFR with Heat Effects 6. Example 8-12 Multiple Reactions in a CSTR

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Professional Reference Shelf 1. Steady State Bifurcation Analysis A. Fundamentals B. Example CD 8-3 Determine the Parameters That Give Multiple Steady States (MSS) Additional Homework Problems CD P8-AB The exothermic reaction A

2B

is carried out in both a plug-flow reactor and a CSTR with heat exchange. You are requested to plot conversion as a function of reactor length for both adiabatic and nonadiabatic operation, as well as to size a CSTR. [2nd Ed. P8-16] CD P8-B B Use bifurcation theory (Section 8.6.5 on the CD-ROM) to determine the possible regions for multiple steady states for the gas reaction with the rate law:

[2nd Ed. P8-26] CD P8-C B In this problem, bifurcation theory (CD-ROM Section 8.6.5) is used to determine if multiple steady states are possible for each of three types of catalyst. [2nd Ed. P8-27] CD P8-DB In this problem, bifurcation theory (CD-ROM Section 8.6.5) is used to determine the regions of multiple steady states for the autocatalytic reaction: A+B

2B

[2nd Ed. P8-28] CD P8-EC This problem concerns the SO 2 reaction with heat losses. [2nd Ed. P8-33] CD P8-F C This problem concerns the use of interstage cooling in SO 2 oxidation. [2nd Ed. P8-34(a)]



CD P8-GB This problem is a continuation of the SO 2 oxidation example problem. Reactor costs are considered in the analysis cooling. [2nd Ed. P8-34(b and c)] CD P8-HB Parallel reactions taking place in a CSTR with heat effects. [1st Ed. P9-21] CD P8-IB This problem concerns multiple steady states for the second-order, reversible, liquid-phase reaction [Old exam problem] CD P8-J B Series reactions take place in a CSTR with heat effects. [1st Ed. P9-23] CD P8-KB A drug intermediate is produced in a batch reactor with heat effects. The reaction sequence is: 2A + B C+A+B

C+D E+D

The desired product is C. CD P8-LB In the multiple steady state for A

B

the phase plane of C A vs. T shows a separatrix. [2nd Ed. P8-22] CD P8-NB A second-order reaction with multiple steady states is carried out in different solvents. CD P8-OC Multiple reactions 2B A C 2A + B are carried out adiabatically in a PFR. CDP8-P B An exothermic 2nd order reversible reaction is carried out in a packed bed reactor.





CRE -- Chapter Nine-Objectives

9

Unsteady-State Nonisothermal Reactor Design After completing Chapter 9 of the text and associated CD-ROM material the reader will be able to: Analyze batch reactors and semibatch not operated isothermally. Analyze the start up of nonisothermal CSTRs. Analyze perturbations in temperature and presence for CSTRs being operated at steady state and describe under what conditions the reactors can be unsafe (safety). Describe the effects of adding a controller to a CSTR. Analyze multiple reactions in batch and semibatch reactors not operated isothermally.

BEGIN



CRE -- Chapter Nine

9

Learning Resources 1. Summary Notes for Lectures 17 and 18 Summary Notes for Lecture 35B 4. Solved Problems A. Example CD 9-1 Startup of a CSTR B. Example CD 9-2 Falling Off the Steady State C. Example CD 9-3 Proportional Integral (PI) Control

Living Example Problems The following examples can be accessed through the Software Toolbox. 1. Example 9-1 Adiabatic Batch Reactor 2. Example 9-2 Safety in Chemical Plants with Exothermic Reactions 3. Example 9-3 Heat Effects in a Semibatch Reactor 4. Example 9-4 Startup of a CSTR 5. Example 9-5 Falling Off the Steady State 6. Example 9-6 Integral Control of a CSTR 7. Example 9-7 Proportional Integral Control of a CSTR 8. Example 9-8 Multiple Reactions in a Semibatch Reactor

Professional Reference Shelf file:///H:/html/09chap/html/nine.htm[05/12/2011 16:54:49]

CRE -- Chapter Nine

1. Intermediate Steps in the Adiabatic Batch Reactor Derivation 2. Approach to Steady-State Phase-Plane Plots and Trajectories of Concentration versus Temperature 3. Unsteady Operation of Plug Flow Reactors

Additional Homework Problems CD P9-AB The production of propylene glycol (discussed in Examples 8-4, 9-4, 9-5, 9-6, and 9-7) is carried out in a semibatch reactor. [2nd Ed. P8-14] CD P9-B C Reconsider problem P9-14 when a PI controller is added to the coolant stream.


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CRE -- Chapter Ten-Objectives

10

Catalysis and Catalytic Reactors After completing Chapter 10 of the text and associated CD-ROM material the reader will be able to: Define a catalyst, a catalytic mechanism and a rate limit step. Describe the steps in a catalytic mechanism and how one goes about deriving a rate law and a mechanism and rate limiting step consistent with the experimental data. Size isothermal reactors for reactions with LangmuirHinschelwood kinetics. Discuss the different types of catalyst deactivation and the reactor types and describe schemes that can help offset the deactivation. Analyze catalyst decay and conversion for CSTRs and PFRs with temperaturetime trajectories, moving bed reactors, and straight through transport reactors. Describe the steps in Chemical Vapor Deposition(CVD). Analyze moving bed reactors that are not operated isothermally.

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CRE -- Chapter Ten-Objectives



CRE -- Chapter Ten

10

Learning Resources 1. Summary Notes for Lectures 19 and 20 Summary Notes for Lectures 21 and 22 Summary Notes for Lectures 23 and 24 3. Interactive Computer Module A. Heterogeneous Catalysis 4. Solved Problems A. Example CD10-1 Analysis of Heterogeneous Data [Class Problem, Winter 1997] B. Example CD10-2 Least-Squares to Determine Rate Law Parameters k, KT, and KB (Example 6-2 in 2nd Edition) C. Example CD10-3 Hydrodemethylation of Toluene in a PBR without Pressure Drop [2nd Ed. Example 6-3] D. Example CD10-4 Cracking of Texas Gas-Oil in a STTR [2nd Ed. Example 6-5] Living Example Problems The following examples can be accessed through the Software Toolbox. 1. Example 10-5 Catalyst Decay in a Fluidized Bed Modeled as a CSTR 2. Example 10-6 Catalytic Cracking in a Moving-Bed Reactor 3. Example 10-7 Decay in a Straight Through Transport Reactor Professional Reference Shelf 1. Hydrogen Adsorption A. Molecular Adsorption B. Dissociative Adsorption 2. Catalyst Poisoning in a Constant Volume Batch Reactor 3. Differential Method of Analysis to Determine the Decay Law 4. Etching of Semiconductors A. Dry Etching B. Wet Etching

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CRE -- Chapter Ten

C. Dissolution Catalysis Additional Homework Problems CDP10-AA Suggest a rate law and mechanism for the catalytic oxidation of ethanol over tantalum oxide when the adsorption of ethanol and oxygen takes place on different sites. [2nd Ed. P6-17] CDP10-BB Analyze the data for the vapor-phase esterification of acetic acid over a resin catalyst at 118 C. CDP10-CB Silicon dioxide is grown by CVD according to the reaction SiH2 Cl 2 (g) + 2N2 0(g)

SiO2 (s) + 2N2 (g) + 2HCL(g)

Use the rate data to determine the rate law, reaction mechanism, and rate law parameters. [2nd Ed. P6-13] CDP10-DB The autocatalytic reaction A + B 2B is carried out in a moving-bed reactor. The decay law is firstorder in B. Plot the activity and the concentrations of A and B as a function of catalyst weight. CDP10-EB Determine the rate law and rate law parameters for the wet etching of an aluminum silicate. CDP10-FB Titanium films are used in decorative coatings as well as wear-resistant tools because of their thermal stability and low electrical resistivity. TiN is produced by CVD from a mixture of TiCl 4 and NH 3 TiN. Develop a rate law, a mechanism, and a rate-limiting step, and evaluate the rate law parameters. CDP10-GC The decomposition of cumene is carried out over a LaY zeolite catalyst, and deactiviation is found to occur by coking. Determine the decay law and rate law, and use these to design a STTR. [2nd Ed. P6-27] CDP10-HB The dehydrogenation of ethylbenzene is carried out over a Shell catalyst. From the data provided, find the cost of the catalyst required to produce a specified amount of styrene. [2nd Ed. P6-20] CDP10-I B A second-order reaction over a decaying catalyst takes place in a moving-bed reactor. [Final Exam, Winter 1994] CDP10-J B A first-order reaction A

B + C takes place in a moving-bed reactor.

CDP10-KB For the cracking of normal paraffins (P n ), the rate has been found to increase with increasing temperature up to a carbon number of 15 (i.e., n < 16) and to decrease with increasing temperature for a carbon number greater than 16. [J. Wei, Chem. Eng. Sci., 51, 2995 (1996)]


CRE -- Chapter Ten

CDP10-LB The formation of CH 4 from CO and H2 is studied in a differential reactor. CDP10-M B The reaction A + B

C + D is carried out in a moving-bed reactor.

CDP10-NA Determine the rate law and mechanism for the reaction A + B

C.

CDP10-OB Determine the rate law from data where the pressures are varied in such a way that the rate is constant. [2nd Ed. P6-18] CDP10-PB Determine the rate law and mechanism for the vapor phase dehydration of ethanol. [2nd Ed. P6-21] CDP10-QA Second order reaction and zero order decay in a batch reactor. CDP10-RB First order decay in a moving bed reactor for the series reaction A

B

C



CRE -- Chapter Eleven-Objectives

11

External Diffusion Effects on Heterogeneous Reactions After completing Chapter 11 of the text and associated CD-ROM material the reader will be able to: Define the mass transfer coefficient, explain what it is function of and how it is measured or calculated. Analyze PBRs in which mass transfer limits the rate of reaction. Discuss how one goes form a region mass transfer limitation to reaction limitation. Describe how catalyst monoliths and wire gauze reactors are analyzed. Apply the shrinking core model to analyze catalyst regeneration.

BEGIN



CRE -- Chapter Eleven

11

Learning Resources 1. Summary Notes for Lectures 27 and 28 2. Solved Problems A. Example CD11-1 Calculating Steady State Fluxes B. Example CD11-2 Relating the Fluxes W A, BA, and J A C. Example CD11-3 Diffusion Through a Stagnant Gas Professional Reference Shelf 1. Mass Transfer Limited Reactions on Metallic Surfaces A. Catalyst Monoliths (Catalytic Converter for Autos) B. Wire Gauzes Additional Homework Problems CDP11-AA An isomerization reaction that follows Langmuir-Hinshelwood kinetics is carried out on a monolith catalyst. [2nd Ed. P10-11] CDP11-BB A parameter sensitivity analysis is required for this problem in which an isomerization is carried out over a 20-mesh gauze screen. [2nd Ed. P10-12] CDP11-CC This problem examines the effect on temperature in a catalyst monolith. [2nd Ed. P10-13] CDP11-DB A second-order catalytic reaction is carried out in a catalyst monolith. [2nd Ed. P10-14] CDP11-EC Fracture acidizing is a technique to increase the productivity of oil wells. Here acid is injected at high pressures to fracture the rock and form a channel that extends out from the well bore. As the acid flows through the channel, it etches the sides of the channel to make it larger, and thus less resistant to the flow of oil. Derive equations for the concentration profile of acid and the channel width, each as a function of distance from the well bore. [2nd Ed. P10-15] CDP11-FC The solid-gas reaction of silicon to form SiO is an important process in microelectronics fabrication. The file:///H:/html/11chap/html/eleven.htm[05/12/2011 16:54:51]

CRE -- Chapter Eleven

2

oxidation occurs at the Si-SiO2 interface. Derive an equation for the thickness of the SiO2 layer as a fucntion of time. [2nd Ed. P10-17] CDP11-GB Mass transfer limitations in CVD processing to product material with ferroelectric and pezoelectric properties. [2nd Ed. P10-17] CDP11-HB Calculate multicomponent properties. [2nd Ed. P10-17] CDP11-I B Application of the shrinking core model to FeS2 rock samples in acid mine drainage. [2nd Ed. P10-18]


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CRE -- Chapter Twelve-Objectives

12

Diffusion and Reaction in Porous Catalysts After completing Chapter 12 of the text and associated CD-ROM material the reader will be able to: Define the Thiele modules and the effectiveness factor. Describe the regions of reaction limitations and internal diffusion limitations and the conditions that affect them. Determine which resistance is controlling in a slurry reactor. Analyze trickle bed reactors. Analyze fluidized bed reactors. Describe the operation of a CVD Boat Reactor.

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CRE -- Chapter Twelve

12

Learning Resources 1. Summary Notes for Lectures 27 and 28 Summary Notes for Lectures 29 and 30 Professional Reference Shelf 1. Trickle Bed Reactors A. Fundamentals B. Limiting Situations C. Evaluating the Transport Coefficients 2. Fluidized Bed Reactors A. Overview B. Mechanics of Fluidized Beds C. Descriptive Behavior of the Kunii-Levenspiel Bubbling Bed Model D. Mass Transfer in Fluidized Beds E. Reaction in a Fluidized Bed F. Mole Balances on the (1) Bubble, (2) Cloud, and (3) Emulsion G. Solution to the Balance Equations for a First Order Reaction Example CD12-3 Catalytic Oxidation of Ammonia H. Limiting Situations Example CD12-4 Calculating the Resistances Example CD12-5 Effect of Particle Size on Catalyst Weight for a Slow Reaction Example CD12-6 Effect of Catalyst Weight for a Rapid Reaction I. Summary 3. CVD Boat Reactors A. Fundamentals B. Examples Additional Homework Problems CDP12-AB file:///H:/html/12chap/html/twelve.htm[05/12/2011 16:54:52]


Determine the catalyst size that gives the highest conversion in a packed bed reactor. CDP12-B D Determine the temperature profiles to achieve a uniform thickness. [2nd Ed. P11-18] CDP12-C B Explain how varying a number of the parameters in the CVD boat reactor will affect the wafer shape. [2nd Ed. P11-19] CDP12-DB Determine the wafer shape in a CVD boat reactor for a series of operating conditions. [2nd Ed. P11-20] CDP12-EC Model the buildup of a silicon wafer on parallel sheets. [2nd Ed. P11-21] CDP12-F C Rework CVD boat reactor accounting for the reaction: SiH 2 + H2 SiH 4 [2nd Ed. P11-22] CDP12-GB Hydrogenation of an unsaturated organic is carried out in a trickle bed reactor. [2nd Ed. P127] CDP12-HH The oxidation of ethanol is carried out in a trickle bed reactor. [2nd Ed. P12-9] CDP12-J B The hydrogenation of aromatics in a napthenic lube oil distillate takes place in a trickle bed reactor. CDP12-KC Compare the models discussed in the article with the Kunii-Levenspiel model--e.g. list major difference, the advantages, disadvantages, and strong and weak points of each model. CDP12-LC Reconsider example problem 12-2, using the correlations of C. Chavarie and J. R. Grace [Ind. Engrg Chem. Fundamentals, Vol. 14. No. 2 pp. 75-79 (1975)] CDP12-M A If the temperature were increased from 273°K to 546°K, the minimum fluidization velocity of the gas would approximately:

(a) Increase by a factor of a) 1.4, b) 2.8, c) 2.0. (b) Decrease by a factor of a) 1.4, b) 2.8, c) 2.0. file:///H:/html/12chap/html/twelve.htm[05/12/2011 16:54:52]


(c) Remain the same. (d) Cannot be calculated from the information given. (e) None of the above. CDP12-NA At the very top of the column, what is the rate of reaction in the emulsion phase in [gmole/(cm3 of

bubble)(s)] as predicted by the Kunii-Levenspiel model? CDP12-OB When a fluidized column is operated at a superficial velocity of 10 cm/sec which is 5 times the minimum fluidization velocity, it was found the bubble size at the mid-point in the column was 10 cm and that bubbles occupied exactly 10.6% of the column. CDP12-P B The irreversible gas phase isomerization reaction

CDP12-QB When operated at twice the minimum fluidization velocity 86.5% conversion is realized in a of 0.5. It is proposed to increase the pressure from 1 fluidized bed reactor which has a atm. to 4.0 atm. and decrease the temperature from 427°C to 227°C. CDP12-R B The irreversible catalytic isomerization CDP12-S B The production of methane from carbon monoxide and hydrogen over a 1/8´ nickel catalyst is carried out at 500°F. CDP12-TB Using tracer studies to characterize fluidized bed reactors. CDP12-UB The hydrogenation of aromatics in a napthenic lube oil distillate takes place in a trickle bed reactor. CDP12C-1 Using values of the minimum fluidization velocity provided by Kurian and Raja Rao (Ind. J. of Tech., 8, 275 (1970)).

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CRE -- Chapter Thirteen-Objectives

13

Distributions of Residence Times for Chemical Reactors After completing Chapter 13 of the text and associated CD-ROM material the reader will be able to: Define a residence time distribution RTD [E(t), F(t)] and the mean residence time. Determine E(t) form tracer data. Write the RTD functions (E(t), F(t), I(t)) for ideal CSTRs, PFRs, and laminar flow reactors. Predict conversions from RTD data using the segregation and maximum mixedness models. Predict effluent concentrations for multiple reactions using the segregation and maximum mixedness models.

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CRE -- Chapter Thirteen

13

Learning Resources 1. Summary Notes for Lectures 31 and 32 2. Web Modules A. The Attainable Region Analysis 4. Solved Problems A. Example CD13-1 Calculate the exit concentrations for the series reaction

using the segregation model and the maximum mixedness model. B. Example CD13-2 Determination of the effect of variance on the exit concentrations for the series reaction

Living Example Problems The following examples can be accessed through the Software Toolbox. 1. Example 13-8 Using Software to Make Maximum Mixedness Model Calculations 2. Example 13-9 RTD and Complex Reactions Reactor with Asymmetric RTD Segregation Model, Maximum Mixedness Model Reactor with Bimodal RTD Maximum Mixedness Model, Segregation Model Professional Reference Shelf file:///H:/html/13chap/html/thirteen.htm[05/12/2011 16:54:53]


1. Attainable Region Analysis 2. Comparison of Conversion for Segregation and Maximum Mixedness Models for Reaction Orders Between 0 and 1 Additional Homework Problems CD P13-AC After showing that E(t) for two CSTRs in series having different values is

you are aksed to make a number of calculations. [2nd Ed. P13-11] CD P13-B B Determine E(t) and from data taken, form a pluse test in which the pulse is not perfect and the inlet concentration varies with time. [2nd Ed. P13-15] CD P13-C B Derive the E(t) curve for a Bingham plastic flowing in a cylindrical tube. [2nd Ed. P13-16]. CD P13-DB The order of a CSTR and PFR in series is investigated for a third-order reaction. [2nd Ed. P13-10] CD P13-EB Review the Muphree pilot plant data when a second-order reaction occurs in the reactor. [1st Ed. P13-15] CD P13-F A Gasoline shortages in the United States have produced long lines of motorists at service stations. CD P13-GB Vary the parameters to learn their effects on conversion. CD P13-HB The reactions described in Problem 6-16 are to be carried out in the reactor whose RTD is described in Example 13-7.

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CRE -- Chapter Fourteen-Objectives

14

Models for Nonideal Reactors After completing Chapter 14 of the text and associated CD-ROM material the reader will be able to: Describe the tanks-in-series and dispersion one parameter models. Describe how to obtain the mean residence time and variance to calculate the number of tanks-in-series and the Peclet number. Calculate the conversion for a first order reaction taking place in a tubular reactor with dispersion Describe how to use combinations of ideal rectors to model a real reactor and how to use tracer data to determine the model parameters.

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CRE -- Chapter Fourteen

14

Learning Resources 1. Summary Notes for Lectures 33 and 34 4. Solved Problems A. Example CD14-1 Two CSTRs with Interchange Professional Reference Shelf 1. Derivation of Equation for Taylor-Aris Dispersion 2. Real Reactor Modeled as an Ideal CSTR with Exchange Volume Additional Homework Problems CDP14-AC A real reactor is modeled as:

[2nd Ed. P14-5] CD P14-B B A batch reactor is modeled as:

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CRE -- Chapter Fourteen

[2nd Ed. P14-13] CDP14-C C Develop a model for a real reactor with RTD data obtained from a step input. [2nd Ed. P1410] CDP14-DB Calculate Da and X from sloppy tracer data. [2nd Ed. P14-6] CDP14-EB Use RTD data from Oak Ridge National Laboratory to calculate the conversion from the tanks-in-series and dispersion models. [2nd Ed. P14-7]


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CD-ROM

D

CD.1 Using Semilog Plots in Rate Data Analysis Semilog graph paper is used when dealing with either exponential growth or decay, such as y 5 bemx

(CD-1)

For the first-order elementary reaction A → products which is carried out at constant volume, the rate of the disappearance of A is given by 2dCA --------------- 5 kCA dt

(CD-2)

When t 5 0: CA 5 CA0 , where the units of CA are g mol/dm3, t is expressed in minutes, and k is expressed in reciprocal minutes. Integrating the rate equation, we obtain CA - 5 2kt ln -------CA0

(CD-3)

We wish to determine the specific reaction rate constant, k. A plot of ln CA versus t should produce a straight line whose slope is 2k. We may eliminate the calculation of the log of each concentration data point by plotting our data on semilog graph paper. The points in Table CDD-1 are plotted on the semilog graph shown in Figure CDD-1. 1

2

CD-ROM TABLE CDD-1

t (min) CA (gmol/dm3 )

0

2

4

8

14

2.0

1.64

1.38

0.95

0.60

10 8

CA (g mol/liter)

6 5 4 3 2 ∆y

1.0 0.8 0.6 0.5 0.4

∆x

0.3 0.2

0.1 0

2

4

6

8

10

12

14

16

t (min) Figure CDD-1

Fogler/PrenHall/CDD.1 S/S

Algebraic Method Draw the best straight line through your data points. Choose two points on this line, t1 and t2, and the corresponding concentrations CA1 and CA2 at these times: CA1 - 5 2kt1 ln -------CA0

CA2 - 5 2kt2 ln -------CA0

ln CA2 2 ln CA1 5 2k (t2 2 t1 )

(CDD-4)

3

CD-ROM

Rearranging yields ln CA2 2 ln CA1 ln ( CA1 / CA2 ) - 5 -----------------------------k 5 2----------------------------------t2 2 t1 t2 2 t1

(CDD-5)

When t 5 8, CA 5 1.05; when t 5 12, CA 5 0.75. Substituting into Equation (CDD-5) gives us 0.336 ln ( 1.05 / 0.75 ) k 5 ---------------------------------- 5 -------------4 min ( 12 2 8 ) min 5 0.084 min21 Graphical Technique In the preceding example, we had CA - 5 2kt ln -------CA0

(CDD-6)

Dividing by 2.3, we convert to log base 10: ln ( CA / CA0 ) 2kt ---------------------------- 5 log ( CA / CA0 ) 5 --------2.3 2.3 The slope of a plot of log CA versus time should be a straight line with slope 2k/2.3. Referring to Figure CDD-1, we draw a right triangle with the acute angles located at points CA 5 1.6, t 5 2.8 and CA 5 0.7, t 5 12.8. Next, the distances x and y are measured with a ruler. These measured lengths in y and x are 1.35 and 4.65 cm, respectively: 1 cycle Dy 5 21.35 cm 3 ----------------- 5 20.35 cycle 3.9 cm 14 min Dx 5 4.65 cm 3 ----------------- 5 9.7 min 6.7 cm 20.35 slope 5 --------------- 5 20.0361 9.7 k 5 22.3 (slope) 5 22.3 ( 20.0361 ) min21 5 0.083 min21 A modification of the algebraic method is possible by drawing a line on semilog paper so that the dependent variable changes by a factor of 10. From Equation (CDD-5) in the form ln ( CA1 / CA2 ) k 5 -----------------------------t2 2 t1 2.3 log ( CA1 / CA2 ) 5 ------------------------------------------t2 2 t1

(CDD-7)

choose the points (CA1 , t1 ) and (CA2 , t2 ) so that CA2 5 0.1CA1 : 2.3 2.3 log 10 k 5 ------------------------ 5 -------------t2 2 t1 t2 2 t1 This modification is referred to as the decade method.

(CDD-8)

MATLAB

E

Susan A. Fugett MATLAB ® version 5 is a very powerful tool useful for many kinds of mathematical tasks. For the purposes of this text, however, MATLAB 5 will be used only to solve four types of problems: polynomial curve fitting, system of algebraic equations, system of ordinary differential equations, and nonlinear regression. This appendix serves as a quick guide to solving such problems. The solutions were all prepared using the Student Edition of MATLAB 5. Please note that the MATLAB 5 software must be purchased independently of the CD-ROM accompanying this book.

E.1 A Quick Tour When MATLAB is opened, the command window of the MATLAB Student Edition appears: To get started, type one of these commands: helpwin, helpdesk, or demo EDU» You may then type commands at the EDU» prompt. Throughout this appendix, different fonts are used to represent MATLAB input and output, and italics are used to explain function arguments.

1

2

CD-ROM

E.1.1 MATLAB’s Method

MATLAB can range from acting as a calculator to performing complex matrix operations and more. All of MATLAB’s operations are performed as matrix operations, and every variable is stored in MATLAB’s memory as a matrix even if it is only 1x1 in size. Therefore, all MATLAB input and output will be in matrix form. E.1.2 Punctuation

MATLAB is case sensitive and recognizes the difference between capital and lowercase letters. Therefore, it is possible to work with the variables “X” and “x” at the same time. The semicolon and period play very important roles in performing calculations using MATLAB. A semicolon placed at the end of a command line will suppress restatement of that output. MATLAB will still perform the command, but will not display the answer. For example, type beta=1+4 and MATLAB will display the answer beta =5, but if you then type alpha = 30/2;, MATLAB will not tell you the answer. To see the value of a variable, simply type the name of the variable, alpha and MATLAB will display its value, alpha =15. The command who can also be used to view a list of current variables: Your variables are: alpha beta The period is used when element-by-element matrix multiplication is performed. To perform standard matrix multiplication of two matrices, say “A” and “B,” type A*B. To multiply every element of matrix “A” by 2 type A*2. However, to multiply every element of “A” with the corresponding element of “B,” one must type A.*B. This element-by-element matrix multiplication will be used for the purposes of this text. To learn more about MATLAB, type demo at the command prompt; to see a demo about matrix manipulations, type matmanip. E.1.3 Help

MATLAB has an extensive on-line help program that can be accessed through the help command, the lookfor command, and by the helpwin command (or by choosing help from the menu bar). By typing “help topic,” for example help log, MATLAB will give an explanation of the topic. LOG Natural logarithm. LOG(X) is the natural logarithm of the elements of X. Complex results are produced if X is not positive. See also LOG2, LOG10, EXP, LOGM. It is likely that in many instances you will not know the exact name of the topic for which you need help. (By typing helpwin, you will open the help window, which houses a list of help topics.)

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The lookfor command can be used to search through the help topics for a key word. For example you could type lookfor logarithm and receive the following list: LOGSPACE LOG LOG10 LOG2 BETALN GAMMALN LOGM

Logarithmically spaced vector. Natural logarithm. Common (base 10) logarithm. Base 2 logarithm and dissect floating point number. Logarithm of beta function. Logarithm of gamma function. Matrix logarithm.

from which the search can be narrowed. Please note that all built-in MATLAB commands are lower case, although in help they are displayed in uppercase letters. It is strongly recommended that students take time to explore the demo before attempting to solve problems using MATLAB. E.1.4 M-files

Many of the commands in MATLAB are really a combination of commands and manipulations that are stored in an m-file. Users can also write their own m-files with their own commands and data. The m-files are simply text files that have an “m” extension (e.g., example1.m). The name of the file can then be called upon later to execute the commands in the m-file as though they were being entered line-by-line by the user at the EDU» prompt. The m-file saves time by relieving the user of the need to type lines of commands over and over and by enabling him or her to change values of one or more variables easily and repeatedly.

E.2 Examples Examples of each of the four types of problems listed above will now be explained. Please refer to the examples in the book that were solved using POLYMATH. It may be wise to type the command clear before starting any new problems to clear the values from all variables in MATLAB’s memory. E.2.1 Polynomial Curve Fitting: Example 2-3

In this example, a third-order polynomial is fit to conversion-rate data.

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Step 1: First, the data have to be entered as matrices by listing them between brackets, leaving a space between each entry. X=[0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.85]; ra=[0.0053 0.0052 0.005 0.0045 0.004 0.0033 0.0025 0.0018 0.00125 0.001]; Step 2: Next, the function polyfit is used to fit the data to a third order polynomial. To learn more about the function polyfit, type help polyfit at the command prompt. p=polyfit(X,ra,3) (matrix of coefficients 5 polyfit(ind. variable, dep. variable, order of polynomial) p =0.0092

-0.0153

0.0013

0.0053

The coefficients are arranged in decreasing order of the independent variable of the polynomial, therefore, ra 5 0.0092X 3 2 0.0153X 2 1 0.0013X 1 0.0053 Please note that the typical mathematical convention for ordering coefficients is y 5 a0 1 a1 x 1 a2 x 2 1 ??? 1 an x n whereas, MATLAB returns the solution ordered from an to a0 . Step 3: Next, a variable “f” is assigned to evaluate the polynomial at the data points (i.e., “f” holds the “ra” values calculated from the equation of the polynomial fit.) Since “X” is a 1x10 matrix, “f” will also be 1x10 in size. f=polyval(p,X); f=polyval(matrix of coefficients, ind. variable) To learn more about the function polyval, type help command prompt.

polyval at the

Step 4: Finally, a plot is prepared to show how well the polynomial fits the data. plot(X,ra,'o',X,f,'-') plot(ind. var., dep. var., ‘symbol’, ind. var., f, ‘symbol’) where ‘symbol’ denotes how the data are to be plotted. In this case, the data set is plotted as circles and the fitted polynomial is plotted as a line. The following commands label and define the scale of the axes. xlabel('X'); ylabel('ra "o", f "-"'); axis([0 0.85 0 0.006]); xlabel(‘text’); ylabel(‘text’); axis([xmin xmax ymin ymax]) Please refer to help plot for more information on preparing plots.

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x The variance, or the sum of the squares of the difference between the actual data and the values calculated from the polynomial, can also be calculated. variance=sum((ra-f).^2) variance =1.6014e-008 The command (ra-f)creates a matrix the same size as “ra” and “f” and contains the element-by-element subtraction of “f” from “ra .” Every element of this new matrix is then squared to create a third matrix. Then, by summing all of the elements of this third matrix, the result is a 1x1 matrix, a scalar, equal to the variance. E.2.2 Solving a System of Algebraic Equations: Example 6-7

In this example a system of three algebraic equations 1/2

1/2

0 5 C H 2 C H0 1 ( k 1C H C M 1 k 2C H C X ) t 1/2

0 5 C M 2 C M0 1 k 1C H C M t 1/2

1/2

0 5 (k1CH CM 1 k2CH C X ) t 2 C X is solved for three variables, CH , CM , and CX . Step 1: To solve these equations using MATLAB the constants are declared to be symbolic, the values for the constants are entered in the equations and the equations are entered as eq1, eq2, and eq3 in the following form: (eq1=symbolic equation).

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It is very important that the three variables, CH , CM , and CX , be represented by variables of only one character in length (e.g., h, m, and x)

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h 5 CH

CH0 5 .021

k1 5 55.2

m 5 CM

CM0 5 .0105

k2 5 30.2

x 5 CX

t 5 0.5

syms h m x; eq1=h-.021+(55.2*m*h^0.5+30.2*x*h^0.5)*0.5; eq2=m-0.0105+(55.2*m*h^0.5)*0.5; eq3=(55.2*m*h^0.5-30.2*x*h^0.5)*0.5-x; Step 2: Next, to solve this system of equations, we type S=solve(eq1,eq2,eq3); The answers can be displayed by typing the following commands: S.h ans = .89435804499169139775064976230242e-2 S.m ans = .29084696757170701507538493259810e-2 S.x ans = .31266410984827736759987989710624e-2 Therefore, CH 5 0.00894, CM 5 0.00291, and CX 5 0.00313. E.2.3 Solving a System of Ordinary Differential Equations: Example 4-7

In Example 4-7, a system of two differential equations and one supplementary equation d ( X ) rate ------------ 5 ---------- ; d (W ) fa0

d ( y) ( 1 1 eps ? X ) ------------ 5 2alpha ----------------------------- ; 2y d (W )

1 1 eps ? X f 5 ------------------------y

was solved using POLYMATH. Using MATLAB to solve this problem requires two steps: (1) Create an m-file containing the equations, and (2) Use the MATLAB ode45 command to numerically integrate the equations stored in the m-file created in step 1. Part 1: Solving for X and y Step 1: To begin, choose New from the File menu and select M-file. A new text editor window will appear; the commands of the m-file are to be written there. Step 2: Write the m-file. The m-file for this example may be divided into four parts.

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Part 1: The first part contains only comments and information for the user or future users. Each comment line begins with a percent sign since MATLAB ignores the rest of a line following %. Part 2: The second part is the function command which must be the first line in the m-file that is not a comment line. This command assigns a new function to the name of the m-file. The new function is composed of any combination of existing commands and functions from MATLAB. The information and commands that define the new function must be saved in a file whose name is the same as that of the new function. Part 3: The third part of the m-file contains all other information and auxiliary equations used to solve the differential equations. It may also include the global command that allows the value for variables to be passed into or out of the m-file. Part 4: The final part of the m-file contains the differential equations to be solved. MATLAB requires that the variables of the ODEs be the elements of a single column vector. Therefore, a vector x is defined such that, for N variables, x=[var1; var2; var3; ...; varN] or x(1)=var1, x(2)=var2, x(N)=varN. In the case of Example 4-7, var1=X and var2=y. Step 3: Save the m-file under the name “ex4_7.m.” This file must be saved in a directory in MATLAB's path. The path is the list of places MATLAB looks to find the files it needs. To see the current path, to temporarily add a directory to the path, or to permanently change the path, use the pathtool command. Step 4: To see the m-file we type type ex4_7 This command tells MATLAB to type the m-file named “ex4_7.m.” Step 5: Now to solve the problem, the initial conditions need to be entered from the command window. A matrix called “ic” is defined to hold the initial conditions of x(1) and x(2), respectively, and “wspan” is used to define the range of the independent variable. ic=[0;1]; wspan = [0 60]; Step 6: The global command is also repeated from the command window. global eps kprime Step 7: Finally, we will use the ode45 built-in function. This function numerically integrates the set of differential equations saved in an m-file. [w,x]=ode45('ex4_7',wspan,ic); [ind. var., dep. var.] = ode45(‘m-file’, range of ind. variable, initial conditions of dep. variables)

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Lines beginning with % are comments and are ignored by MATLAB. The comment lines are used to explain the variables in the m-file. This line assigns the function xdot to the m-file ex4_7.m (in this case w is the independent variable and x is the dependent variable). This line tells MATLAB to allow the value for the variables "eps" and "kprime" to be passed outside the m-file. These lines provide important information necessary to solve the problem.

% % % % % % %

Part 1 "ex4_7" m-file to solve example 4-7 x(1)=X x(2)=y xdot(1)=dX/dW, xdot(2)=dy/dW

%Part 2 function xdot=ex4_7(w,x) global eps kprime %Part 3 kprime=0.0266; eps=-0.15; alpha=0.0166; rate=kprime*((1-x(1))/(1+eps*x(1)))*x(2); fa0=1; %Part 4 xdot(1, :)=rate/fa0; xdot(2, :)=-alpha*(1+eps*x(1))/(2*x(2)); These lines are the equations for the ODEs to be solved. MATLAB requires that the variables of the ODE's be assigned to one column vector. Therefore, a vector x is defined such that x(1)=X and x(2)=y. Also, xdot is the derivative of x.

For more information, type help ode45 at the command prompt. Part 2: Evaluating Variables not Contained in the Solution Matrix Step 1: We want to solve for “f,” which is not contained in the solution matrix, “x,” but is a function of part of the solution matrix. To see the size of the matrix “x,” we type size(x). This returns the following: ans = 57 2. Therefore, “x” is a 57 by 2 matrix of the form: X (1) y(1) x1( 1 ) x2( 1 ) X (2) y(2) x1( 2 ) x2( 2 ) x 5 x1( 3 ) x2( 3 ) 5 X ( 3 ) y( 3 ) Ú Ú Ú Ú x 1( 57 ) x 2( 57 ) X ( 57 ) y( 57 ) Step 2: Next we need to write the equation for “f” in terms of the “x” matrix. Using MATLAB notation, x(1:z,1:y) represents x(row 1:row n, column 1:column n) rows 1 through z and columns 1 through y of X=x(1:57,1:1) = x(1:57,1) y=x(1:57,2:2) = x(1:57,2) the matrix “x.” Similarly, x(1:57,1) represents all the rows in the first column of the “x” matrix, which in our case is X. Similarly x(1:57,2) defines the second column, y.

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The notation x( : , 1) also defines all the rows in the first column of the “x” matrix. This is usually more convenient than sizing the matrix, but at times, only part of the solution matrix may be needed. For example, you may want to plot only part of the solution. So, we can write the formula (f=(1+eps*X)/y) Multiplication and division signs in the following way: are preceded by a period to denote element-by-element operations as described in the Quick Tour. (The operation is performed on every element in the matrix.)

f=(1+eps.*x(:,1))./x(:,2);

And we can write the formula for “rate” as follows: rate=kprime.*((1-x(:,1))./(1+eps.*x(:,1))) .*x(:,2); Note: This is why we used the global command. We needed the values for “eps” and “kprime” to solve for “rate” and “f.” Step 3: A plot can then be made displaying the results of the computation. To plot “X”, “y,” and “f” as a function of “w”: plot(w,x,w,f); plot(ind. var., dep. var., ind. var., dep. var.); title('Example 4.7');xlabel('w (lb)');ylabel('X,y,f') Since the solution matrix “x” contains two sets of data (two columns) and “f” contains one column, the plot should display three lines. Example 4–7 3.5

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To plot the rate: plot(w,rate);title('Example 4.7');xlabel('w (lb)');ylabel('rate');

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Example 4–7 0.03

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w (lb) E.2.4 Solving a System of Ordinary Differential Equations: Example 4-8

To review what you learned about Example 4-7, please examine Example 4-8. type ex4_8 % % % % % %

"ex4_8" m-file to solve example 4.8 x(1)=X x(2)=y xdot(1)=dX/dz, xdot(2)=dy/dz

function xdot=ex4_8(z,x) Fa0=440; P0=2000; Ca0=0.32; R=30; phi=0.4; kprime=0.02; L=27; rhocat=2.6; m=44; Ca=Ca0*(1-x(1))*x(2)/(1+x(1)); Ac=pi*(R^2-(z-L)^2); V=pi*(z*R^2-1/3*(z-L)^3-1/3*L^3);

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G=m/Ac; ra=-kprime*Ca*rhocat*(1-phi); beta=(98.87*G+25630*G^2)*0.01; W=rhocat*(1-phi)*V; xdot(1,:)=-ra*Ac/Fa0; xdot(2,:)=-beta/P0/x(2)*(1+x(1)); Now, from the command window enter ic=[0;1]; zspan = [0 54]; [z,x]=ode45('ex4_8',zspan,ic); plot(z,x);title('Example4.8');xlabel('z(dm)') ;ylabel('X,y'); axis([0 54 0 1.2]) Example 4–8

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In this example, rate-pressure data are fit to four rate equations to evaluate the rate constants. These fits are then compared to determine the best rate equation for the data. To accomplish this, an m-file is required to compute the least-squares regression for the data. Only part (a) of Example 5-6 will be demonstrated here. The rate equation for part (a) is kP E P H ra 5 ---------------------------------------------1 1 K A P EA 1 K E P E

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Step 1: Write the m-file. The structure of this m-file is very similar to the m-file for Example 4-7. Please refer to Example 4-7 for comparison and further explanation. The function "f" is assigned to the m-file "rate_a" f must have an initial value. Therefore, it is given a value of 0 before the “for” loop to initiate the sum at zero.

% "rate_a" % m-file to perform least-squares regression % x(1)=k; x(2)=Ke; x(3)=Ka function f=rate_a(x) global ra pe pea ph2 n f=0; for i=1:n f=f+(ra(i)-(x(1)*pe(i)*ph2(i))/(1+x(3)*pea(i)+x(2)*pe(i)))^2; end This "for" loop calculates the square of the difference between the actual rate and the proposed rate equation. The result of the loop is the sum of the squares we are trying to minimize and is saved in the variable "f." This equation will be different for each rate law. n

f 5 ^ ( ra 2 ra )2 actual calculated 1

Step 2: From the command window, the global command is repeated. In this case it allows the values for variables to be passed into the m-file. global ra pe pea ph2 n Step 3: The data are entered. ra=[1.04 3.13 5.21 3.82 4.19 2.391 3.867 2.199 0.75]; pe=[1 1 1 3 5 0.5 0.5 0.5 0.5]; pea=[1 1 1 1 1 1 0.5 3 5]; ph2=[1 3 5 3 3 3 5 3 1]; Also, the value for “n” is assigned. Since there are nine data points, n=9; Step 4: To perform the least-squares regression for the data, the fmins command is used to find the values of the constants that minimize the value of “f.” Type help fmins for more information. xo=[1 1 1]; x=fmins('rate_a',xo) dep. variable = fmins(‘m-file’,[matrix of initial guesses])

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The solution: x =3.3479

2.2111

0.0428

Therefore, k 5 3.35; KE 5 2.21; and KA 5 0.043. Step 5: To see how close the solution fits the data, look at the sum of the squares to see the final value of “f” for the solution in step 4. Assign to the variable “residual” the final value of “f,” residual=rate_a(x) residual =0.0296 The sum of the squares (s 2) is 0.0296.

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H

H.1 AIChE National Student Chapter Competition This exercise was given as an open-ended problem to 160 students (working in groups of 4) in the University of Michigan Chemical Reaction Engineering class Winter Term 1997. However, the students were to only design the experiment and not build the experiment as is the case with the National Competition. Each group of students presented their design on poster boards to an outside panel of judges. The same judging criteria applied for the posters as for the National Competition. MEMORANDUM TO: Student Chapters RE: National Student Chapter Competition For the past several years we have all seen the esprit de corps, excitement, and learning that has been generated among undergraduates from engineering disciplines engaging in national competitions. The civil engineers have the “concrete canoe race,” the mechanical engineers the “egg-drop competition,” and there in the interdisciplinary “solar car race.” Many students, faculty, and practicing engineers would like to give chemical engineers a similar experience, one that would educate others about our profession and receive similar publicity (e.g., newspapers, perhaps even TV coverage). It has been suggested that the latter would be more probable if the competition involved topical issues: environment, energy reduction, world-wide food production, and the like. In any case, safety should be a primary concern (e.g., no explosive or toxic chemicals). 1

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To celebrate the newest division of the American Institute of Chemical Engineers, the Chemical Reaction Engineering Division, the first competition will be concerned with chemical reaction engineering. THE 1998 COMPETITION

Design, build, and operate an apparatus for an undergraduate laboratory experiment that demonstrates chemical reaction engineering principles and that is novel or perhaps strives to do the improbable (e.g., won't a concrete canoe sink?). The experiment should be bench scale and of the type currently found in undergraduate laboratory courses. It is also possible that the experiment could be used for a lecture demonstration. The experiment should cost less than $500 in purchased parts to build. The first year’s, competition could include experiments that would either produce a product (e.g., yogurt or something that would find use in the feeding of starving nations) or demonstrate how an environmental problem might be solved (e.g., wetlands to degrade toxic chemicals). The winners (perhaps second place also) of the regional competition will be invited to bring their experiments to the annual AIChE meeting, where the national winners will be selected. General Mills has agreed to sponsor the competition and the following prizes will be awarded to the student chapters: 1st prize 2nd prize 3rd prize

$2000 $1000 $500

In addition, a description of the winning experiment will be published in Chemical Engineering Education. The first regional competition will be held at the 1998 Student Chapter Regional Conferences and the finals at the 1998 annual meeting in Miami Beach, Florida. The rules and judging criteria are attached. Rules 1. “Design, build, and operate an apparatus for an undergraduate laboratory experiment that demonstrate, the principles of reaction engineering principles and that is novel or perhaps strives to do the improbable (e.g, won't a concrete canoe sink?).” The experiment should be bench scale and of the type currently found in most undergraduate laboratory courses. It is also acceptable that the experiment be of the type that would be used for a lecture demonstration. 2. The experiments should encompass “simplicity/ease of communication to nontechnical people.” The process should be one easily understood by people outside the profession. Either the object of the process, such as the manufacture of yogurt, the importance of the

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project, such as feeding many people from increasingly scarce resources, or the process itself should be easily communicated to people without a background in chemical engineering. Media coverage (newspapers/television/radio) is one way to show success. The competition will be conducted on the honor system. The faculty and graduate students cant act only as sounding boards to student queries. The faculty cannot be idea generators for the project. The student chapter advisor or department chair must write a cover letter stating to the best of his or her knowledge that students have abided by the rules. Students who work on the project must also sign a statement stating that they have abided by the rules. The competition is to be a team competition with at least 20% of the team being composed of members from each of the junior and senior chemical engineering classes. The minimum number of participants is 5 and the maximum is 15 per university. Associated measuring equipment (e.g., pH meter) must be of the type that is readily available at most universities through department ownership or borrowing from other departments in the university. Purchased parts must cost less than $500. This price does not include a PC for data acquisition, or associated measuring or other (e.g., pumps, fittings, vessels) equipment that exists in most chemical engineering undergraduate laboratories. The experiments will be displayed at the regional meeting. A poster board should accompany the apparatus as well as a 5- to 10-page report describing how the idea for the experiment was generated, the underlying principles, the experimental procedure, and sample results. In the event that the apparatus may not be physically brought to the meeting, videotape or other means may be used to assist understand the experiment. The top one or two winners of the regional student chapters will be eligible to compete in the finals to be held at the annual meeting. Safety regarding assembling and operating the experiment must be addressed. The student chapter advisor or department chair at the host chapter at the regional conference will select a panel of three judges. The judges can be from industry, faculty, or students. The judges cannot be affiliated with any organization that has an entry. The number of winners selected to go to the finals will depend the number of regional entries. If there are six or fewer entries, one winner will be selected to advance to the national competition. If there are seven or more entries, two winners will be selected. The decision of the regional and national judges shall be final.

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Judging Criteria 1. Creativity/novelty/originality. 2. Statement of the principle to be demonstrated and clarity in demonstrating that principle. 3. Proper description of the safety issues associated with building and operating the experiment. 4. Simplicity/ease of communication/media coverage. 5. Quality of communication: Introduction: How was the idea generated? What principle does the experiment demonstrate and why is it important? Discussion: Explain the fundamentals. Procedure: Discuss safety concerns. Results: Describe what you found. 6. Opportunity for subsequent laboratory groups to study different variables or outcomes using the same apparatus. 7. Ease, desirability, and feasibility of being replicated by another student chapter. 8. Physical appearance. 9. Participation (more than 16 hours) by someone who is not a chemical engineering major) (3 points for each non-chemical engineering major) and/or participation by chemical engineering sophomores (2 points for each sophomore) (10 points maximum).

40 points 30 points 15 points 15 points 10 points

10 points 10 points 10 points

10 points 150 points

H.2 Effective Lubricant Design* Background Lubricants are often applied at the interface between rubbing surfaces to reduce friction and prevent wear by disallowing direct surface to surface contact. An automobile engine has many contacting metal parts, such as the pistons and cylinders, and the cam lobes and cam followers. Without adequate lubrication, the sliding metal parts within the engine would wear appreciably, leading to engine failure. A typical consumer may expect to drive more than 100,000 miles before experiencing severe engine problems resulting from wear within the engine. To meet this expectation, lubricant manufacturers, in close collaboration with automobile manufacturers, continue to develop improved lubricant formulations. Lubricants are formulated by blending a base oil with additives to yield a mixture with the desirable physical and chemical properties dictated by the application environment. Base oils are typically derived from petroleum and * Problem provided by General Motors Research Laboratories, Warren, Michigan.

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are complex mixtures of aliphatic and aromatic hydrocarbons. However, some lubricants are blended using synthetic base oils. Examples of synthetic base oils include esters, polyphenyl ethers, polyalphaolefins, and perfluoroalkylethers. Lubricant additives are classified according to their function, including antioxidants, viscosity index (VI) improvers (to maintain desired viscosity over a wide temperature range), antiwear additives, friction modifiers, dispersants, detergents, pour point depressants, and antifoaming agents. A fully formulated lubricant typically consists of 80–90% base oil and 10–20% additives. Lubricant development requires an understanding of the specific problems and needs associated with lubrication such as: the need for automotive lubricants with enhanced oxidation stability, antiwear properties, and physical properties under severe operating condition. A kinetics model may be used to try and predict the oxidative degradation behavior of lubricants under differing conditions. Lubricant Degradation Model: A fresh lubricant may have physical properties ideally suited to its application, but as the lubricant degrades its physical properties can change markedly. This transformation can lead to increased friction and wear at lubricated surfaces. After significant degradation takes place, sludge and varnish deposits may form on lubricated surfaces to further hinder the smooth operation of lubricated components. There is general agreement in the literature that under normal service conditions a major portion of lubricant degradation is due to oxidation of the lubricant base oil. Consequently, a great deal of lubricant research has focused on base oil degradation and the inhibition of oxidation through the use of antioxidant additives. The oxidation of a lubricant base oil follows the hydroperoxide chain mechanism for hydrocarbon oxidation. Some of the major steps of this mechanism are listed in Reactions (1)–(10). Low Temperatures, No Antioxidants   INITIATION ( 2 ) I?1 RH → R? 1 HI  k0

( 1 ) I2 → 2I? ki

kP1

  PROPAGATION ( 4 ) RO2? 1 RH → ROOH 1 R? 

( 3 ) R? 1 O2 → RO2? kP2

 kt ( 5 ) RO2? 1 RO2? → INACTIVE PRODUCTS  TERMINATION  High Temperatures, No Antioxidants  ki ( 6 ) ROOH → RO? 1 ?OH  INITIATION  3

P ( 7 ) RO? 1 RH → ROH 1 R?  PROPAGATION  kP ( 8 ) ?OH 1 RH → HOH 1 R? 

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Low and High Temperatures With Antioxidants k

A1 ( 9 ) RO2? 1 AH → ROOH 1 A? kA2 ( 10 ) A? 1 RO2? → inactive products

Also hydroperoxide decomposing antioxidants can transform hydroperoxides to stable products as shown in Reaction (11) below: k

A3 ( 11 ) ROOH 1 A? → inactive products

This antioxidant is effective provided the products are stable, and the rate is much faster than reaction (6). An example of a hydroperoxide decomposing antioxidant is phenothiazine. In the absence of an antioxidant (or after antioxidant additive depletion), significant quantities of hydroperoxides may accumulate as a result of extensive base oil oxidation. Resulting secondary oxidation reactions may occur, leading to the formation of alcohols, ketones, carboxylic acids, and esters. Extensive oxidation can also lead to the formation of high-molecular-weight material which may form deposits on lubricated surfaces. A dramatic increase in viscosity generally results from extensive oxidation of a lubricant, which may also lead to poor lubricant performance. Clearly, extensive lubricant oxidation leads to a rapid deterioration of lubricant effectiveness, and at the point of antioxidant depletion, an automobile engine lubricant should generally be considered ineffective and should be replaced with fresh lubricant. Problem Statement 1. Consider the possible attributes of currently available, general purpose engine lubricants. List them in what you consider their order of importance. (You may wish to examine sales displays, advertisements, cans/bottles of motor oil, etc. for help with this information.) 2. If you could design an ideal engine lubricant to totally dominate the marketplace, what characteristics would you give it? 3. In the future, a priority will most likely be to make automobile maintenance easier for the owner. What kind of creative ideas/inventions can you conceive of for streamlining oil change maintenance for the consumer? After you’ve generated some ideas (by brainstorming perhaps) critique them from the standpoint of practicality, cost, availability of technology, etc. 4. Degradation by oxidation is a major cause for having to replace engine lubricants. Using your knowledge of reaction kinetics, analyze the degradation of engine lubricants due to oxidation using the reactions shown in the introduction (i.e. find an expression for the rate of degradation). Consider the following four cases. In the absence of antioxidants, examine base oil (RH) degradation a) at low temperatures (258C)

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b) In c) d)

at high temperatures (1008C) the presence of antioxidants, examine base oil (RH) degradation at low temperatures (258C) at high temperatures (1008C)

Estimated Values for the Kinetic Parameters (258C) [ RH ]0 5 3 M

[ I2 ]0 5 0.10 M

k 0 5 1026 ( Eact 5 7.5)

k A1 5 103 ( Eact 5 10)

k P2 5 1.58 ( Eact 5 8.5)

k i3 5 1026 ( Eact 5 9)

k t 5 107 ( Eact 5 0)

k A2 5 103 ( Eact 5 9)

k A3 5 3.33 3 1023 ( Eact 5 10)

The units on the above rate constants are sec21 for first order and dm3 /mole ? sec for second order. The units for the activation energies are kcal/mole. You may assume for the purposes of this investigation that 10% conversion of the base oil (RH) is the point at which the lubricant will have to be replaced. From your analysis, what kind of recommendations can you make for the improvement of the engine lubricants? If you were making suggestions to the R&D division of a lubricant manufacturer, what would you have them investigate to make the most impact on the retardation of oil degradation by oxidation? What kind of experiments should they do? What are the drawbacks of this type of kinetic model for oil degradation? Do you have a better modeling suggestion? 5. Some suggestions that have been turned in to the suggestion box of Synthoil, an up and coming, newly formed lubricant manufacturer, are: a) New cars should be equipped up with a feed and bleed system for the oil. Every so often, a quart of used oil should be drained and a quart of new oil should be added to the automobile. This should lengthen the necessary time between complete oil changes and save the consumer money. b) We should design and market an inhibitor feed system for automobiles that would allow us to maintain a minimum inhibitor concentration in the engine oil, thereby protecting it from excessive oxidative breakdown. As head of the R&D division of this progressive company, it is your job to investigate the technical feasibility of these suggestions and report on them at the next Board of Directors meeting. Investigate one of these suggestions, or substitute an equally good one of you own and investigate it. Be creative! Problem Information Synthoil Cost Data for Evaluation Purposes: Complete Oil Change - $29.95 (includes 5 qts. oil, filter, and labor) Quart of Oil - $1.50 (typical engine 5 5 qt. capacity) Inhibitor (Antioxidant, AH) - $0.10/gram, Approx. MW 5 100 g

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Additional Information Foulers H. Scott. Elements of Chemical Reaction Engineering. 2nd Ed., Englewood Cliffs, N.J.: Prentice-Hall, 1992. Given as an OEP at The University of Michigan, Winter 1991.

H.3 Peach Bottom Nuclear Reactor* Background The Peach Bottom nuclear reactor, located in Georgia, has been built and is almost operational. The reactor is a boiling water nuclear reactor that produces 1100 MW of power and cost approximately 2 billion dollars to build. The effluent from the reactor contains cooling air and isotopes of Krypton and Xenon. The major constituents of the stream are Kr-83m and Xe-135 which are present in the exit gas at concentrations of 3.19 3 106 mCi/dm3 and 1.4 3 1023 mCi/dm3, respectively. The volumetric gas flow rate exiting the reactor is 2.75 m3 /hr at a temperature of 308C. The Nuclear Regulatory Commission (NRC) issued emission limits for Kr-83m and Xe-135 in the Federal Register, Vol. 56, No. 98, Part VI, dated Tuesday, May 21, 1991. The emission standards as issued are .05 mCi/dm3 for Kr-83m and 7.0 3 1025 mCi/dm3 for Xe-135.

Peach Bottom Nuclear Reactor

Effluent 2.75 m3/hr Air Kr-83m Xe-135

* Problem developed by Susan Stagg, University of Michigan, from a problem suggested by Octave Levenspiel, Oregon State University.

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Problem Statement Propose two or more solutions that will enable the Peach Bottom nuclear reactor to go on line with the 2.75 m3 /hr of exit gas meeting the NRC emission standards. Please show all calculations and include any diagrams necessary to thoroughly explain your solutions. Additional information and properties for Kr-83m and Xe-135 can be obtained from the Handbook of Chemistry and Physics. Problem Information Ci is the symbol for a curie. A curie is the unit of radioactivity equivalent to 3.70 3 1010 disintegrations per second. This unit is named after Marie Curie. Effluent data: Constituent in Effluent Kr-83m Xe-135

Concentration (mCi/dm3) 3.19 3 106 1.4 3 1023

NRC Emission Standard (mCi/dm3) 0.05 7.0 3 1025

Additional Information Fogler, H. Scott. Elements of Chemical Reaction Engineering. 2nd Ed., Englewood Cliffs, N.J.: Prentice-Hall, 1992. Lide, David R. CRC Handbook of Chemistry and Physics. Ann Arbor: CRC Press, 1993. Given as OEP at The University of Michigan, Winter 1994.

H.4 Underground Wet Oxidation* Background Several of your company’s chemical processes generate aqueous waste streams containing a large number of hazardous compounds that are presently being destroyed by incineration. The Chief Executive Officer (CEO) of your company saw the attached article in a local journal. He asked the Director of the Engineering Service Division (ESD) if the technology would be useful for treating the aqueous waste streams from your plant. After assuring the CEO that he would investigate the possibilities, the ESD Director asked the Manager of the Reaction Engineering Group to check it out. The manager, who is also your supervisor, handed the assignment to you. Problem Statement Your mission is to evaluate the technology, size a reactor system, and specify appropriate operating conditions for oxidizing the components of the aqueous waste streams. The Engineering Economics group will then compare * Problem developed by Professor Phillip E. Savage, University of Michigan.

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the costs of your artesian with incineration to assess the relative financial merits of using the new technology. A few questions to provide some initial direction in your evaluation are given below. • At what temperature and pressure should the reactor operate? • Is an underground reactor better than the conventional above-ground reactor? • What safety considerations do you need to include in your design for this high-temperature, high-pressure process involving hazardous chemicals? • Can you ethically recommend this technology to your management? Is it sufficiently proven? • How confident are you that your reactor will be able to destroy the hazardous chemicals and meet the design specifications? • Are the products of incomplete oxidation also hazardous? • What material should be used to construct the reactor? Will corrosion be a problem? • Should the reactor operate isothermally, adiabatically, or with heat transfer? Problem Information To complete your work you will need more information than the article provides. Fortunately, your company’s Technical Service Division can conduct experiments for you (for a fee, of course). To request their services, you need only send the Technical Services Manager a written memo explaining what you want them to do. They will let you know the cost and time required to do the work. Then, if you still want the work done, they will provide the results. Additional Information: Fogler, H. Scott. Elements of Chemical Reaction Engineering. 2nd Ed., Englewood Cliffs, N.J.: Prentice-Hall, 1992. Given as an OEP at The University of Michigan, Winter 1990. (Note: The following journal article is based upon one found in the December 7, 1988 edition of the New York Times.)

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UNDERGROUND WET OXIDATION OF WASTE MATERIAL Oxygen

Clean Water Wet Waste Material

Five Layer View

Reaction

When wet waste material is mixed with oxygen under high pressures, the wet oxidation process produces sterile ash and clean water. The process is ideal for degrading sewage sludge and other waste material. However, frequently the high pressure requires special equipment and large process plants. It has been suggested that gravity might provide the high pressures, as indicated by the above set-up. The reaction is carried out 5,000 ft. underground. The waste material and oxygen are transported by pipes to the bottom of the vessel. Falling waste material provides the needed excess pressure. The reaction typically occurs at approximately 550 degrees Fahrenheit and 2,000 psi. The products are drawn to the surface via a third pipe.

H.5 Hydrodesulfurization Reactor Design* Background Just as you arrive at work one morning, your supervisor, Dr. Jones, says he needs to speak with you and your design group. He seems concerned about something, so you locate your group members and hurry into his office. As the

* Problem developed by John T. Santini, Jr., University of Michigan.

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last member of your group shuffles in, you turn to your supervisor and ask, “O.K, we are all here. What did you want to talk to us about?” “Well, one of the processes recently proposed by the Process R&D group produces a by-product stream consisting of nearly pure benzothiophene. Because benzothiophene contains sulfur and an aromatic ring, we cannot vent this stream into the atmosphere. The engineers in Catalyst Development believe that we can use a hydrodesulfurization reaction to convert benzothiophene into ethylbenzene with a cobalt-molybdenum catalyst supported on alumina. If we can design a reactor to do this efficiently, we could sell the ethylbenzene as a commodity chemical. The other product, hydrogen sulfide, could be sent to the sulfur treatment facilities in Building 12.” “I know that your group’s specialty is reactor design, so I’m assigning the hydrodesulfurization reactor project to you. I would like a progress report in three weeks and a final design report four weeks after that. I’ve compiled a list of items that you should include in each of these reports. I know it has been a while since most of you designed a reactor from start to finish, so I’ve included a partial list of references that may help you. They will be especially helpful with the selection of the materials of construction. I know this assignment is open-ended and requires a lot of engineering judgment, but just remember to use your common sense and BE CREATIVE! Any questions?” “No. We’ll get started right away,” you reply as you and your group leave the office. Problem Statement As your supervisor told you, the engineers in Catalyst Development think that benzothiophene could be converted to ethylbenzene by a hydrodesulfurization reaction. Before you commit the company’s time and money to design a reactor for this reaction, you may want to attempt to verify that the production of ethylbenzene is economically feasible. In other words, are the products worth more than the reactants and energy required to make them? If you discover the answer is no, you may have saved your company thousands of dollars in design fees. Research this issue and discuss your findings in your progress report. (See the Additional Information section for some references that may help in answering these questions.) If you knew that your supervisor supported the design of the new reactor and you discovered that producing ethylbenzene from benzothiophene was not cost effective, how would you inform your supervisor of this? The progress report may consist of a maximum of five pages, excluding figures and appendices. In the progress report, be sure that your group provides support for your choice of: • reactor (i.e. PER, PER, fluidized CSTR, etc.), • adiabatic vs. isothermal reactor operation, • reactor temperatures and pressures (Hint—single phase reactions are less complicated than multiple phase reactions), • feed ratio of hydrogen to benzothiophene, • effluent conditions and compositions, and • the weight of catalyst required.

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Also include: • a qualitative discussion of the effect that operating conditions and the method of operation have on capital and operating costs, • justification for any assumptions made, and • appendices summarizing your calculations. The final design report may consist of a maximum of ten pages, excluding figures and appendices. In the final design report, you should provide support for your choice of: • materials of construction for the reactor (Hint—is the material susceptible to accelerated corrosion due to the presence of sulfur or hydrogen?) • reactor shape and dimensions, and • reactor wall thickness. Also include: • support for any changes in the initial reactor design presented in your progress report, • any environmental or safety concerns that may be relevant to your design, • diagram of the reactor, • justification for any assumptions made, and • appendices summarizing ALL calculations. Problem Information The reaction is:

+

3H2

+ H2S

S Benziothiophene (Thianapthene)

Ethylbenzene

In past research studies, the proposed reaction has been run in the vapor phase ith reactor temperatures of 2408-3008C, total pressures of 2–30 atm, and hydrogen to benzothiophene feed ratios of 4:1 to 9:1. These experiments resulted in the development of the rate law given below. (Note: You may assume that the rate law holds for conditions outside this range. Therefore, you are in no way constrained to using these operating ranges and should use them only as guides.)

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The rate law is: k B K B K H2 P B P H2 2 r9B 2 ----------------------------------------------------------------------------------------------------P H2 S   1 1 ( K H2 PH2 )0.5 1 K H2S ----------  1 ( K B PB ) P H2   where mol k B ( 2608C ) 5 6.65 3 1025 --------------------gcat ? sec

mol k B ( 300 8C ) 5 1.80 3 1024 --------------------gcat ? sec

K B ( 260 8C ) 5 19.3 atm21

K B ( 3008C ) 5 9.90 3 1022 atm21

K H2 ( 2608C ) 5 0.358 atm21

K H2 ( 3008C ) 5 1.84 3 1023 atm21

K H2S ( 2608C ) 5 211

K H2S ( 3008C ) 5 1.082

(Note: The heat of adsorption was estimated at 280 kcal/mol for all three species) Catalyst Properties: Particle Diameter: For a packed bed reactor:

For a fluidized CSTR:

0.08 cm f 5 porosity 5 0.30 a 5 pressure drop parameter 5 0.34 kg21 f 5 porosity 5 0.75 a 5 pressure drop parameter 5 0.005 kg21

Feed (pure benzothiophene before the addition of hydrogen): mol FB0 5 20 --------hr T 0 5 entering temperature 5 2608C T melt 5 melting temperature at 1 atm 5 328C T boil 5 boiling temperature at 1 atm 5 221 8C References Girgis, M.J. and Gates, B.C. “Reactivities, Reaction Networks and Kinetics in High-Pressure Catalytic Hydroprocessing.” Ind. Eng. Chem. Res., 30, 2021–2058, [1991]. Van Parijs, I.A., Hosten, L.H., Froment, G.F. “Kinetics of Hydrodesulfurization on a CoMo/g-Al2O3 Catalyst. 2. Kinetics of the Hydrogenolysis of Benzothiophene.” Ind. Eng. Chem. Prod. Res. Dev., 25, 437–443, [1986].

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Additional Information Economic Sources: Chemical Marketing Reporter. New York: Schnell Publishing Co., Inc. Oil, Paint, and Drug Reporter. New York: Schnell Publishing Co., Inc. Reactor Codes: Rules for Construction of Pressure Vessels. A.S.M.E. Boiler and Pressure Vessel Code (Section VIII). July 1, 1980. Yokell, S. “Understanding the Pressure Vessel Codes.” Chemical Engineering. May 12, 1986. pp. 75–85. Selection of Materials: Kirby, G.N. “How to Select Materials.” Chemical Engineering. Nov. 3, 1980. pp. 86–131. Kirby, G.N. “Corrosion Performance of Carbon Steel.” Chemical Engineering. March 12, 1979. pp. 72–84. Peters, M.S. and Timmerhaus, K.D. Plant Design and Economics for Chemical Engineers. 4th Ed. New York: McGraw-Hill, l99l. Schillmoller, C.M. “Solving High-Temperature Problems in Oil Refineries and Petrochemical Plants.” Materials Engineering. January 6, 1986. pp. 83–87. Thermodynamic and Physical Property Data: Lide, David R. CRC Handbook of Chemistry and Physics. Ann Arbor: CRC Press, 1993. Perry, R.H., Green, D.W., Maloney, J.O. Chemical Engineers’ Handbook, 6th ed. New York: McGraw-Hill, 1984.

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H.6 Continuous Bioprocessing* Background In some biochemical processes, it is desired to use continuous rather than batch processing for economic and efficiency reasons. An example of a continuous process is given in Figure 1.

growth media feed

supernatant (cell free)

Animal cells suspended in media

ρ ≈ ρ cells media Figure 1

Continuous Bioreactor.

The animal cells are in a suspension (they may or may not be on beads) in the reactor unit. When the growth medium is fed in, the cells begin to produce the desired product. The exit stream (supernatant) is composed of this product along with any unused growth media. It is critical that the animal cells are not removed with the supernatant, but are retained in solution. It is also important to keep the reactor well-stirred without exerting a large shear stress on the cells, since this may kill or damage them. In addition, the product flow rate from the reactor must not be fast (the space time is of the order of 0.5 to 2 days). Problem Statement Your problem is to design a reactor for an actual process of your choice which will meet the above specifications (see the journal references below for hints and examples). Consider the aspects of mixing, separation, and kinetics. Additional Information Beck, C., Stiefel, H., Stinnett, T. “Cell-Culture Bioreactors,” Chemical Engineering, February 16, 1987, pp. 121–129. Miller, R., Melick, M. “Modeling Bioreactors,” Chemical Engineering, February 16, 1987, pp. 112–120. * Problem provided by The Upjohn Company, Kalamazoo, Michigan.

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Fogler, H. Scott. Elements of Chemical Reaction Engineering. 2nd Ed., Englewood Cliffs, N.J.: Prentice-Hall, 1992. Rubinstein, M.F., Tools for Thinking and Problem-Solving, Englewood Cliffs, New Jersey, Prentice-Hall, 1986.

H.7 Methanol Synthesis* Background Kinetic models based on experimental data are being used more frequently in the chemical industry for the design of catalytic reactors, but the modeling process itself can influence the final reactor design and its ultimate performance by incorporating different interpretations of experimental design into the basic kinetic models. Model Reaction. The reaction for the synthesis of methanol is 2H2 1 CO → CH3OH This reaction is commercially significant and chemically simple, and the thermodynamic properties of the chemical species are well known. The mechanism assumed here is complex enough to make some sophistication necessary for the analysis, but it is too simple to be really true. We assumed a chemical mechanism of medium complexity, comprised of several elementary reaction steps, for the synthesis of methanol. The data were generated for the overall reaction as it would occur in a backmixed, gradientless, experimental reactor at realistic reaction conditions. The final data set is from a statistically designed, central composite set of simulated experiments, to which 5% random error was added. It comprises a total of 27 simulated results (see Table 1). Problem Statement The primary purpose of this model is to develop kinetic modeling methods and approaches. We have included the reactor simulation part primarily to afford a realistic basis for the comparison of different kinetic models. The design of the reactor to be simulated, the thermodynamic, transport, and physical properties data to be used, and the reaction conditions to be assumed are specified in Tables 2 and 3. The reactor is a commercially realistic, plant-scale, shell-and-tube reactor, suitable for the synthesis of methanol. However, its actual design, its reaction conditions, and its performance will be different from those of any existing commercial methanol process. Simulate the shell-and-tube reactor at specified conditions, using a simple, one-dimensional, plug-flow, pseudohomogeneous, nonisothermal reactor model. Further, investi-

* Problem presented by J. Berty, S. Lee, F. Szeifert, and J. Cropley, at the International Workshop on Kinetic Model Development, A.I.Ch.E. Meeting, Denver, CO, August 1983. (With Permission)

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gate the effect of different coolant temperatures. In all calculations, assume that the ideal gas law applies. With this in mind, the following tasks should be completed. 1) Develop a kinetic model for the synthesis of methanol from the set of synthetic rate data shown in Table 1. 2) Simulate a plant-scale catalytic reactor at specified reaction conditions, using your kinetic model. The design of the reactor, the reaction conditions, and necessary thermodynamic and physical property data are given in Tables 2 and 3. 3) Summarize your results in the format shown in Table 4. Then plot the results with temperature on the y-axis and distance on the x-axis. 4) Suggest a cooling water temperature to be used. Problem Information TABLE 1.

DATA

FOR

KINETIC ANALYSIS Partial Pressure (kPa)

Experiment

Rate (mol/m3 ? s)

Temp. (K)

Methanol

CO

Hydrogen

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

6.573 4.819 6.270 4.928 10.115 7.585 9.393 7.124 1.768 1.177 1.621 1.293 2.827 2.125 2.883 2.035 4.030 3.925 3.938 10.561 1.396 2.452 5.252 3.731 3.599 5.085 3.202

495 495 495 495 495 495 495 495 475 475 475 475 475 475 475 475 485 485 485 500 470 485 485 485 485 485 485

1013 1013 1013 1013 253 253 253 253 1013 1013 1013 1013 253 253 253 253 507 507 507 507 507 1520 172 507 507 507 507

4052 4052 1530 1530 4052 4052 1530 1530 4052 4052 1530 1530 4052 4052 1530 1530 2533 2533 2533 2533 2533 2533 2533 4862 1276 2533 2533

8509 5906 8509 5906 8509 5906 8509 5906 8509 5906 8509 5906 8509 5906 8509 5906 7091 7091 7091 7091 7091 7091 7091 7091 7091 9330 5369

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REACTOR, CATALYST, AND PROCESS CONDITIONS SIMULATION REACTOR CONDITIONS

TABLE 2.

FOR

Reactor Type: shell and tube Tubes: 3000, 38.1 mm i.d. 3 12 m Coolant: boiling water is on shell side; assume coolant temperature constant at 483 K Heat-transfer coefficient (overall): assume 631 W/m2 ? K Catalyst Description Shape: Approximately spherical Diameter: 2.87 mm Effective catalyst bed void fraction: 40% Diffusional resistance: may be ignored Process Conditions Feed gas: Composition: 70 mol% H2 ; 30 mol% CO Space Velocity: 10,000 standard cubic meters per hour per cubic meter of reactor volume Reactor inlet pressure: l0.13 MPa Reactor inlet temperature: 473 K Reactor coolant temperature: 483 K (constant)

TABLE 3.

PHYSICAL PROPERTY

AND

THERMODYNAMIC INFORMATION

Prandtl number of gas: 0.70 (assume constant) Heat capacity of gas: 29.31 J/g mol ? K (assume constant) Viscosity of gas: 1.6 3 1025 Ps ? s (assume constant) Heat of reaction: 297.97 kJ/mol methanol formed Thermodynamic equilibrium constant: (T in K) 3921 log10 K eq 5 ------------ 2 7.971 log10 T 1 0.002499 T T 2 ( 2.953 3 1027 ) T 2 1 10.2

TABLE 4.

( dimensionless )

RESULTS

Authors: Shell-side temperature (K)

483

Maximum tube-side temperature Location from inlet of max. Temp. Outlet temperature Outlet methanol concentration Same as fraction of equilib. value Production rate (kg/hr)

Additional Information Fogler, H. Scott. Elements of Chemical Reaction Engineering. 2nd Ed., Englewood Cliffs, N.J.: Prentice-Hall, 1992.

Use of Computational Chemistry Software Packages

J

Thermodynamic properties of molecular species that are used in reactor design problems can be readily estimated from thermodynamic data tabulated in standard reference sources such as Perry’s Handbook or the JANAF Tables. Thermochemical properties of molecular species not tabulated can usually be estimated using group contribution methods. Estimation of activation energies is, however, much more difficult due to the lack of reliable information on transition state structures, and the data required to carry out these calculations is not readily available. Recent advances in computational chemistry and the advent of powerful easy-to-use software tools have made it possible to estimate important reaction rate quantities (such as activation energy) with sufficient accuracy to permit incorporation of these new methods into the reactor design process. Computational chemistry programs are based on theories and equations from quantum mechanics, which until recently could only be solved for the simplest systems such as the hydrogen atom. With the advent of inexpensive high-speed desktop computers, the use of these programs in both engineering research and industrial practice is increasing rapidly. Molecular properties such as bond length, bond angle, net dipole moment, and electrostatic charge distribution can be calculated. Additionally, reaction energetics can be accurately determined by using quantum chemistry to estimate heats of formation of reactants, products, and also for transition state structures. Examples of commercially available computational chemistry programs include SPARTAN, developed by Wavefunction, Inc. (http://www.wavefun.com) and Cerius2, from Molecular Simulations, Inc. (http://www.msi.com). The following example utilizes SPARTAN 4.0 to estimate the activation energy for a nucleophilic substitution reaction (SN2). The calculations cited below were performed on an IBM 43–P RS–6000 UNIX workstation. 1

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J.1 Example: Estimation of Activation Energies Synthesis of chemical-grade ethanol can be carried out by a nucleophilic substitution reaction using hydroxide ion as the nucleophile and a haloethane as the substrate. For this problem we will investigate the activation energy for the reaction: C2 H5Cl 1 OH2 → C2 H5OH 1 Cl2 This non-catalytic reaction takes place in water at moderate temperatures. In order to investigate the energetics of this reaction, SPARTAN 4.0 will be used to estimate the data required to carry out the following calculations: a) the heat of formation of all reactants and products (including the ionic species) and the overall enthalpy change for the reaction: DH Rx 5 ∑ H f8 ( Products ) 2 ∑ H f8 ( Reactants ) b) the activation energies for the forward and reverse reactions based on an assumed model for the transition state species. From organic chemistry, we learn that the mechanism for SN2 (substitution, nucleophilic, bi-molecular) reactions involves back-side attack of the nucleophile at the carbon atom where a suitable leaving group (such as a halogen atom) is attached. The entering and leaving groups are presumed to be simultaneously bonded in axial positions at the carbon atom where the reaction takes place, hence requiring a particular geometry for the transition state that accommodates this arrangement. The following approximations can be used to obtain a reasonable structure for the transition state species for this and other SN2 reactions: * assume that the geometry of the carbon atom where the nucleophilic attack takes place is trigonal-bipyramidal; H

Cl

H—C—C H

H H

OH

* arrange the entering and leaving groups in axial positions at this carbon atom (e.g. 1808 apart) * bond distances for the entering and leaving groups will be constrained to be somewhat larger than is typical for normal covalent bonds signifying the simultaneous formation and cleavage of bonds for the attacking hydroxide ion and the leaving group We first set up the calculation for the transition state species using the following step-by-step procedure.

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1. Start SPARTAN, click on File/New to enter the Builder. Due to the non-standard geometry at the reaction center, the Expert Builder will have to be used (click on Expert under the Fragment Panel). 2. Click on carbon ( C ) in the periodic table and choose the normal tetrahedral geometry for the carbon atom not involved in the reaction —x— . To place the atom in the workspace, move the mouse pointer to the builder page and click once (left mouse button). 3. With carbon still selected as the atom, select trigonal-bipyramidal geometry from the bonding geometry menu o—x

and click on any

free valance (shown in yellow). 4. To complete the transition state structure, click on chlorine as the atom and select terminal geometry ( —x ) and click on an axial free valance for the carbon atom where the reaction is to take place. Now select oxygen with angular geometry, —x , and click on the other axial free valance shown on the screen. Hydrogen atoms can now be added at all other free valences, but SPARTAN assumes that any unsatisfied free valance is saturated and automatically assigns a hydrogen atom to each of the yellow bonds. The following schematic should be shown on the screen Cl O 5. Click on Minimize in the builder menu in the Fragment Panel. This performs a crude energy minimization calculation. 6. We now wish to constrain (i.e. set) the bonds at the reaction center to values that are somewhat larger than would be characteristic of C—O and C—Cl covalent bonds in normal molecules. In the builder, perform the following steps: • Click on Geometry/Constrain Distance and click on one of the two bonds involved in the reaction • The current C—O and C—Cl bond distance is shown (depending on which bond you selected) in the Constrain Distance dialog box: values of about 1.77 Å for C—Cl and 1.42 Å for C—O are typical. • Click on Constrain Distance in the dialog box, and delete the current entry with the backspace key. Enter a value that is about 50% longer than normal ( , 2.5 Å works well). Click on OK: a pink cylinder will be shown around the bond to indicate that it will be held constant for subsequent calculations. 7. Click on Minimize to perform a crude energy minimization of the constrained transition state structure, and then click on OK and save the result (File/Save—Yes/supply a name then Quit ). This will bring you out of the builder into the main SPARTAN window. The appearance of the molecule can be changed by selecting different renderings

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(ball and wire, ball and spoke, tube, etc.) in the Model pull-down menu: experiment with some models to find one that you like best. The molecule can be rotated and translated using the middle and right mouse buttons, respectively. 8. We are now ready to carry out some quantum chemistry calculations in order to first examine the feasibility of our assumed transition state structure, and then calculate the energy (heat of formation) of the transition state. To determine if we have a good approximation for the transition state for this reaction, carry out the following calculations: * Select Setup/Semi-empirical Title: SN2 Transition State * Task 5 Single-point Energy * Model 5 AM1 (a compromise between accuracy and computational speed) * Solvent 5 Water (C–T) * Charge 5 21 {Note: the charge on the transition state is equal to the summation of the formal charges of all atoms in the molecule. The formal charge on the carbon atom where the reaction is occurring is Formal Charge = # valence electrons 2 # non-bonding electrons 2 1/2 (# bonding electrons) 5 4 2 0 2 1/2 (10) 5 21}) * Multiplicity 5 1 {Note: the spin multiplicity is the number of unpaired electrons plus 1. The transition state has no unpaired electrons.} * The Constraints button in the Semi-empirical dialog window at the right should be clicked off for this calculation. * Save * Select Setup/Properties and click on Frequency, then Save. In this step we are requesting that SPARTAN calculate the vibrational spectrum of our assumed transition state species. * Select Setup/Submit to start the calculation: click on OK when the dialog box appears. Depending on the speed of your computer, the requested calculation may take several minutes (5 or more). * When the calculation is complete, click OK to remove the dialog box and select Display/Vibration. From quantum mechanics we know that the normal mode vibration of a transition state species will be an imaginary frequency. Examine the list of vibrational frequencies: there should be at least one imaginary frequency, and several (3 in this case) may be listed. To determine which imaginary frequency corresponds to the reaction that we are studying, we will use the animation utility in SPARTAN to visualize the vibrational motion. * Click on the imaginary frequency around 357. * Click on Display and examine the action shown. Change the default value from 7 frames to 25 frames to slow down the animation. Does this correspond to the reaction? No, this frequency corresponds to a “wag” of the methyl group.

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* Examine the other imaginary frequencies (e.g. 112) until you identify the one that corresponds to the reaction of interest—normally this will be the imaginary frequency with the largest value (around 562 in this case). For our assumed transition state the imaginary frequency at about 112 has the correct bonds involved, but the action shown indicates an elimination reaction rather than substitution. Animate the 562 normal mode and convince yourself that this frequency corresponds to the correct reaction coordinate. 9. We will now use SPARTAN to calculate the heat of formation of the transition state * Select Setup/Semi-empirical * Task 5 Transition Structure (a change from previous setup) * Model 5 AM1 * Solvent 5 Water (C–T) * Click on Save and Save again (i.e. twice). This procedure searches for the saddle point energy. * Select Setup/Submit to start the calculation. Click on OK to clear the dialog box. SPARTAN is now using quantum mechanics to search the potential energy surface looking for a global saddle point that defines the transition state (including solvation effects). This calculation will take several minutes. Click on OK when the dialog box appears indicating that the job has finished. 10. Retrieve the calculated heat of formation. Select Display/Properties/ Energy. A value close to 2121.835 kcal/mol should be obtained. Select Display/Vibration only one imaginary frequency should appear. Animate the frequency to be sure that it is the correct one. 11. Build (File/New) and Minimize both ethyl chloride and ethanol using either the entry or expert builders. Save the structures, and in the main window determine the heat of formation for each molecule using the following steps: * Setup/Semi-empirical * Task 5 Geometry Optimization, (a change from previous setup) * Model 5 AM1 * Solvent 5 Water (C–T) * Charge 5 0 (a change from previous setup) * Multiplicity 5 1 * Save * Setup/Submit Click on OK to clear the dialog box. This calculation will be quicker than the transition state search. Click on OK when the dialog box appears indicating that the calculation has finished, and retrieve the estimated heat of formation (Display/Properties/Energy). Values very close to the following should be obtained: H f8 for ethanol 5 269.164 kcal/mol H f8 for ethyl chloride 5 226.649 kcal/mol

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12.

13.

14.

15.

Note: Because of the solution technique and local minima, slightly different values may result. Build the Cl2 ionic species in the expert builder by selecting chlorine with terminal geometry ( —x ) and then deleting the free valence. Select Delete Atom and then click on the very top of the bond from the Cl atom and then enter a period, i.e. “.” from the keyboard. Do not minimize the chlorine atom (there is nothing to minimize!), save, and exit to the main SPARTAN window. Build the OH2 ion by selecting oxygen in the expert builder with terminal geometry ( —x ). Again, do not minimize: save, and exit to the main SPARTAN window. Calculate the heats of formation for each of the two ionic species. The following commands should be used to set up and execute the calculations: * Setup/Semi-empirical * Task 5 Single Point Energy (a change from previous setup) * Model 5 AM1 * Solvent 5 Water (C–T) * Charge 5 21 (a change from previous setup) * Multiplicity 5 1 * Save * Setup/Submit Click on OK to clear the dialog box indicating that the job has been submitted. These calculations will take only a few seconds as the geometry of the ion is not being changed to find a global energy minimum. Values close to the following should result: H f8 for OH2 5 2122.206 kcal/mol H f8 for Cl2 5 2114.669 kcal/mol

16. Use the heats of formation to calculate the heats of reaction and activation energy. ∑ H f8 ( Reactants ) 5 H f8 ( C2H5Cl ) 2 H f8 ( OH2 ) 5 2 26.65 1 2122.2 5 2148.85 kcal / mol ∑ H f8 ( Products ) 5 H f ( C2H5OH ) 1 H f8 ( Cl2 ) 5 2 69.16 1 2144.67 5 2183.83 kcal / mol For the forward reaction EA 5 H f8 ( Transition State ) 2 ∑ H f8 ( Reactants ) 5 2 121.85 2 ( 2148.85 ) 5 27 kcal / mol

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For the reverse reaction EAr 5 H f ( Transition State ) 2 ∑ H f8 ( Products ) 5 2 121.85 2 ( 2183.83 ) 5 62 kcal / mol D H Rx 8 5 ∑ H f8 ( Products ) 2 ∑ H f8 ( Products ) 5 2 183.83 2 ( 2148.85 ) 5 235 kcal / mol 17. The complete reaction energy diagram is shown below. TS TS

–121.85

E Faf Af R –148.85

P

–183.83 Reactants

Products Reaction Coordinate

[Example provided by Professor Robert M. Baldwin, Chemical Engineering Department, Colorado School of Mines in Golden, Colorado] CDPApp.J-1A

Redo Example Appendix J.1. (a) Choose different methods of calculation such as using a value of 2.0Å to constrain the C—Cl and C—O bonds. (b) Choose different methods to calculate the potential energy surface. Compare the Ab Initio to the semi-empirical method. (c) Within the semi-empirical method compare the AM1 and PM3 models.

CRE -- Kinetics Challenge 1

Kinetics Challenge 1 -- Quiz Show Concepts

Time Reference Description

Definitions of rates of reactions. Types of reactors. General mole balances for batch reactors, CSTR's and PFR's. 29 minutes ± 10 minutes Fogler: Chapter 1 This module allows students to test their knowledge about general mole balance equations, reaction rate laws, and different types of reactions and reactors. Individual students will find themselves going head-to-head against computer opponents in an interactive game with timed responses. Twenty multiple-choice questions are selected from a pool of approximately 100 possible questions, so the game will be different every time. The questions fall under four main categories: mole balance, reactions, rate laws, and reactor types; and there are five difficulty levels within each category. Each correct answer will earn the student a given number of points; the more difficult the question, the higher the point values.

The student has one minute to choose the correct answer. The module responds to the student's choice, either reinforcing the reasoning for a correct answer, or immediately clarifying a misunderstanding if an incorrect answer is entered. If no response is entered within the time limit, or if an incorrect response is entered, the points are lost, and one of the computer competitors tries to answer the question:

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The competitor who last answered a question correctly gets to pick the next category and degree of difficulty. (Note that this will not necessarily always be the student). In addition to regular questions, one question is randomly assigned as the "Double Challenge" in which the student has the option of betting some or all of his/her points. After all twenty questions have been answered, the contestants with positive scores go on to the "Final Challenge" quesion, in which they are also allowed to bet points.

Grade Base

The game score is the number of accumulated points, including gains or losses from the Double Challenge (if applicable) and the Final Challenge. For the performance scores, the student is given 3 points for every correct answer in the 100-300 point range, and 7 points for each 400-500 point question. The Final Challenge is worth 8 points. 75 points are needed to achieve mastery of the module.

Comments

Students have used this module as review material before an exam, to ensure that they have a solid grasp of the basics of reaction kinetics. Some professors have also made use of it in recitation sections, inviting student volunteers to enter responses, then discussing any conceptual misunderstandings that might be discovered.

Installation

Instructions for installing the ICMs and for using the ICMs are available.

Return to Chapter One


CRE -- Staging

Staging -- Reactor Sequencing Optimization Game Concepts


Concentration as a function of conversion CSTR vs. PFR volume-conversion relationships Effect of changing order of reactor placement on final conversion. 35 minutes ± 10 minutes Fogler: Chapter 2 This module presents the student with a set of five reactors (CSTRs and PFRs) and asks him/her to connect them in series. The goal is to maximize the product flowrate for a given reaction, while maintaining a minimum conversion of 75%. The student is provided with a graph of -FAo/rA vs X, and the reactors' volumes are specified. The student may arrange the reactors in any order, and he/she may also vary the inlet flowrate. Each arrangement of reactors may be tested using a simulator that provides instant feedback for any change in reactor order or inlet flowrate.

The student may at any time access a reference section that reviews the derivation of the design equations for PFRs and CSTRs, clarifying the change in conversion down a PFR, and the wellmixedness of the CSTR:

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CRE -- Staging

The reactor optimization simulator can also be run independently of the scenario. This allows the professor to present the student with a variety of open-ended problems to be investigated using the simulator. Grade Base

The student’s score is based on the conversion achieved, as well as the total flow rate of material produced.

Comments

This module makes use of a scenario to increase the level of interest of the student. In the scenario, the student must generate a sufficient amount of an antidote of high enough purity to help Mr. Hyde get back to his Dr. Jeckyll persona.

Installation


Return to Chapter Two

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Kinetics Challenge 2 -- Quiz Show Concepts


Arrhenius equation Stoichiometry tables Rate laws 36 minutes ± 16 minutes Fogler: Chapter 3 This module focuses on rate laws and stoichiometry, allowing the students to master the elements of the stoichiometric table:

The interactive portion of the module is similar to that in Kinetic Challenge 1. Students can choose from four categories (reactants, products, rate law, potpourri) and four levels of difficulty (200-1,000 points). Each question has four multiple-choice answers, and students have a limited amount of time to make a response:



The goal of this module is to provide the students with practice setting up stoichiometric relationships, so that they will avoid mistakes, such as expressing the reaction rate law for an irreversible reaction as if it were reversible, or using the ideal gas law for liquid-phase reactions.

Grade Base

The game score is the number of accumulated points, including gains or losses from Double Challenge and Final Challenge. For the performance scores, the student is given 3 points for every correct answer in the 200-600 point range, and 7 points for the 800-1,000 point questions. The Final Challenge question is worth 8 points.

Comments

Students report that they find this module very useful as a review before the first examination. In some instances, some of the text strings in the last few questions and the Final Challenge question will become garbled, and the computer will lock up at the end of the module. We have been unable to determine the circumstances that give rise to this effort. Some students comment that the one minute time limit does not allow them enough time to derive the required expressions. It is helpful to suggest to students that they examine the four options available and choose the correct answer by process of elimination, based on the information provided in the problem statement, rather than trying to derive the expressions.

Installation


Return to Chapter Three


CRE -- Murder Mystery

Murder Mystery: CSTR Volume Algorithm Concepts Time Reference Description

Isothermal CSTR reactor design Problem solving and analysis 32 minutes ± 10 minutes Fogler: Chapter 4 The goal of this module is to allow students to practice the CSTR design algorithm:

In the interactive portion of the module the student must solve a murder mystery, with the aid of Sir Nigel Ambercrombie, an English investigator. It seems that overnight there was a slight irregularity in conversion for the CSTR at the Nutmega Spice Company:

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It is feared that one of the employees may have been murdered by a fellow employee, and the dead body left within the reactor. By analyzing the conversion data, gathering information about the plant's personnel, and using their knowledge of CSTR design, students must determine the identity of both the murderer and the victim. Help may be obtained by questioning the suspects:

Grade Base

Successful solution of the murder mystery with a minimum of assistance.

Comments

Students have always enjoyed the murder mystery scenario in this module.

Installation




Return to Chapter Four


CRE -- Tic Tac

Tic Tac: Isothermal Reactor Design - Ergun, Arrhenius, and Van't Hoff Equations Concepts Time Reference Description

Interaction of the Ergun, Arrhenius, and Van't Hoff equations and other considerations in isothermal reactor design. 33 minutes ± 9 minutes Fogler: Chapter 4 This module allows the student to examine nine reactor design problems, and investigate the effect of varying reactor parameters on process performance. The problems are organized as in a tic-tactoe board. The reactors covered by these problems include PFRs, CSTRs, packed bed reactors and semi-batch reactors:

The student must master the concepts in enough squares to successfully win the tic-tac-toe game (Three adjacent squares horizontally, vertically, or diagonally). Each problem allows the student the opportunity to examine the effect of a specified operational parameter on reactor performance, using simulators:

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CRE -- Tic Tac

After performing the “experiments” the student proceeds to answer the questions that examine the effects observed. These effects can be explained through the Ergun, Arrhenius, and Van'’t Hoff equations In many cases, competing effects are highlighted. The square is "won" by answering two out of the three questions correctly. Grade Base

Grade based on mastery of concepts within each square, and successful completion of the tic-tactoe game.

Comments

Some students have found the questions in this module to be slightly above their current level of understanding. They have mentioned, however, that the process was helpful in exploring these concepts.

Installation


Return to Chapter Four

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CRE -- Ecology

Ecology -- A Wetlands Problem Concepts Time Reference Description

Collection and analysis of rate data Ecological engineering concepts 49 minutes ± 21 minutes Fogler: Chapter 5 The student, as an employee of a company trying to meet environmental regulatory agency standards, must sample concentration data for a toxic material found in a wetlands channel between a chemical plant upstream and a protected waterway to analyze the rate of decay of the toxic material.

The wetlands are modeled as a PFR. The student must first develop the necessary reactor design equation for a PFR, then start collecting data. This concentration data (which includes experimental error) is then analyzed in various ways (polynomial fit of the data followed by differentiation of the resulting equation, difference equations, etc.) to determine the rate law, as well as the rate constants and reaction order. Students must determine which points are to be excluded from the analysis (if any) and which points may be resampled:

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CRE -- Ecology

The student then analyzes this information and submits a memo with the requested parameters:

This information is reviewed by the boss, who evaluates the parameter values and makes recommendations. Grade Base

Based on correct determination of rate law parameters.

Comments

This module is useful in exposing the student to experimental error and the dangers of using curve-fitting tools without discretion. It also exposes the student to “real world” applications of reaction kinetics. To introduce an element of levity, the student performs the analysis in a "Mr. Potatohead" persona. Students have reported enjoying this.

Installation


Return to Chapter Five

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CRE -- Heat Effects 1

Heat Effects 1 -- Basketball Challenge Concepts Time Reference Description

Effect of parameter variations on operation of a nonisothermal CSTR 36 minutes ± 14 minutes Fogler: Chapter 8 This module allows students to investigate the effect of parameter variations on the operation of a nonisothermal CSTR. An extensive review section derives the energy balance for the CSTR, and also describes the terms in the mole balance that are temperature dependent:

A simulator is also included in the review section. This allows the student to vary parameters and to observe the effects on the conversion-temperature relationships as described by both the mole balance and the energy balance. The parameters that may be varied include: feed flowrate, feed temperature, the reversibility/irreversibility of the reaction, heat of reaction, heat exchanger area, and heat transfer fluid temperature. The operating conditions can be determined from the intersection(s) of the mole balance and the energy balance:

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The module can also be run in an interactive mode, in which the scenario takes the student to a basketball tournament. They have the choice of two-point and three-point questions:

A simulator is available to help the student answer the three-point questions. Grade Base

A grade is only assigned in the interactive mode. The student is given a “shooting percentage” for the t-o point and thr-p point questions, as well as an overall shooting percentage. A shooting percentage greater than 85% demonstrates mastery of the module.

Comments

This module requires a lot of memory to run.

Installation




Return to Chapter Eight



Heat Effects 2 -- Effect of Parameter Variation on a PFR Concepts Time Reference Description

Effect of parameter variations on operation of a nonisothermal plug flow reactor 61 minutes ± 36 minutes Fogler: Chapter 8 This simulation allows the student to explore the effects of various parameters on the performance of a non-isothermal plug flow reactor. The student may choose from eight simulations that span all combinations of exothermic/endothermic reactions and reversible/irreversible reactions, as well as a simulation that takes pressure drop into account. The variable parameters include the heat transfer coefficient, the inlet reactant flowrate, the diluent flowrate, the inlet temperature, and the ambient temperature:

The results of the simulator may be analyzed in the form of plots of concentration, conversion, or temperature as a function of reactor volume. The module may also be run in an interactive mode, in which the student must achieve specific goals (e.g. achieve a given conversion without exceeding a given temperature within the reactor), in order to get to the center of the reactor complex. The review section includes a derivation of the energy balance equation for a PFR:



Grade Base

In the interactive mode, mastery is based on the correct solution of two consecutive problems (e.g.: arriving at the center of the reactor complex).

Comments

We have used the simulator portion of this module as a tool in a group problem solving exercise. Students had to vary various parameters and explain their observations, then use the newly gained insight to optimize a system. A sample assignment, for System 2 in the “individual problems” menu, follows. Since the assignment was to be completed within a one hour class period, explicit instructions and suggestions for parameter values were given.

Sample Assignment

You are to investigate how some important reactor parameters affect the conversion and the temperature profiles down a tubular reactor. You will be told which parameter to vary, then asked to explain the results you observe. In each case, in addition to a general statement ("increase UA"), you will be given a set of optional reactor conditions to use, in the order in which they appear in the left-hand side of the simulation screen: (UA, Ta, Fio, Fao, To). You may use these conditions if you wish, or pick your own for your investigation. GETTING STARTED Choose "5. individual problems" from the main menu , then choose problem 2. “Endothermic irreversible." Once the F-key bar at the bottom shows up, you may want to hit "F2" for a short description of each of the components of the simulator. Things to keep in mind, once you are running the simulator: • To change the step size in varying the parameters: 1(smallest step size) 10 (largest one) • To delete a run so its curve doesn't show up on the grap, -select it and hit backspace. • Not sure what keys to hit -Press I for Information. EFFECTS OF HEAT EXCHANGE To analyze the effect of the heat exchanger on the reaction, compare the conversion and the temperature profiles with and without heat exchange: Set the y-axis to temperature - Choose temperature with the arrow keys.



Select your first run - Hit until the yellow selector box is around the UA box Perform a run with UA equal to 0 (e.g., UA=0, Ta=300, Fio=10, Fao=10, To=300) How does the temperature change with volume down the reactor with no heat exchange for an endothermic irreversible reaction? Select a second run - Use the arrows to select the blue run. Perform a new run with a higher UA (e.g. UA=250, Ta=300, Fio=10, Fao=10, To=300) How does the temperature change down the reactor with heat exchange for an endothermic irreversible reaction? Set the y-axis to conversion - Press "A" for axes How do the conversion profiles for the cases with and without heat exchange compare? EFFECTS OF FLOW PARAMETERS Perform a new run with no inerts. (e.g. UA=250, Ta=300, Fio=0, Fao=10, To=300) Now perform a run with a higher reactant rate. (e.g. UA=250, Ta=300, Fio=0, Fao=20, To=300) How does the presence of inerts affect the results from the previous question? Perform a run with inerts (e.g. UA=250, Ta=300, Fio=10, Fao=20, To=300). Compare the temperature profiles for the above three cases. APPLICATION Given your new-found intuition, try to get the highest conversion given the limitation that the reactor temperature (at ALL positions within the reactor), must be between 250-300 K. An easily achieved value is 0.50. The highest conversion found so far is 0.711. Turn in the conditions you used (UA,Ta,Fio, Fao,To) as well as the conversion obtained, and a few sentences explaining what you learned. SUMMARY Write a paragraph (1/2 to 1 page) describing the effects of heat exchange on the reaction and the effects of changes in the reactant flowrate and the inert flowrate on conversion and temperature profiles for the tubular reactor. Include sketches illustrating the trends and the equations necessary to predict the results. Based on these results can you predict what would happen in an exothermic, irreversible reaction? How about reversible reactions? Installation


Return to Chapter Eight


CRE -- Heterogeneous Catalysis

Hetcat: Heterogeneous Catalysis Concepts Time Reference Description

Derivation of catalytic rate equations based on experimental data Selection of reaction mechanism and rate-limiting step that support the rate equation 33 minutes ± 13 minutes Fogler: Chapter 10 The review section of this module reviews the essential elements of heterogeneous catalysis:

The student must derive the rate equation for a given reactive system by analyzing the rate data obtained in a differential reactor. The student must choose which experiments to run, that is, the entering pressures of each species and total flow rate. In order to obtain the dependence of the rate equation on the pressure of a given species, the student must select which of the points are to be included in a plot of reaction rate vs. species partial pressure. Given the requested plot, the student must determine the form of the dependence of the rate law on the pressure of the given species:

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Once all dependencies have been established, the student must decide which rate law parameters can be determined, through judicious plotting of the experimental data. The review section also outlines the derivation of the governing equations of heterogeneous catalysis:

Grade Base

Based merely on completion of the module, i.e. derivation of the reaction rate expression.

Comments

Students reported that this module was very helpful to them in preparing to do the homework problems from the textbook. This module requires a large amount of memory to run.

Installation




Return to Chapter Ten


CRE -- Installing the ICMs

Installing the ICMs A Note about the ICMs The interactive computer modules (ICMs) were written for a DOS environment; therefore, you will need to have a PC to use them. Our apologies to Mac owners, who either don't have PC-cards installed, or whose PC-emulator software is slow or unstable. The ICMs may be converted to a more platform-independent format in the future, but for now, PCs or PC-emulators are the only options. So, if you have a PC, you can run the ICMs directly from the CD-ROM. See the page on using the ICMs for more information.

Installation Procedure For your convenience, the interactive computer modules have also been included on the CD-ROM in two compressed formats: (1) as a zip file (newmods.zip) and (2) as a self-extracting, executable file (newmods.exe), which might require you to have a copy of WinZip (a free shareware utility for Windows 95) to run it. This is so you can run the ICMs without needing a CD-ROM for every computer. They are located in the ICMs\Files directory (and also in the Html\DOS_Mod\Files directory) on your CD-ROM.

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Using newmods.exe Do the following: 1. Using Windows Explorer, select your CD-ROM drive and locate the newmods.exe program file. As mentioned above, it should be in the ICMs\Files directory. 2. Double-click on the newmods.exe file to run this self-extracting installation program.



3. This will bring up a pop-up window that looks like this:

You'll notice that the "Unzip To Folder:" textfield already has C:\Newmods listed as the location to which it will unzip files. (You may change this "unzip to" location if you do not want install the ICMs to this default directory.) 4. Click the Unzip button to have the program automatically install the Interactive Computer Modules for you. 5. After the files unzip themselves, click the Close button to exit the self-extractor program. 6. Select the Newmods directory that was created on your C: drive.



At this point, you should see a number of files and directories in there. The important files are the ones with names like mystery.bat and kinchal1.bat. These are known as batch files. You can doubleclick one of these batch files to run an ICM. NOTE: Each ICM has its own batch file and a directory that houses the "guts" of the ICM. Do not open the directory. Use the batch file to run the corresponding ICM. 7. See the page on using the ICMs for more information.

Using newmods.zip Do the following: 1. Using Windows Explorer, create a directory on your hard drive called Newmods. For example, select your C: drive. Then under File, select New -- Folder. Type in Newmods as the folder name. 2. Select your CD-ROM drive and open the ICMs folder. 3. Open the Files folder and select the newmods.zip file. 4. Unzip the newmods.zip file into the Newmods folder that you created in step one. There are two programs that I recommend. The first is PKUNZIP and the second is



WinZip. Both should be available as shareware from one of the major shareware sites, such as SHAREWARE.COM or DOWNLOAD.COM. WinZip is available from its own site at www.winzip.com. 5. Select the Newmods directory that you created earlier. At this point, you should see a number of files and directories in there. The important files are the ones with names like mystery.bat and kinchal1.bat. These are known as batch files. You can double-click one of these batch files to run an ICM. NOTE: Each ICM has its own batch file and a directory that houses the "guts" of the ICM. Do not open the directory. Use the batch file to run the corresponding ICM. 6. See the page on using the ICMs for more information.


CRE -- Using the ICMs

Using the ICMs You can run the interactive computer modules directly from the CD-ROM (PCs only) by accessing the ICMs directory (or the Html\DOS_Mod\Newmods directory) on your CD-ROM, and then following the instructions, below. Otherwise, if decide to run the ICMs from your hard-drive, but you have not yet installed them on your computer, then read the information on installing the ICMs.

Once you're set, do the following: 1. Make sure you are in the directory that contains the ICMs. For example, if you're running the interactive computer modules from your CD-ROM, then you can use Windows Explorer to go to the ICMs directory (or the Html\DOS_Mod\Newmods directory).

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If you installed the interactive computer modules on your computer's hard drive, in the directions on the installation page you were asked to create a directory on your hard drive called Newmods. 2. Run the batch file (file extension .bat) for the ICM you wish to use. In the Windows Explorer program of Windows 95, double-click on the batch file that you want to run, or at the DOS Prompt, type the batch file name and then hit ENTER (a.k.a. RETURN). For example, say you want to run the Mystery Theater module. In Windows 95, you would double-click on the mystery.bat file to run that ICM.



In DOS, you would type mystery.bat and then hit ENTER to run the ICM. 3. Follow the directions within each module. (The right arrow key is used to advance to new material, unless otherwise noted.) 4. Each ICM will instruct you on how to exit that particular module.

List of Interactive Computer Modules Module Name

Ecology Heat Effects 1 Heat Effects 2 Heterogeneous Catalysis Kinetics Challenge 1 Kinetics Challenge 2 Murder Mystery Staging Tic Tac file:///H:/html/dos_mod/using.htm[05/12/2011 16:55:03]

File Name ecology.bat heatfx1.bat heatfx2.bat hetcat.bat kinchal1.bat kinchal2.bat mystery.bat staging.bat tictac.bat



Polymath Short Course

Polymath Short Course by Dieter Andrew Schweiss

The following mini-course on Polymath will provide you with the bare minimum that you need to know, in order to use Polymath to look at the Living Example Problems. For more complete instructions on using Polymath, please be sure to review the official Polymath Manual.

1. Starting Polymath A. Windows 95/NT Procedure 1. Using Windows Explorer, switch to the Polymat4 directory on your CD-ROM.

2. Double-click on the Polymath batch file (Polymath.bat) to run Polymath.

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NOTE: Polymath will run in its own, full-screen DOS Window. To be able to follow the step-by-step instructions of the Polymath Short Course and use Polymath at the same time, you will have to make frequent use of the ALT-ENTER keys to alternate between the Polymath Short Course and Polymath itself.

B. DOS Procedure 1. If you are already in DOS (or in a DOS window), then move to the next step. Otherwise, open a DOS window by double-clicking on the DOS Prompt. In Windows 95/NT, the DOS Prompt should be available as an icon on your desktop or in your Start menu. 2. Switch to Polymat4 directory. This will either be the one on your CD-ROM, or the one Polymath is installed to on your hard drive (if you installed Polymath). 3. Type polymath at the command prompt and hit ENTER to run Polymath. NOTE: Polymath will run in its own, full-screen DOS Window. To be able to follow the step-by-step instructions of the Polymath Short Course and use Polymath at the same time, you will have to make frequent use of the ALT-ENTER keys to alternate between the Polymath Short Course and Polymath itself. Page 1 | Page 2 | Page 3 | Page 4 | Page 5



Polymath Main | Using Polymath | Installing Polymath | Short Course


Installing Polymath

Installing Polymath The following files have been included on your CD-ROM in the Polymat4\Files\poly402 directory:

These files will allow you to install Polymath 4.0.2 on your hard drive. Please read the Polymath 4.0.2 Manual (the mpoly402.pdf file) to learn how to use Polymath. The library.zip and the library.exe files are used to install the Living Example Problems, discussed below. In addition to the installation instructions that follow, please read the readme.txt file, since it also contains information about installing Polymath. The installation files for Polymath 4.1 are also on the CD (in the Polymat4\Files\poly41 directory).

Why Two Versions of Polymath? This is the second printing of the text and the CD. Polymath 4.1 has better printing features than Polymath 4.0.2, but it was not available for the first printing of the CD. We could have replaced Polymath 4.0.2 with Polymath 4.1, but we

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Installing Polymath

decided to make both versions available instead.

1. Installing Polymath This is the Polymath 4.0.2 installation procedure. The Polymath 4.1 installation procedure is similar, yet much simpler, so refer to the Polymath 4.1 readme.txt file, if you plan to install Polymath 4.1 instead. Since Polymath is a DOS-based program, directions have been written so that you may use either Windows 95/NT or a DOS Prompt to install Polymath. The installation instructions for Windows 95/NT are given below. The DOS installation instructions are located on a separate page.

A. Windows 95/NT Procedure 1. Using Windows Explorer, select your CD-ROM drive and switch to the directory containing the Polymath files. As mentioned above, they should be in the Polymat4\Files\poly402 directory. 2. Double-click on the install.exe file to run the Polymath installation program.

A DOS window will open, since Polymath is a DOS-based program. 3. The installation program will ask you the following questions: a. Enter drive and directory for POLYMATH [C:\POLYMAT4] :


Installing Polymath

Polymath is suggesting that you install to a default directory called Polymat4. Please select this directory by simply hitting the RETURN key. NOTE: It is important for you to install Polymath in the default directory, because you will also have to install the library files for the Living Example Problems in this directory. b. Is this a network installation? [N] Type N when you're asked this question, even if you're running your computer on a network. Trust me. c. Please select the type of output device you prefer [1]: Polymath wants to know if you have a (1) printer or (2) plotter. The default response is that you have a printer, which is what we assume you will select, too. d. Please select the type of printer you have: There should be a long list of printers here. If you don't see your printer in the list, we suggest you select the one that comes closest. NOTE: If you have trouble printing with with Polymath (especially if you print on a network), then contact the authors of Polymath directly. Their contact information should be in the Polymath Manual. e. Please select the mode you want output in [1]: You'll be offered a list of printing options to choose from; the options are similiar, yet different for each printer. We suggest you choose a portrait (or full page) option for printing code or tables, and a landscape option for printing graphs. f. Please select the port your printer is connected to [1]: You'll be offered a list of printing ports to choose from; most people have printers on LPT1. 4. Polymath should now be able to run on your machine. Close the DOS window and switch to the Polymat4 directory on your hard drive. 5. Double-click on the Polymath batch file (Polymath.bat) to run Polymath.


Installing Polymath

IMPORTANT: You will need to follow the instructions for installing the library files before you can access the Polymath code for the Living Example Problems. 6. Now read the instructions on using Polymath or take the Polymath Short Course.

2. Installing the Library Files The Polymath code for the Living Example Problems is contained in a file called library.zip, which is located in the Polymat4\Files directory on your CD-ROM. For your convenience, a second file called library.exe is in that same directory. This program will automatically install the Polymath code for the Living Example Problems in the default Polymath directory on your hard drive (C:\Polymat4).

A. Windows 95/NT Procedure 1. Using Windows Explorer, select your CD-ROM drive and locate the library.exe program file. As mentioned above, it should be in the Polymat4\Files\poly402 directory. 2. Double-click on the library.exe file to run this self-extracting installation program.


Installing Polymath

3. This will bring up a pop-up window that looks like this:

You'll notice that the "Unzip To Folder:" textfield already has C:\polymat4 listed as the location to which it will unzip files. (You may change this "unzip to" location if you didn't install Polymath to its default directory.) 4. Click the Unzip button to have the program automatically install the library files for you. 5. After the library files unzip themselves, click the Close button to exit the self-extractor program. You may now access the Living Example Problems using Polymath. Polymath Main | Using Polymath | Installing Polymath | Short Course


Installing Polymath


Using Polymath

Using Polymath Polymath Manual Polymath is easy to use -- once you know how. Before you use Polymath for the first time, take a look at the Polymath Manual for instructions on how to run and generally use Polymath: Polymath 4.0.2 Manual in PDF format. Polymath 4.1 Manual in PDF format. NOTE: You will need the Adobe Acrobat Reader Plug-in to read this file. (See the CD-ROM Introduction for more information.) If you just can't wait to dig into Polymath, then take a look at the Polymath Short Course. It won't teach you everything, but it will get you started.

Why Two Versions of Polymath? This is the second printing of the text and the CD. Polymath 4.1 has better printing features than Polymath 4.0.2, but it was not available for the first printing of the CD. We could have replaced Polymath 4.0.2 with Polymath 4.1, but we decided to make both versions available instead.

Accessing the Polymath Examples Before you begin, refer to the Polymath Manual for specific instructions on using Polymath. More-detailed instructions for using Polymath and opening the Living Example Problems are available in the Polymath Short Course. Some of the examples are for use with Polymath's ordinary differential equation (ODE) solver, while others can be used with the non-linear regression solver. If you follow the link to a specific example problem to see the code for it, the code will give you an idea as to which tool to use to open the example problem. You can tell the difference between the two types of code, in the way the data is presented: For instance, Example 5-6 holds Polymath code for a non-linear regression. The code has data in a tabular format, since you enter regression data in a table. By comparison, Example 4-10 holds Polymath code for the ODE solver. The code presents differential equations and constants as a list of equations, since that is how the information would be entered in the ODE solver.

file:///H:/html/toolbox/polymath/using.htm[05/12/2011 16:55:06]

Using Polymath

To actually load a Living Example Problem: 1. Select the correct option from Polymath's main menu. (You'll probably select either option 1 or option 4 most of the time.) 2. On the next screen, hit F9 for file options. 3. At the TASK MENU, hit F9 again for LIBRARY OPTIONS. 4. Select the example you want to load from the list. (NOTE: The files in the list will depend on the solver you're using.) 5. When the example loads, you will be instructed to hit any key to look at the example problem, so do that. 6. Now you can play around with the example as you see fit. See the Polymath Manual for further instructions. Confused? Please take a look at the Polymath Short Course for a more-detailed explanation of the information given here, as well as a step-by-step example of how to navigate through Polymath. Polymath Main | Using Polymath | Installing Polymath | Short Course

file:///H:/html/toolbox/polymath/using.htm[05/12/2011 16:55:06]

POLYMATH VERSION 4.02

USER-FRIENDLY NUMERICAL ANALYSIS PROGRAMS - SIMULTANEOUS DIFFERENTIAL EQUATIONS - SIMULTANEOUS ALGEBRAIC EQUATIONS - SIMULTANEOUS LINEAR EQUATIONS - POLYNOMIAL, MULTIPLE LINEAR AND NONLINEAR REGRESSION for IBM and Compatible Personal Computers THIS MANUAL AND SOFTWARE ARE TO ACCOMPANY Elements of Chemical Reaction Engineering - 3rd Edition by H. Scott Fogler and published by Prentice Hall

POLYMATH Internet Site http://www.polymath-software.com Users are encouraged to obtain the latest general information on POLYMATH and its use from the above Internet site. This will include updates on this version and availability of future versions.

Copyright  1998 by M. Shacham and M. B. Cutlip This manual may be reproduced for educational purposes by licensed users. IBM and PC-DOS are trademark of International Business Machines MS-DOS and Windows are trademarks of Microsoft Corporation

i -2 PREFACE

POLYMATH 4.0 PC

POLYMATH LICENSE AGREEMENT The authors of POLYMATH agree to license the POLYMATH materials contained within this disk and the accompanying manual.pdf file to the owner of the Prentice Hall textbook Elements of Chemical Reaction Engineering, Third Edition, by H. Scott Fogler. This license is for noncommercial and educational uses exclusively. Only one copy of this software is to be in use on only one computer or computer terminal at any one time. One copy of the manual may be reproduced in hard copy only for noncommercial educational use of the textbook owner. This individual-use license is for POLYMATH Version 4.02 and applies to the owner of the textbook. Permission to otherwise copy, distribute, modify or otherwise create derivative works of this software is prohibited. Internet distribution is not allowed under any circumstances. This software is provided AS IS, WITHOUT REPRESENTATION AS TO ITS FITNESS FOR AND PURPOSE, AND WITHOUT WARRANTY OF ANY KIND, EITHEREXPRESS OR IMPLIED, including with limitation the implied warranties of merchantability and fitness for a particular purpose. The authors of POLYMATH shall not be liable for any damages, including special, indirect, incidental, or consequential damages, with respect to any claim arising out of or in connection with the use of the software even if users have not been or are hereafter advised of the possibility of such damages. HARDWARE REQUIREMENTS POLYMATH runs on the IBM Personal Computer and most compatibles. A floating-point processor is required. Most graphics boards are automatically supported. The minimum desirable application memory is 450 Kb plus extended memory for large applications. POLYMATH works with PC and MS DOS 3.0 and above. It can also execute as a DOS application under Windows 3.1, Windows 95, and Windows NT. It is important to give POLYMATH as much of the basic 640 Kb memory as possible and up to 1024 Kb of extended memory during installation. A variety of drivers are provided for most printers and plotters. Additionally some standard graphics file formats are supported. POLYMATH 4.0 PC

PREFACE i-3

TABLE OF CONTENTS - POLYMATH PAGE INTRODUCTION POLYMATH OVERVIEW..................................................................... 1 MANUAL OVERVIEW.......................................................................... 1 INTRODUCTION................................................................................ 1 GETTING STARTED.......................................................................... 1 HELP.................................................................................................... 1 UTILITIES........................................................................................... 1 APPENDIX........................................................................................... 1 DISPLAY PRESENTATION................................................................. 1 KEYBOARD INFORMATION............................................................. 1 ENTERING VARIABLE NAMES........................................................ 1 ENTERING NUMBERS......................................................................... 1 MATHEMATICAL SYMBOLS............................................................ 1 MATHEMATICAL FUNCTIONS........................................................ 1 LOGICAL EXPRESSIONS.................................................................... 1 POLYMATH MESSAGES...................................................................... 1 HARD COPY........................................................................................... 1 GRAPHICS.............................................................................................. 1 -

1 2 2 2 2 2 2 3 3 4 4 5 5 6 6 6 6

GETTING STARTED HARDWARE REQUIREMENTS......................................................... POLYMATH SOFTWARE .................................................................... INSTALLATION TO INDIVIDUAL COMPUTERS & NETWORKS. FIRST TIME EXECUTION OF POLYMATH..................................... EXITING POLYMATH PROGRAM.....................................................

2 2 2 2 2

-

1 1 1 2 3

HELP MAIN HELP MENU............................................................................... ACCESSING HELP BEFORE PROBLEM ENTRY.......................... ACCESSING HELP DURING PROBLEM ENTRY........................... CALCULATOR HELP........................................................................... UNIT CONVERSION HELP.................................................................

3 3 3 3 3

-

1 2 2 3 3

UTILITIES CALCULATOR....................................................................................... 4 CALCULATOR EXPONENTIATION............................................... 4 AVAILABLE FUNCTIONS................................................................ 4 ASSIGNMENT FUNCTIONS............................................................. 4 CALCULATOR EXAMPLES............................................................. 4 UNIT CONVERSION............................................................................ 4 PREFIXES FOR UNITS...................................................................... 4 UNIT CONVERSION EXAMPLE...................................................... 4 i-4 PREFACE POLYMATH 4.0

- 1 - 1 - 1 - 3 - 3 - 5 - 5 - 6 PC

PROBLEM STORAGE........................................................................... FILE OPERATIONS............................................................................ LIBRARY OPERATIONS..................................................................... LIBRARY STORAGE.......................................................................... LIBRARY RETRIEVAL...................................................................... PROBLEM OUTPUT AS PRINTED GRAPHICS............................. SAMPLE SCREEN PLOT.................................................................... OPTIONAL SCREEN PLOT................................................................ PRESENTATION PLOT....................................................................... PROBLEM OUTPUT TO SCREEN AND AS PRINTED TABLES.. PROBLEM OUTPUT AS DOS FILES.................................................. PROBLEM OUTPUT AS GRAPHICS FILES.....................................

4 - 7 4 - 7 4 - 8 4 - 8 4 - 8 4 - 9 4 - 9 4-10 4-10 4- 10 4-11 4-11

DIFFERENTIAL EQUATIONS SOLVER QUICK TOUR.......................................................................................... DIFFERENTIAL EQUATION SOLVER............................................. STARTING POLYMATH.................................................................... SOLVING A SYSTEM OF DIFFERENTIAL EQUATIONS.............. ENTERING THE EQUATIONS.......................................................... ALTERING THE EQUATIONS.......................................................... ENTERING THE BOUNDARY CONDITIONS................................. SOLVING THE PROBLEM................................................................. PLOTTING THE RESULTS................................................................. EXITING OR RESTARTING POLYMATH....................................... INTEGRATION ALGORITHMS.......................................................... TROUBLE SHOOTING......................................................................... SPECIFIC ERROR MESSAGES.......................................................... NONSPECIFIC ERROR MESSAGES.................................................

5 5 5 5 5 5 5 5 5 5 5 5 5 5

-

1 1 1 2 3 4 4 5 6 7 8 9 9 9

ALGEBRAIC EQUATIONS SOLVER QUICK TOUR.......................................................................................... NONLINEAR ALGEBRAIC EQUATION SOLVER......................... STARTING POLYMATH.................................................................... SOLVING ONE NONLINEAR EQUATION...................................... SOLVING A SYSTEM OF NONLINEAR EQUATIONS.................. EXITING OR RESTARTING POLYMATH....................................... SELECTION OF INITIAL ESTIMATES FOR THE UNKNOWNS METHOD OF SOLUTION.................................................................... TROUBLE SHOOTING.........................................................................

6 6 6 6 6 6 6 6 6

-

1 1 1 3 5 7 7 7 8

POLYMATH 4.0 PC

PREFACE i-5

LINEAR EQUATIONS SOLVER QUICK TOUR.......................................................................................... LINEAR EQUATION SOLVER......................................................... STARTING POLYMATH.................................................................... SOLVING FIVE SIMULTANEOUS EQUATIONS............................ EXITING OR RESTARTING POLYMATH.......................................

7 7 7 7 7

REGRESSION QUICK TOUR......................................................................................... REGRESSION PROGRAM.................................................................. STARTING POLYMATH.................................................................... QUICK TOUR PROBLEM 1................................................................ RECALLING SAMPLE PROBLEM 3.... ............................................ FITTING A POLYNOMIAL................................................................ FITTING A CUBIC SPLINE................................................................ EVALUATION OF AN INTEGRAL WITH THE CUBIC SPLINE... MULTIPLE LINEAR REGRESSION.................................................. RECALLING SAMPLE PROBLEM 4................................................. SOLVING SAMPLE PROBLEM 4...................................................... TRANSFORMATION OF VARIABLES............................................. NONLINEAR REGRESSION.............................................................. EXITING OR RESTARTING POLYMATH....................................... SOLUTION METHODS.........................................................................

8 8 8 8 8 8 8 8 8 8 8 8 8 8 8

APPENDIX INSTALLATION AND EXECUTION INSTRUCTIONS................... DOS INSTALLATION......................................................................... DOS EXECUTION............................................................................... WINDOWS 3.1 INSTALLATION....................................................... WINDOWS 3.1 EXECUTION.............................................................. WINDOWS 95 INSTALLATION........................................................ WINDOWS 95 EXECUTION............................................................... INSTALLATION QUESTIONS............................................................ "OUT OF ENVIRONMENT SPACE" MESSAGE............................. CHANGING PRINTER SELECTION.................................................. PRINTING FOR ADVANCED USERS................................................ PRINTING TO STANDARD GRAPHICS FILES...............................

9 9 9 9 9 9 9 9 9 9 9 9

i-6 PREFACE

-

1 1 1 3 4

- 1 - 1 - 1 - 1 - 2 - 4 - 5 - 6 - 7 - 7 - 8 -11 -13 -15 -15

-

1 1 1 2 2 3 3 3 7 8 8 9

POLYMATH 4.0 PC

INTRODUCTION POLYMATH OVERVIEW POLYMATH is an effective yet easy to use computational system which has been specifically created for professional or educational use. The various programs in the POLYMATH series allow the user to apply effective numerical analysis techniques during interactive problem solving on a personal computer. Whether you are student, engineer, mathematician, scientist, or anyone with a need to solve problems, you will appreciate the ease in which POLYMATH allows you to obtain solutions. Chances are very good that you will seldom need to refer to this manual beyond an initial reading because POLYMATH is so easy to use. With POLYMATH, you are able to focus your attention on the problem at hand rather than spending your valuable time in learning how to use or reuse the program. You are encouraged to become familiar with the mathematical concepts being utilized in POLYMATH. These are discussed in most textbooks concerned with numerical analysis. The available programs in POLYMATH include: - SIMULTANEOUS DIFFERENTIAL EQUATION SOLVER - SIMULTANEOUS ALGEBRAIC EQUATION SOLVER - SIMULTANEOUS LINEAR EQUATION SOLVER - POLYNOMIAL, MULTIPLE LINEAR AND NONLINEAR REGRESSION Whether you are a novice computer user or one with considerable computer experience, you will be able to make full use of the programs in POLYMATH which allow numerical problems to be solved conveniently and interactively. If you have limited computer experience, it will be helpful for you to read through this manual and try many of the QUICK TOUR problems. If you have considerable personal computer experience, you may only need to read the chapters at the back of this manual on the individual programs and try some of the QUICK TOUR problems. This manual will be a convenient reference guide when using POLYMATH. POLYMATH 4.0 PC

INTRODUCTION 1-1

MANUAL OVERVIEW This manual first provides general information on features which are common to all of the POLYMATH programs. Particular details of individual programs are then presented. Major chapter topics are outlined below: INTRODUCTION The introduction gives an overview of the POLYMATH computational system and gives general instructions for procedures to follow when using individual POLYMATH programs. GETTING STARTED This chapter prepares you for executing POLYMATH the first time, with information about turning on the computer, loading POLYMATH, and making choices from the various menu and option screens. HELP On-line access to a general help section is discussed. UTILITIES This chapter discusses features that all programs have available. These include a scientific calculator and a convenient conversion for units and dimensions. This chapter discusses saving individual problems, data and/or result files on a floppy or hard disk. It also describes the use of the problem library for storing, retrieving and modifying problems on a disk. Options for the printing and plotting of results are explained. The remaining chapters of the manual present a QUICK TOUR of each individual POLYMATH program and are organized according to the following subsections: 1. PROGRAM OVERVIEW This subsection gives general details of the particular program. 2. QUICK TOUR You can use this subsection to see how easy it is to enter and solve a problem with a particular POLYMATH program. APPENDIX Detailed installation instructions and additional output options are presented for advanced users. 1-2 INTRODUCTION

POLYMATH 4.0 PC

DISPLAY PRESENTATION Throughout this manual, a full screen is indicated by a total enclosure:

An upper part of screen is contained within a partial enclosure:

A lower part of screen is shown by a partial enclosure:

An intermediate part of a screen is given between vertical lines:

The option box is given by:

KEYBOARD INFORMATION When using POLYMATH, it is not necessary to remember a complex series of keystrokes to respond to the menus, options, or prompts. The commands available to you are clearly labeled for easy use on each display. Normally the keystrokes which are available are given on the display as indicated on the PROBLEM OPTIONS display shown below.

POLYMATH 4.0 PC

INTRODUCTION 1-3

USING THE KEY symbol is used to indicate In this manual as in POLYMATH, the the carriage return key which is also called the enter key. Usually when you are responding to a menu option, the enter key is not required. However, when data or mathematical functions are being entered, the enter key is used to indicate that the entry is complete. SHIFTED KEYPRESSES Some options require that several keys be pressed at the same time. This is indicated in POLYMATH and in this manual by a dash between the keys such as a ⇑ F8 which means to press and hold the ⇑ or "shift" key, then press the F8 function key and finally release both keys. THE EDITING KEYS Use the left and right arrow keys to bring the cursor to the desired position, while editing an expression. Use the Del key to delete the character (Back Space) key to delete the first character to above the cursor or the the left of the cursor. Typed in characters will be added to the existing expression in the first position left to the cursor. BACKING UP KEYS Press either the F8 or the Esc key to have POLYMATH back up one program step. ENTERING VARIABLE NAMES A variable may be called by any alphanumeric combination of characters, and the variable name MUST start with a lower or upper case letter. Blanks, punctuation marks and mathematical operators are not allowed in variable names. Note that POLYMATH distinguishes between lower and upper case letters, so the variables 'MyVar2' and 'myVar2' are not the same. ENTERING NUMBERS All numbers should be entered with the upper row on the key board or with the numerical keypad activated. Remember that zero is a number from the top row and not the letter key from the second row. The number 1 is from the top row while letter l is from the third row.

1-4 INTRODUCTION

POLYMATH 4.0 PC

The results of the internal calculations made by POLYMATH have at least a precision of eight digits of significance. Results are presented with at least four significant digits such as xxx.x or x.xxx . All mathematical operations are performed as floating point calculations, so it is not necessary to enter decimal points for real numbers. MATHEMATICAL SYMBOLS You can use familiar notation when indicating standard mathematical operations. Operator +

Meaning addition subtraction multiplication division power of 10

Symbol + * / x.x10a

Entry + x * -: / x.xea x.xEa (x.x is numerical with a decimal and a is an integer) exponentiation rs r**s or r^s MATHEMATICAL FUNCTIONS Useful functions will be recognized by POLYMATH when entered as part of an expression. The arguments must be enclosed in parentheses: ln (base e) abs (absolute value) sin arcsin sinh log (base 10) int (integer part) cos arccos cosh exp frac (fractional part) tan arctan tanh POLYMATH 4.0 PC

exp2(2^x) round (rounds value) sec arcsec arcsinh exp10 (10^x) sign (+1/0/-1) csc arccsc arccosh sqrt (square root) cbrt (cube root) cot arccot arctanh INTRODUCTION 1-5

LOGICAL EXPRESSIONS An "if" function is available during equation entry with the following syntax: if (condition) then (expression) else (expression). The parentheses are required, but spaces are optional. The condition may include the following operators: > greater than < less than >= greater than or equal <= less than or equal == equals <> does not equal | or & and The expressions may be any formula, including another "if" statement. For example: a=if(x>0) then(log(x)) else(0) b=if (TmaxT) then (maxT) else (T)) POLYMATH MESSAGES There are many POLYMATH messages which may provide assistance during problem solving. These messages will tell you what is incorrect and how to correct it. All user inputs, equations and data, are checked for format and syntax upon entry, and feedback is immediate. Correct input is required before proceeding to a problem solution. HARD COPY If there is a printer connected to the computer, hard copy of the problem statements, tabular and graphical results etc. can be made by pressing F3 key wherever this option is indicated on the screen. Complete screen copies can also be made if the DOS graphics command is used before entering POLYMATH and your printer accepts this graphics mode. Problem statements and results can be also printed by saving them on a file and printing this file after leaving POLYMATH. GRAPHICS POLYMATH gives convenient displays during problem entry, modification and solution. Your computer will always operate in a graphics mode while you are executing POLYMATH. 1-6 INTRODUCTION

POLYMATH 4.0 PC

GETTING STARTED This chapter provides information on the hardware requirements and discusses the installation of POLYMATH. HARDWARE REQUIREMENTS POLYMATH runs on IBM compatible personal computers from the original IBM PC XT to the latest models with Pentium chips. Most graphics boards are automatically supported. The minimum application memory requirement is 560Kb, and a new feature uses extended memory when it is available. POLYMATH works with PC and MS DOS 3.0 and above. POLYMATH can be a DOS application under Windows 3.1 and Windows 95. Calculations are very fast since the floating point processor is used when it is available. POLYMATH SOFTWARE The complete set of POLYMATH application programs with a general selection menu is available on a single 3-1/2 inch 1.44 Mb floppy in compressed form. It is recommended that a backup disk be made before attempting to install POLYMATH onto a hard disk. Installation is available via an install program which is executed from any drive. INSTALLATION TO INDIVIDUAL COMPUTERS AND NETWORKS POLYMATH executes best when the software is installed on a hard disk or a network. There is a utility on the POLYMATH distribution disk which is called "install". Detailed installation instructions are found in the Appendix of this manual. Experienced users need to simply put the disk in the floppy drive, typically A or B. Type "install" at the prompt of your floppy drive, and press return. Follow the instructions on the screen to install POLYMATH on the particular drive and directory that you desire. Note that the default drive is "C:" and the default directory is called "POLYMAT4". Network installation will require responses to additional questions during installation. Latest detailed information can be found on the INSTALL. TXT and README.TXT files found on the installation disk. The PRINTSET program which can be used to change the printer specifications without completely reinstalling POLYMATH is discussed in the Appendix. POLYMATH 4.0 PC

GETTING STARTED 2-1

FIRST TIME EXECUTION OF POLYMATH The execution of POLYMATH is started by first having your current directory set to the subdirectory of the hard disk where POLYMATH version 4.0 is stored. This is assumed to be C:\POLYMAT4 C:\POLYMAT4 > Execution is started by entering "polymath" at the cursor C:\POLYMAT4 > polymath and then press the Return ( ) key. The Program Selection Menu should then appear:

The desired POLYMATH program is then selected by entering the appropriate letter. You will then taken to the Main Program Menu of that particular program. Individual programs are discussed in later chapters of this manual. GETTING STARTED 2-2

POLYMATH 4.0 PC

EXITING POLYMATH PROGRAM The best way to exit POLYMATH is to follow the instructions on the program display. However, a Shift F10 keypress (⇑F10) will stop the execution of POLYMATH at any point in a program and will return the user to the Polymath Program Selection Menu. THIS ACTION WILL DELETE THE EXISTING PROBLEM so be sure to store your problem as an individual file or in the library before exiting the program in this manner. A query is made to determine if the user really wants to end the program in this manner while losing the current problem. This ⇑ F10 keypress is one of the few POLYMATH commands which is not always indicated in the various Display Menus. It is worth remembering.

POLYMATH 4.0 PC

GETTING STARTED 2-3

HELP MAIN HELP MENU Each individual POLYMATH program has a detailed help section which is available from many points in the program by pressing F6 when indicated. The Help Menu allows the selection of the topic area for specific help as shown below for the Differential Equation Solver:

For example, pressing "a" gives a discussion on entering the equations.

3-1 HELP

POLYMATH 4.0 PC

Once the current topic is completed, the Help Options Menu provides for additional options as shown below:

The ⇑ - F8 option to return to the program will take you to the display where you originally requested HELP ACCESSING HELP BEFORE PROBLEM ENTRY The Main Help Menu is reached during the startup of your POLYMATH program from the Main Menu as shown below and from the Problem Options Menu by pressing F6.

ACCESSING HELP DURING PROBLEM ENTRY When you are entering a problem, the HELP MENU is available from the Problem Options Menu. This will allow you to obtain the necessary help and return to the same point where the HELP MENU was originally requested. As an example, this access point is shown in the Problem Options Menu shown on the next page. POLYMATH 4.0 PC

HELP 3-2

CALCULATOR HELP A detailed discussion of the POLYMATH Calculator is given in Chapter 4 of this manual. The Calculator can be accessed from by pressing F4 from any point in a POLYMATH program.

An F6 keypress brings up the same page help which provides a brief instruction inside the Calculator window. UNIT CONVERSION HELP The Unit Conversion Utility is discussed in Chapter 4 of this manual. There is no on-line help for this utility.

3-3 HELP

POLYMATH 4.0 PC

UTILITIES CALCULATOR A sophisticated calculator is always available for use in a POLYMATH program. This calculator is accessed by pressing the F4 key . At this time a window will be open in the option box area which will give you access to the calculator.

CALCULATOR: Enter an expression and press to evaluate it. Press to leave or press F6 for information

The POLYMATH calculator allows you to enter an expression to be evaluated. After the expression is complete, press to have it calculated. You may then press again to clear the expression, or you may edit your expression using the standard editing functions. When you wish to leave the calculator, just press the F8 or Esc key. CALCULATOR EXPONENTIATION Numbers may also be entered in scientific notation. The calculator will recognize E or e as being equivalent to the notation *10**. Either ** or ^ indicates general exponentiation. For example, the following three expressions are equivalent for a particular value of A: 4.71*10**A = 4.71eA = 4.71*10^A AVAILABLE FUNCTIONS A number of standard functions are available for use in the calculator. The underlined portion of the following functions is all that is required provided that all arguments are enclosed in parentheses. The arguments may themselves be expressions or other functions. The nesting of function is allowed. ln ( ) or alog ( ) = natural logarithm to the base e log ( ) or alog10 ( ) = logarithm to the base 10 exp ( ) = exponential (ex) exp2 ( ) = exponential of 2 (2x) exp10 ( ) = exponential of 10 (10x) sqrt ( ) = square root abs ( ) = absolute value POLYMATH 4.0 PC

UTILITIES 4-1

int ( ) or ip ( ) = integer part frac ( ) = or fp ( ) = fractional part round ( ) = rounded value sign ( ) = returns + 1 or 0 or -1 N! = factorial of integer part of number N (this only operates on a number) sin ( ) = trigonometric sine with argument in radians cos ( ) = trigonometric cosine with argument in radians tan ( ) = trigonometric tangent with argument in radians sec ( ) = trigonometric secant with argument in radians csc ( ) = trigonometric cosecant with argument in radians cot ( ) = trigonometric cotangent with argument in radians arcsin ( ) = trigonometric inverse sine with result in radians, alternates arsin ( ) and asin ( ) arccos ( ) = trigonometric inverse cosine with result in radians, alternates arcos ( ) and acos ( ) arctan ( ) = trigonometric inverse tangent with result in radians, alternate atan ( ) arcsec ( ) = trigonometric inverse secant with result in radians arccsc ( ) = trigonometric inverse cosecant with result in radians arccot ( ) = trigonometric inverse cotangent with result in radians sinh ( ) = hyperbolic sine cosh ( ) = hyperbolic cosine tanh ( ) = hyperbolic tangent arcsinh ( ) = inverse hyperbolic sine arccosh ( ) = inverse hyperbolic cosine arctanh ( ) = inverse hyperbolic tangent You should note that the functions require that their arguments be enclosed in parentheses, but that the arguments do not have to be simple numbers. You may have a complicated expression as the argument for a function, and you may even nest the functions, using one function (or an expression including one or more functions) as the argument for another.

4-2 UTILITIES

POLYMATH 4.0 PC

ASSIGNMENT FUNCTIONS The assignment function is a way of storing your results. You may specify a variable name in which to store the results of a computation by first typing in the variable name, then an equals sign, then the expression you wish to store. For example, if you wish to store the value of sin (4/3) 2 in variable 'a', you would enter: a = sin (4/3)**2 Variable names must start with a letter, and can contain letters and digits. There is no limit on the length of the variable names, or on the number of variables you can use. You can then use the variable 'a' in other calculations. These variables are stored only as long as you remain in the current POLYMATH program. Please note that all stored values are lost when the particular program is exited. Calculator information is not retained during problem storage. CALCULATOR EXAMPLES Example 1. In this example the vapor pressure of water at temperatures of 50, 60 and 70 o C has to be calculated using the equation: log10 P = 8.10765 – 1750.29 235.0 + T For T = 50 the following expression should be typed into the calculator: 10^(8.10675 - 1750.29 / (235+50)) CALCULATOR: Enter an expression and press to evaluate it. Press to leave or press F6 for information.

Pressing brings up the desired answer which is 92.3382371 o mm Hg at 50 C. To change the temperature use the left arrow to bring the cursor just right to the zero of the number 50, use the (BkSp or delete) key to erase this number and type in the new temperature value.

POLYMATH 4.0 PC

UTILITIES 4-3

Example 2. In this example the pressure of carbon dioxide at temperature of T = 400 K and molal volume of V = 0.8 liter is calculated using the following equations: P = RT – a V – b V2

Where

2 2 a = 27 R Tc Pc 64

b = RTc 8 Pc

R = 0.08206, Tc = 304.2 and Pc = 72.9. One way to carry out this calculation is to store the numerical values to store in the named variables. First you can type in Pc = 72.9 and press this value as shown below.

Pc=72.9 CALCULATOR: Enter an expression and press to evaluate it. =72.9.

After that you can type in Tc = 304.2 and R = 0.08206. To calculate b, you must type in the complete expression as follows: b=R*Tc/(8*Pc) CALCULATOR: Enter an expression and press to evaluate it. =0.0428029012

The value of a is calculated in the same manner yielding a value of 3.60609951. Finally P can be calculated as shown:

P=R*400/(0.8-b)-a/(0.8*0.8) CALCULATOR: Enter an expression and press to evaluate it. =37.7148168

4-4 UTILITIES

POLYMATH 4.0 PC

UNIT CONVERSION A utility for unit conversion is always available for use within a POLYMATH program. Unit Conversion is accessed by pressing F5 wherever you desire. This will result in the following window in the option box area: Type the letter of the physical quantity for conversion. a) Energy b) Force c) Length d) Mass e) Power f) Pressure g) Volume h) Temperature F8 or ESC to exit

The above listing indicates the various classes of Unit Conversion which are available in POLYMATH. A listing of the various units in each class is given below: ENERGY UNITS: joule, erg, cal, Btu, hp hr, ft lbf, (liter)(atm), kwh FORCE UNITS: newton, dyne, kg, lb, poundal LENGTH UNITS: meter, inch, foot, mile, angstrom, micron, yard MASS UNITS: kilogram, pound, ton (metric) POWER UNITS: watts, hp (metric), hp (British), cal/sec, Btu/sec, ft lbf /sec PRESSURE UNITS: pascal, atm, bar, mm Hg (torr), in Hg, psi [lbf /sq in] VOLUME UNITS: cu. meter, liter, cu. feet, Imperial gal, gal (U.S.), barrel

(oil), cu. centimeter

TEMPERATURE UNITS: Celsius, Fahrenheit, Kelvin, Rankine

PREFIXES FOR UNITS It is convenient to also specify prefixes for any units involved in a Unit Conversion. This feature provides the following prefixes: deci 10 -1 hecto 10 2

centi 10 -2 kilo 10 3

POLYMATH 4.0 PC

milli 10 -3 mega 10 6

micro 10 -6 giga 10 9

deka 10

UTILITIES 4-5

UNIT CONVERSION EXAMPLE Suppose you want to convert 100 BTU's to kilo-calories. First you should access the Unit Conversion Utility by pressing F5. This will bring up the following options Type the letter of the physical quantity for conversion. a) Energy b) Force c) Length d) Mass e) Power f) Pressure g) Volume h) Temperature F8 or ESC to exit

Press "a" to specify an Energy conversion: From units: Type in a letter (F9 to set a prefix first) a. joule b. erg c. cal d. Btu f. ft lbf g. (liter)(atm) h. kwh

e. hp hr

Type a "d" to specify Btu: From units : Btu To units: a. joule b. erg c. cal f. ft lb, g. (liter)(atm) h. kwm

(F9 for a prefix) d. Btu e. hp hr

Use F9 to indicate a Prefix: Press the number of the needed prefix or F9 for none. 1) deci 10 -1 2) centi 10 -2 3) milli 10 -3 4) micro 10 -6 2 3 5) deka 10 6) hecto 10 7) kilo 10 8) mega 106 9) giga 109

Please indicate kilo by pressing the number 7. From units: Btu a. joule b. erg f. ft lbf g. (liter)(atm)

To units: kilo c. cal d. Btu h. kwh

e. hp hr

Complete the units by pressing "c" for calories. Indicate the numerical value to be 100 and press enter: From units: Btu Numerical value: 100 100.00 Btu = 25.216 kilo-cal

4-6 UTILITIES

To units: kilo-cal

POLYMATH 4.0 PC

PROBLEM STORAGE POLYMATH programs can be stored for future use as either DOS files or in a "Library" of problems. The Library has the advantage that the titles are displayed for only the problems for the particular POLYMATH program which is in use. Both the DOS files and the Library can be placed in any desired subdirectory or floppy disk. In both cases, only the problem and not the solution is stored. The storage options are available from the Task Menu which is available from POLYMATH programs by pressing either F9 from the Main Menu or ⇑ F8 from the Problem Options Menu.

FILE OPERATIONS A current problem can be saved to a DOS file by selecting "S" from the Task Menu. The desired directory and DOS file name can be specified from the window given below:

Note that the path to the desired directory can also be entered along with the file name as in "A:\MYFILE.POL" which would place the DOS file on the Drive A. A previously stored problem in a DOS file can be loaded into POLYMATH from the Task Menu by selecting "L". A window similar to the one above will allow you to load the problem from any subdirectory or floppy disk. An F6 keypress gives the contents of the current directory for help in identifying the file name for the desired problem. POLYMATH 4.0 PC

UTILITIES 4-7

LIBRARY OPERATIONS The Library is highly recommended for storing problems as the titles of the problems are retained and displayed which is a considerable convenience. Also, only the problems for the particular POLYMATH program in current use are displayed. The Library is accessed from the Task Menu by pressing F9 as shown below:

If there is no current Library on the desired subdirectory or floppy disk, then a Library is created. LIBRARY STORAGE The Library Options menu allows the current POLYMATH problem to be stored by simply entering "S". The title as currently defined in the active problem will be displayed. The user must choose a file name for this particular problem; however, it will then be displayed along with the Problem title as shown above. LIBRARY RETRIEVAL The Library Options window allows the current POLYMATH problem to be recalled by first using the cursor keys to direct the arrow to the problem of interest and then entering "L". A window will confirm the library retrieval as shown below:

Problems may be deleted from the Library by using the arrow to identify the problem, and then selecting "D" from the Library Options menu. Users are prompted to verify problem deletion. 4-8 UTILITIES

POLYMATH 4.0 PC

PROBLEM OUTPUT AS PRINTED GRAPHICS One of the most useful features of POLYMATH is the ability to create graphical plots of the results of the numerical calculations. The command to print graphical output is F3. The first step in printing graphical output is to display the desired output variables. The POLYMATH programs allow the user to make plots of up to four variables versus another variable. An example which will be used to demonstrate plotting is the Quick Tour Problem 1 from the next chapter. Here the POLYMATH Differential Equation Solver has produced a numerical solution to three simultaneous ordinary differential equations. The calculations are summarized on a Partial Results display which has the following Display Options Menu:

SIMPLE SCREEN PLOT The selection of "g" from the Display Options Menu allows the user to select desired variables for plotting. A plot of variables A, B, and C versus the independent variable t can be obtain by entering "A,B,C" at the cursor and pressing the Return key ( ).

The resulting graph is automatically scaled and presented on the screen.

POLYMATH 4.0 PC

UTILITIES 4-9

OPTIONAL SCREEN PLOT The selection of "g" from the Display Options Menu with the entry of "B/A" results in B plotted versus A. This demonstrates that dependent variables can be plotted against each other. PRESENTATION PLOT A simple plot can be printed directly or it can be modified before printing by using options from the Graph Option Menu shown below:

This menu allow the user to modify the plot before printing as desired to obtain a final presentation graphic with specified scaling and labels. PROBLEM OUTPUT TO SCREEN AND AS PRINTED TABLES The Display Options Menu also allows the user to select tabular output from the Partial Results Display by pressing "t":

This is shown below for the same entry of "A, B, C" for the Quick Tour Problem 1 from the next chapter on differential equations.

The output shown above gives variable values for the integration interval at selected intervals. The maximum number of points is determined by the numerical integration algorithm. Output variable values for a smaller number of points are determined by interpolation. A Screen Table can be printed by using F3. 4-10 UTILITIES

POLYMATH 4.0 PC

PROBLEM OUTPUT AS DOS FILES The output from many of the POLYMATH programs can also be stored for future use as DOS files for use in taking results to spreadsheets and more sophisticated graphics programs. Typically this is done after the output has been sent to the screen. This is again accomplished with option "d" from the Display Options Menu.

This option take the user to a display where the name and location of the DOS data file is entered:

Please note that the user can change the drive and the directory to an desired location. One the location is indicated and the file name is entered, the desired variable names must be provided and the number of data points to be saved. The file shown below was created as shown for the request of "A,B,C" and 10 data points for Quick Tour Problem 1 from the next chapter: t 0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3

A B C 1 0 0 0.74081822 0.19200658 0.067175195 0.54881164 0.24761742 0.20357094 0.40656966 0.24127077 0.35215957 0.30119421 0.21047626 0.48832953 0.22313016 0.17334309 0.60352675 0.16529889 0.13797517 0.69672595 0.12245643 0.10746085 0.77008272 0.090717953 0.082488206 0.82679384 0.067205513 0.062688932 0.87010556 0.049787068 0.047308316 0.90290462

Note that the separate columns of data in this DOS output file are separated by tabs which is suitable format for input to various spreadsheet or graphics programs. POLYMATH 4.0 PC

UTILITIES 4-11

PROBLEM OUTPUT AS GRAPHICS FILES Advanced POLYMATH users can direct their printed output to many standard graphics files for direct use in word processing, desktop publishing, etc. This is accomplished though special polymath.bat files which direct any printer output to specified graphics files with user-defined file names. Details of this option are found in Chapter 9 of this manual. A typical example would be to create the following output in a TIFF file for inclusion in a written reporting using word processing or desktop publishing. The problem is again the for Quick Tour Problem 1 from the next chapter. Note that the figure below has utilized the title and axis definition options.

The modified bat file which lead to the above TIFF image was ECHO OFF SET BGIPATH=C:\POLYMAT4\BGI SET PM_WORKPATH=C:\POLYMAT4 SET PM_PRINTER=_TIF,0,FILE:E:\PMOUT+++.TIF POLYMENU CLS and the first tiff image was stored as PMOUT001.TIF in C:\POLYMAT4 directory. See Chapter 9 for more information. 4-12 UTILITIES

POLYMATH 4.0 PC

DIFFERENTIAL EQUATION SOLVER QUICK TOUR This section is intended to give you a very quick indication of the operation of the POLYMATH Differential Equation Solver Program. DIFFERENTIAL EQUATION SOLVER The program allows the numerical integration of up to 31 simultaneous nonlinear ordinary differential equations and explicit algebraic expressions. All equations are checked for syntax upon entry. Equations are easily modified. Undefined variables are identified. The integration method and stepsize are automatically selected; however, a stiff algorithm may be specified if desired. Graphical output of problem variables is easily obtained with automatic scaling. STARTING POLYMATH To begin, please have POLYMATH loaded into your computer as detailed in Chapter 2. Here it is assumed that your computer is set to the hard disk subdirectory or floppy drive containing the POLYMATH package. At the prompt (assumed C:\POLYMAT4 here), you should enter "polymath" C:\POLYMATH4 > polymath then press the Return ( ) key. The Program Selection Menu should then appear, and you should enter "1" to select the Simultaneous Differential Equation Solver. This should bring up the Main Program Menu:

POLYMATH 4.0 PC

DIFFERENTIAL EQUATIONS 5 -1

Now that POLYMATH is loaded, please press F6 and then the letter "a" to get information on "Entering the equations". The first page of the Help Section should be on your screen as shown here:

Please press F8 to return from the Help to the program, and then press the Enter key ( )to continue this Quick Tour example. SOLVING A SYSTEM OF DIFFERENTIAL EQUATIONS Let us now enter and solve a system of three simultaneous differential equations: d(A) / d(t) = - kA (A) d(B) / d(t) = kA (A) - kB (B) d(C) / d(t) = kB(B) In these equations, the parameter kA is to be constant at a value of 1.0 and the parameter kB is to be constant at the value of 2.0. The initial condition for dependent variable "A" is to be 1.0 when the initial value of the independent variable "t" is zero. The initial conditions for dependent variables "B" and "C" are both zero. The solution for the three differential equations is desired for the independent variable "t" between zero and 3.0. Thus this problem will be entered by using the three differential equations as given above along with two expressions for the values for kA and kB given by: kA = 1.0; kB = 2.0 5-2 DIFFERENTIAL EQUATIONS

POLYMATH 4.0 PC

ENTERING THE EQUATIONS The equations are entered into POLYMATH by first pressing the "a" option from the Problem Options Menu. The following display gives the first equation as it should be entered at the arrow. (Use the Backspace key, to correct entry errors after using arrow keys to position cursor.) Press the ) to indicate that the equation is to be entered. Don't worry Return key ( if you have entered an incorrect equation, as there will soon be an opportunity to make any needed corrections. d(A)/d(t)=-ka*A_

The above differential equation is entered according to required format which is given by: d(x)/d(t)=an expression where the dependent variable name "x" and the independent variable name "t" must begin with an alphabetic character and can contain any number of alphabetic and numerical characters. In this Quick Tour problem, the dependent variables are A, B and C for the differential equations, and the independent variable is t. Note that POLYMATH variables are case sensitive. The constants kA and kB are considered to be variables which can be defined by explicit algebraic equations given by the format: x=an expression In this problem, the variables for kA and kB will have constant values. Note that the subscripts are not available in POLYMATH, and in this problem the variable names of ka and kb will be used. Please continue to enter the equations until your set of equations corresponds to the following: Equations: → d(A)/d(t)=-ka*A d(B)/d(t)=ka*A-kb*B d(C)/d(t)=kb*B ka=1 kb=2

As you enter the equations, note that syntax errors are checked prior to being accepted, and various messages are provided to help to identify input errors. Undefined variables are also identified by name during equation entry. POLYMATH 4.0 PC

DIFFERENTIAL EQUATIONS 5-3

ALTERING THE EQUATIONS with no After you have entered the equations, please press equation at the arrow to go to the Problem Options display which will allow needed corrections:

The Problem Options Menu allows you to make a number of alterations on the equations which have been entered. Please make sure that your equations all have been entered as shown above. Remember to first indicate the equation that needs altering by using the arrow keys. When all equation are correct, press ⇑ F7 (keep pressing shift while pressing F7) to continue with the problem solution. ENTERING THE BOUNDARY CONDITIONS At this point you will be asked to provide the initial values for the independent variable and each of the dependent variables defined by the differential equations. Enter initial value for t _

Please indicate this value to be the number "0" and press Return. The next initial value request is for variable "A". Please this value as the value "1." Enter initial value for A 1_ 5-4 DIFFERENTIAL EQUATIONS

POLYMATH 4.0 PC

The initial values for B and C will be requested if they have not been previously entered. Please enter the number "0" for each of these variables. Next the final value for t, the independent variable, will be requested. Set this parameter at "3": Enter final value for t 3_

As soon as the problem is completely specified, then the solution will be generated. However, if you corrected some of your entries, then you may need to press ⇑ F7 again to request the solution. Note that a title such as "Quick Tour Problem 1" could have been entered from the Problem Option Menu. SOLVING THE PROBLEM The numerical solution is usually very fast. For slower computers, an arrow will indicate the progress in the independent variable during the integration. Usually the solution will be almost instantaneous. The screen display after the solution is given below:

POLYMATH 4.0 PC


Another Return keypress gives the partial Results Table which summarizes the variables of the problem as shown below:

The Partial Results Table shown above provides a summary of the numerical simulation. To display or store the results you can enter "t" (tabular display), "g" (graphical display), or "d" (storing the results on a DOS file). This Table may be printed with the function key F3. PLOTTING THE RESULTS Let us now plot the variable from this Problem 1 by entering "g" for a graphical presentation. When asked to type in the variable for plotting, please enter the input indicated below at the arrow: Type in the names of up to four (4) variables separated by commas (,) and optionally one 'independent' variable preceded by a slash(/). For example, myvar1, myvar2/timevar

A, B, C __

A Return keypress ( ) will indicate the end of the variables and should generate the graphical plot on the next page of the specified variables A, B, and C versus t, the independent variable, for this example.

5-6 DIFFERENTIAL EQUATIONS

POLYMATH 4.0 PC

Suppose that you want to plot variable B versus variable A. Select the option "g" from the Display Options Menu and enter B/A when asked for the variable names. Type in the names of up to four (4) variables separated by commas (,) and optionally one 'independent' variable preceded by a slash(/). For example, myvar1, myvar2/timevar

B/A

This will results in a scaled plot for variable B versus variable A. This concludes the Quick Tour problem using the Differential Equation Solver. If you wish to stop working on POLYMATH, please follow the exiting instructions given below. EXITING OR RESTARTING POLYMATH A ⇑ - F10 keypress will always stop the operation of POLYMATH and return you to the Program Selection Menu. THIS ACTION WILL DELETE THE EXISTING PROBLEM. The program can be exited or restarted from the Program Selection Menu. POLYMATH 4.0 PC


INTEGRATION ALGORITHMS The program will first attempt to integrate the system of differential equations using the Runge-Kutta-Fehlberg (RKF) algorithm. A detailed discussion of this algorithm is given by Forsythe et al.* This algorithm monitors the estimate of the integration error and alters the step size of the integration in order to keep the error below a specified threshold. The default values for both relative and absolute (maximal) errors are less than 10-10. If this cannot be attained, then the absolute and relative errors are set as necessary to 10-7 and then to 10-4. If it is not possible to achieve errors of 10-4, then the integration is stopped, and the user is given a choice to continue or to try an alternate integration algorithm for stiff systems of differential equations. Under these circumstances, the system of equations is likely to be "stiff" where dependent variables may change in widely varying time scales, and the user is able to initiate the solution from the beginning with an alternate "stiff" integration algorithm The algorithm used is the semi-implicit extrapolation method of Bader-Deuflhard**, and the maximal errors are again started at 10-10. When the integration is very slow, the F10 keypress will allow the selection of the stiff algorithm, and the problem will be solved from the beginning.

* Frosythe, B. E., M. A. Malcolm, and C. B. Moler, Computer Methods for Mathematical Computation, Prentice-Hall, Englewood Cliffs, NJ, 1977. ** Press, W. H., P. B. Flannery, S. A. Teukolsky, and W. T. Vetterling, Numerical Recipes, 2nd Ed., Cambridge University Press, 1992, pp. 735739.


POLYMATH 4.0 PC

TROUBLE SHOOTING SPECIFIC ERROR MESSAGES Most error messages given by POLYMATH are self-explanatory, and they suggest the type of action which should be taken to correct the difficulty. NONSPECIFIC ERROR MESSAGES "Circular dependency detected." This message appears during the inputting of equations when the equations are not all explicit. For example, an attempt to define y=z/x when z has been previously defined will cause this error message to appear. This version of POLYMATH Differential Equations Solver can only solve variables which can be explicitly expressed as a function of other variables. "The expression ... is undefined at the starting point." This common problem can be solved by starting the integration from t=eps where eps is a very small number and t represents the independent problem variable. "Solution process halted due to a lack of memory." This message may result when the default Runge-Kutta-Fehlberg algorithm is used for a stiff system of differential equations, and thus very small step sizes are taken. Consequently, a large number of data points for possible plotting of the results. Use the F10 to stop the integration and switch to the stiff algorithm. If the message persists, then take the following steps to resolve the difficulty: (1) If you are running under Windows, make sure the PIF for POLYMATH specifies 640K of conventional memory and 1024K or more of XMS(but see item (3) below). (2) Remove other memory-resident programs from your computer. (3) Reduce the number of equations. This is most easily accomplished by introducing the numerical values of the constants into the equations, instead of defining them separately. Due to a limitation of some versions of HIMEM, POLYMATH may not be able to access most of the XMS in the number of equations exceeds approximately 24. (The exact number depends on the computer's configuration.) (4) Reduce the integration interval. POLYMATH 4.0 PC


"Solution process halted because it was not going anywhere." This message usually appears when the problem is very stiff, and the default RKF algorithm is used for integration. The stiff algorithm should be used, or the interval of integration should be reduced. If the error message persists, there are probably errors in the problem setup or input. Please check for errors in the basic equation set, the POLYMATH equation entry, and the numerical values and the units of the variables.

.


POLYMATH 4.0 PC

ALGEBRAIC EQUATION SOLVER QUICK TOUR This chapter is intended to give a very brief discussion of the operation of the POLYMATH Nonlinear Algebraic Equation Solver. NONLINEAR ALGEBRAIC EQUATION SOLVER The user can solve up to a combination of 32 simultaneous nonlinear equations and explicit algebraic expressions. Only real (non-complex) roots are found. All equations are checked for correct syntax and other errors upon entry. Equations can be easily be modified, added or deleted. Multiple roots are given for a single equation. STARTING POLYMATH To begin, please have POLYMATH loaded into your computer as detailed in Chapter 2. Here it is assumed that your computer is set to the hard disk subdirectory or floppy drive containing the POLYMATH package. At the prompt (assumed C:\POLYMAT4 here), you should enter "polymath" C:\POLYMAT4>polymath then press the Return ( ) key. The Main Program Menu should then appear: The Program Selection Menu should then appear, and you should enter "2" to select the Simultaneous Algebraic Equation Solver.

POLYMATH 4.0 PC

NONLINEAR ALGEBRAIC EQUATIONS 6-1

Once that POLYMATH is loaded, please utilize the Help Menu by pressing F6 and then the letter "a" to obtain details on in order to learn how to input the equations. The first page of this Help Section is given below:

This Help Section gives detailed information for entering the nonlinear and auxiliary equations. Press F8 to return from the Help Section to the program, and then press the Enter key ( ) to enter an equation for the first Quick Tour example. The Problem Options Menu at the bottom of your display allows entry of equations with the keypress of "a". Now you are ready for the first problem.

6-2 NONLINEAR ALGEBRAIC EQUATIONS

POLYMATH 4.0 PC

SOLVING ONE NONLINEAR EQUATION The first nonlinear equation to be solved as Quick Tour Problem 1 is: x2 -5x + 6 = 0 The solution is to be obtained over the range of x between 1 and 4. This equation is entered into POLYMATH using the equation entry guidelines where the equation is to be zero at the solution. The following display gives the equation as it should be entered at the arrow: (Use the or the delete key to erase entered characters. Standard DOS editing is available at the cursor.)

f(x)=x^2-5*x+6_ The format for the above equation for f(x) is that the left side of the equation will be equal to zero when the solution has been obtained. The variable which is to be determined is set as an argument within the parentheses for the function f( ). Thus in this case, the variable is x and the function to be solved as being zero is x2-5x+6. Also note that in POLYMATH one way of entering x2 is x^2. An alternative entry is x**2. After you have correctly typed the equation at the arrow, please press once to enter it and then again to end equation entry. This should result in the Problem Options Menu at the bottom of the display and the equation at the top:

The Problem Options Menu indicates which options are now available for you to carry out a number of tasks. In this case, the problem should be complete, and these options for the equation at the arrow will not be needed. If an equation needed to be changed, then you would enter a "c" at the above display. (The arrow is moved by the arrow keys on the keyboard.) POLYMATH 4.0 PC


Once you have the equation entered properly, please press ⇑F7 to solve the problem. You will then be asked to provide the interval over which you wish to find solutions for the equation. This interval is only requested during the solution of a single nonlinear equation.

Please indicate the xmin to be 1 and press ( and press ( ).

). Then indicate xmax to be 4

The entire problem is then display above the Problem Options Menu:

For this single equation, the solution is presented graphically over the search range which you indicated. The solution is where the function f(x) is equal to zero. POLYMATH has the ability to determine multiple solutions to a single equation problem, and the first of two solutions is shown below:


POLYMATH 4.0 PC

Press enter (

) for the second solution.

A shift-return (⇑

) will return you to the Problem Options Display.

SOLVING A SYSTEM OF NONLINEAR EQUATIONS Next, you will solve two nonlinear equations with two unknowns. To enter this new set of equation press ⇑ F8 for a new problem, then press followed by "y" to enter a new problem. The equations that will be solved are:

v CAf – CA1 v CA1 – CA2 2 and k CA2 = V V where k = 0.075; v = 30; CAf = 1.6 ; CA2 = 0.2CAf . Thus there are two unknowns: CA1 and V. 2 k CA1 =

To solve this system of equations, each nonlinear equation must be rewritten in the form f(x) = (an expression that is to have the value of zero at the solution). The appropriate forms for these equations are: v CAf – CA1 2 f CA1 = kCA1 – V v CA1 – CA2 2 f V = kCA2 – and V All equations can be entered into POLYMATH as shown below. Note that each of the problem unknowns (CA1 and V) should appear once and only once inside the brackets in the left of the equal sign. The unknown variable may not be in that particular equation. POLYMATH just needs to know the variable names that you are using in your problem. The explicit algebraic equations may be entered directly. Please enter the equations as given below. The order of the equation is not important as POLYMATH will order the equations during problem solution. Equations f(Ca1)=k*Ca1^2-v*(Caf-Ca1)/V f(V)=k*Ca2^2-v*(Ca1-Ca2)/V k=0.075 Caf=1.6 v=30 Ca2=0.2*Caf

Press ⇑F7 to solve this system of equations. POLYMATH 4.0 PC


For two or more nonlinear equations, POLYMATH requires an initial estimate to be specified for each unknown.

While the solution method used is very robust, it often will not be able to find the solution if unreasonable initial estimated are entered. In this example, physical considerations dictate that CA1 must be smaller than CA0 and larger than CA2. So please enter initial estimate of Ca1 as 1.0. As for V, any positive value up to about V=3900 can be a reasonable estimate. Please use the initial value of 300 for V in this Quick Tour example. After entering the initial values, this example problem should be:

Please press ⇑-F7 to solve the problem. The solution process will start and its progress will be indicated on the screen by an arrow moving along a ruler scale. For most computers, the solution is so fast that the display of the iterations in the numerical solution to a converged solution will not be seen. When visible, the arrow indicates how far from zero the function values are at a particular stage of the solution on a logarithmic scale. Details are given in the Help Section by pressing F6. The results are given after any keypress as shown below:

Please note that the values of the various nonlinear equation functions (nearly zero) are given along the with values of all the problem variables. 6-6 NONLINEAR ALGEBRAIC EQUATIONS

POLYMATH 4.0 PC

This concludes the Quick Tour problem using the Simultaneous Nonlinear Algebraic Equation Solver. If you wish to stop working on POLYMATH, please follow the exiting instructions given below. EXITING OR RESTARTING POLYMATH A ⇑ - F10 keypress will always stop the operation of POLYMATH and return you to the Program Selection Menu. THIS ACTION WILL DELETE THE EXISTING PROBLEM. The program can be exited or restarted from the Program Selection Menu. SELECTION OF INITIAL ESTIMATES FOR THE UNKNOWNS The solution algorithm requires specification of initial estimates for all the unknowns. Generally speaking, closer initial estimates have a better chance of converging to the correct solution. If you wish to solve only a single nonlinear equation, the program will plot the equation so that the location of the roots (if any) can be seen. The program will then show the roots. If no roots exist in the chosen range, the plot will indicate what range should be explored to have the nonlinear function f( ) cross zero. When several equations are to be solved, the selection of the initial values is more complicated. First, the user should try to find the limiting values for the variables using physical considerations. (For example: The mole or mass fraction of a component can neither be negative nor greater that 1; the temperature of cooling water can be neither below freezing nor above boiling; etc.) Typical initial estimates are taken to be mid range. Users should be particularly careful no to select initial estimates where some of the functions may be undefined. (For example, f(xa)=1/(xa-xb)+... is undefined whenever xa=xb; f(xb)=log(1-xb) is undefined whenever xb>=1; etc.) The selection of such initial estimates will stop the POLYMATH solution, and an error message will be displayed. METHOD OF SOLUTION For a single nonlinear equation, the user must specify an interval in which the real root(s) can be found. The program will first attempt to locate points or regions where the function is undefined inside this interval*. If the equations are too complicated for determination of discontinuity points, a warning message is issued. ______________________ *For details of the method, see Shacham, O. and Shacham, M., Acm. Trans. Math. Softw., 16 (3), 258-268 (1990). POLYMATH 4.0 PC


The function is plotted and smaller intervals in which root(s) are located by a sign change of the function. The Improved Memory Method*, which employs a combination of polynomial interpolation and bisection, is used to converge to the exact solution inside those intervals. Iterations are stopped when the relative error is <10-10. Numerical pertubation is used to calculate needed derivatives. For solving a system equations, the bounded Newton-Raphson (NR) method is used. The NR direction is initially used, but the distance is limited by the possible discontinuities **. The progress in iteration will be either the full NR step or close to the first point of discontinuity in that direction. The logarithm of the Euclidean norm of the function residuals is displayed during solution. Norm values below 10-5 are not shown. Iterations end when the norm of the relative error is < 10-10. TROUBLE SHOOTING Most error messages are self-explanatory and suggest needed action. Other less obvious error messages and suggested actions are given below: "Circular dependency detected." This message appears while inputting equations when an attempt is made to define a variable as a function of another variable that was already defined as a function of the new variable. For example, attempting to define y=z/x, when z is already defined as a function of y, will cause this error. This can be prevented by writing the equation in a implicit form: f(y)=y-z/x. "Solution process halted because ..." This message for simultaneous nonlinear equations suggests that: 1. The system contains equations which are very nonlinear. 2. The initial estimates are too far from the solution. or 3. The problem has no solution. Equations can be made less nonlinear by eliminating division by unknowns (multiplying both sides of the equation by the expression that contains the unknowns). Often, substitution and reordering can bring the system of equations to a form that only one equation is implicit. Realistic initial estimates can often help achieve a solution. Finally, nonconvergence may indicate that the set of equations set has no solution. Check the problem setup in this case and pay particular attention to the units. __________________ *For details of this method, see Shacham, M., Computers & Chem. Engng., 14 (6), 621-629 (1990) **For details of the method, see Shacham, O. and Shacham, M., Acm. Trans. Math. Softw., 16 (3), 258-268 (1990). 6-8 NONLINEAR ALGEBRAIC EQUATIONS

POLYMATH 4.0 PC

LINEAR EQUATION SOLVER QUICK TOUR This chapter is intended to give a very brief discussion of the operation of the POLYMATH Linear Equation Solver. LINEAR EQUATION SOLVER The user can solve up to a combination of 32 simultaneous linear equations. The equations are entered in matrix form. STARTING POLYMATH To begin, please have POLYMATH loaded into your computer as detailed in Chapter 2. Here it is assumed that your computer is set to the hard disk subdirectory containing the POLYMATH package. At the prompt (assumed C:\POLYMAT4 here), you should enter "polymath" C:\POLYMAT4>polymath then press the Return ( ) key. The Main Program Menu should then appear: The Program Selection Menu should then appear, and you should enter "3" on the keyboard to select the Linear Equation Solver.

POLYMATH 4.0 PC

LINEAR ALGEBRAIC EQUATIONS 7-1

Once that POLYMATH is loaded, please utilize the Help Menu by pressing F6 for information regarding the use of the Linear Equation Solver. This Help Section is given below:

This Help Section gives detailed information for entering a system of linear equations. Press any key to return from the Help Section to the program, and your display should be at the Main Menu for the Linear Equation Solver. (An alternate command to reach the Main Menu is the ⇑ F10 keypress.) To begin the first Quick Tour example, please press the Return key ( ) from the Main Menu. This will give the Task Menu as shown below:

7-2 LINEAR ALGEBRAIC EQUATIONS

POLYMATH 4.0 PC

SOLVING FIVE SIMULTANEOUS LINEAR EQUATIONS A typical problem for simultaneous linear equations is given below for the variables x1 through x5: x1 + 0.5 x2 + 0.333333 x3 + 0.25 x4 + 0.2 x5 = 0.0 0.5 x1 + 0.333333 x2 + 0.25 x3 + 0.2 x4 + 0.166667 x5 = 1.0 0.333333 x1 + 0.25 x2 + 0.2 x3 + 0.166667 x4 + 0.142857 x5 = 0.0 0.25 x1 + 0.2 x2 + 0.166667 x3 + 0.142857 x4 + 0.125 x5 = 0.0 0.2 x1 + 0.166667 x2 + 0.142857 x3 + 0.125 x4 + 0.111111 x5 = 0.0 The above problem in stored as a Sample Problem in POLYMATH. To recall the above problem, press F7 from the Task Menu of the Linear Equation Solver. Then select problem number "2" to obtain the Problem Options Menu shown below:

Solve this system of equations by pressing ⇑ F7 which should yield the results and the Display Options Menu on the next page. Remember that this keypress combination is accomplished by pressing and holding the Shift key and then pressing the F7 function key.

POLYMATH 4.0 PC


Lets explore making changes to this system of equations. This is accomplished by first pressing ⇑ to "make changes" to the problem. Use the arrow keys to take the highlighted box to the top of the "b" of constants for the equation. Please delete the 0 and enter 1.0 in this box which corresponds to changing the first linear equation to: x1 + 0.5 x2 + 0.333333 x3 + 0.25 x4 + 0.2 x5 = 1.0 This involves using the arrow key and pressing the return key ( highlighted box is in the desired location as shown.

) when the

Then enter the new value at the cursor:

Please solve the problem by pressing ⇑F7. The results are shown below:

This concludes the Quick Tour Problem for Simultaneous Linear Equations. When you are ready to leave this program and return to the Program Selection Menu, use the ⇑ F10 keypress which is discussed below. EXITING OR RESTARTING POLYMATH A ⇑ F10 keypress will always stop the operation of POLYMATH and return you to the Program Selection Menu. THIS ACTION WILL DELETE THE EXISTING PROBLEM. The program can be exited or restarted from the Program Selection Menu. 7-4 LINEAR ALGEBRAIC EQUATIONS

POLYMATH 4.0 PC

REGRESSION QUICK TOUR This chapter is intended to give a very brief overview of the operation of the POLYMATH Polynomial, Multiple Linear and Nonlinear Regression program. REGRESSION PROGRAM This program allows you to input numerical data into up to 30 columns, with up to 100 data points in each column. The data can be manipulated by defining expressions containing the names of previously defined columns. Relationships between different variables (columns of data) can be found using polynomial, multiple linear and nonlinear regression as well as cubic spline interpolation. Fitted curves can be interpolated, differentiated and integrated. Graphical output of the fitted curves and expressions is presented, and a statistical analysis of the parameters found during the regressions is given. STARTING POLYMATH To begin, please have POLYMATH loaded into your computer as detailed in Chapter 2. Here it is assumed that your computer is set to the hard disk subdirectory or floppy drive containing the POLYMATH package. At the prompt (assumed C:\POLYMAT4 here), you should enter "polymath" C:\POLYMAT4 > polymath ) key. The Program Selection Menu should then then press the Enter ( appear, and you should enter "4" on the keyboard to select the Polynomial, Multiple Linear and Nonlinear Regression program. This should bring up the Main Program Menu as given in the next page. In order to save time in entering data points during this quick tour, we will use sample problems which have been stored in POLYMATH. Press F7 to access the Sample Problems Menu from the Main Program Menu. QUICK TOUR PROBLEM 1 Let us consider a fairly typical application of the Regression Program in which some data are available. When these data are fitted to a polynomial

POLYMATH 4.0 PC

REGRESSION 8 -1

within POLYMATH, the polynomial expression has the form: P(x) = a0 + a1x + a2x2 +... + anxn where y is the dependent variable, x is the independent variable, and the parameters are a0 ...an. Variable "n" here represents the degree of the polynomial. In POLYMATH, the maximum degree which is shown is 5. The above polynomial expression gives a relationship between the dependent variable and the independent variable which is obtained by determining the parameters according to a least squares objective function. Data points are usually available which give x and y values from which the parameters a0... an can be determined.

RECALLING SAMPLE PROBLEM 3 After pressing F7 at the Main Program Menu, the Sample Problems Menu should appear on your screen as shown on the next page.. The sample problem to be discussed should be retrieved by pressing "3" on the keyboard. This will result in the Problem Options Display which includes 10 data points of x and y as shown on the next page.

8 -2 REGRESSION

POLYMATH 4.0 PC

POLYMATH 4.0 PC

REGRESSION 8 -3

FITTING A POLYNOMIAL The Problem Options Menu includes problem editing, library, printing, help and solution options. To fit a polynomial to the data of Y versus X you should select the "⇑ F7 to fit a curve or do regression" option. After pressing ⇑ F7 the following "Solution Options" menu appears:

After pressing "p" (lower case), you should be asked for the name of the independent variable's column, as shown below:

You should enter a capital "X" (upper case) as name of the independent variable and press . The same question regarding the dependent variable will be presented. Please enter a capital "Y" (upper case) at the arrow. The following display should appear:

8-4 REGRESSION

POLYMATH 4.0 PC

On this display the coefficients of the polynomial P(x), up to the fifth order are shown together with the value of the variance. One of the polynomials is highlighted by having a box around it. This is the lowest order polynomial, such that higher order polynomial does not give significantly better fit. The same polynomial is also plotted versus the experimental data. Other polynomials can be highlighted and plotted by pressing a number between 1 and 5. There are many additional calculations and other operations that can be carried out using the selected polynomial. Please make sure the highlighted box is on the 4th degree polynomial. Let us find the value of X for Y = 10. To do that you should press "y" and enter after the prompt regarding the value of Y: "10". The following display results:

The resultant X values are shown both graphically and numerically. For Y = 10 there are two X values, X = 1.36962 and X = 5.83496. FITTING A CUBIC SPLINE We will now fit a cubic spline to these data of Sample Problem 3. Please press F8 two times to return to the Problems Options Menu. Then press ⇑ F7 to "fit a curve or do regression". The Solutions Options Menu should appear. POLYMATH 4.0 PC

REGRESSION 8-5

Enter "s" (lower case) for a cubic spline followed by "X" and then "Y". The following display should present the results:

EVALUATION OF AN INTEGRAL WITH THE CUBIC SPLINE Please take options "i" and request the initial value for the integration to be "1" at the arrow:

Press integration.

and then enter "6" at the arrow for the find value of the

Press to have the resulting integration shown on the next display with both graphical and numerical results:

8-6 REGRESSION

POLYMATH 4.0 PC

Please press any key to end this Sample Problem 3. MULTIPLE LINEAR REGRESSION It will often be useful to fit a linear function of the form: y(x) = a0 + a1x1 + a2 x2 +... + anxn where x1, x2, ..., xn are n independent variables and y is the dependent variable, to a set of N tabulated values of x1,i, x2,i, ... and y (xi). We will examine this option using Sample Problem 4. RECALLING SAMPLE PROBLEM 4 First exit to the main title page by pressing ⇑ F10. Press F7 to access the Sample Problems Menu, and select problem number 4 by pressing "4". (The problem display is shown on the next page).

POLYMATH 4.0 PC

REGRESSION 8-7

SOLVING SAMPLE PROBLEM 4 After you press ⇑ F7 "to fit a curve or do regression", the following Solution Options Menu should appear:

This time press "l" (lower case letter "l") to do "linear regression". You will be prompted for the first independent variable (column) name.

Please type in "X1" at the arrow and press . You will be prompted for the 2nd independent variable. Enter "X2" as the second independent variable name and press once again. A prompt for the 3rd independent variable will appear. You should press here without typing in anything else, since there are no additional independent variables. 8 -8 REGRESSION

POLYMATH 4.0 PC

At the prompt for the dependent variable (column) name shown below you should type "Y" and press .

Once the calculations are completed, the linear regression (or correlation) is presented in numerical and graphical form.

Please note that the correlation the equation for variable "Y" has the form of the linear expression: Y = a0 + a1X1 + a2X2 where a0 = 9.43974, a1 = -0.1384 and a2 = 3.67961. This graphical display of Sample Problem 4 presents the regression data versus the calculated values from the linear regression. The numerical value of the variance and the number of the positive and negative residuals give an indication regarding the validity of the assumption that Y can be represented as linear function of X1 and X2. The results in this case indicate a good fit between the observed data and the correlation function.

POLYMATH 4.0 PC

REGRESSION 8-9

The Display Options Menu allows the user to use an "s" keypress "to save results in a column". This refers to saving the calculated value of Y from the linear regression to the Problems Options Display under a column name provided by the user. The "r" keypress from the Display Options Menu give a statistical residual plot as shown below:

The "F9" keypress from the Display Options Menu give a statistical summary:

The confidence intervals given in the statistical summary are very useful in interpreting the validity of the linear regression of data. This concludes Sample Problem 4 which illustrated multiple linear regression.

8-10 REGRESSION

POLYMATH 4.0 PC

TRANSFORMATION OF VARIABLES A nonlinear correlation equation can be often brought into a linear form by a transformation of the data. For example, the nonlinear equation: Y = a0 X1a 1 X2a 2 can be linearized by taking logarithm of both sides of the equation: ln Y = ln a0 + a1 ln X1 + a2 ln X2. To demonstrate this option please recall Sample Problem 5. To do this, please press ⇑ F10 to get to the Main Program Menu, F7 to access the Sample Problems Menu and select Sample Problem number 5. This should result in the Problem Option Display below:

In this display X1, X2 and Y represent the original data, the variables (columns) lnX1, lnX2 and lnY represent the transformed data. You can see the definition of ln X1 , for example, by moving the cursor (the highlighted box), which located in row number 1 of the first column, into the box containing "lnX1" (using the arrow keys) and press .

POLYMATH 4.0 PC

REGRESSION 8-11

The following window is brought up:

Note that the expression in the right hand side of the column definition equation must be a valid algebraic expression, and any function arguments used in the expression should be enclosed within parentheses. Since we do not want to change this expression, please press to close the window. Now press ⇑ F7 to do regression, then "l" to do linear regression. Type in "lnX1" as the name of the first independent variable, "lnX2" as the name of the second independent variable and "lnY" as the name of the dependent variable. The results should be displayed as shown below:

All of the statistical analyses are available for the transformed variable. Please note that the results indicate that the equation for variable "Y" can be written as: Y = a0 X1a 1 X2a 2 where a0 = exp (-0.666796) = 0.5133, a1 = 0.986683 and a2 = -1.95438. This concludes the transformation of variable and the multiple linear regression for Sample Problem 5. 8-12 REGRESSION

POLYMATH 4.0 PC

NONLINEAR REGRESSION It is often desirable to fit a general nonlinear function model to the independent variables as indicated below: y(x) = f(x1, x2, ..., xn; a0, a1, ..., am) In the above expression, x1, x2, ..., xn are n independent variables, y is the dependent variable, and a0, a1, ..., am are the model parameters. The data are represented by a set of N tabulated values of x1,i, x2,i, ... and y(xi). The regression adjusts the values of the model parameters to minimize the sum of squares of the deviations between the calculated y(x) and the data y(xi). The nonlinear regression capability of POLYMATH allows a general nonlinear function to be treated directly without any transformation. Lets return to Sample Problem 5 and this time treat the model for Y directly where Y = a0 X1a 1 X2a 2 . Please recall Sample Problem 5. From the Problem Options Display press ⇑ F7 and then enter "R" (upper case R) to "Do nonlinear regression." The user is then prompted to:

The user can then enter the model equation using any of the variables from the columns of the Problem Options Display and any unknown parameters (maximum of five) which are needed. For this example, please enter

Thus in this problem, the unknown parameters are k, alpha, and beta. The next query for the user is to supply initial estimates for each of the unknown parameter in turn:

It is good practice to provide good initial parameter estimates from either reasonable physical/chemical model values or from a linearized treatment of the nonlinear model. In this example however, please set all initial guesses for the parameters as unity, "1.0". Then POLYMATH will provide a summary of the problem on the Regression Option Display as shown on the next page.

POLYMATH 4.0 PC

REGRESSION 8-13

The Regression Options Menu gives several useful options for model changes and alterations to initial parameter guesses; however, please press ⇑F7 to solve this problem. The program search is shown to the user and the converged solution is indicated below:

There are a number of options from the Display Options Menu (not shown here). Perhaps the most useful is the "statistical analysis" which is given on the next page. 8-14 REGRESSION

POLYMATH 4.0 PC

This concludes the Quick Tour section dealing with nonlinear regression and the Chapter on the Polynomial, Multiple Linear and Nonlinear Regression Program. Remember, when you wish to stop POLYMATH, please follow the exiting instructions given below. EXITING OR RESTARTING POLYMATH A ⇑ F10 keypress will always stop the operation of POLYMATH and return you to the Program Selection Menu. THIS ACTION WILL DELETE THE EXISTING PROBLEM. The program can be exited or restarted from the Program Selection Menu. SOLUTION METHODS When fitting a polynomial of the form P(x) = a0 + a1x + a2x2 +...+anxn to N points of observed data, the minimum sum of square error correlation of the coefficients a0, a1, a2...an can be found by solving the system of linear equation (often called normal equations): POLYMATH 4.0 PC

REGRESSION 8-15

XT XA = XT Y where

Y=

y y1 .2 . . y

A=

N

ao a1 . . . an

x0 X=

1 x0 2 x0 N

x1

1 x1 2 x1 N

. . . . . . . . . . . .

xn

1 xn 2 xn N

and where y1, y2...yN are N observed values of dependent variable, and x1, x2...xN are N observed values of the independent variable. Multiple linear regression can also be expressed in the same form except that the matrix X is redefined as follows:

X=

1

x1,1

x2,1 . . . x n,1

1

x1,2

x2,2 . . . x n,2

1

x1,N x2,N . . . x n,N

where xi,j is the j-th observed value of the i-th independent variable. When polynomial or multiple linear regression are carried out without the free parameter (a0), the first element in vector A and the first column in matrix X must be removed. In POLYMATH the normal equations are solved using the GaussJordan elimination. It is indicated in the literature that direct solution of normal equations is rather susceptible to round off errors. Practical experience has should this method to sufficiently accurate for most practical problems. The nonlinear regression problems in POLYMATH are solved using the Levenberg-Marquardt method. A detailed description of this method can be found, for example, in the book by Press et al.*

*Press, W. H., P. B. Flannery, S. A. Teukolsky, and W. T. Vetterling, Numerical Recipes, 2nd Ed., Cambridge University Press, 1992, pp. 678683. 8-16 REGRESSION

POLYMATH 4.0 PC

APPENDIX INSTALLATION FROM THE CRE-99 CD-ROM ACCOMPANYING "ELEMENTS OF CHEMICAL REACTION ENGINEERING" AND EXECUTION INSTRUCTIONS FOR VARIOUS OPERATING SYSTEMS This Appendix provides complete instructions for the installation and execution of POLYMATH for the DOS, Windows 3.1 and Windows 95 operating systems. Detailed information is provided for advanced options. Latest updates are on the README.TXT and INSTALL.TXT files. DOS INSTALLATION Place the CD-ROM in an appropriate drive. This is usually drive D. First set your current directory to that of the CD-ROM and indicate the location of the POLYMATH installation software. For example, if the drive to be used is D, then insert “D:” at the cursor and press Enter. Then enter “cd\html\toolbox\polymath\files”. At the cursor enter “dir” to verify that you have the following POLYMATH installation files in that directory: PMUNZIP.EXE BGI.EXE INSTALL.EXE README.TXT INSTALL.TXT POLYMATH installation is initiated by typing “install” at the cursor the drive and subdirectory containing the above files. For a CD-ROM as drive D, this would be: D:\html\toolbox\polymath\files>install The installation program will ask a number of questions. These are listed later on this file. Please refer to the details of these questions as needed. Follow the instructions on the screen. Note that the initial recommendation for the Printer Output setting is option 1. Half Page, Low Resolution. When the installation is complete, press Enter. DOS EXECUTION 1. Change your directory to the POLYMATH directory by entering "C:" and then "cd\polymat4" at the cursor. Substitute your specified location if you did not use the default directory location. 2. Enter "polymath" at the cursor. 3. Always end POLYMATH by exiting from the Program Selection Menu. POLYMATH 4.0 PC

APPENDIX 9-1

WINDOWS 3.1 INSTALLATION 1. Put the CD-ROM in the appropriate drive such as “D:”. 2. Double click on the Main icon in the Program Manager window. 3. Double click on the MS-DOS Prompt icon. 4. Change the directory to the CD-ROM drive to where POLYMATH is stored. For example, if the drive to be used is D, then insert “D:” at the cursor and press Enter. Then enter “cd\html\toolbox\polymath\files”. 5. At the D:\html\toolbox\polymath\files> prompt, enter “dir”. 6. The following five files should be listed: PMUNZIP.EXE, INSTALL.EXE, INSTALL.TXT, BGI.EXE, and README.TXT. 7. At the D:\html\toolbox\polymath\files\poly402> prompt, enter “install” 8. Follow the instructions on the screen. Please note that the initial recommended setting for Output is option 1. Half Page, Low Resolution. 9. At the DOS prompt enter “exit” to return to Windows CONTINUE STEPS BELOW TO CREATE A PIF FILE FOR POLYMATH ONLY IF POLYMATH DOES NOT EXECUTE PROPERLY 10. From the Program Manager Window click on Main. 11. From the Main Window double click on PIF Editor. 12. Please enter the following in the PIF Editor: Program Filename: POLYMATH.BAT Window Title: POLYMATH 4.0 Startup Directory: C:\POLYMAT4 Video Memory: Low Graphics Memory Requirements: -1 -1 EMS Memory: 0 1024 XMS Memory: 1024 1024 Display Usage: Full Screen 13. From File use Save As “POLYMAT4.PIF”. WINDOWS 3.1 EXECUTION* 1. From the File options in the Program Manager Window, select Run. 2. The Command Line for your POLYMATH location should be entered such as “c:\polymat4\polymath”. 3. Click on “OK” 4. If POLYMATH does not run properly, then create a PIF file starting with Step 10 as given above. 5. Always end POLYMATH by exiting from the Program Selection Menu. * A Program Group and a Program Item can be created under Windows to allow POLYMATH to be executed conveniently from the desktop. 9-2 APPENDIX

POLYMATH 4.0 PC

Windows 95 Installation 1. Put the installation disk in your floppy drive. 2. Click on the Start button. 3. Click on the My Computer icon. 4. Double click on the CD-ROM icon indicating cre-99. Then continue double clicking on html, toolbox, polymath, files, and poly402 5. Double click on the Install.exe file 6. Follow the directions on the screen. Please note that the initial recommended setting for Output is option 1. Half Page, Low Resolution. 7. When installation is complete, press Enter. 8. Close the DOS window. Windows 95 Execution* 1. Click on the Start button. 2. Click on the Run icon. 3. Enter “c:\polymat4\polymath” 4. Always end POLYMATH by exiting from the Program Selection Menu. * A Program Group and a Program Item can be created under Windows 95 to allow POLYMATH to be executed conveniently from the desktop. INSTALLATION QUESTIONS 1. Enter drive and directory for POLYMATH [C:\POLYMAT4] : ==> The default response is indicated by the contents of the brackets [...] which is given by pressing Enter key. The full path (drive and directory) where you wish the POLYMATH program files to be stored must be provided here. If the directory does not exist, then the installation procedure will automatically create it. NOTE: Network clients will need read and execute permission for this directory and its subdirectories. This procedure does not provide the needed permissions. 2. Is this a network installation? [N] If you are installing POLYMATH on a stand-alone computer, take the default or enter "N" for no and GO TO 5. on the next page of this manual. If you are installing on any kind of network server, answer "Y" and continue with the installation. POLYMATH 4.0 PC

APPENDIX 9-3

3. What will network clients call [POLYdir]? ===> _ This question will only appear if you answered "Y" to question 2 to indicate a Network installation. Here "POLYdir" is what was provided in question 1. On some networks, the clients "see" server directories under a different name, or as a different disk, than the way the server sees them. This question enables POLYMATH to print by indicating where the printerdriver files are located. They are always placed in subdirectory BGI of the POLYMATH directory by the installation procedure. During run time, they must be accessed by the client machines, thus POLYMATH must know what the client's name is for the directory. 4. Enter drive and directory for temporary print files [C:\TMP]: ===> _ This question will only appear if you answered "Y" to question 2 to indicate a Network installation. Depending on the amount of extended memory available, and the type of printer is use, POLYMATH may need disk workspace in order to print. Since client machines are not normally permitted to write on the server disk, you are requested to enter a directory where files may be written. The temporary print files are automatically deleted when a print is completed or cancelled. 5. The POLYMATH installation program now copies all of the needed files according to your previous instructions. This may take some time as the needed files are compressed on the installation diskette. 6. Please select the type of output device you prefer [1]: _ 1. Printer 2. Plotter ===> _ Your answer here will take you to an appropriate selection menu. 7 A. POLYMATH INSTALLATION - printer selection Please select the type of printer you have: 1. Canon Laser Printer 2. Canon BJ 200 3. Canon BJC 600 4. HP LaserJet II 9-4 APPENDIX

POLYMATH 4.0 PC

5. HP LaserJet III 6. HP LaserJet IV 7. HP LaserJet III or IV (HPGL/2) 8. Epson 9-pin Dot Matrix Printer 9. Epson 24-pin Dot Matrix Printer 10. Epson 9-pin Dot Matrix Printer (color) 11. Epson 24-pin Dot Matrix Printer (color) 12. Epson Stylus 13. Epson Color Stylus 14. IBM Proprinter 15. IBM Proprinter X24 16. IBM Quietwriter 17. Kodak Diconix 18. OkiData Dot Matrix Printer (native mode) 19. PostScript printer 20. Toshiba 24-pin Dot Matrix Printer (native mode) 21. Xerox CP4045/50 Laser Printer (USA) 22. Xerox CP4045/50 Laser Printer (International) 23. HP DesignJet 24. HP DeskJet (Black & White) 25. HP DeskJet 500C (Color) 26. HP DeskJet 550C (Color) 27. HP DeskJet 1200C 28. HP PaintJet 29. HP PaintJet XL300 30. HP ThinkJet ===> _ Type in the number of your printer (or a type of printer that your printer can emulate) and press Enter. 7 B. POLYMATH INSTALLATION - plotter selection Please select the type of plotter you have: 1. HP 7090 Plotter 2. HP 7470 Plotter 3. HP 7475 Plotter 4. HP 7550 Plotter 5. HP 7585 Plotter 6. HP 7595 Plotter 7. Houston Instruments DMP/L Plotters ===> _ POLYMATH 4.0 PC

APPENDIX 9-5

Type in the number of your plotter (or a type of plotter that your plotter can emulate) and press Enter. Once you have selected either a printer or a plotter, you will be asked to select the mode for your output. 8. POLYMATH INSTALLATION - printer/ plotter mode selection Please select the mode you want output in [1]: 1. Half Page, Low Resolution 2. Half Page, Medium Resolution 3. Half Page, High Resolution 4. Landscape, Low Resolution 5. Landscape, Medium Resolution 6. Landscape, High Resolution 7. Portrait, Low Resolution 8. Portrait, Medium Resolution 9. Portrait, High Resolution ===> _ Generally, the fastest printing with adequate resolution is option (1), which is also the default. This choice is recommended. It is easy to change your printer/plotter selection and mode by using the PRINTSET program which is installed in the POLYMATH directory. 9. POLYMATH INSTALLATION - printer/ plotter port selection Please select the port your printer/plotter is connected to [1] : 1. LPT1 2. LPT2 3. COM1 4. COM2 5. LPT3 ===> _ The default port, LPT1, is the first parallel port on most personal computers and usually is connected to the printer. For network installations, these are usually "logical" printer names, but they work just as well. If your network uses other printer logical names, please read the Appendix section entitled "PRINTING FOR ADVANCED USERS" which starts on the next manual page.

9-6 APPENDIX

POLYMATH 4.0 PC

10. POLYMATH Execution The POLYMATH program can now be executed by first changing the directory of your personal computer to the one where the program has been installed. Then you should enter "polymath" at the cursor as shown below and press Enter: C:\POLYMAT4> polymath_ Windows users must use POLYMATH as a DOS program. Advanced users may wish to place the polymath program in the "path" statement of the autoexec.bat file so that POLYMATH is more easily available from any cursor. "OUT OF ENVIRONMENT SPACE" MESSAGE If you receive this message or if you are having difficulty in printing from POLYMATH, then follow ONE of the following instruction set for your particular operating system. Window 95, Windows 98, or Windows NT 4.0 1. Open a DOS prompt window (if yours opens full-screen, hit Alt+Enter to get a window). 2. Click on the "Properties" button at the top. 3. At the end of the Cmd line, add the text "/e:2048" (if you already have this, then change the existing number to a higher number in increments of 1024). Windows 3.x and Windows 95 1. Open a DOS prompt window. 2. At the C:\ prompt type "CD/WINDOWS, and press Enter. 3. Type "EDIT SYSTEM.INI" and press Enter. 4. Locate a line that reads "[NonWindows App]". 5. Make sure that this section contains the following entry: "CommandEnvSize=2048". 6. Save the modified file. 7. Reboot your computer. 8. Adjust the size upwards in increments of 1024 as necessary. DOS or Windows 3.1 1. Open a DOS prompt window. 2. Type "SET" at the prompt. POLYMATH 4.0 PC

APPENDIX 9-7

3. Look for the the line that displays the value of the COMSPEC environment variable. If COMSPEC is set to C:\COMMAND.COM then add the following line to the CONFIG.SYS file: SHELL=C:\COMMAND.COM C:\ /E:2048 /P If COMSPEC is set to C:\DOS\COMMAND.COM then add this line to the CONFIG.SYS file: SHELL=C:\DOS\COMMAND.COM C:\DOS /E:2048 /P 4. Reboot your computer. 5. Adjust the size of E: in Step 3 upwards in increments of 1024 as necessary. CHANGING PRINTER SELECTION The printer selection may be changed without a complete reinstallation of POLYMATH by using the PRINTSET utility program which is stored on the directory containing POLYMATH. This program is executed by entering "printset" while in the POLYMATH directory. If there are several printers to be used with POLYMATH, then different POLYMATH.BAT files can be created for each printer. The batch file for printers is discussed in the following section. PRINTING FOR ADVANCED USERS POLYMATH printing is accomplished using GRAF/DRIVE PLUS which is a trademark for copyrighted software from Fleming Software. The printer setup is controlled by three environmental variables which are set in the POLYMATH.BAT file. This file is created by the INSTALL procedure, and it can be subsequently be modified by the PRINTSET utility or with an editor. CAUTION: This file should only be altered with an editor by advanced users. A typical POLYMATH.BAT file is shown below: ECHO OFF SET BGIPATH=C:\POLYMAT4\BGI SET PM_WORKPATH=C:\POLYMAT4 SET PM_PRINTER=_LJ3R,0,LPT1 POLYMENU CLS 9-8 APPENDIX

POLYMATH 4.0 PC

In this file, the three environmental variables are: BGIPATH - This is the directory in which the printer driver "BGI" files for POLYMATH reside. It is usually the BGI subdirectory of the POLYMATH directory. This variable is not normally changed. PM_WORKPATH - This is the directory in which temporary print files are stored. PM_PRINTER - This variable has the following form: ,, This variable in the previous POLYMATH.BAT file is defined as: PM_PRINTER=_LJ3R,0,LPT1 where _LJ3R indicates the printer_type as HP LaserJet III, 0 (the number zero) indicates the page_format as half page with low density print, and LPT1 indicates the printer_port as LPT1. PM_PRINTER Options for and A detailed listing of the and options is given in Table 1 at the end of this Appendix. PM_PRINTER Options for The may be any of the following: LPT1, LPT2, LPT3, COM1, COM2 or other physical/logical device names. PRINTING TO STANDARD GRAPHICS FILES It is possible to have all output which is "printed" by POLYMATH to be saved as various graphics files for use in word processing, desktop publishing, etc. This involves specialized use of the PM_PRINTER variable which is not available during the INSTALL procedure or the PRINTSET program for printer modification. All printing to graphics files requires the creation of a special batch file for this purpose. POLYMATH 4.0 PC

APPENDIX 9-9

A typical file for this purpose which is arbitrarily called POLYGRAP.BAT is shown below: ECHO OFF SET BGIPATH=C:\POLYMAT4\BGI SET PM_WORKPATH=C:\POLYMAT4 SET PM_PRINTER=_TIF,0,FILE:C:\GRAPHS\PMOUT+++.TIF POLYMENU CLS In the above file, the PM_PRINTER variable _TIF indicates the printer_type as a Tagged Image Format (TIFF), 0 indicates low resolution with two colors, and FILE:C:\GRAPHS\PMOUT+++.TIF indicates the location and name of the resulting graphics file. PM_PRINTER Options for and for Graphics Files A detailed listing of the and options for graphics file output is given in Table 2 at the end of this Appendix. PM_PRINTER Options for for Graphics Files This variable can be defined as FILE: when the output is to be to a file and not a port. (Note that if is a logical device name, such as LPT3, the output will be printed.) may be a particular filename or a general template, including '+' signs where a sequence number is to be written. The above example will create a sequence of files named PMOUT001.TIF, PMOUT002.TIF, etc., stored in directory C:\GRAPHS .

POLYMATH 4.0 PC

APPENDIX 9-10

Table 1.1 - PM_PRINTER OPTIONS

Output Page /Resolution /Color


Epson-compatible 9-pin Dot Matrix, and IBM Proprinter _FX

0

HalfLo

5

LandHi

1

HalfMed

6

FullLo

2

HalfHi

7

FullMed

3

LandLo

8

FullHi

4

LandMed

Epson-compatible 24-pin Dot Matrix _LQ

0

HalfLo

5

LandHi

1

HalfMed

6

FullLo

2

HalfHi

7

FullMed

3

LandLo

8

FullHi

4

LandMed

Epson-compatible 9-pin Dot Matrix Printer (color) _CFX

0

HalfLoC

3

LandMedC

1

HalfMedC

4

FullLoC

2

LandLoC

5

FullMedC

Epson-compatible 24-pin Dot Matrix Printer (color) _CLQ

9-11 APPENDIX

0

HalfLoC

3

LandMedC

1

HalfMedC

4

FullLoC

2

LandLoC

5

FullMedC

POLYMATH 4.0 PC




0

HalfLo

5

LandHi

1

HalfMed

6

FullLo

2

HalfHi

7

FullMed

3

LandLo

8

FullHi

4

LandMed

0

HalfLo

5

LandHi

1

HalfMed

6

FullLo

2

HalfHi

7

FullMed

3

LandLo

8

FullHi

4

LandMed

2

Full

2

Full

IBM Proprinter X24 _PP24

IBM Quietwriter _IBMQ

OkiData Dot Matrix Printer (native mode) _OKI92

0

Half

1

Land

Toshiba 24-pin Dot Matrix Printer (native mode) _TSH

0

Half

1

Land

POLYMATH 4.0 PC

APPENDIX 9-12




LaserJet II, LaserJet III, LaserJet IV, DeskJet (black cartridge), Canon laser _LJ _LJ3R _LJ4 _DJ _Canon

0

HalfLo

5

LandHi

1

HalfMed

6

FullLo

2

HalfHi

7

FullMed

3

LandLo

8

FullHi

4

LandMed

DeskJet 500C (color cartridge), and DeskJet 550C _DJC _DJC550

0

HalfLoC8

5

LandHiC8

1

HalfMedC8

6

FullLoC8

2

HalfHiC8

7

FullMedC8

3

LandLoC8

8

FullHiC8

4

LandMedC8

0

HalfLoC2

8

FullLoC8

1

LandLoC2

9

HalfHiC8

2

FullLoC2

10

LandHiC8

3

HalfHiC2

11

FullHiC8

4

LandHiC2

12

HalfLoC16

5

FullHiC2

13

LandLoC16

6

HalfLoC8

14

FullLoC16

7

LandLoC8

PaintJet _PJ

9-13 APPENDIX

POLYMATH 4.0 PC




0

Half

6

HalfC16

1

Land

7

LandC16

2

Full

8

FullC16

3

HalfGR

9

HalfC256

4

LandGR

10

LandC256

5

FullGR

11

FullC256

0

DraftPL

2

DraftPLB

1

LQPL

3

LQPLB

_HP7470

0

DraftPL

1

LQPL

_HP7475 _HP7550

0

DraftPL

4

DraftPLr

1

LQPL

5

LQPLr

2

DraftPLB

6

DraftPLBr

3

LQPLB

7

LQPLBr

0

DraftPL

5

LQPLC

1

LQPL

6

DraftPLD

2

DraftPLB

7

LQPLD

3

LQPLB

8

DraftPLE

4

DraftPLC

9

LQPLE

PostScript Printers _PS

Hewlett-Packard Plotters _HP7090

_HP7585 _HP7595

POLYMATH 4.0 PC

APPENDIX 9-14




0

HalfLo

5

LandHi

1

HalfMed

6

FullLo

2

HalfHi

7

FullMed

3

LandLo

8

FullHi

4

LandMed

Canon BJ 200, and Epson Stylus _BJ200 _ESCP2

Canon BJC600, and Epson Color Stylus _BJC600 _ESCP2C

0

HalfLoC8

5

LandHiC8

1

HalfMedC8

6

FullLoC8

2

HalfHiC8

7

FullMedC8

3

LandLoC8

8

FullHiC8

4

LandMedC8

HPGL/2 (DeskJet 1200C, PaintJet XL300, and DesignJet) _HGL2

9-15 APPENDIX

0

HalfC2

5

FullC8

1

LandC2

6

HalfC256

2

FullC2

7

LandC256

3

HalfC8

8

FullC256

4

LandC8

POLYMATH 4.0 PC




0

LoResC16

3

LoResC2

1

MedResC16

4

MedResC2

2

HiResC16

5

HiResC2

0

ColorMode

1

MonoMode

0

LoResC2

2

HiResC2

1

MedResC2

Zsoft PCX _PCX

Windows BMP _BMP GEM IMG _IMG

Tagged Image Format (TIFF) compressed and uncompressed _TIF _UTIF

0

LoResC2

1

MedResC2

2

HiResC2

1

MonoMode

1

MonoMode

Computer Graphics Metafile (ANSI) _CGM

0

ColorMode

0

ColorMode

AutoCad DXF _DXF

Video Show (ANSI NAPLPS) _VSHO

0

ColorMode

Word Perfect Graphics _WPG

0

POLYMATH 4.0 PC

ColorMode

APPENDIX 9-16




0

ColorMode

1

MonoMode

ColorMode

1

MonoMode

ColorMode

1

MonoMode

Windows Metafile _WMF

Adobe Illustrator PostScript _AI

0

Color QuickDraw (PICT) _PCT

9-17 APPENDIX

0

POLYMATH 4.0 PC

POLYMATH VERSION 4.1 Provides System Printing from Windows 3.X, 95, 98 and NT

USER-FRIENDLY NUMERICAL ANALYSIS PROGRAMS - SIMULTANEOUS DIFFERENTIAL EQUATIONS - SIMULTANEOUS ALGEBRAIC EQUATIONS - SIMULTANEOUS LINEAR EQUATIONS - POLYNOMIAL, MULTIPLE LINEAR AND NONLINEAR REGRESSION for IBM and Compatible Personal Computers Internet: http://www.polymath-software.com Users are encouraged to obtain the latest general information on POLYMATH and its use from the above Internet site. This will include updates on this version and availability of future versions. Michael B. Cutlip and Mordechai Shacham, the authors of POLYMATH, have prepared this software to accompany the book entitled Problem Solving in Chemical Engineering with Numerical Methods. This book is a companion book for students and professional engineers who want to utilize the POLYMATH software to effectively and efficiently obtain solutions to realistic and complex problems. Details on this Prentice Hall book, ISBN 0-13-862566-2, can be found at www.polymathsoftware.com or at www.prenhall.com.


i -2 PREFACE

POLYMATH 4.1 PC

POLYMATH LICENSE AGREEMENT The authors of POLYMATH agree to license the POLYMATH materials contained within this disk and the accompanying mpoly41.pdf file to the owner of the Prentice Hall textbook Elements of Chemical Reaction Engineering, by H. Scott Fogler. This license is for noncommercial and educational uses exclusively. Only one copy of this software is to be in use on only one computer or computer terminal at any one time. One copy of the manual may be reproduced in hard copy only for noncommercial educational use of the textbook owner. This individual-use license is for POLYMATH Version 4.1 and applies to the owner of the textbook. Permission to otherwise copy, distribute, modify or otherwise create derivative works of this software is prohibited. Internet distribution is not allowed under any circumstances. This software is provided AS IS, WITHOUT REPRESENTATION AS TO ITS FITNESS FOR AND PURPOSE, AND WITHOUT WARRANTY OF ANY KIND, EITHEREXPRESS OR IMPLIED, including with limitation the implied warranties of merchantability and fitness for a particular purpose. The authors of POLYMATH shall not be liable for any damages, including special, indirect, incidental, or consequential damages, with respect to any claim arising out of or in connection with the use of the software even if users have not been or are hereafter advised of the possibility of such damages. HARDWARE REQUIREMENTS POLYMATH runs on the IBM Personal Computer and most compatibles. A floating-point processor is required. Most graphics boards are automatically supported. The minimum desirable application memory is 450 Kb plus extended memory for large applications. POLYMATH works with PC and MS DOS 3.0 and above. It can also execute as a DOS application under Windows 3.1, Windows 95, Windows 98, and Windows NT. It is important to give POLYMATH as much of the basic 640 Kb memory as possible and up to 2048 Kb of extended memory during installation. This version of POLMATH supports only the Windows printers that are available to your operating system. POLYMATH 4.1 PC

PREFACE i-3


1 2 2 2 2 2 2 3 3 4 4 5 5 6 6 6 6


2 2 2 2 2

-

1 1 1 2 3


3 3 3 3 3

-

1 2 2 3 3


- 1 - 1 - 1 - 3 - 3 - 5 - 5 - 6 PC


4 - 7 4 - 7 4 - 8 4 - 8 4 - 8 4 - 9 4 - 9 4-10 4-10 4- 10 4-11 4-11


5 5 5 5 5 5 5 5 5 5 5 5 5 5

-

1 1 1 2 3 4 4 5 6 7 8 9 9 9


6 6 6 6 6 6 6 6 6

-

1 1 1 3 5 7 7 7 8

POLYMATH 4.1 PC

PREFACE i-5


7 7 7 7 7


8 8 8 8 8 8 8 8 8 8 8 8 8 8 8

- 1 - 1 - 1 - 1 - 2 - 4 - 5 - 6 - 7 - 7 - 8 -11 -13 -15 -15

9 9 9 9 9 9 9 9

-

1 1 2 2 2 3 3 3

9 9 9 9 9

-

4 4 6 6 7

APPENDIX INSTALLATION AND EXECUTION INSTRUCTIONS................... WINDOWS 3.X INSTALLATION....................................................... WINDOWS 3.X EXECUTION............................................................. WINDOWS 3.X SHUTDOWN............................................................. USING PRINT METAFILES IN DOCUMENTS IN WINDOWS 3.X. WINDOWS 95, 98, AND NT INSTALLATION.................................. WINDOWS 95, 98, AND NT EXECUTION........................................ WINDOWS 95, 98, AND NT SHUTDOWN........................................ USING PRINT METAFILES IN DOCUMENTS FOR WINDOWS 95, 98, AND NT......................................................... INSTALLATION QUESTIONS............................................................ ADDITIONAL NETWORK INSTALLATION NEEDS..................... TROUBLESHOOTING.......................................................................... "OUT OF ENVIRONMENT SPACE" MESSAGE.............................

i-6 PREFACE

-

1 1 1 3 4

POLYMATH 4.1 PC


INTRODUCTION 1-1


POLYMATH 4.1 PC







POLYMATH 4.1 PC

INTRODUCTION 1-3


1-4 INTRODUCTION

POLYMATH 4.1 PC



Symbol + * / x.x10a



LOGICAL EXPRESSIONS An "if" function is available during equation entry with the following syntax: if (condition) then (expression) else (expression). The parentheses are required, but spaces are optional. The condition may include the following operators: > greater than < less than >= greater than or equal <= less than or equal == equals <> does not equal | or & and The expressions may be any formula, including another "if" statement. For example: a=if(x>0) then(log(x)) else(0) b=if (TmaxT) then (maxT) else (T)) POLYMATH MESSAGES There are many POLYMATH messages which may provide assistance during problem solving. These messages will tell you what is incorrect and how to correct it. All user inputs, equations and data, are checked for format and syntax upon entry, and feedback is immediate. Correct input is required before proceeding to a problem solution. HARD COPY If there is a printer connected to the computer, hard copy of the problem statements, tabular and graphical results etc. can be made by pressing F3 key wherever this option is indicated on the screen. This version of POLYMATH allows printing from the Windows printers, and the Windows meta files can also be used in various documents. See pages 9-2 and 9-4 for more details. Problem statements and results can be also printed by saving them as a file and printing this file after leaving POLYMATH. GRAPHICS POLYMATH gives convenient displays during problem entry, modification and solution. Your computer will always operate in a graphics mode while you are executing POLYMATH. 1-6 INTRODUCTION

POLYMATH 4.1 PC

GETTING STARTED This chapter provides information on the hardware requirements and discusses the installation of POLYMATH. HARDWARE REQUIREMENTS POLYMATH 4.1 runs on IBM compatible personal computers that support the Windows-based operating systems. Most graphics boards are automatically supported. The minimum application memory requirement is approximately 450Kb, and extended memory is used when it is available. POLYMATH 4.1 runs as a DOS application under Windows 3.X, Windows 95, Windows 98, and Windows NT. Printing is accomplished by a separate program operating under Windows. This allows the printing from POLYMATH to be done by any printer that can be used with the particular Windows operating system. POLYMATH SOFTWARE The complete set of POLYMATH application programs with a general selection menu is available on a single 3-1/2 inch 1.44 Mb floppy in compressed form. It is recommended that a backup disk be made before attempting to install POLYMATH onto a hard disk. Installation is available via an install program which is executed from any drive. INSTALLATION TO INDIVIDUAL COMPUTERS AND NETWORKS POLYMATH executes best when the software is installed on a hard disk or a network. There is a utility on the POLYMATH distribution disk which is called "install". Detailed installation instructions are found in the Appendix of this manual. Experienced users need to simply put the disk in the floppy drive, typically A or B. Type "install" at the prompt of your installation floppy drive, and press return. Follow the instructions on the screen to install POLYMATH on the particular drive and directory that you desire. Note that the default drive is "C:" and the default directory is called "POLYMAT4". Network installation will require responses to additional questions during installation. Latest detailed information can be found on the README.TXT file found on the installation disk.

POLYMATH 4.1 PC

GETTING STARTED 2-1



POLYMATH 4.1 PC

EXITING POLYMATH PROGRAM The best way to exit POLYMATH is to follow the instructions on the program display. However, a Shift-¡F10 keypress (⇑F10) will stop the execution of POLYMATH at any point in a program and will return the user to the Polymath Program Selection Menu. THIS ACTION WILL DELETE THE EXISTING PROBLEM so be sure to store your problem as an individual file or in the library before exiting the program in this manner. A query is made to determine if the user really wants to end the program in this manner while losing the current problem. This ⇑ F10 keypress is one of the few POLYMATH commands which is not always indicated in the various Display Menus. It is worth remembering.

POLYMATH 4.1 PC

GETTING STARTED 2-3



3-1 HELP

POLYMATH 4.1 PC




HELP 3-2



3-3 HELP

POLYMATH 4.1 PC




UTILITIES 4-1


4-2 UTILITIES

POLYMATH 4.1 PC

ASSIGNMENT FUNCTIONS The assignment function is a way of storing your results. You may specify a variable name in which to store the results of a computation by first typing in the variable name, then an equals sign, then the expression you wish to store. For example, if you wish to store the value of sin (4/3) 2 in variable 'a', you would enter: a = sin (4/3)**2 Variable names must start with a letter, and can contain letters and digits. There is no limit on the length of the variable names, or on the number of variables you can use. You can then use the variable 'a' in other calculations. These variables are stored only as long as you remain in the current POLYMATH program. Please note that all stored values are lost when the particular program is exited. Calculator information is not retained during problem storage. CALCULATOR EXAMPLES Example 1. In this example the vapor pressure of water at temperatures of 50, 60, and 70 o C has to be calculated using the equation: log10 P = 8.10765 – 1750.29 235.0 + T For T = 50 the following expression should be typed into the calculator: 10^(8.10675 - 1750.29 / (235+50)) CALCULATOR: Enter an expression and press to evaluate it. Press to leave or press F6 for information.


POLYMATH 4.1 PC

UTILITIES 4-3


Where

2 2 a = 27 R Tc Pc 64

b = RTc 8 Pc






4-4 UTILITIES

POLYMATH 4.1 PC







POLYMATH 4.1 PC



deka 10

UTILITIES 4-5



e. hp hr






e. hp hr


4-6 UTILITIES

To units: kilo-cal

POLYMATH 4.1 PC




UTILITIES 4-7




POLYMATH 4.1 PC




POLYMATH 4.1 PC

UTILITIES 4-9





POLYMATH 4.1 PC




A B C 1 0 0 0.74081822 0.19200658 0.067175195 0.54881164 0.24761742 0.20357094 0.40656966 0.24127077 0.35215957 0.30119421 0.21047626 0.48832953 0.22313016 0.17334309 0.60352675 0.16529889 0.13797517 0.69672595 0.12245643 0.10746085 0.77008272 0.090717953 0.082488206 0.82679384 0.067205513 0.062688932 0.87010556 0.049787068 0.047308316 0.90290462


UTILITIES 4-11

PROBLEM OUTPUT AS GRAPHICS FILES This 4.1 version of POLYMATH creates Windows Meta files (WMFs) that can be printed or entered into documents. These files are normally printed directly by special MetaFile Print programs for Windows 3.X (MFP16) or for Windows 95, 98, and NT (MFP32) . If these special print programs are not running, the generated WMFs are found in the SPOOL subdirectory under the POLYMAT4 directory. Thus the user can place these file directly into word processors and desktop publishing software. Details of this option are found in Chapter 9 of this manual. A typical example would be to create the following output in a WMF file for inclusion in a written reporting using word processing or desktop publishing. The problem is again the for Quick Tour Problem 1 from the next chapter. Note that the figure below has utilized the title and axis definition options.

4-12 UTILITIES

POLYMATH 4.1 PC


POLYMATH 4.1 PC




POLYMATH 4.1 PC

ENTERING THE EQUATIONS The equations are entered into POLYMATH by first pressing the "a" option from the Problem Options Menu. The following display gives the first equation as it should be entered at the arrow. (Use the Backspace key, to correct entry errors after using arrow keys to position cursor.) Press the ) to indicate that the equation is to be entered. Don't be Return key ( concerned if you have entered an incorrect equation, as there will soon be an opportunity to make any needed corrections. d(A)/d(t)=-ka*A_







POLYMATH 4.1 PC



POLYMATH 4.1 PC




A, B, C __



POLYMATH 4.1 PC


B/A






POLYMATH 4.1 PC

TROUBLE SHOOTING SPECIFIC ERROR MESSAGES Most error messages given by POLYMATH are self-explanatory, and they suggest the type of action which should be taken to correct the difficulty. NONSPECIFIC ERROR MESSAGES "Circular dependency detected." This message appears during the inputting of equations when the equations are not all explicit. For example, an attempt to define y=z/x when z has been previously defined will cause this error message to appear. This version of POLYMATH Differential Equations Solver can only solve variables which can be explicitly expressed as a function of other variables. "The expression ... is undefined at the starting point." This common problem can be solved by starting the integration from t=eps where eps is a very small number and t represents the independent problem variable. "Solution process halted due to a lack of memory." This message may result when the default Runge-Kutta-Fehlberg algorithm is used for a stiff system of differential equations, and thus very small step sizes are taken. Consequently, a large number of data points for possible plotting of the results. Use the F10 to stop the integration and switch to the stiff algorithm. If the message persists, then take the following steps to resolve the difficulty: (1) If you are running under Windows, make sure the PIF for POLYMATH specifies 640K of conventional memory and 1024K or more of XMS(see the Appendix for more details). (2) Remove other memory-resident programs from your computer. (3) Reduce the number of equations. This is most easily accomplished by introducing the numerical values of the constants into the equations, instead of defining them separately. (4) Reduce the integration interval.

POLYMATH 4.1 PC



.


POLYMATH 4.1 PC


POLYMATH 4.1 PC





POLYMATH 4.1 PC











POLYMATH 4.1 PC

Press enter (


A shift-return (⇑











POLYMATH 4.1 PC




POLYMATH 4.1 PC


POLYMATH 4.1 PC





POLYMATH 4.1 PC



POLYMATH 4.1 PC



) when the




POLYMATH 4.1 PC


POLYMATH 4.1 PC

REGRESSION 8 -1



8 -2 REGRESSION

POLYMATH 4.1 PC

POLYMATH 4.1 PC

REGRESSION 8 -3




8-4 REGRESSION

POLYMATH 4.1 PC



REGRESSION 8-5



Press integration.



8-6 REGRESSION

POLYMATH 4.1 PC


POLYMATH 4.1 PC

REGRESSION 8-7




POLYMATH 4.1 PC




POLYMATH 4.1 PC

REGRESSION 8-9




8-10 REGRESSION

POLYMATH 4.1 PC



POLYMATH 4.1 PC

REGRESSION 8-11




POLYMATH 4.1 PC





POLYMATH 4.1 PC

REGRESSION 8-13



POLYMATH 4.1 PC


REGRESSION 8-15

XT XA = XT Y where

Y=

y y1 .2 . . y

A=

N

ao a1 . . . an

x0 X=

1 x0 2 x0 N

x1

1 x1 2 x1 N

. . . . . . . . . . . .

xn

1 xn 2 xn N


X=

1

x1,1

x2,1 . . . x n,1

1

x1,2

x2,2 . . . x n,2

1

x1,N x2,N . . . x n,N



POLYMATH 4.1 PC

APPENDIX INSTALLATION FROM THE CD-ROM ACCOMPANYING "ELEMENTS OF CHEMICAL REACTION ENGINEERING." EXECUTION INSTRUCTIONS FOR VARIOUS OPERATING SYSTEMS This Appendix provides complete instructions for the installation and execution of POLYMATH 4.1 for Windows 3.X, Windows 95, Windows 98, and Windows NT operating systems. Detailed information is also provided for advanced options. Latest updates will be on the README.TXT file. WINDOWS 3.X INSTALLATION 1. Put the CD-ROM in the appropriate drive. 2. Double click on the Main icon in the Program Manager window. 3. Double click on the MS-DOS Prompt icon. 4. Change the directory to the CD-ROM drive on which POLYMATH is stored. For example, if the drive to be used is D, then insert “D:” at the cursor and press Enter. Then enter "cd\html\toolbox\polymath\files\poly41". 5. At the D:\html\toolbox\polymath\files\poly41> prompt, enter “dir”. 6. There should be three files: Readme.txt, Install.exe, and Pmunzip.exe 7. At the D:\html\toolbox\polymath\files\poly41> prompt, enter “install” 8. Follow the instructions on the screen. 9. At the D:\html\toolbox\polymath\files\poly41> prompt enter “exit” to return to Windows CONTINUE STEPS BELOW TO CREATE A PIF FILE FOR POLYMATH ONLY IF POLYMATH DOES NOT EXECUTE PROPERLY 10. From the Program Manager Window click on Main. 11. From the Main Window double click on PIF Editor. 12. Please enter the following in the PIF Editor: Program Filename: POLYMATH.BAT Window Title: POLYMATH 4.0 Startup Directory: C:\POLYMAT4 Video Memory: Low Graphics Memory Requirements: -1 -1 EMS Memory: 0 1024 XMS Memory: 1024 1024 Display Usage: Full Screen 13. From File use Save As “POLYMAT4.PIF”. POLYMATH 4.1 PC

APPENDIX 9-1

WINDOWS 3.X EXECUTION* This 4.1 version of POLYMATH is a DOS program, but it creates Windows Meta files (WMFs) that can be printed or entered into documents. A special MetaFile Print program for Windows 3.X (MFP16) prints only POLYMATH WMF files that are generated upon printing requests from within POLYMATH. These generated WMFs are normally placed in the SPOOL subdirectory under the POLYMAT4 directory. Thus the user must first run the MFP16 Windows program before POLYMATH to insure that all requests to Print will be printed on the Windows printer. The MFP16 program produces a small blue printer icon on the bottom right of the desktop. A right mouse click on this icon gives the options which can be executed by a left mouse click. 1. Click on the File options in the Program Manager Window, select Run. 2. Enter the Command Line for the Metafile Print program such as “c:\polymat4\mfp16.exe” 3. The Command Line for your POLYMATH location should then be entered such as “c:\polymat4\polymath.bat”. 4. Click on “OK” 5. If POLYMATH does not run properly, then create a PIF file starting with Step 10 as given in the preceding Windows 3.X Installation section. 6. Always end POLYMATH by exiting from the Program Selection Menu. * A Program Group and a Program Item can be created under Windows to allow POLYMATH.bat and MFP16.exe to be executed conveniently from the desktop. WINDOWS 3.X SHUTDOWN 1. POLYMATH can be terminated at any time by pressing Shift-F10 to return to the Program Selection Menu and then by pressing F8 to exit. 2. The MFP16 program can be terminated by a right click with the mouse on the small blue printer icon followed by a left mouse click on "terminate." USING PRINT METAFILES IN DOCUMENTS FOR WINDOWS 3.X The WMFs files will accumulate in the SPOOL subdirectory of POLYMAT4 when the MFP16 program is not running, These WMF files, designated by TEMP00X.WMF, can be copied and inserted as files within word processing and desktop publishing software. MS Word, for example, 9-2 APPENDIX

POLYMATH 4.1 PC

will conveniently give a small preview of each selected file so that the content will be known prior to insertion. Execution of MFP16.exe will allow you to print all or delete all remaining files in SPOOL. Thus POLYMATH output can be selectively inserted into documents or printed or both. Windows 95, 98, and NT Installation 1. Put the CD-ROM in the appropriate drive, for example Drive D. 2. Minimize all Windows until Desktop appears. 3. Double click on the My Computer icon. 4. Double click on the CD-ROM icon. Then continue double clicking in turn on html, toolbox, polymath, files, and poly41. 5. Double click on the Install icon. 6. Follow the directions on the screen. 7. When installation is complete, press Enter. 8. Close the DOS window. Windows 95, 98, and NT Execution* This 4.1 version of POLYMATH is a DOS program, but it creates Windows Meta files (WMFs) that can be printed or entered into documents. A special MetaFile Print program for Windows 95, 98, and NT (MFP32) prints only POLYMATH WMF files that are generated upon printing requests from within POLYMATH. These generated WMFs are normally placed in the SPOOL subdirectory under the POLYMAT4 directory. When the user executes the POLYMATH.bat file, the MFP32.exeWindows program is also launched. POLYMATH runs in its own Window by the MFP32.exe program only produces a small blue printer icon on the bottom right of the desktop. A right mouse click on this icon gives the options which can be executed by a left mouse click. Program execution is accomplished by the following steps: 1. Click on the Start button. 2. Click on the Run icon. 3. Start POLYMATH by entering the storage location for POLYMATH and specify the program as "polymath.bat" such as "c:\polymat4\polymath.bat".

POLYMATH 4.1 PC

APPENDIX 9-3

4. The POLYMATH program will initiate both the POLYMATH software and the MFP32.exe program as discussed previously. If the MFP32.exe was not active at POLYMATH startup, then you may need to click on the POLYMATH window to continue with POLYMATH. * Icons for both the Printer Utility and POLYMATH can be placed on the desktop by using Windows Explorer to find both the Mfp32.exe and Polymath.bat files in the POLYMATH directory. A right mouse click on each of these files followed by a left mouse click on "Create Shortcut" can create a "Shortcut to ..." which can be dragged to the desktop. WINDOWS 95, 98, AND NT SHUTDOWN 1. POLYMATH can be terminated at any time by pressing Shift-F10 to return to the Program Selection Menu and then by pressing F8 to exit. 2. The MFP32.exe program can be terminated by a right click with the mouse on the small blue printer icon followed by a left mouse click on "terminate." USING PRINT METAFILES IN DOCUMENTS FOR WINDOWS 95, 98, AND NT The WMFs files will accumulate in the SPOOL subdirectory of POLYMAT4 when the MFP32.exe program is not running, These WMF files, designated by TEMP00X.WMF, can be copied to word processing and desktop publishing software and inserted as files. MS Word will give a small preview of each selected file so that the content will be known. Execution of MFP32.exe will allow you to print all or delete all remaining files in SPOOL. Thus POLYMATH output can be selectively inserted into documents or printed or both. INSTALLATION QUESTIONS (DETAILS) 1. Enter drive and directory for POLYMATH [C:\POLYMAT4]: ==> The default response is indicated by the contents of the brackets [...] which is given by pressing Enter key. The full path (drive and directory) where you wish the POLYMATH program files to be stored must be provided here. If the directory does not exist, then the installation procedure will automatically create it.

9-4 APPENDIX

POLYMATH 4.1 PC

NOTE: Network clients will need read and execute permission for this directory and its subdirectories. This procedure does not provide the needed permissions. 2. Is this a network installation? [N] If you are installing POLYMATH on a stand-alone computer, take the default or enter "N" for no and GO TO 5. on this list. If you are installing on any kind of network server, answer "Y" and continue with the installation. 3. What will network clients call [POLYdir]? ===> _ This question will only appear if you answered "Y" to question 2 to indicate a Network installation. Here "POLYdir" is what was provided in question 1. On some networks, the clients "see" server directories under a different name, or as a different disk, than the way the server sees them. This question enables POLYMATH to print by indicating where the printerdriver files are located. They are always placed in subdirectory BGI of the POLYMATH directory by the installation procedure. During run time, they must be accessed by the client machines, thus POLYMATH must know what the client's name is for the directory. 4. Enter drive and directory for temporary print files [C:\TMP]: ===> _ This question will only appear if you answered "Y" to question 2 to indicate a Network installation. Depending on the amount of extended memory available, and the type of printer is use, POLYMATH may need disk workspace in order to print. Since client machines are not normally permitted to write on the server disk, you are requested to enter a directory where files may be written. The temporary print files are automatically deleted when a print is completed or cancelled. 5.

POLYMATH INSTALLATION Which version of MS Windows are you using [2]: 1. Windows 3.X 2. Windows 95 3. Windows 98 4. Windows NT ===> _

POLYMATH 4.1 PC

APPENDIX 9-5

6. The POLYMATH installation program now copies all of the needed files according to your previous instructions. This may take some time as the needed files are compressed on the installation diskette. Additional Network Installation Needs: 1. POLYMATH requires the creation of the directory C:\TMP\SPOOL. 2. There must be an MFP.INI file on each computer (including the server) connected to the network that is going to run POLYMATH. In this file, a spool directory on the local disk should be specified. This directory has to exist. Any Windows Meta File (*.wmf) that is placed into this directory by the Print Utility (either MFP16.exe or MFP32.exe) will get printed and deleted, so it should not be in a directory in general use. A sample MFP.INI file follows: [location] SpoolPath=C:\TMP\SPOOL TROUBLESHOOTING Please look for the most recent troubleshooting hints at www.polymath-software.com (a) If you are attempting to print on Windows 3.X and you get the message "abnormal program termination", then try running POLYMATH again. Additional messages such as "out of memory" or "missing batch file" indicates that you must remove some RAM resident programs. If all else fails, then use the DOS printing version 4.02 of POLYMATH. (b) When POLYMATH says "Could not begin printing: I/O port error", the SPOOL subdirectory does not exist. Please make sure it exists. See the following part (c). (c) If POLYMATH indicates "print successful" and nothing comes out of the printer then... - Use Print Manager to check to see that there is no problem with the printer. If there is a job in the queue, this means that POLYMATH and MFP did their part, and the problem is in Windows, Print Manager, or the printer itself. 9-6 APPENDIX

POLYMATH 4.1 PC

- Check that MFP16.exe is running for Windows 3.X or that MFP32.exe is running for Windows 95, 98, and NT. There should be a little printer icon visible at the bottom right of the desktop. If not, then run the proper program (File->Run ... C:\POLYMAT4\MFP16.exe for Windows 3.X ; File->, Run ... C:\POLYMAT4\MFP32.exe for Windows 95, 98, and NT) - MFP.INI file doesn't exist or doesn't match the POLYMATH.BAT file. Please check the POLYMATH.BAT file for a line similar to PM_PRINTER=_WMF,0,FILE:C:\POLYMAT4\SPOOL\TEMP+++.WMF where the directory specified in the preceding line, "C:\POLYMAT4\SPOOL" should be the same as it appears in the MFP.INI file. If they don't match, the you must change one of the other, shut down MFP, exit POLYMATH, and start both again.

"OUT OF ENVIRONMENT SPACE" MESSAGE If you receive this message or if you are having difficulty in printing from POLYMATH, then follow ONE of the following instruction sets for your particular operating system. Window 95, Windows 98, or Windows NT 4.0 1. Open a DOS prompt window (if yours opens full-screen, hit Alt+Enter to get a window). 2. Click on the "Properties" button at the top. 3. At the end of the Cmd line, add the text "/e:2048" (if you already have this, then change the existing number to a higher number in increments of 1024). Windows 3.x and Windows 95 1. Open a DOS prompt window. 2. At the C:\ prompt type "CD/WINDOWS, and press Enter. 3. Type "EDIT SYSTEM.INI" and press Enter. 4. Locate a line that reads "[NonWindows App]". 5. Make sure that this section contains the following entry: "CommandEnvSize=2048". 6. Save the modified file. 7. Reboot your computer. 8. Adjust the size upwards in increments of 1024 as necessary. POLYMATH 4.1 PC

APPENDIX 9-7

DOS or Windows 3.1 1. Open a DOS prompt window. 2. Type "SET" at the prompt. 3. Look for the line that displays the value of the COMSPEC environment variable. If COMSPEC is set to C:\COMMAND.COM then add the following line to the CONFIG.SYS file: SHELL=C:\COMMAND.COM C:\ /E:2048 /P If COMSPEC is set to C:\DOS\COMMAND.COM then add this line to the CONFIG.SYS file: SHELL=C:\DOS\COMMAND.COM C:\DOS /E:2048 /P 4. Reboot your computer. 5. Adjust the size of E: in Step 3 upwards in increments of 1024 as necessary.

POLYMATH 4.1 PC

APPENDIX 9-8

using.htm

Using and Modifying the M-Files

Using the M-Files 1. Run MATLAB. 2. At the Command Window prompt, type p=path and press ENTER. MATLAB's path (the list of places MATLAB looks for m-files) will be listed. 3. To add the m-files on the CD-ROM to MATLAB's path, type path(p,' [your cdrom drive letter]:\Html\Toolbox\Matlab\m-files') and press ENTER. For example, if your CD drive is drive D, then you would type: path(p, 'D:\html\toolbox\matlab\m-files') NOTE: This command will only temporarily change the path. To permanently change the path, type pathtool at the command prompt. A new window will open. Add the [your cd-rom drive letter]:\Html\Toolbox\Matlab\m-files directory to the list of path directories and click "Save Settings." (If you are using a copy of MATLAB owned by your college or university, you may not have sufficient access to modify the path permanently.) 4. Run Microsoft Word. 5. Open the Html\Toolbox\Matlab\Word directory on the CD-ROM. You will see a number of folders labeled CH#. 6. Open the folder corresponding to the chapter of the example problem you wish to examine. 7. Open the file for that example problem.

Additional Information These files were created using the Notebook interface with Word. The Word files contain examples that have been solved using the m-files. The text comes in three different colors: file:///H:/html/toolbox/matlab/using.htm[05/12/2011 16:55:11]

using.htm

Lines in green represent commands you should enter at the command prompt. Lines in blue contain output from MATLAB, Lines in black contain descriptive text. We highly recommend using the Word file for each example as a guideline (print a copy if necessary). Retype the text from the green lines (one by one) at the MATLAB command prompt. These commands define the boundary conditions and the initial conditions necessary to solve the example problems. They also show you how to make graphs, so you can display the results of the calculations. To change the boundary conditions and/or the initial conditions of a problem, simply change the values in the green input commands. To change the differential equations and/or their supporting equations, you will have to edit the m-files themselves. (You will have to copy the m-files to your computer's hard drive to edit them, since you won't be able to save them on your CD-ROM.)

Modifying the M-Files The m-files are in the Html\Toolbox\Matlab\m-files directory on the CD-ROM. 1. In MATLAB, open the m-file for the example you wish to examine. 2. Select all of the text in the m-file and copy it. 3. Under the File menu in MATLAB, choose New and then M-File. An M-File Editor window will appear. 4. Paste the text into this window. 5. Make changes to the equations you want to modify. 6. Save the file under a new name (e.g., temp1.m) to your computer's hard drive or to a floppy disk. Remember that this file must also be saved into a directory contained in MATLAB's path. (See steps 2 and 3 from Using the M-Files, above.) Also, the "function" command in the m-file should be changed to match the new filename. 7. Enter the command lines as before, using the Word file as a guide. For all commands that use the m-file name, be sure to replace the old m-file name with the name of the file you just created (e.g., temp1.m).

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Lectures 1 and 2

Lectures 1 and 2

Chemical Identity (Chapter 1) * A chemical species is said to have reacted when it has lost its chemical identity. The identity of a chemical species is determined by the kind, number, and configuration of that species' atoms. Three ways a chemical species can lose its chemical identity: 1. decomposition 2. combination 3. isomerization

Reaction Rate (Chapter 1) The reaction rate is the rate at which a species looses its chemical identity per unit volume. The rate of a reaction can be expressed as the rate of disappearance of a reactant or as the rate of appearance of a product. Consider species A:

r A = the rate of formation of species A per unit volume -r A = the rate of a disappearance of species A per unit volume r B = the rate of formation of species B per unit volume For a catalytic reaction, we refer to -r A ' , which is the rate of disappearance of species A on a per mass of catalyst basis. NOTE: dCA /dt is not the rate of reaction Consider species j: r j is the rate of formation of species j per unit volume r j is a function of concentration, temperature, pressure, and the type of catalyst (if any) r j is independent of the type of reaction system (batch, plug flow, etc.) r j is an algebraic equation, not a differential equation We use an algebraic equation to relate the rate of reaction, -r A , to the concentration of reacting species and to the file:///H:/html/course/lectures/one/index.htm[05/12/2011 16:55:12]

Lectures 1 and 2

temperature at which the reaction occurs [e.g. -r A = k(T)C A 2 ]. Would you like to see an How about doing a

of some more rates of reaction? on what you've learned?

General Mole Balance Equation (Chapter 1)

Mole Balance on Different Reactor Types The GMBE applied to the four major reactor types (and the general reaction,

Reactor

Differential

Algebraic

Batch

CSTR

PFR

PBR

file:///H:/html/course/lectures/one/index.htm[05/12/2011 16:55:12]

Integral

):

Lectures 1 and 2

Conversion (Chapter 2) The conversion of species A in a reaction is equal to the number of moles of A reacted per mole of A fed. Batch

Flow

Design Equations (Chapter 2) The following design equations are for single reactions only. Design equations for multiple reactions will be discussed later.

Reactor Mole Balances in Terms of Conversion Reactor

Differential

Algebraic

Integral

Batch

CSTR

PFR

PBR

Reactor Sizing (Chapter 2) Given -r A as a function of conversion, one can size any type of reactor. The volume of a CSTR and the volume of a PFR can be represented as the shaded areas in the Levenspiel Plots shown below:


Lectures 1 and 2

Numerical Evaluation of Integrals (Appendix A.4) The integral to calculate the PFR volume can be evaluated using a method such as Simpson's One-Third Rule (pg 925):

NOTE: The intervals ( ) shown in the sketch are not drawn to scale. They should be equal.

Simpson's One-Third Rule is one of the more common numerical methods. It uses three data points. Other numerical methods (see Appendix A, pp 924-926) for evaluating integrals are: 1. Trapezoidal Rule (uses two data points) 2. Simpson's Three-Eighth's Rule (uses four data points) 3. Five-Point Quadrature Formula

Reactors in Series (Chapter 2) Given -r A as a function of conversion, one can also design any sequence of reactors:


Lectures 1 and 2

Relative Rates of Reaction aA + bB

cC + dD

*

All chapter references are for the 3rd Edition of the text Elements of Chemical Reaction Engineering. Back to the top of Lectures 1 and 2.

Jump to Lecture(s): 1 & 2 | 3 & 4 | 5 & 6 | 7 & 8 | 9 & 10 11 & 12 | 13 & 14 | 15 & 16 | 17 & 18 | 19 & 20 21 & 22 | 23 & 24 | 25 & 26 | 27 & 28 | 29 & 30 31 & 32 | 33 & 34 | 35A | 35B | 36 & 37 | 38 & 39


Lectures 1 and 2 © 1999 Prentice-Hall PTR Prentice Hall, Inc. ISBN 0-13-531708-8 Legal Statement


Thoughts on Problem Solving: Algorithm

1. Write out the problem statement. Include information on what you are to solve, and consider why you need to solve the problem.

2. Make sure you are solving the real problem as opposed to the perceived problem. Use techniques such as "Finding out Where the Problem Came From," "The Duncker Diagram," "The Explore Phase," etc. to check to see that you define and solve the real problem. Recast the problem statement if necessary.

3. Draw and label a sketch. Define and name all variables and/or symbols. Show numerical values of variables if known.

4. Identify and name A. Relevent principles, theories and equations B. Systems and subsytems C. Dependent and independent variables D. Knowns and unknowns E. Inputs and outputs F. Necessary (missing) information

file:///H:/html/probsolv/closed/alg/alg.htm[05/12/2011 16:55:14]


5. List assumptions and approximations involved in solving the problem. Question the assumptions and then state which ones are the most reasonable for your purposes.

6. Check to see if the problem is either under-specified or over-specified. If it is under-specified, figure out how to find the missing information. If over-specified, identify the extra information that isn't needed.

7. Relate problem to a similar problem or experience (compare to an example problem in lecture or in the book).

8. Use an algorithm (e.g. reaction engineering) A. Mole Balance B. Rate Laws i. Kinetic ii. Transport C. Stoichiometry i. Gas or liquid ii. Pressure drop D. Combine

E. Energy Balance



F. Evaluate

9. Develop/derive/integrate and/or manipulate an equation or equations from which the desired variable can be determined.

10. Substitute numerical values and calculate the desired variable. Check your units at each step in the solution to find possible errors.

11. Examine and evaluate the answer to see it makes sense. Is it reasonable, considering the problem statement? Does it consider safety and ethical issues? See how this algorithm can be applied to Examples in Chemical Reaction Engineering.


Thoughts on Problem Solving: Getting Unstuck

1. Persevere! Keep trying different strategies and stay open to creative ideas. Try not to get frustrated.

2. Be more active in the solution!! A. Ask yourself questions about the problem. Is this problem a routine one? What data are missing? What equations can I use? Explore the problem. B. Draw sketches of what you think that the solution should look like. (e.g. temperature-time curve). C. Write equations. D. Keep track of your progress.

3. Re-focus on the fundamentals. Review the textbook and lecture material. Look for similar examples. Study the examples given. Change what is given in the example and what is asked, then try to see how it might relate to the problem you are addressing.

4. Break the problem into parts. Analyze the parts of the problem. Concentrate on the parts of the problem you understand and that can be solved.

file:///H:/html/probsolv/closed/unstuck/unstuck.htm[05/12/2011 16:55:15]


5. Verbalize the problem to yourself and others. Describe...

A. what the problem is B. what you have done, and C. where you are stuck

6. Paraphrase. Re-describe the problem. Think of simpler ways to describe the problem. Ask other classmates to describe the problem to you in their own words.

7. Use a heuristic or algorithm. The algorithm for closedended problems on this site is a good start, and others may be available to you.

8. Look at extreme cases that could give insight and understanding. For instance: What happens if x = 0? x = infinity?

9. Simplify the problem and solve a limiting case. Break up the problem into simpler pieces and solve each piece by itself. Find a related but simpler example and work from there.



10. Try substituting numbers to see if a term can be neglected.

11. Try solving for ratios to drop out parameters that are not given. You may find that you don't need to find some parameters because they cancel out!

12. Look for hidden assumptions or for what information you have forgotten to use. After reading each phrase or sentence of the problem statement, ask yourself if any assumptions can be inferred from that phrase.

13. Alternate working forward towards a solution and backwards from a solution you assumed. Working backwards may at least give you clues as to how you should approach the problem while working forward.

14. Take a break. Incubate. Let your subconscious work on the problem while you do something else, file:///H:/html/probsolv/closed/unstuck/unstuck.htm[05/12/2011 16:55:15]


like exercising, talking to friends, or just relaxing! Sometimes all you need is a break to achieve that final breakthrough!

15. Brainstorm. Think of different approaches to the problem, no matter how strange. Guess the solution to the problem and then check the answer.

16. Check again to make sure you are solving the right problem. Double-check all of your values, assumptions, and approaches. Make sure you haven't missed anything and that you are looking for the correct solution.

17. Try using a different strategy. There is usually more than one way to solve a problem, and you may find a method that you haven't considered is much easier than the one you're working on currently.

18. Ask for help! There are many resources you may go to for additional instruction or ideas. Instructors can usually steer you in the right direction and clarify your understanding of the problem. If allowed, your classmates may be the biggest source of help, since they usually utilize many different approaches and can relate



to your approach.

Back to Closed-Ended Problems


CRE -- Problem 1-A

Learning Resource/Additional Homework Problem CDP1AA

A 200-dm3 constant-volume batch reactor is pressurized to 20 atm with a mixture of 75% A and 25% inert. The gas-phase reaction is carried out isothermally at 227 C.

V = 200-dm 3 P = 20 atm T = 227 C

a. Assuming that the ideal gas law is valid, how many moles of A are in the reactor initially? What is the initial concentration of A? b. If the reaction is first order:

Calculate the time necessary to consume 99% of A. c. If the reaction is second order:

file:///H:/html/01chap/html/ahp01-a.htm[05/12/2011 16:55:16]

CRE -- Problem 1-A

Calculate the time to consume 80% of A. Also calculate the pressure in the reactor at this time if the temperature is 127 C. Solution



CRE -- Reactor Photos

REACTOR PHOTOS One of the biggest complaints, that chemical engineering students have, is that they never get to see the equipment that they're spending so much time designing. Heat exchangers, pumps, containment vessels, etc. all remain abstract "little black boxes," and reactors are the worst culprits, because they're used in almost every homework assignment (at least in reaction engineering). So, we've taken the liberty of including pictures of some common (and not-so-common) industrial reactors on our pages. Industrial Reactor Photos

Reactors Module An interactive module on chemcial engineering reactors was created by Sam Catalano, a student at the University of Michigan working with Professor Susan Montgomery in the Department of Chemical Engineering. Check out some screenshots from the Reactors Module. The module is available in both Windows 95 and Macintosh formats. Contact Professor Susan Montgomery, if you would like more information about this module.


file:///H:/html/01chap/html/reactors/reactors.htm[05/12/2011 16:55:16]

Problem 1-A Solution

Learning Resource/Additional Homework Problem CDP1-A Solution a. How many moles of A are in the reactor initially? What is the initial concentration of A?

If we assume ideal gas behavior, then calculating the moles of A initially present in the reactor is quite simple. We insert our variables into the ideal gas equation:

Knowing the mole fraction of A (yAo ) is 75%, we multiply the total number of moles (NTo) by the yA:

The initial concentration of A (CAo ) is just the moles of A divided by the volume:

b. Time (t) for a 1st order reaction to consume 99% of A.

With both 1st and 2nd order reactions, we will begin with the mole balance:

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There is no flow in or out of our system, and we will assume that there is no spatial variation in the reaction rate. We are left with:

Knowing the moles per volume (NA/V) is concentration (CA), we then define the reaction rate as a function of concentration:

First Order Reaction This is the point where the solutions for the different reaction orders diverge. Our first order rate law is:

We insert this relation into our mole balance:

and integrate:



Knowing C A=0.01 C Ao and our rate constant (k=0.1 min -1), we can solve for the time of the reaction:

c. Time for 2nd order reaction to consume 80% of A and final pressure (P) at T = 127 C.

Second Order Reaction Our second order rate law is:

We insert this relation into our mole balance:

and integrate:

We can solve for the time in terms of our rate constant (k = 0.7) and our initial concentration (CAo ):

To determine the pressure of the reactor following this reaction, we will again use the ideal gas law. First, we determine the number of moles in the reactor:



Now, we calculate the new pressure using the ideal gas law:

Back



01prob.htm

Additional Homework Problems CDP1AA

A 200-dm 3 constant-volume batch reactor is pressurized to 20 atm with a mixture of 75% A and 25% inert. The gas-phase reaction A

B+C

is carried out isothermally at 227°C. (a) Assuming that the ideal gas law is valid, how many moles of A are in the reactor initially? What is the initial concentration of A? (b) If the reaction is first order:

calculate the time necessary to consume 99% of A (c) If the reaction is second order:

calculate the time to consume 80% of A. Also calculate the pressure in the reactor at this time if the temperature is 127°C. CDP1BA

A novel reactor used in special processing operations is the foam (liquid + gas) reactor (Figure CDP1-B). Assuming that the reaction occurs only in the liquid phase, derive the differential general mole balance equation (1-4) in terms of

(Hint: Start from a differential mole balance.)

Figure CDP1-B Foam reactor. [2nd Ed. P1-10

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01prob.htm



Aspen Plus

Aspen PlusTM Creating and Simulating Chemical Reaction Models with Aspen Plus TM WELCOME to the ASPEN PLUS TM Pages! ASPEN PLUS TM is a software package designed to allow a user to build a process model and then simulate the model without tedious calculations. ASPEN PLUS TM can be used for a wide variety of chemical engineering tasks. For example, it can execute tasks as simple as describing thermodynamic properties of an ethanol and water mixture, or as complex as predicting the steady-state behavior of a full-scale petrochemical plant. This web site, however, will introduce ASPEN PLUS TM as a handy tool for simulating reaction engineering scenarios, such as designing and sizing reactors, predicting reaction conversions, and understanding reaction equilibrium behavior. So, get to know ASPEN PLUS TM by following the outline below. It will surely enhance your understanding of chemical reaction phenomena and the engineering principles behind them!

I. Introduction II. Accessing ASPEN PLUS TM III. Creating a Reaction Engineering Process Model A. Building a Process Flowsheet B. Entering Process Conditions IV. Running the Process Model A. Interpreting the Results B. Changing Process Conditions and Rerunning the Model V. Example Problems A. 8-6: Adiabatic Production of Acetic Anhydride B. 8-7: Operation of a PFR with Heat Exchanger VI. Other Need-to-Knows A. Saving your Process Model B. Printing your Process Model C. Changing Names of Streams and Unit Operations D. Changing Units of Parameters E. Exiting ASPEN PLUS TM VII. Credits

file:///H:/html/toolbox/aspen/index.htm[05/12/2011 16:55:18]

Thoughts on Problem Solving: Closed-Ended Problems

Closed Ended Problems are the type with only one right answer. These are the same types of problems that are usually found at the end of chapters in textbooks, and they reinforce concepts learned in the corresponding chapter. Algorithm - you can consistently solve difficult problems by following this simple heurisitic. (Includes links to solved example problems: PFR/CSTR and SREP.) Getting Unstuck - these tips can help you overcome mental barriers in problem solving.

file:///H:/html/probsolv/closed/cep.htm[05/12/2011 16:55:19]

Thoughts on Problem Solving: Open-Ended Problems

Open-ended problems are those which have many solutions or no solutions for the problem as defined. The solutions to these problems usually involves the use of all the skills discussed in Bloom's Taxonomy. First Steps in Solving Open-Ended Problems Examples of Previous OEPs Pharmacokinetics of Cobra Bites Bloom's Taxonomy

file:///H:/html/probsolv/open/oep.htm[05/12/2011 16:55:19]

Thoughts on Problem Solving Ten Types of Home Problems

1. Problems that require simple calculations, or formula substitution. These problems will bolster confidence, they are often limited to the knowledge and comprehension levels of Bloom's Taxonomy. Homogeneous Example 1 Heterogeneous Example 1

2. Problems that require intermediate calculations or manipulations. This type of problem may seem unsolvable at first glance, causing the student to go back and re-read the text and lecture notes. One needs to know what laws apply to make these calculations; consequently, this is level 3 in Bloom's Taxonomy application. Homogeneous Example 2 Heterogeneous Example 2

3. Problems that are over-specified so the student has to decide which data and conditions are relevent. Homogeneous Example 3 Heterogeneous Example 3

4. Problems that are under-specified so the student has to consult other information sources in order to complete the problem. Homogeneous Example 4 Heterogeneous Example 4

The Following Types of Problems will Receive Greater Emphasis in the Digital Age 5. "What if..." problems that promote discussion Homogeneous Example 5 Heterogeneous Example 5

6. Problems, or parts of problems (i.e. extensions), that are Open-ended. Homogeneous Example 6 Heterogeneous Example 6

7. Problems where the student must explore the situation by varying operating conditions or parameters. Here, the student may need to create techniques or guidelines to learn whether or not the solution is reasonable. Ordinary differential equation (ODE) solvers can be used to explore the problem. Homogeneous Example 7 Heterogeneous Example 7

8. Problems that challenge assumptions. The student can use ODE solvers or process simulators to redo the problem, changing the assumptions to learn the effects on the answer. Homogeneous Example 8 Heterogeneous Example 8

9. Problems where groups of students work on different parts (or the same part) then come together for discussion. The discussion provides the opportunity for the students to explore different points of view, as well as interact and increase their interpersonal skills. Homogeneous Example 9 Heterogeneous Example 9

10. Problems that develop life-long learning. These problems address the issue of teaching the students to learn on their own. With the explosion of knowledge that is occuring, it will be essential that they be able to learn material independently (i.e.. life-long learning skills). Homogeneous Example 10 Heterogeneous Example 10

file:///H:/html/probsolv/tentypes/index.htm[05/12/2011 16:55:20]

review.htm

Strategies for Creative Problem Solving by H. Scott Fogler and Steven E. LeBlanc

Book Summary Every individual possesses creative skills of one type or another. Honing these skills to a razor-sharp level requires work, and this book was designed to help problem-solvers improve their street smarts. Authors H. Scott Fogler and Steven E. LeBlanc provide a systematic approach to problem solving that helps guide readers through the solution process and the generation of alternative solution pathways. Specifically, Strategies for Creative Problem Solving presents techniques and guidelines that will help readers to: identify the real problem, effectively explore the constraints, plan a robust approach, carry it through to a viable solution, and then evaluate what has been accomplished.

Awards Authors H. Scott Fogler and Steven E. LeBlanc were chosen to receive the prestigious 1996 American Society of Engineering Education Meriam/Wiley Distinguished Author Award, in recognition of ground-breaking Strategies for

file:///H:/html/probsolv/strategy/psgen.htm[05/12/2011 16:55:21]

review.htm

Creative Problem Solving book.

Other Learn more about the authors Read a review of the book How can I purchase a copy?


file:///H:/html/probsolv/strategy/psgen.htm[05/12/2011 16:55:21]

Elements of Chemical Reaction Engineering, Updates -- 1st Printing

Typos in the First Printing The following are typographical errors that have been found after the book was printed, but before the CD-ROM was manufactured. PREFACE page xxvi: The virtual reality module is not on the CD but can be found at http://www.engin.umich.edu/labs/vrichel TABLE OF CONTENTS page xi: 8.5.3 Operating Conditions should read: 8.5.4 Operating Conditions CHAPTER 1 page 30: P1-18A, change "predictor-prey" to "predator-prey" relationships CHAPTER 2 page 52: FA0 is missing in the equation just above Eqn. (E2-6.2) and the upper bound of the integral should be 0.8, not 8. Should read: page 61 (top of the page): The upper bound of the integral in Equation (S2-4) should be 0.8, not 8. CHAPTER 3 page 117: P3-9 Part (a), end of sentence should read, "...and rate law parameters for P3-7(a)." CHAPTER 4

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page 128: Figure 4-2. There are four typos: a) Half way down the figure, the subzeros should be the opposite of what they are on P and T. Should read:

and

b) Change 3 to 4, i.e., 3. COMBINE should read: 4. COMBINE c) Equal sign missing. Should read: d) Last equation in the figure, there is an unwanted X before the [ln( )] term. Should read: page 146 (add to the margin): "From our discussion of reactor staging in Chapter 2, we could have predicted that the series arrangement would have given the higher conversion." page 177: a) Eqn. (2-20). Delete minus signs on c and d Should read: b) Table 4-5 1) Batch reactor balance on B: CA should be CB Should read: 2) The left hand side of the PFR and PBR mole balances should be multiplied by v0 Should read: PFR: PBR: page 191: There is an extra vo in Equation 4-58 -- delete one of them. The equation should read dCB/dt=rb+vo(CB0-CB)/V



page 195: Delete the V on the left hand side of Equation 4-65. Should read: page 206: Problem 2(d) should read (recall that alpha is a function of the particle diameter and Po). page 209: Problem 4-7(b) should read (Ans. 967 dm 3 ) page 219: Problem P4-31, change parameter values for K C and kCB . Should read:

CHAPTER 6 page 327: Problem 6-13(b), the answers should read (Ans.: CA = 0.61, CB = 0.79, CF = 0.25, CD = 0.45) CHAPTER 7 page 381: The y-axis on Figure E7-5.1(c) should read D not mu D page 418: Add What if ki >> kp and R1 = I o page 419: P7-21B, add the word styrene Line 3 should read "...in styrene polymerization for..." In the Library (F9) of the Polymath section of the CD Living Example Problems "Example 7-3" should read "Example 7-2 part b" CHAPTER 8 page 434: Equation (8-23) should have a "dT" after the brackets in the heat capacity term. page 443: The Fa in the box under V specified: should read Fao (i.e. entering molar feed rate).



page 450: Table E8-5.2 should read, Example 8-4 Non-adiabatic Reactor page 461: There is a 2 missing in Equation (T8-3.7). It should read Cc=2CaoX(To/T) page 498: just above Equation (8-77), replace (8-74) by (8-76). It should read: "...we divide Equation (8-67) by Equation (8-76) to obtain..." page 505: Example 8-12, delete the word "adiabatically" in the second line of the problem statement. It should read: "...take place in a 10 dm 3 CSTR..." CHAPTER 9 page 545: Equation E9-2.12, replace 142 with 35.85. It should read: "=35.85(448-298)" page 547: Second equation from the bottom of the page replace 1433 with 8600. It should read: "=4.48x105 kcal/min + 8600 kcal/min" page 577: Problem P9-11, add Cp(coolant) = 18 cal/mol/K page 578: P9-16(c) should read: "...versus time up to 6 hours and then..." CHAPTER 11 page 722: complete reference is AIChE J 44, p.1933(1998) CHAPTER 12 page 764: The eff. diff. should be 1.82 not 1.89 page 766: The eff. diff. should be 1.82 not 1.89 page 767: The units for rho_b are g/m3 , not g/cm3 CHAPTER 13 page 838: The last line just above Figure 13-13 the words early and late are reversed. It should



read: "The extremes of late and early mixing are referred to as complete segregation and maximum mixedness, respectively." In the margin, replace the word "one" with the word "are". It should read: "...conversion are:" page 844 (at the top of the page): Table E13-6.1 should read Table E13-2.1 Figure E13-6.1 should read Figure E13-2.1 page 852: Eqn. (13-71) "-" missing on right hand side. Should read: page 853: the second term of Equation E13-8.4, E2. Replace the second term (+ 1.180 ) with (+ 1.3618 ) The POLYMATH program is correct! page 872: Add to the margin: "A Model Must: 1. Fit the data 2. Be able to extrapolate theory and experiment 3. Have realistic parameters" CHAPTER 14 page 888: Eqn. (14-36). Change "+" to "-" in the term

Should read: page 908: Pe=Ul/De should read Pe=Ul/Da page 910: Problem 14-2(e), s = 0.01 cm2 /s Should read:

= 0.01 cm2 /s

Problem 14-3: "Maze and Blue" Should read: "Maize and Blue"



APPENDIX page 945: the rate constant above the A + BC ---> (ABC)* should be k1, not k2 page 946: Either equation (G-15) should read: -rA = kKc*(A)(BC), or (G-14) needs to use (B) instead of (BC) to be consistent. page 947: in the top portion, the equation should have a kb, not a k in it, so that it reads: q = (SIGMA)q(i) e^(eps(i) / kbT) page 947: Drop the subscript B in K B. It should read: "...where K is Boltzmann's constant."



Elements of Chemical Reaction Engineering, Updates -- Which Printing?

Which Printing Do I Have? If you do not know which printing you have, check the publication information on the reverse side of the title page. If you see Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 then you have the first printing. If you see Printed in the United States of America 10 9 8 7 6 5 4 3 2 then you have the second printing.


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Elements of Chemical Reaction Engineering, FAQs

Frequently Asked Questions (FAQs)

Chapter 1 1. In the formula for CSTR, if the rate of reaction is not constant and is dependent on the concentration, should we take to mean? the integral mean? Because the reactor is well-mixed, the concentrations, temperature, and rate of reaction are the same throughout the reactor volume, including the exit point. Consequently, the concentrations, temperature, and rate of reaction in the reactor are all evaluated at the exit conditions of the CSTR. 2. What is the difference between packed bed and fluidized bed reactors? Packed Bed Reactor: Catalyst particles are packed in a tube Fluidized Bed Reactor: Analogous to a CSTR with catalyst particles, see Figure 10-16 (p.620). 3. Why does a semi-batch reactor have better temperature control than a batch reactor? One can also control the feed rate of one of the reactants as well as control the heat exchanger. 4. Why would you choose to have CSTRs and PFRs in series? It is what reactors you might have on hand that you could connect together to obtain a high conversion.

Chapter 2 1. How would the problem involving three reactors in series in Chapter 2 change if there were sidestreams between? One can't use the definition of total conversion up to a point because the reactant is fed to the stream between reactors. One must work in terms of molar flow rates when writing the mole balances. 2. Is there ever a time when a CSTR will have a lower volume than a PFR for the same conversion and flow rates? Yes, for some adiabatic reactions. 3. In the 3-reactor series, (CSTR, PFR, CSTR) why wouldn't you just use the PFR to have the least volume overall to achieve the best conversion? You would if you had a PFR large enough. We are assuming that you have these reactors available for your use. See ICM-Staging.

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4. For the three reactors in series example, can a PFR use liquid like the 2 CSTRs? Yes. 5. If you had three reactors and two were CSTR's and the other was a PFR, would the PFR be placed at the end to minimize volume? Yes, in most instances when the reaction is isothermal and the curve of (1/-rA) increases monotonically (i.e. no valleys or mountains) with X. 6. Does

(for aA + bB

cC + dD) only hold for first order reactions?

No! This relationship has only to do with stoichiometry and nothing to do with rate laws. It holds for reactions of ANY order. 7. How do you use the equation to model reactors in parallel? See Example 4-2 (p.142).

Chapter 3 1. What is the frequency factor and where can we get values for it? What is it dependent on? Generally the frequency factor is independent of temperatures, however on occasion it can be a weak function of temperature. See p.944 2. Why is the limiting reactant our basis of calculation? One could calculate a NEGATIVE concentration otherwise. See Example 3-5 (p.90). 3. What is the relationship between the K in chemistry (A + B <--> C) and the k in the rate laws? KC is an equilibrium constant, and k is specific rate constant. k has units of time, K does not. 4. How does the k (specific reaction rate) depend on pressure, or does it? ONLY in very very rare instances at very high pressures such as, 6000 atm, for liquid phase reactions is k a function of pressure. See p.220 and CD-ROM on critiquing what you read. 5. What is the frequency factor, A, in the Arrhenius Equation; I want to know what it's physical meaning is and/or what it is a frequency factor. Arrhenius Equation is k = Ae-E/RT The frequency factor, A, is the coefficient of the exponential term. It has the same units as k. It is related to the number of collisions between molecules. See p.942 and 943. 6. What does the overall order of the power law model indicate? One can classify reactions by their overall order of reaction.



7. Who determines all the rate laws? These can be found in the literature, journals, books, tables, etc., see the footnote on p.75. They can also be determined in the laboratory. See Ch.5

Chapter 4 1. In solving problems for this class, is there ever a case where you need more steps than the mole balance, rate law, stoichiometry, and combining? (When do you deviate from this algorithm?) We will always use this basic algorithm, just add to these steps to it, e.g. the energy balance. We will not deviate from these first four steps. 2. What specifically causes a CSTR in series to have a higher conversion than a CSTR in parallel? The CSTR is always operating a the lowest concentration, the exit concentration. When say two CSTRs are in series, the first operates at a higher concentration, therefore the rate is greater, therefore the conversion is greater. The second reactor in series builds on the conversion in the first reactor. The conversion in the parallel scheme is the same as the conversion to the first reactor to the series scheme. See Figure p.50 and Example 4-2. 3. When are reactors in parallel used since it seems as though reactors in series would always achieve higher conversion? The PBRs in parallel are sued where there would otherwise be a large pressure drop in one long reactor (or identically several PBRs connected in series). 4. Is it possible to have a pressure drop for a liquid phase reaction, as is possible for a gas phase reaction? You can have a pressure drop in liquid phase systems, but it does not affect the reaction rate because liquids are virtually incompressible and therefore the concentration does not change with pressure. 5. Since two equal CSTR in series give a higher conversion than two in parallel, are reactors in parallel ever used to increase conversion? Not for a CSTR, only a PFR with DP. 6. The Damköhler (Da) number a. What is the Damköhler number? See p.138 of text. b. How is the Damköhler (Da) number defined for a reaction (A + B C) when the reaction is first order in A, first order in B, but second order overall? Just substitute the rate law evaluated entrance to the reactor, -r A0, [e.g. definition . See p.138. file:///H:/htmlmain/faqs.htm[05/12/2011 16:55:24]

] into the


c. Is Da always indicative of certain conversion? Yes, for irreversible reactions. d. How does defining an extra variable, the Damköhler number, save us time and confusion, as opposed to solving without it? When should it be used? It serves as rule of thumb. When Da < 0.1 then X < 10% and when Da >10 then X >90%. See margin note p.138 of Ch.4. 7. Is the only way to determine a rate constant, k, by experimentation? Yes. 8. Why do they use a batch reactor to determine k if they are going to be using CSTR in actual industrial process? Batch experiments are most always easier to take data to measure k. 9. In what cases would you use the order of magnitude reaction times other than to check k values that you calculate? When you are short on time and want to get quick engineering estimates. 10. At some of the polymer plants and refineries I've visited, a huge problem is fouling of the reactors. The plant workers would sometimes have to go into the reactors to break through the solids/sludge that adhere to the reactor walls. I imagine this solid build up leads to a drastic volume decrease. So, how do we take into account the change of volume and it's detrimental effect to conversion? Good point. It would however, catalyst decay by fouling is usually more important. See Ch.10. 11. Is rinsing the reactor with water after a batch ample cleaning, or are chemical cleans necessary in between batches? It depends, if there are no side reactions, a chemical clean is probably not necessary. Also the larger the reactor the greater the cleaning time. 12. Please clarify the method for deriving the rate law expressed with partial pressures. The conversion used when studying catalytic reactions is that the rate law is developed in terms of partial pressure, e.g.

.

Just use ideal gas law to relate to concentrations CA and CB

writing partial pressures in terms of conversion



13. How accurate is the perfect mixing assumption in dealing with CSTRs? It seems kind of far fetched that the entire CSTR is at the same concentration as the exit. True but these are ideal CSTRs, andnon ideal reactors and the perfect mixing assumption is discussed and modeled in Ch.13. Once we understand ideal reactors (perfectly mixed), we can easily model non-ideal reactors. 14. In a PFR or CSTR reactor, wouldn't the reaction still be happening in the pipe that the products leave through? Why does the reactor just magically stop occurring when the contents leave the reactor? Is there a good way to model a CSTR that is not perfectly mixed? It does continue to react to some extent. It depends on the temperature in the pipe! However, these are ideal reactors. Different models for Non-ideal reactors are discussed in Chapters 13 and 14. 15. Can you compare space time for a flow reactor to the time spent in a batch reactor for the purposes of measuring conversion? Sometimes. See Lecture 6 on the CD-ROM. 16. We understand that for a CSTR, the conversion X increases as residence time increases. We were unsure as to what the relationship is for a PFR? Same is true

for PFR. If you increase

and you increase X.

17. Is the conversion equation always the same no matter the order of reaction? No! This equation is only for first order. 18. Is the assumption that there are no radial gradients a good one for most tubular reactors? When is it not valid? It is quite a good assumption for turbulent flow. It is not valid for Laminar flow - see Ch.13 p.831. 19. How did you develop equation E4-9.8 on page 180? i.e. FB = 2 (F A0 - FA )

For every two moles of B are formed, one mole of A is consumed.



20. When trying to determine the optimal catalyst size for the internal diffusion limited case, do we always use the relation

or are there any other relations that can be used.

You can only use this relationship in the internal diffusion limited regime shown in Figure 12-5, with p.750. For a first order reaction (12-35)

or lumping all the constants not involving particle size into "a"

then k2 = k1

then

. This

for Pb. 4-23 one must use all of Figure 12-5. It can be

shown that Figure 12-5 can be represented as

21. When accounting for pressure drop in a membrane reactor, does the same method as we would use with a PFR apply? Yes,

22. Do we only use the

form of the PBR design equation for membrane reactors (IMRCF)?

23. In the text it states that for developing the design equation for a PBR when X << 1 you can use the relationship

Neglecting X

For what values of X is this valid?



24. Is there a transition regime for the engineering analysis we performed, if so, what is the relations? Yes, in the transition region one must use full equations for 25. For what particle size (if any) does the porosity change (or void) to make a calculation difference? It changes with packing of catalyst, but usually very little for the PBR and catalysts commonly used. 26. Why can't you write the membrane equations in terms of conversion X? Because you can't relate CB and X. 27. In the chapter we are given total cycle times excluding reaction for a batch polymerization. Is there a similar standard for a PBR when the catalyst must be removed and are there similar standards for flow systems that experience coking on the walls and need cleaning eventually. The onstream time for flow systems is much much greater than the down time for cleaning and repair. 28. Would there ever by a case in which a membrane reactor would be used for the reverse? I.e. if H 2 , for example, was used as a reactant in a reaction, would one ever want to run concentrated H 2 along the sides of the reactor and let it diffuse into the reaction zone? Yes, especially when O2 is one of the reactants. 29. In a membrane reactor, how can se quantify the equilibrium point on the graph of F vs. V? In this region (i.e. after the knee) the reaction is in equilibrium and the rate of removal of B is what limits the overall rate of reaction. 30. How can you assume that a semi-batch system has constant density? (p.199). Most liquids do not change density during the course of the reaction. 31. On page 199, why does Is it just for CD ?

and not

not like in all of the other examples?

In Case I, it was for a batch system, thus no flow, and that is why there is no vot term. If it was a semi-batch system with immediate evaporation:



Updates The following are typographical errors that have been found after the book was printed , but before the CD-ROM was manufactured. Please check the following web site for additional typographical errors. http://www.engin.umich.edu/~cre TABLE OF CONTENTS 1) page xi: 8.5.3 Operating Conditions should read 8.5.4 Operating Conditions CHAPTER 2 1) page 52: FA0 is missing in the equation just above Eqn. (E2-6.2) 0 .8 dX Should read: V2 = FA0 ∫ 0 −r A CHAPTER 4 1) page 128: Figure 4-2. There are four typos: a) Half way down the figure the subzeros are miss placed on P and T. P T P T Should read: v = v 0 (1 +εX ) 0 and V = Vo (1 + εX) 0 P T0 P T0 b) Change 3 to 4, i.e. 3. COMBINE Should read: 4. COMBINE P c) Equal sign mising. Should read: y = P0 d) Last equation in the figure, there is an unwanted X before the [ln( )] term. v  1  Should read: V = 0 (1+ε ) ln −εX k  1−X  2) page 177: a) Eqn. (2-20). Delete minum signs on c and d r r r r Should read: A = B = C = D − a −b c d b) Table 4-5 1) Batch reactor balance on B CA should be CB , i.e. should read dCB a = rA dt b 2) The left hand side of the PFR and PBR mole balances should be multipled by v 0 , e.g. should read dC dC b PFR v0 A = rA and v0 B = rA dV dV a dCA dCB b PBR v0 = rA and v0 = r dW dW a A

3) page 195: Delete the V on the left hand side of Eqn. (4-65).  C C  Should read: −rA = k CA CB − C D   KC  4) page 219: Problem P4-31, change parameter values for K C and k CB. Should read: K C = 0.01 mol2 dm6 k CB = 40 min -1 CHAPTER 6 1) page 327: Problem 6-13, part (b), Answers should read: (Ans.: CA = 0.61, CB = 0.79, CF = 0.25, CD = 0.45) CHAPTER 13 1) page 838: The last line just above Figure 13-13 the words early and late are reversed. It should read “The extremes of late and early mixing are referred to as complete segregation and maximum mixedness respectively.” dX −rA 2) page 852: Eqn. (13-71) “–” missing on right hand side. Should read = dt CA0 3) page 853: Eqn. (E13-8.4) replace the second term for E2 , (+ 1.180 e–6 λ 2) with (+ 1.3618 e –6 λ 2) The POLYMATH program is correct! CHAPTER 14  kC 1) page 888: Eqn. (14-36) Change “+” to “–” in the term + A   U  Da d2 CA dCA kCA Should read: − − =0 U dz2 dz U 2) page 910: Problem 14-2(e), (s = 0.01 cm 2 /s should read:: µ = 0.01 cm 2 /s

Sample Syllabus for a 3 Credit Course

ChE 381 University of Illinois Chemical Rate Processes and Reactor Design Schedule Professor Richard Braatz

Date

Topic

Reading

8/26

Introduction, basic mole balances

App A, Ch 1

8/28

P1-6, P1-7, and P1-2(b)s

8/31

Conversion in ideal reactors

9/2

P2-4, P2-7s, and P2-8

9/4

Basic chemical kinetics

9/7

Labor Day - no class

9/9

Stoichiometry

9/11

P3-10, P3-12, and CDP3-E

9/14

Ch 2

Homework Due

P1-3, P1-5, CDP1-8, P1-8(a)s

Ch 3.1

Ch 3.2-3.4

P2-3, P2-12s, P3-4, P3-6(d)

Reactor choice, sequences of reactors

Ch 4.1-4.3

P3-7, CDP3-D

9/16

Pressure drop in reactors

Ch 4.4-4.5

9/18

Simultaneous reaction and separation

Ch 4.6-4.7

9/21

P4-9s, P4-22, and P4-26

9/23

Review for Exam I

9/25

Exam I (Ch 1-4)

9/28

Analysis of rate data

Ch 5

9/30

Basic catalysis

Ch 10.1-10.5 P5-3, P5-7

10/2

Catalyst deactivation

Ch 10.6-10.8

10/5

P10-3, P10-12s, and CDP10-L or (P10-18)

10/7

Polymerization

Ch 7.1-7.3

10/9

Enzymatic reactions

Ch 7.4

P4-7s, P4-23s, P4-27

P10-4, P10-14

10/12 P7-5, Example 7-2, and CDP7-C 10/14 Review of thermochemistry file:///H:/htmlmain/ill_syll.htm[05/12/2011 16:55:25]

Ch 8.1-8.2

P7-2(a), P7-3, P7-4, P7-11


10/16 Nonisothermal reactors

Ch 8.3

10/19 Equilibrium conversion

Ch 8.4

10/21 Unsteady-state operation

Ch 9

10/23 Reactor stability

Ch 8.7

10/26 P8-13, P9-7, and P8-15 10/28 Multiple Reactions

Ch 6

P8-4, P8-5(a,b,c,d), P8-6

10/30 P6-5, P6-6, and P6-14 11/2

Review for Exam II

P6-4, P6-7, P6-8

11/4

Exam II (Ch 5-10)

11/6

Diffusion effects

Ch 11.1-11.3

11/9

Shrinking core model

Ch 11.4-11.5

11/11 P11-3 and P11-4 11/13 Diffusion and reaction in porous catalysts

Ch 12.1-12.3 P11-2(a,b), P11-5

11/16 Effectiveness factors

Ch 12.4-12.5

11/18 Packed bed reactors, limiting cases

Ch 12.6-12.7

11/20 Chemcial vapor deposition

Ch 12.8

11/23 P12-5 and P12-7 11/25 Slurry and trickled bed reactors (qualitative) Ch 12

P12-3, P12-4s, P12-6

11/27 Thanksgiving - no class 11/30 Bioreactors

Ch 7.5

12/2

P7-25 and CDP7-G

12/4

Residence time distributions

Ch 13.1-13.4 P7-24, P7-27

12/7

Micromixing

Ch 13.5-13.6

12/9

P13-12, P13-13

12/11 Nonideal reactors - last day of class

Ch 14

s = Suggestion

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P13-8, CDP13-A


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UM Course Syllabus, 4 Credit Hours

ChE 344: CHEMICAL REACTION ENGINEERING at the University of Michigan (4 Credit Hours) Instructor: Office Hours: TA's: TA Office Hours:

Professor H. Scott Fogler Tu and Th, 3:30 - 4:30 and as needed Dieter Andrew Schweiss and Aleksander Franz M and W, 3 - 5:30, 3158 Dow (Podbielniak Lounge)

F, 2:30-4:00, 3158 Dow (Podbielniak Lounge) Text: Elements of Chemical Reaction Engineering ( 3rd ed.), Prentice Hall, Upper Saddle River, NJ (1998). Recommended: Strunk and White, The Elements of Style. Lecture Date

Topic

Finish Reading

Hand In

1

Th Jan 8

Introduction POLYMATH Mole Balances

2

Tu Jan 13

Design Equations Rate vs. X Plots Summary Notes for Lectures 1 and 2

Chapter 1 Appendix A

1-1, 1-2 (a,b), 1-3, 1-8(a,b), 1-17, 1-18

3

Th Jan 15

Reactor Staging

Chapter 2

ICM: Kinetics Challenge I ICM: Staging 2-2, 2-6, 2-7

4

Tu Jan 20

Rate Laws Stoichiometry Summary Notes for Lectures 3 and 4

Chapter 3

3-1(a,b), 3-6, 3-9, 1-20

5

Th Jan 22

Expressing CA as a Function of X or of (F A / FT )

Chapter 3

ICM: Kinetics Challenge II 3-2(a), 3-10, 3-13, 3-21

6

Tu Jan 27

Isothermal Reactor Design Irreversible Reactions Summary Notes for Lectures 5 and 6

Chapter 4

ICM: Mystery Theater 4-6, 4-7

7

Th Jan 29

Reversible Reactions

Chapter 4

4-2(e), 4-12, 4-13

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Pressure Drop

Appendix C

8

Tu Feb 3

Spherical Reactors Membrane Reactors Summary Notes for Lectures 7 and 8

Chapter 4

ICM: Tic Tac 4-2(c,d), 4-23

9

Th Feb 5

Semibatch Reactors

Chapter 4 Integrating Factor handout

Start 4-1

10

Tu Feb 10 Analysis of Rate Data Summary Notes for Lectures 9 and 10

Chapter 5 4-1, 4-2(f,j), 4-26 Appendix A.3

11

Th Feb 12 Multiple Reactions

Chapter 6

5-4(a,b), 5-5, 5-19

12

Tu Feb 17 Multiple Reactions Summary Notes for Lectures 11 and 12

Chapter 6

6-2(d), 6-8

13

Th Feb 19 Multiple Reactions OEP Discussion

Chapter 6

6-12(a,b)

M Feb 23 14

15

Hand in Extra Problems by 5:30 PM at the TA Office Hours

6-12(c)

Tu Feb 24 Heat Effects Energy Balance Summary Notes for Lectures 13 and 14

Chapter 8

6-12(c), 6-17

Tu Feb 24 Mid Term Exam I (Closed Book) Media Union

Chapters 1-6 Crib Sheet

Group 1: 5:30-8:00PM Group 2: 8:15-10:45PM

Th Feb 26 Heat Effects PFR and CSTR Algorithm

Chapter 8

8-2(b-d)

F Feb 27

Sat February 28

Hand in OEP Interim Report at 3168 H.H. Dow Bldg by 3PM ***SPRING BREAK***

Sun March 8

16

Tu Mar 10 Heat Effects Summary Notes for Lectures 15 and 16

Chapter 8

8-5(a,b,c,d,f,g)

17

Th Mar 12 Heat Effects Multiple Steady States

Chapter 8

ICM: Heat FX 1 8-9

18

Tu Mar 17 Unsteady State

Chapter 9

ICM: Heat FX 2



Summary Notes for Lectures 17 and 18

8-20

19

Th Mar 19 Heat Effects in Multiple Reactions

Chapter 9

9-2(a), 9-5

20

Tu Mar 24 Octane Ratings Summary Notes for Lectures 19 and 20

Chapter 10

8-30 (to be worked individually), Triple Point Value

21

Th Mar 26 Catalysis

Chapter 10

ICM: Heterogeneous Catalysis 10-3(a,c)

22

Tu Mar 31 Catalysis CVD Summary Notes for Lectures 21 and 22

Chapter 10

10-8, 10-10 (Hint: use regression on 10-10)

23

Th Apr 2

Catalyst Decay

Chapter 10

10-14

24

Tu Apr 7

Moving Bed Reactors Summary Notes for Lectures 23 and 24

Chapter 10

10-19(b), 10-21

25

Th Apr 9

Bioreactors Summary Notes for Lectures 25 and 26

Chapter 7.5

Th Apr 9

Mid Term Exam II (Open Book) Media Union

Chapter 8, 9, 10

Group 1: 8:15-10:45PM Group 2: 5:30-8:00PM

Tu Apr 21 OEP Judging

Last Day of Classes

East Room Pierpont Commons 2-5:30PM

M Apr 28 Final Exam 11:30AM- 3rd Floor, Media Union 6:00PM

Chapters 1-6, 7.5, 8-10

Group 1: 12 Noon-2:30 PM Group 2: 3:00-5:30PM

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ChE 344 - Course Syllabus

Instructors Professor: Office Hours:

H. Scott Fogler Tu Th 3:30-4:30

Teaching Assistants: Probjot Singh, Barry Wolf, Anuj Hasija TA Office Hours: 3074G DOW, MW 2-6, F 11-12, Sa 11-12 (Sa hours - trial 1st 4 weeks)

Texts The Elements of Chemical Reaction Engineering H. Scott Fogler Problem Solving in Chemical Engineering with Numerical Methods Cutlip & Shacham (recommended) The Elements of Style Strunk & White (recommended)

Schedule January February March April 1) Thursday, January 7 Introduction, POLYMATH, Mole balances Topic: Finish Reading: Preface 2) Tuesday, January 12 Design equations, Rate versus X plots, Reactor staging Finish Reading: Chapter 1, P1-8, Appendix A, Course Guidelines from the Web Hand In: 1-1 (try to keep a notebook of items similar to P11 for each chapter), 1-2(c), 1-3, 1-18, homework group and signed rules sheet ICM: Kin Chal 1 (25 min) Study Problems: 1-10,1-12

Topic:

3) Thursday, January 14

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Topic: Rate laws, Stoichiometry Finish Reading: Chapter 2, CDP2-B, Chapter 3, pp 68-77 Hand In: 2-6, 2-7, 2-11 ICM: Staging (32 min) Class Problem: 1 Study Problems: 2-2,2-5, CDP2-A 4) Tuesday, January 19 Expressing CA as a function of X or (F A /F T) Finish Reading: Chapter 3 Hand In: 3-1(a,b), 3-3, 3-6, 3-7, 3-8, 3-9(a) ICM: Kin Chal 2 Class Problems: 2 Study Problems: 3-10, 3-14, 3-18, CDP3-AB

Topic:

5) Thursday, January 21 Isothermal reactor design Topic: Finish Reading: Chapter 4, pp 125-138 Hand In: 3-9(c,d) ICM: Mystery Theater Class Problems: 3 Study Problems: 4-3 6) Tuesday, January 26 Irreversible reactions Topic: Finish Reading: Chapter 4, pp 138-154 Hand In: 4-2(a), 4-6, 4-8(a) Class Problems: 4 Study Problems: 4-11 7) Thursday, January 28 Reversible reactions, Pressure drop Topic: Class Problems: 5 Finish Reading: Chapter 4, pp 154-182; Appendix C (omit Spherical Reactors, pp 168-173) Study Problems: 4-2, 4-9 Hand In: 4-12 8) Tuesday, February 2 Membrane Reactors Topic: Class Problems: 6 Finish Reading: Chapter 4, pp 182-187



Study Problems: 4-15, 4-19 Hand In: 4-2(b,d,j), 4-13, 4-20 9) Thursday, February 4 Semibatch Reactors 7 Chapter 4, pp 187-197 (omit Sections 4.7.3 and 4.8) Additional Reading: Integrating Factor Study Problems: 4-27, 4-31 Hand In: ICM: Tic Tac Topic: Class Problems: Finish Reading:

10) Tuesday, February 9 Analysis of Rate Data Topic: Class Problems: 8 Finish Reading: Chapter 5 (omit Section 5.3, Example 5-4, Sections 5.5.1 and 5.5.3, and pp 262-267); Appendix A.3; Wetlands Module; P4-17 Hand In: 4-2(f,g), 4-5(e,f), 4-25, 4-26 11) Thursday, February 11 Multiple Reactions Topic: Class Problems: 9 Finish Reading: Chapter 6, pp 282-304 Study Problems: 5-8, 5-20 Hand In: 5-3 ICM: Ecology Part I 12) Tuesday, February 16 Multiple Reactions Topic: Class Problems: 10 Finish Reading: Chapter 6, pp 304-316 Study Problems: 6-4, 6-6, 6-12 Hand In: 5-4, 5-19, 6-8 13) Thursday, February 18 Multiple Reactions, OEP Discussion Topic: Class Problems: 11 Finish Reading: Chapter 6 Study Problems: 6-16, 6-18 Hand In: 6-4(a,b)



14) Tuesday, February 23 Topic: Heat Effects, Energy Balance Finish Reading: Chapter 8, pp 383-406 Hand In: 6-13, 6-20 15) Thursday, February 25 - MIDTERM EXAM I Part I (80%): In-class 8:30 AM (closed book, note, web) Part II (20%): Due 9 AM Feb 26 to Barry Wolf 3106 DOW (open book, note, web) Crib Sheet

SPRING BREAK 16) Tuesday, March 9 CSTR and PFR Algorithms Topic: Class Problems: 12 Finish Reading: Chapter 8, pp 406-459 Study Problems: Hand In: 8-2(d), Interim OEP Report Outline ICM: Heat Effects I - Basketball 17) Thursday, March 11 Interstage Cooling Topic: Class Problems: 13 Finish Reading: Chapter 8, pp 460-473 Study Problems: 8-6, 8-7 Hand In: 8-5(omit e and f) 18) Tuesday, March 16 Multiple Steady States Topic: Class Problems: 14 Finish Reading: Chapter 8, pp 474-498 Study Problems: 8-9, 8-10, 8-12 Hand In: 8-3, 8-8 19) Thursday, March 18 Multiple Reactions, Unsteady State Topic: Class Problems: 15 Finish Reading: Chapter 8, pp 500-510; Chapter 9, pp 534-548 Study Problems: 8-21, 8-22 Hand In: Group Problem 8-20(omit g)



ICM:

Heat Effects II

20) Tuesday, March 23 Unsteady Steady State, Control of Reaction Topic: Class Problems: 16 Finish Reading: Chapter 9, pp 549-572 Study Problems: 8-31 Hand In: 8-30 (value = 1/3 of Midterm Exam) to be worked individually 21) Thursday, March 25 Octane Ratings Topic: Class Problems: Finish Reading: Chapter 10, pp 581-591 Study Problems: 9-2, 9-11 Hand In: 9-2(e) 22) Tuesday, March 30 Catalysis Topic: Class Problems: 17 Finish Reading: Chapter 10, pp 592-620 Study Problems: 10-4, 10-5, 10-7 Hand In: 9-3, 10-3, 10-8 23) Thursday, April 1 CVD, Catalyst Decay Topic: Class Problems: 18 Finish Reading: Chapter 10, pp 620-633 Study Problems: 10-12, 10-13, 10-14 Hand In: ICM: Heterogeneous Catalysis 24) Tuesday, April 6 Catalyst Decay, Moving Bed Reactors Topic: Class Problems: 19 Finish Reading: Chapter 10, pp 634-659 Study Problems: 10-21, 10-24(a) Hand In: 10-10 (use Regression) 25) Thursday, April 8 STTR Topic: Class Problems: 20



Finish Reading: Study Problems: 10-16 Hand In: Group Problems 10-18, 10-21(a,b) 26) Monday, April 12 - MIDTERM EXAM II Group A: 5:30-8:00 Group B: 8:15-10:45

Media Union

27) Tuesday, April 13 - NO CLASS (Work on OEP) 28) Thursday, April 15 - NO CLASS (Work on OEP) 29) Tuesday, April 20 - OEP POSTER PRESENTATIONS (3-6 pm) 26) Thursday, April 22 - FINAL EXAM 10:30 AM


Media Union

About the CD

ABOUT THE CD-ROM This CD-ROM is a companion to the book Elements of Chemical Reaction Engineering, 3rd Edition, by H. Scott Fogler. It contains web browser-driven content intended to supplement the information in the book. We suggest that you use Netscape NavigatorTM 3.x or greater or Netscape CommunicatorTM 4.x or greater to view the content of this CD. We have included the latest version of Netscape NavigatorTM on this CD. Also included on this CD are executable software intended to be installed and run directly from your computer. The software included are Windows® 95 and/or DOS 3.1 specific and unfortunately, do not include macintosh versions. There are instructions and links to this software from the CD.

System Requirements: Windows® 95/NT, or Power Macintosh (cannot run executable modules) CD-ROM Drive, 4x or higher Java-enabled, HTML browser (latest version of NetscapeTM is included) We recommend 36 MB of ram and at least 16 MB of available space on your hard drive to install the software. Prentice Hall does not offer technical support for this software. However, if there is a problem with the media, you may obtain a replacement copy by e-mailing us with your problem at [email protected].


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Problem 2-A

Learning Resource CDP2-A For the irreversible gas-phase reaction:

the following correlation was determined from laboratory data (the initial concentration of A is 0.2 g mol/L):

The volumetric flow rate is 5 m 3/s. a. Over what range of conversions are the plug-flow reactor and CSTR volumes identical? b. What conversion will be achieved in a CSTR that has a volume of 90 L? c. What plug-flow reactor volume is necessary to achieve 70% conversion? d. What CSTR reactor volume is required if effluent from the plug-flow reactor in part (c) is fed to a CSTR to raise the conversion to 90%? e. If the reaction is carried out in a constant-pressure batch reactor in which pure A is fed to the reactor, what length of time is necessary to achieve 40 % conversion? f. Plot the rate of reaction and conversion as a function of PFR volume. g. Critique the answers to this problem. Solution



Problem 2-A



02prob.htm

Additional Homework Problems CDP2AB

For the irreversible gas-phase reaction

the following correlation was determined from laboratory data. The initial concentration of A is 0.2 g mol/L.

The volumetric flow rate is 5.0 m 3/s. (a) Over what range of conversions are the plug-flow reactor and CSTR volumes identical? (b) What conversion will be achieved in a CSTR that has a volume of 90 L? (c) What plug-flow reactor volume is necessaryto achieve 70% conversion? (d) What CSTR reactor volume is required if effluent from the plug-flow reactor in part (c) is fed to a CSTR to raise the conversion to 90%? (e) If the reaction is carried out in a constant-pressure batch reactor in which pure A is fed to the reactor, what length of time is necessary to achieve 40% conversion? (f) Plot the rate of reaction and conversion as a function of PFR volume. (g) Critique the answers to this problem. CDP2BA

[2nd Ed. P2-12B ]

You are employed in a small company that is trying to build a plant to produce chemical X. As a result of your low overhead and other factors, you should be able to underprice your competitor significantly. You have sized all the pieces of equipment except the reactor(s). You cannot do this because you don't know the kinetic parameters, nor do you have laboratory equipment to determine them. Time is of the essence and your boss suggests that you take a photograph of your competitor's reactor which is located outside, and use that to size your reactor. He suggests that you hire a plane for aerial photographs or a truck similar to those used to repair telephone lines that could see over the fences (see Figure 2-7). Your boss also suggests you could get


02prob.htm

an estimate of the production rate by monitoring the size and number of trucks shipping the product from the plant. Using your results from Problem 2-17A as a basis, how confident are you that your estimates of the sizes will be reasonably accurate? Do you feel it is ethical to estimate the reactor sizes and production rates in this manner? Make a list of reasons and arguments as to why your boss might feel that it is ethical to request you to do this. If you don't feel that this is ethical, make a list of reasons and arguments as to why it should not be done. Suggest alternative ways to obtain the desired information. You may wish to consult the book The Power of Ethical Management by K. Blanchard and N. V. Peale (New York: Fawcett Crest., 1988) to identify and evaluate the ethical issues. If you feel that the situation above is clearly ethical or unethical, revise the scenario so that it is in a gray area. For example, would it be ethical if the reactor were in full view from the street? What if your boss suggested that you get a tour of the plant with a Boy Scout troop, and try to take pictures and obtain other information (e.g., read gauges) while on the tour at the Annual "Engineering as a Profession Day" three weeks from now? Use Blanchard and Peale's Ethics Check List Questions (Is it legal?, Is it balanced? How will it make me feel about myself? ) to help take the grayness out of the situation and make it black and white. [2nd Ed. P2-18B ] CDP2- Pure A is fed at a volumetric flow rate of 1000 ft3 /h and at a CA 3

concentration of 0.005 lb mol/ft to an existing CSTR, which is connected in series to an existing tubular reactor. If the volume of the CSTR is 1200 ft3 and the tubular reactor volume is 600 ft3, what are the intermediate and final conversions that can be achieved with the existing system? The reciprocal rate is plotted in Figure P2-7 as a function of conversion for the conditions at which the reaction is to be carried out.


02prob.htm

Figure P2-7 CDP2DB

A 400-L CSTR and a 100-L PFR are available to process 1.0 L of feed per second. The feed contains 41% A, 41% B, and 18% inerts. The irreversible gas-phase reaction

is to be carried out at 10 atm and 227°C. The rate of reaction in g mol/L min is given below as a function of conversion:

(a) What is the maximum conversion that can be achieved with these two reactors connected in series? (Ans.: X C,P = 0.445, XP,C. = 0.515.) (b) What would be the overall conversion if two 400-L CSTRs were connected in series for the same feed and operating conditions? (Ans.: X = 0.595.) (c) What would be the overall conversion if two 400-L CSTRs were connected in parallel with half of the feed going to each reactor? (Ans.: X = 0.52.) (d) What is the volume of a single tubular reactor necessary to achieve 60% conversion if the molar feed rate is 2 g mol A/min? (Ans.: V = 180 L) (e) If the total pressure were reduced by a factor of 10, would the conversion increase, decrease, or remain the same? (f) Plot the rate of reaction and conversion as a function of PFR volume. (g) Give a critique of the answers to this problem.



Lectures 3 and 4

Lectures 3 and 4

Rate Laws (Chapter3) * A rate law describes the behavior of a reaction. The rate of a reaction is a function of temperature (through the rate constant) and concentration.

Power Law Model

k is the specific reaction rate (constant) and is given by the Arrhenius Equation: k = Ae-E/RT , where E = activation energy (cal/mol) R = gas constant (cal/mol*K) T = temperature (K) A = frequency factor Example: If the rate law for the non-elementary reaction

is found to be

then the reaction is said to be 2nd order in A, 1st order in B, and 3rd order overall.

Elementary Reactions A reaction follows an elementary rate law if the stoichiometric coefficients are the same as the individual reaction order of each species. For the reaction in the previous example (above), the rate law would be:

Test yourself!

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Lectures 3 and 4

Reversible Reactions The net rate of formation of any species is equal to its rate of formation in the forward reaction plus its rate of formation in the reverse reaction: ratenet = rateforward + ratereverse At equilibrium, ratenet 0 and the rate law must reduce to an equation that is thermodynamically consistent with the equilibrium constant for the reaction. Example: Consider the exothermic, heterogeneous reaction

At low temperature, the rate law for the disappearance of A is

At high temperature, the exothermic reaction is significantly reversible:

What is the corresponding rate law? Let's see. If the rate of formation of A for the forward reaction (A + B

C) is

then we need to assume a form of the rate law for the reverse reaction that satisfies the equilibrium condition. If we assume the rate law for the reverse reaction (C A + B) is

then:

and:

Does this rate law satisfy our requirement that, at equilibrium, it must reduce to an equation that is thermodynamically consistent with KP ? Let's see. From Appendix C we know that for a reaction at equilibrium:


Lectures 3 and 4

At equilibrium, r net

0, so:

Solving for KP gives:

The conditions are satisfied.

Stoichiometric Tables (Chapter 3)

Using stoichiometry, we set up all of our equations with the amount of reactant A as our basis.

Batch System Stoichiometric Table (Chapter 3): Species

Symbol

A

A

B

B

C

C

D

D

Inert

I

Initial


Change

-------

Remaining

Lectures 3 and 4

________

____________

Where:

and

Concentration -- Batch System:

Constant Volume Batch:

[a liquid phase reaction, or a gas phase reaction in a rigid (e.g., steel) reactor]

etc.

Flow System Stoichiometric Table (Chapter 3):

Species

Symbol

Reactor Feed


Change

Reactor Effluent

Lectures 3 and 4

A

A

B

B

C

C

D

D

Inert

I

------________

____________

Where:

and

Concentration -- Flow System:

Liquid Phase Flow System:

etc. If the rate of reaction were


Lectures 3 and 4

then we would have

This gives us -rA = f(X). Consequently, we can use the methods discussed in Chapter 2 to size a large number of reactors, either alone or in series.

Gas Phase Flow System:

etc. Again, these equations give us information about -rA = f(X), which we can use to size reactors.

Algorithm for Isothermal Reactor Design (Chapter 4)

Example: The elementary gas phase reaction


Lectures 3 and 4

takes place in a CSTR at constant temperature (500 K) and constant pressure (16.4 atm). The feed is equal molar in A and B.

Mole Balance

Rate Law Stoichiometry

gas phase, isothermal (T = T0), no pressure drop (P = P 0)

[Why do you suppose C B is a constant, when B is being consumed?]

Combine

Evaluate

Example: The elementary liquid phase reaction


Lectures 3 and 4

is carried out isothermally in a CSTR. Pure A enters at a volumetric flow rate of 25 dm 3 /s and at a concentration of 0.2 mol/dm3 . What CSTR volume is necessary to achieve a 90% conversion when k = 10 dm 3 /(mol*s)?

Mole Balance

Rate Law Stoichiometry

liquid phase (v = vo)

Combine

Evaluate

at X = 0.9,

V = 1125 dm 3 Space Time

*

All chapter references are for the 3rd Edition of the text Elements of Chemical Reaction Engineering.


Lectures 3 and 4

Back to the top of Lectures 3 and 4.




Lectures 5 and 6

Lectures 5 and 6

Using the Algorithm for Isothermal Reactor Design (Chapter 4) * Gas Phase Elementary Reaction

Additional Information only A fed P 0 = 8.2 atm T0 = 500 K

C A0 = 0.2 mol/dm 3 k = 10 dm 3/mol-s vo = 25 dm 3/s

Batch

CSTR

PFR

Gas: V = V0

Gas: T = T0, P = P 0

Gas: T = T0, P = P 0

Mole Balance: Rate Law: Stoichiometry:

(e.g., constant volume steel container)

Combine: file:///H:/html/course/lectures/five/index.htm[05/12/2011 16:55:33]

Lectures 5 and 6

V = 378 dm 3

For X = 0.9:

V = 45.3 dm 3

t = 4.5 s

Reversible Reactions (Chapter 3) To determine the conversion or reactor volume for reversible reactions, one must first calculate the maximum conversion that can be achieved at the isothermal reaction temperature, which is the equilibrium conversion. (See Example 3-8 in the text for additional coverage of equilibrium conversion in isothermal reactor design.)

Equilibrium Conversion, Xe From Appendix C:

Example: Determine Xe for a batch system with constant volume, V = V0 Reaction

Additional Information C A0 = 0.2 mol/dm 3 KC = 100 dm 3/mol

Xe = 0.83

file:///H:/html/course/lectures/five/index.htm[05/12/2011 16:55:33]

Lectures 5 and 6

Example: Determine Xe for a PFR with no pressure drop, P = P0 Given that the system is gas phase and isothermal, determine the reactor volume when X = 0.8 X e.

Reaction

Additional Information C A0 = 0.2 mol/dm 3 KC = 100 dm 3/mol

k = 2 dm 3/mol-min F A0 = 5 mol/min

First calculate Xe:

Xe = 0.89 X = 0.8Xe = 0.711

One could then use Polymath to determine the volume of the PFR. The corresponding Polymath program is shown below.

Using Polymath

Algorithm Steps

Polymath Equations

Mole Balance

d(X)/d(V) = -rA/FA0

Rate Law

rA = -k*((CA**2)-(CB/KC))

Stoichiometry

CA = (CA0*(1-X))/(1+eps*X) CB = (CA0*X)/(2*(1+eps*X))

Parameter Evaluation

eps = -0.5 CA0 = 0.2 k=2 FA0 = 5


Lectures 5 and 6

KC = 100 Initial and Final Values

X0 = 0 V0 = 0 Vf = 500

Polymath Screen Shots Equations Plot of X vs. V Results in Tabular Form A volume of 94 dm 3 (rounding up from slightly more than 93 dm 3 ) appears to be our answer.

Arrhenius Equation (Chapter 3)

1.)

2.)

General Guidelines for California Problems (Chapter 4) file:///H:/html/course/lectures/five/index.htm[05/12/2011 16:55:33]

Lectures 5 and 6

Many of these problems involve an intermediate calculation to determine the final answer.

1. group unknown parameters/values on the same side of the equation example: [unknowns] = [knowns] 2. look for a Case 1 and a Case 2 (usually two data points) to make intermediate calculations 3. take ratios of Case 1 and Case 2 to cancel as many unknowns as possible 4. carry all symbols to the end of the manipulation before evaluating, UNLESS THEY ARE ZERO

*





Problem 3-A

Learning Resource Additional Homework Problems CDP3-AB Data on the tenebrionid beetle whose body mass is 3.3 g show that it can push a 35-g ball of dung at 6.5 cm/s at 27 C, 13 cm/s at 37 C, and 18 cm/s at 40 C. How fast can it push dung at 41.5 C? [B. Heinrich. The Hot-Blooded Insects (Cambridge, Mass.: 1993).]

Solution



Problem 3-B

Learning Resource Additional Homework Problems CDP3-B B For silicon to be used in the manufacture of microelectronic devices, it must contain less than 150 parts per trillion of impurities. This ultrapure silicon can be produced by reacting metallurgical silicon (98% pure) and HCl to form trichlorosilane and other products. This mixture is distilled and then the trichlorosilane is reacted with hydrogen at 1100 C to form polycrystalline silicon, HCl, and other silane products in the following reaction:

Set up a stoichiometric table for the following case: x=1

y=2

z=2

The total pressure may be taken to be 2 atm and the feed is stoichiometric. Sketch the concentrations of each species as a function of conversion. Solution


file:///H:/html/03chap/html/ahp03-b.htm[05/12/2011 16:55:35]

03prob.htm

Additional Homework Problems The elementary reaction

CDP3CB

is taking place only in the gas phase of a square duct. The feed to the duct consists of a gas stream of pure A and a liquid stream of pure B. The flowing liquid B covers the bottom of the duct and evaporates into the gas phase, maintaining its equilibrium vapor pressure throughout the system. The gas phase flows in plug flow. Ignore the volume occupied by liquid B (see Figure CDP3-C).

Figure CDP3-C

(a) Express the rate law solely as a function of conversion, and evaluate numerically all possible symbols. (b) What is the rate of reaction, -r A, when the conversion is 50%? Total pressure (considered constant): 1 atm Value of k: 106 ft3/lb mol s Temperature within the reactor (considered constant): 540°F Vapor pressure of B: 0.25 atm at 540°F Inlet flow rate of A: 1.5 lb mol/ s

(Ans.: 0.174 lb mol/ ft3

s.)


03prob.htm

CDP3DB

The following elementary gas-phase reaction takes place in a constant-pressure isothermal vessel (1 atm, 25°C):

Additional information:

CDP3-EB

Benzylamide is the product obtained from the liquidphase reaction of ammonia and benzoyl chloride:

(a) Taking benzoyl chloride as your basis of calculation, set up a stoichiometric table for a batch system. (b) If the initial mixture consisted solely of ammonia at a concentration of 6 gmol/L and benzoyl chloride at a concentration of 2 g mol/L, calculate the concentrations of ammonia and benzylamide when the conversion is 25%. (c) Taking ammonia as your basis of calculation, explain how your stoichiometric table would be different for a flow system.


03prob.htm



Lectures 7 and 8

Lectures 7 and 8

Pressure Drop in Reactors (Chapter 4) * Analyze the following second order gas phase reaction that occurs isothermally in a PBR:

Mole Balance Must use the differential form of the mole balance to separate variables:

Rate Law Second order in A and irreversible:

Stoichiometry

Isothermal, T = T0

Combine

file:///H:/html/course/lectures/seven/index.htm[05/12/2011 16:55:38]

Lectures 7 and 8

Need to find (P/P0 ) as a function of W (or V if you have a PFR).

Pressure Drop in Packed Bed Reactors (Chapter 4) Ergun Equation

Variable Density

let

Catalyst Weight where

let

then


Lectures 7 and 8

Isothermal Operation

let

then

recall that

notice that The two expressions are coupled ordinary differential equations. We can solve them simultaneously using an ODE solver such as Polymath. For the special case of isothermal operation and epsilon = 0, we can obtain an analytical solution.

Analytical Solution

Combine


Lectures 7 and 8

Solve

Could now solve for X given W, or for W given X.

Engineering Analysis We want to learn how the various parameters (particle diameter, porosity, etc.) affect the pressure drop and hence conversion. We need to know how to respond to "What if…" questions, such as: "If we double the particle size, decrease the porosity by a factor of 3, and double the pipe size, what will happen to D P and X?" To answer these questions we need to see how a varies with these parameters.

Turbulent Flow


Lectures 7 and 8

Compare Case 1 and Case 2. For example, Case 1 might be our current situation and Case 2 might be the parameters we want to change to.

For constant mass flow through the system

= constant

Laminar Flow

POLYMATH Consider the following gas phase reaction carried out isothermally in a packed bed reactor. Pure A is fed at a rate of 2.5 moles/s and with , and a = 0.0002 kg-1 . 2A Mole Balance

Elementary Rate Law

Stoichiometry Gas with T = T0


B

Lectures 7 and 8

A

B/2

POLYMATH will combine everything

________________

Profiles

Spherical Reactors (Chapter 4)


Lectures 7 and 8

Mole Balance

Stoichiometry Cross-sectional Area Pressure Drop Equations

Combine

Polymath will combine for you

Membrane Reactors (Chapter 4) Membrane reactors can be used to achieve conversions greater than the original equilibrium value. These higher


Lectures 7 and 8

conversions are the result of Le Chatelier's Principle; you can remove one of the reaction products and drive the reaction to the right. To accomplish this, a membrane that is permeable to that reaction product, but is impermeable to all other species, is placed around the reacting mixture. Example: The following reaction is to be carried out isothermally in a membrane reactor with no pressure drop. The membrane is permeable to Product C, but it is impermeable to all other species.

For membrane reactors, we cannot use conversion. We have to work in terms of the molar flow rates FA , FB , FC .

Polymath Program Mole Balances

Rate Laws

Stoichiometry


Lectures 7 and 8

Combine

Polymath will combine for you

*





Lectures 9 and 10

Lectures 9 and 10

Semibatch Reactors (Chapter 4) * The reactant that starts in the reactor is always the limiting reactant.

Three Forms of the Mole Balance Applied to Semibatch Reactors: 1. Molar Basis

2. Concentration Basis

3. Conversion

For constant molar feed:

For constant density:

Use the algorithm to solve the remainder of the problem.

Example: Elementary Irreversible Reaction file:///H:/html/course/lectures/nine/index.htm[05/12/2011 16:55:41]

Lectures 9 and 10

Consider the following irreversible reaction:

The combined mole balance, rate law, and stoichiometry may be written in terms of conversion and/or concentration:

Conversion

Concentration

Number of Moles

Polymath Equations: Conversion

Concentration

Moles

d(X)/d(t) = -ra*V/Nao

d(Ca)/d(t) = ra - (Ca*vo)/V

d(Na)/d(t) = ra*V

ra = -k*Ca*Cb

d(Cb)/d(t) = rb + ((Cbo-Cb)*vo)/V d(Nb)/d(t) = rb*V + Fbo

Ca = Nao*(1 - X)/V

ra = -k*Ca*Cb

ra = -k*Ca*Cb

Cb = (Nbi + Fbo*t - Nao*X)/V rb = ra

rb = ra

V = Vo + vo*t

V = Vo + vo*t

V = Vo + vo*t

Vo = 100

Vo = 100

Vo = 100

vo = 2

vo = 2

vo = 2

Nao = 100

Fbo = 5

Fbo = 5

Fbo = 5

Nao = 100

Ca = Na/V

Nbi = 0

Cbo = Fbo/vo

Cb = Nb/V

k = 0.1

k = 0.01

k = 0.01

Na = Ca*V

X = (Nao-Na)/Nao

Polymath Screenshots: Conversion

Concentration

Polymath Equations

Polymath Equations

Summary Table

Summary Table

Conversion vs. Time

Conversion vs. Time

Concentration vs. Time

Concentration vs. Time

file:///H:/html/course/lectures/nine/index.htm[05/12/2011 16:55:41]

Lectures 9 and 10

Volume vs. Time

Volume vs. Time

Equilibrium Conversion in Reversible Reactions (Chapter 4) Consider the following reversible reaction:

Everything is the same as for the irreversible case, except for the rate law:

Where:

See Also: Web Module on Reactive Distillation

Finding the Rate Law (Chapter 5) Consider the following reaction that occurs in a constant volume batch reactor:


Lectures 9 and 10

Mole Balance: Rate Law: Stoichiometry: Combine:

Differential Method (Chapter 5)

Taking the natural log of

The reaction order can be found from a ln-ln plot of:

Three Ways to Determine -dC A/dt (Chapter 5) 1. Graphical


Lectures 9 and 10

2. Polynomial (using Polymath) CA = a o + a 1 t + a 2 t 2 + a 3 t 3 +a 4 t 4

3. Numerical Finite Difference

Non-Linear Least-Squares Analysis

Generally want

to be a minimum.

For concentration-time data,

Integral Method (Chapter 5)

Zero Order

First Order


Second Order

Lectures 9 and 10

*





Wetlands

Wetlands Web Module What Are Wetlands? When most people think of wetlands, they often picture them as big swamps or marshes, like The Everglades in Florida -- places that are inhabited by alligators and other reptiles -- basically, some place where no one would want to live. In some respects, these people are right, but wetlands are much more than just "swamps." Wetlands support complex ecosystems and many species of birds, reptiles, amphibians, and fish call them home. And let's not forget the multitude of plant life you find in wetlands, either!

In addition to supporting plant and animal life, wetlands act as nature's water treatment facilities. In fact, wetlands have been used as low-cost alternatives to human water treatment plants since the early 1900's. For example, the Great Meadows natural wetland near the Concord River in Lexington, MA began receiving wastewater in 1912, and the Brillion Marsh in Wisconsin has received wastewater discharges since1923. The potential of wetlands for water treatment was not fully realized for many years, because wetlands were not monitored for water quality until the 1960's and 1970's. Fortunately, much has changed since then, for in 1994 there were over 300 wetland treatment systems in North America, and over 1000 worldwide. The Des Plaines River Wetlands Project (DPRWP) is a great example of how one can make use of natural resources to clean polluted rivers. Wetlands | DPRWP | Modeling Polymath | References This module is a learning resource for Chapter Four of Elements of Chemical Reaction Engineering, 3rd Ed.

file:///H:/html/web_mod/wetlands/index.htm[05/12/2011 16:55:42]

Wetlands

file:///H:/html/web_mod/wetlands/index.htm[05/12/2011 16:55:42]

Membrane Reactors

Membrane Reactors Chemical Reaction Engineering

Welcome to the Membrane Reactor Web Site! This site: 1. Introduces membrane reactors and their many applications. 2. Develops an algorithm for solving membrane reactor problems. 3. Shows an example of how membrane reactors work. 4. Compares membrane reactors with plug flow reactors. Main | Introduction | Algorithm Example | Comparison | Credits This module is a learning resource for Chapter Four of Elements of Chemical Reaction Engineering, 3rd Ed.

file:///H:/html/web_mod/membrane/index.htm[05/12/2011 16:55:42]

Reactive Distillation

This module is a learning resource for Chapter Four of Elements of Chemical Reaction Engineering. Reactive distillation is used with reversible, liquid phase reactions. Suppose a reversible reaction had the following chemical equation :

For many revesible reactions the equilibrium point lies far to the left and little product is formed :

However, if one or more of the products are removed more of the product will be formed because of Le Chatlier's Principle :

Removing one or more of the products is one of the principles behind reactive distillation. The reaction mixture is heated and the product(s) are boiled off. However, caution must be taken that the reactants won't boil off before the products. For example, Reactive Distillation can be used in removing acetic acid from water. Acetic acid is the byproduct of several reactions and is very usefull in its own right. Derivatives of acetic acid are used in foods, pharmaceuticals, explosives, medicinals and solvents. It is also found in many homes in the form of vinegar. However, it is considered a polutant in waste water from a reaction and must be removed.

By Timothy Mashue and H. Scott Fogler

Introduction

file:///H:/html/web_mod/distill/index.htm[05/12/2011 16:55:43]

Review

Problem

References

CRE -- Problem 4-A

Additional Homework Problems CDP4-AB A rather sinister-looking gentleman sidles up to you one night and in a sibilant whisper asks you to make him some methyl perchlorate. You question his motives because the product of the reaction between solid silver perchlorate and methyl iodide "explodes violently when struck" [M. F. Radies and T. Iredale, J. Phys. Chem., 48, 224 (1944)]. He responds by telling you the truth: he owns a tree stump removal business, and he needs cheap explosives. The legal route has occurred to him, but his funds are quickly depleting. You're not too comfortable with this situation, but times are hard, and you need the money. What you don't need, however, is to destroy your laboratory. Therefore, you decide to make the material in batch in a benzene solution, give the sinister stranger the product still in the benzene, and let him figure out how to get the methyl perchlorate out. You use a vessel containing 30 dm 3 of solution, starting with 0.7 M CH 3I and 0.5 M AgClO 4 concentrations. How long will it take you to convert 98% of the silver perchlorate? (You handle the silver perchlorate very carefully, since it, too, detonates with disturbing frequency when struck, jarred, or annoyed in other fashions.) Additional Data: CH 3I + AgClO 4

CH 3ClO 4 + AgI

where

at 298 K in benzene, k = 0.00042 (dm3/mol)3/2 /s. Solution



Thoughts on Problem Solving: Sample Registration Exam Problem

This problem was taken from a Professional Engineer's Exam. 1. Problem Statement: A homogeneous liquid phase second-order irreversible reaction

is to be carried out in a tubular (plug-flow) reactor. The PFR has a volume of 5.33 ft3 . It is proposed that to increase conversion, a stirred reactor (CSTR) of 100-gal capacity be installed in series with, and immediately upstream of, the PFR. Calculate the new conversion. 2. Real Problem:

Will a CSTR upstream of a PFR improve conversion? Which variables have the same values for both of the systems? Answer

Which variables need to be found in the PFR system in order to have enough data to work on the CSTR & PFR system? Answer 3. Sketch:

Figure 1. Schematic of the original design with PFR. see equations associated with sketch

file:///H:/html/probsolv/closed/alg/cre-exam/srp/index.htm[05/12/2011 16:55:45]


Figure 2. Schematic of the CSTR and PFR system It is necessary to break the problem into 2 parts: 1. Original design with PFR-determine the problem parameters (e.g. k) from the data given 2. New design with CSTR upstream of PFR - use the parameters for the new design Back to Chemical Reaction Engineering Examples

file:///H:/html/probsolv/closed/alg/cre-exam/srp/index.htm[05/12/2011 16:55:45]

French Menu Analogy

French Menu Analogy This image illustrates the analogy between a menu at a posh French restaurant and the algorithm for isothermal reactor design. Bon appetit!

file:///H:/html/04chap/html/link01.htm[05/12/2011 16:55:45]

French Menu Analogy


file:///H:/html/04chap/html/link01.htm[05/12/2011 16:55:45]

Learning Resources - Chapter 4

Learning Resources

Algorithm for Gas-Phase Reaction when conversion is not used as a variable.

file:///H:/html/04chap/html/04lr.htm[05/12/2011 16:55:46]

Learning Resources - Chapter 4



04prof.htm

Professional Reference Shelf 4.1 Time to Reach Steady State for a First-Order Reaction in a CSTR

To determine the time to reach steady-state operation of a CSTR, we begin with the general mole balance equation applied to a well-mixed CSTR (e.g., Figure CD4-13a):

Utilizing the definitions of F

have

(CD4-43) A

and N A, we

Mole balance

(CD443A)

Conversion does not have any meaning in startup because one cannot separate moles reacted from moles accumulated. Consequently, we must use concentration as the variable in our balance equation. For liquid-phase reactions V =V0 and for a constant overflow, v =v0. After dividing by v0 and replacing V/v0 by the space time , we find that

Stoichiometry

(CD4-44)

Analytical solutions to predict the effluent concentration as a function of time can be found only for zero- and first-order reactions. Using the rate law for a firstorder reaction (A B),

Rate law

(CD4-45)

and dividing by , we arrive at the differential equation

file:///H:/html/04chap/html/04prof.htm[05/12/2011 16:55:47]

04prof.htm

with the initial condition C A = C Ai at t = 0, where C Ai is the initial concentration of A in the CSTR. This first-order differential equation can readily be solved by Laplace transform or with the integrating factor, IF:

Analytical technique

Equation (CD4-45) then becomes

integrating and then dividing by gives

Using the initial condition to eliminate the constant of integration, K, we find

Concentration as a function of time for CSTR startup

(CD445A)

Equation (CD4-45A) gives concentration as a function of time from either startup or an upset occurring at t = 0. For the case of C Ai = 0, we will say that steady state is virtually achieved when the effluent concentration reaches 99% of its steadystate value, C As . At steady state:

(CD4-46)

(CD4-47)

Letting t s represent the time when steady state is virtually achieved, we have


04prof.htm

Combining equations (CD4-46) and (CD4 47) gives us

(CD4-48)

Solving for the time to reach steady state yields

For C A = 0.99C As ,

Time to approach steady state

(CD4-49)

For slow first-order reactions, the time for C A to reach 99% of C As is t s = 4.6 . For fast reactions, the time to reach steady state is 4.6/k. For most first-order systems, steady state is usually achieved in three to four space times or less.



04prof2.htm

Professional Reference Shelf

4.4.2 Recycle Reactors

Recycle reactors are used when the reaction is autocatalytic, or when it is necessary to maintain nearly isothermal operation of the reactor or to promote a certain selectivity (see Section 5.6.6). They are also used extensively in biochemical operations. To design recycle reactors, one simply follows the procedure developed in this chapter and then adds a little additional bookkeeping. A schematic diagram of the recycle reactor is shown in Text Figure 4-15.

The recycled stream is drawn off at point Q and merged with the fresh feed at Point P. We shall define the recycle parameter R as the moles recycled per mole of product removed at point Q.

Two conversions are usually associated with recycle reactors: the overall conversion, X0, and the conversion per pass, Xs :

Two conversions: Xs and X0

(4-88)

(4-89)

The only new twist in calculating reactor volumes or conversions for a recycle reactor is a mole balance at the stream intersections (points P and Q) to properly express the species concentrations as a function of conversion. Consider the gas-phase reaction

file:///H:/html/04chap/html/04prof2.htm[05/12/2011 16:55:48]

04prof2.htm

occuring in our reactor. Let X be the conversion of A in the reactor per mole of A fed to the reactor. The design equation is

Then: Design equation:

Rate law:

with Stoichiometry:

1. From the definition from the overall conversion, we can define FA3 and FB3 leaving the system,

(CD4-90)

(CD4-91)

From the definition for conversion per pass, we can define F A2 and FB3 leaving the reactor,

(CD4-92)

(CD4-93)

2. From the definition for the recycle parameter, R, we can define FAR and FBR and the total molar flow rate in the recycle stream, FtR

(CD4-94)


04prof2.htm

(CD4-95)

where

(CD4-96)

3. From the balance on the stream intersections, we have

(CD4-97)

(CD498) (CD499)

(CD4100)

Relating the molar flow rates in the various streams

(CD4101) (CD4102) (CD4103) (CD4104)

(CD4105)


04prof2.htm

(CD4106)

The volumetric flow rate in the reactor, , is related to the volumetric flow rate entering the reactor by

(CD4-107)

where X is the number of moles of A reacted per mole of A entering the reactor, and is defined by

(CD4108)

The molar flow rate of A within the reactor is

(CD4109) (CD4-110)

The concentrations of A and B are

(CD4-111)


04prof2.htm

These equations for concentration are substituted into the rate law, which is in turn substituted into the design equation and integrated. For a first-order reaction in A and in B,

Recycle reactor volume

(CD4-114)

(CD4-115)

where

The relationship between the overall conversion and the conversion per pass can be found by equating FA2 from Equations (CD4-107) and (CD4106):

Then using Equation (CD4-97) and simplifying, we have


(CD4-116)

04prof2.htm



04prof3.htm


4.3 Critiquing Journal Articles

Your textbooks after your graduation will be, in part, the professional journals that you read. As you read the journals, it is important that you study them with a critical eye. You need to learn if the author's conclusion is supported by the data, if the article is new or novel, if it advances our understanding, and to learn if the analysis is current. To develop this technique, one of the major assignments used in the graduate course in chemical reaction engineering at the University of Michigan for the past 20 years has been an in-depth analysis and critique of a journal article related to the course material. Significant effort is made to ensure that a cursory or superficial review is not carried out. The students are asked to analyze and critique ideas rather than ask questions such as: Was the pressure measured accurately? They have been told that they are not required to find an error or inconsistency in the article to receive a good grade, but if they do find such things, it just makes the assignment that much more enjoyable. Beginning with Chapter 4, a number of the problems at the end of each chapter in this book are based on the students' analyses and critiques of journal articles and are designated with a C (e.g., P4C-1). These problems involve the analysis of journal articles that may have minor or major inconsistencies. Analysis is the process of breaking a problem down into recognizable elements (i.e., pieces) such that meaningful relationships exist between the elements. Consequently, the first step in analyzing a problem is to identify all the elements of the problem and clarifying them.

Finally, after classifying the elements into different groups, we need to formulate relationships between the elements within a group and between groups. The heart of the analysis process is the careful comparison of similarities and differences of the elements. What distinction accounts for the


04prof3.htm

differences? What is the reason for the distinction? Look for increases or decreases between sequential elements in a series. Seek correspondences between diverse ideas. In analyzing and critiquing journal articles, students are asked to identify elements of the article and classify them according to key ideas, key assumptions, key theories, key experiments, and key results. Students then continue to analyze the article by asking such questions as: 1. What do you believe was the most difficult hurdle for the authors to overcome? 2. What is the novelty or innovation in this work? 3. How will the results or techniques described in this work be useful to other investigators? 4. Can the author's results be explained by another theory or hypothesis? 5. Why are the last data points always off the curve predicted by the theoretical model? All challenges by students of ideas and conclusions of the articles must be supported by calculation or pertinent literature. When reading the literature, students look for experimental points that fall off the theoretical curve and for statements or equations that are not in a familiar form and do not look quite right. For example, the article analyzed for Problem P4C-1 discusses the effect of pressure on a liquidphase reaction of the type A+B

C

The rate equation for the batch reaction was written in terms of the mole fraction of A, y, for an equal molar feed containing an inert I:

The results showed that the specific reaction rate ky increased by a factor of 250 as the pressure was increased to 6000 atm. A mole balance on species A for this irreversible elementary reaction is


(PS4-1)

(PS4-2)

04prof3.htm

where ka is the specific reaction rate based on activities aA and aB . Substituting for

in Equation (PS4-2) we will find

(PS4-3)

where C T0is the initial total liquid-phase concentration. Comparing equations (PS3-2) and (PS3-3), we have

Consequently, it may be possible to explain all or part of the increase in the specific reaction rate by the fact that the initial concentration, C T0, increases as pressure increases, especially for a pressure increase of 6000 atm. By learning the liquid compressibility of this or an analogous liquid, one may carry out calculations to see if this challenge is substantiated.

Before the students write the final copies of their critiques, they are asked to outline the challenges they intend to pursue. Many students state that they will check the mechanism proposed by the author, check the assumptions made (such as neglecting pressure drops), rederive the basic material balance equation, and so on. Unless there is evidence that something looks a little "fishy," these types of statements are not sufficiently penetrating, as a deeper level of questioning is required in a critique. If an assumption is to be challenged, it should be one that would affect the maximum sensitivity of the experiment. For example, if a 10-fold increase in the pressure would only result in a 2% increase in the rate of reaction, it doesn't make much sense to redo the pressure-drop calculations for a packedbed reactor. On the other hand, if a temperature rise of 2 or 3 degrees causes the reaction rate to increase significantly, it would be logical to check the isothermality of the reaction bed. Finally, one should refrain from proposing questions relating to


04prof3.htm

the authors' experimental planning and techniques, since one would not be able to resolve that without going to the laboratory and actually taking the data. However, suggesting key experiments that would resolve conflicting theories or reaction mechanisms is encouraged.



04prob.htm

Additional Homework Problems (Ecological engineering) Several researchers have examined the feasibility of using wetlands to clean up high volumes of polluted water. Experiemental wetland three (EW3) at the Des Moines Experimental Wetlands site in Illinois has a volume of 15,000,000 dm 3 and an inflow of 70,000 dm 3 of water per house from the Des Plaines Rier. The outflow eventually returns to the river. During late spring, the river of the herbicide atrazine. water typically contains However, the maximum contaminant level (MCL) under the As a first federal Drinking Water Act is approximation treat EW3 as a perfectly mixed CSTR and assume that atrazine decomposition is first order with k= 0.0025 h-1.

CDP4-BB

(a) First consider that EW3 contains water but no atrazine at the time the flow fromthe river is diverted. Plot C Aas a function of t for the case where the outflow is kept equal to inflow. After what time does C A reach 99% of its steady-state value? Is it below the MCL? (b) Next consider the case where EW3 initially contains neither atrazine nor water. Outflow is kept at 50,000 dm 3/h for 750 h. after which it becomes 70,000 dm 3/h. Plot the concentration as a function of time and explain how it differs from the plot in part (a). Plot the number of moles of atrazine in EW3 versus time. Why is N A increasing while C Aversus time is decreasing? (c) EW3 is operating initially at the steady-state conditions found in part (a). Suppose that weekly thunderstorms periodically increase the amount of atra zine leached into the Des Plaines River, thereby increasing the concentration in the inflow to EW3. Concentration as a function of time is given by

where the trigonometric argument is in radians. Plot C A0 and C Aversus t on the same graph. Does the outflow exceed the MCL at any time? Do C A0 and C A reach their maximums and minimums at the same times? Does C Aever exceed CA0 ? How can this be? file:///H:/html/04chap/html/04prob.htm[05/12/2011 16:55:51]

04prob.htm

(d) EW3 is operating initially at the steady-state conditions found in part (a). Suppose that there is a drought lasting 1000 h during which the evaporative flux of water from EW3 is 10,000 dm 3/h. The overall water balance is such that the wetland volume remains constant, however. No atrazine leaves via evaporation. What does the concentration profile look like? How can this be explained?

CDP4-CB

The cells in your body need to obtain nutrients, hormones, growth factors, and other molecules present in very low concentrations in the fluid around them. To avoid engulfing a large quantity of this fluid and then intracellularly separating useful from useless molecules, the cells possess what are known as receptors on their surface. These receptors are able to bind interesting molecules or ligands with high affinity, thus capturing molecules for the cell's use (Figure CDP4-C). (a) You are growing 106 cells/mL in a T flask containing 10 mL of media. Each cell has 105 receptors on its surface. The association rate constant for the binding of ligands to receptors is 106 M-1 min -1. Calculate the time for 50% of the receptors to bind ligands if you add ligands at a concentration of 10-7 M. Assume irreversible binding and perfect mixing. (b) Show that the ligand concentration in part (a) is sufficiently in excess so that the binding could be considered pseudo-first order.


04prob.htm

Figure CDP4-C

(c) The binding of ligand to receptor is actually a reversible reaction. For the binding of your ligand to receptors, the dissociation rate constant kr is 0.1 min -1 . Using the approximation justified in part (b) and assuming perfect mixing, calculate the percentage of receptors bound 5 min after you add the ligand to the media. (J. Linderman, University of Michigan)

[2nd Ed. P4-34]

CDP4-DB

(Batch bromination of p-chlorophenyl isopropyl ether) You are in charge of the production of specialty chemicals for your organization and an order comes in for 3 lb of brominated p-chlorophenyl isopropyl ether. You decide to use the technique reported by Bradfield et al. [J. Chem. Soc., 1389 (1949)], who carried out the reaction in 75% acetic acid at 68°F. You have a batch reactor that holds 5 gal (0.670 ft3) of a reacting mixture that can be used. Starting out with a mixture that contains 0.002 lb mol (0.34 lb) of pchlorophenyl isopropyl ether and 0.0018 lb mol (0.288 lb) of bromine in the 5 gal, you decide to run 10 batches of the mixture to 65% conversion of the p-chlorophenyl isopropyl ether. This procedure will give the desired 3 lb. How long will each batch take? Additional information: Kinetics (from Bradfield et al.):


04prob.htm

Reaction: where A is p-chlorophenyl isopropyl ether, B is bromine, and C is monobrominated product. Rate Law: Specific reaction rates at 68° F: k1 = 1.98 ft3/lb mol min k2 = 9.2 x 103 (ft3/lb mol)2 min -1 [2nd Ed. P4-29]

CDP4-EB

A liquid organic substance, A, contains 0.1 mol % of an impurity, B, which can be hydrogenated to A:

The material is purified by hydrogenation as a liquid in a continuous well-mixed reactor at 100°C. The feed rate of the liquid is constant at 730 lb/h. The reactor holds 50 gal of liquid, at 500 psig, and the amount of B in the product levels out at 0.001 mol %. What will be the concentration of B in the product if the hydrogen pressure is held at 300 psig? Assume that the reaction behaves as though it were first order with respect to both B and H2, that is, in batch,

Assume perfect gas laws and Henry's law. Also assume the following properties:


04prob.htm

[2nd Ed. P4-15]

CDP4-FA

The gas-phase reaction is to be carried out in an isothermal plug-now reactor at 5.0 atm. The mole fractions of the feed streams are A = 0.20, B = 0.50, and inerts = 0.30. (a) What is the steady-state volumetric flow rate at any point in the reactor if the pressure drop due (1 to fluid friction can be ignored? [Ans.: 0.2X).] (b) What are the expressions for the concentrations of A, B, and D as a function of conversion at any point along the reactor? (c) What is the feed concentration (units: mol/ dm3) of A if the feed temperature is 55°C? (d) Determine how large the plug-flow reactor must be to achieve a conversion (based on A) of 0.70 if the temperature in the reactor is uniform (55°C), the volumetric feed rate is 50 dm3/min, and the rate law at 55°C is -r = 2.5 CA(1/2) C B kmol/m 3 min

(Ans.: V = 50.21 dm3.)

(e) Plot the concentrations, volumetric flow rate, and conversion as a function of reactor length. The reactor diameter is 7.6 cm. (f) How large would a CSTR have to be to take the effluent from the PF reactor in part (d) and achieve a conversion of 0.85 (based on the feed of A to the plug-flow reactor) if the temperature of the CSTR is 55°C? (g) How many 1-in.-diameter pipe tubes, 20 ft in length, packed with a catalyst, are necessary to


04prob.htm

achieve 95% conversion of A starting with the original stream? Plot the pressure and conversion as a function of reactor length. The particles are 0.5 mm in diameter and the bed porosity is 45%. (h) Calculate the PFR size to achieve 70% of the equilibrium conversion and the CSTR size necessary to raise the conversion of the PFR effluent to 85% of the equilibrium conversion if their temperatures were uniform at 100°C. The activation energy for the reaction is 30 kJ/mol, and the reaction is reversible with an equilibrium constant at 100°C of 10 (m3/kmol)1/2 . (Ans: VPFR = 8.56 dm3, VCSTR= 6.45 dm3) [2nd Ed. P4-8]

CDP4-G B

You are designing a reactor system for carrying out the constant-density liquid-phase reaction

which has the rate law

(a) What system (i.e., type and arrangement) of flow reactors, either one alone or two in series, would you recommend for continuous processing of a feed of pure A in order to minimize the total reactor volume? (90% conversion of A is desired.) (b) What reactor size(s) should be used? (c) Plot the conversions and concentrations of A and B as a function of plug-flow reactor volume. Additional information: k 1 = 10.0 (lb-mol/ft 3 ) 0.5 h-1 k 2 = 6.0 ft 3 /lb mol feed = 100 lb mol/h of pure A CAO = 0.25 lb mol/ft 3


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total pressure = 2000 kPa [2nd Ed. P4-14]

CDP4-H A

The removal of nitrogen from organic compounds (i.e., hydrodenitrogenation) is an important industrial reaction. We consider the first reaction step of this liquid-phase process in which 5,6-benzoquinoline (species A) is reacted at 100°C in a solution saturated with hydrogen (2500 psig) (i.e., excess H2) [Ind. Eng. Chem., 28, 139 (1989)]. The following data were obtained at 100°C in a batch reactor using sulfided CoMo as a catalyst at a concentration of 20 g/dm 3.

Also note that the rate at 110° C is approximately four times the rate at 80°C. Verify that the reaction is pseudofirst order in 5,6-benzoquinoline and determine the specific reaction rate. It has been learned that the specific reaction rate is directly proportional to the catalyst concentration (i.e., first order). It is proposed to double the catalyst concentration and drop the temperature to 90°C. Plot the conversion expected under these conditions as a function of time and compare with the data above. [2nd Ed. P4-7]

CDP4-I B

The liquid-phase reaction is carried out in a semibatch reactor. The reactor volume is 1.2 m 3. The reactor initially contains 5 mol of B at a concentration of 0.015 kmol/m 3. A at an aqueous concentration of 0.03 kmol/m 3 is fed to the reactor at a rate of 4 dm 3/min. The reaction is first order in A and half order in B with a specific reaction rate of k = 6 (m3/kmol)1/2 / min. The activation energy is 35 kJ /mol. The feed rate to the 3


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reactor is discontinued when the reactor contains 0.53 m of fluid. (a) Plot the conversion, volume, and concentration as a function of time. Calculate the time necessary to achieve: (b) 97% conversion of A. (c) 59% conversion of B. (d) The reaction temperature is to be increased from 25°C to 70°C and the reaction is to be carried out isothermally. At this temperature the reaction is reversible with an equilibrium constant of 10 (m3/kmol)1/2 . Plot the conversion of A and B and the equilibrium conversion of A as a function of time. (e) Repeat part (d) for the case when reactive distillation is occurring. Study the effect of the evaporation rate on conversion.

[2nd Ed. P4-27]

CDP4-JB

The irreversible liquid-phase acid-catalyzed isomerization reaction

is carried out isothermally in a semibatch reactor (Figure CDP4-J). A 2 M solution of H 2SO 4 is fed at a constant rate of 5 dm 3/ min to a reactor that initially contains no sulfuric acid. The initial volume of pure A solution in the reactor is 100 dm 3. The concentration of pure A is 10 mol / dm 3. The reaction is first order in A and first order in catalyst concentration, and the specific reaction rate is 0.05 dm 3/mol min -1. The catalyst, of course, is not consumed during the reaction.


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Figure CDP4-J

(a) Determine both the number of moles of A and of H and of H2SO 4 in the reactor and the concentration of A and of H2SO 4 as a function of time. (b) Obtain an analytical solution for the number of moles of A, N A , and the concentration of A, C A , as a function of time. What are the concentrations of A and of and of H2SO 4 after 30 min? (c) If the reaction is reversible with K C = 2.0, plot the concentration of A and C as a function of time. (d) How would your answers change if a 2 M solution of A were fed to a 2 M solution by H2SO 4 that had an initial volume of 100 dm 3? (e) Rework this problem where A is fed at a concentration of 2 M and 5 dm 3/min to 200 dm of 2 M H2SO 4 . If the reaction is first order in A and zero-order in H2O and the specific reaction rate is 0.05 min -1, what is the concentration of A after 30 min? If the reaction is reversible, with K C = 2.0, plot the equilibrium conversion and the concentrations of A and C as a function of time.

(Hint: Try to use the mole balance expressed in terms of N A .)

CDP4-

In many industrial processes where the conversion per


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KB

pass through the reactor is low, it may be advantageous to use a recycle reactor (Figure CDP4-K). Here a significant portion of the exit stream is recycled back through the reactor. Calculate the overall conversion

that can be achieved in a 2-m 3 plug-flow reactor when the irreversible, isothermal first order gas-phase reaction

is carried out at 500°C and 5 m 3 feed of gas is recycled for every cubic meter of fresh.

Figure CDP4-K


[1st Ed. P4-28]

CDP4-LB

The elementary gas-phase isomerization reaction

is carried out in a packed-bed recycle reactor. The recycle ratio is 5 mol recycled per mole taken off in the exit stream. For a volumetric flow rate of 10 dm 3/s through the reactor (Figure CDP4-L), the corresponding pressure gradient (assumed constant) in the reactor is 0.0025 atm/ m. The flow in the reactor is turbulent. What overall conversion can be achieved in a reactor that is 10 m in length and 0.02 m 2 in cross-sectional area? Additional information:


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C A0 =0.01 mol /dm3 (P0= 2 atm). The volumetric flow rate of fresh reactant to the reactor is 0 = 10 dm3/s. The specific reaction rate is k = 0.25 s -1.

[1st Ed. P4-22]

Figure CDP4-L

CDP4MB

Consider the recycle reactor system shown in Figure CDP4-M, where the elementary irreversible gas-phase is carried out isothermally at reaction 570°C and 1 atm. A part of the condenser unit is heated to reactor temperature and recycled to reactor inlet. The pressure drop in the conduits can be assumed to be negligible. Estimate the reactor volume for 50% converstion of the feed FA0 .

Additional information: FA0 = 1 kmol/h FB0= 1 kmol/h k = 100 m 3/kmol h R=4 Condenser exit temperature = 45°C Vapor pressure of C at 45°C = 0.2 atm


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Figure CDP4-M (Courtesy of H. S. Shankar)

CDP4-NB

Radial-flow reactors can be used to good advantage for exothermic reactions with high heats of reaction. The high radial velocities at the entrance to the reactor are useful in reducing hot spots within the reactor. As fluid moves out into the reactor, the velocity, U, varies inversely with r:

where U0 is the velocity (m/s) at the inlet radius, R irreversible, gas-phase reaction R0. Consider the elementary, irreversible, gas-phase reaction

carried out in a radial-flow reactor similar to the one shown in Figure CDP4-N. Derive an equation for conversion as a function of radius carried out in a radialflow reactor similar to the one shown in Figure P4-38.

(a) Derive an equation for conversion as a function of radius for isothermal operation neglecting pressure drop. (b) Plot X as a function of r for the case when the pressure drop is significant with a 5 0.07 kg-1. (c) Vary the reaction and reactor parameter values and write a paragraph describing your findings. What parameter effects the results the file:///H:/html/04chap/html/04prob.htm[05/12/2011 16:55:51]

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most?

Figure CDP4-N

Suggested parameter values: , FA0 = 10 mol/min, C A0 = 0.1 mol/dm 3,

h = 0.4 dm, R0 = 0.1 dm, R1 = 10 dm Bulk density of catalyst 2000 g/dm3 [2nd Ed. P4-31]

CDP4OB

The growth of a bacterium, B, is to be carried out in excess nutrient:

The growth rate for this bacteria is best described by a logistic growth model:

Where C B is the cell concentration (g/dm3) and C MAX are constant rate law parameters.

and

(1) Plot the cell concentration as a function of time in a 10-dm 3 batch reactor. (2) If the reaction is to be carried out in a 10-dm3 CSTR,


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what is the exit cell concentration for a volumetric flow rate of 0.5 dm 3/h. Vary the volumetric flow rate to arive at a plot of exit cell concentration as a function of space time. On the same figure, plot the total numbr of cells exiting the reactor (i.e. C B ) as a function of space time. Additional information: Initial cell concentration in feed 10-6 g/dm 3, C Bmax = 5 x 10-3 g/ dm 3.

= 0.5 h< -1,

[2nd Ed. P4-35]

CDP4-PB

A bimolecular (elementary) second-order reaction , takes place in a homogeneous liquide system. Reactants and products are mutually soluble, and the volume change as a result of reaction is negligible. Feed to a tubular (plug-flow) reactor that operates essentially isothermally at 260°F consists of 210 lb/h of A and 260 lb/h of B. Total volume of the reactor is 5.33 ft3, and with this feed rate, 50% of compound A in the feed is converted. It is proposed that to increase conversion, a stirred reactor of 100-gal capacity be installed in series with, and immediately upstream of, the tubular reactor. If the stirred reactor operates at the same temperature, estimate the conversion of A that can be expected in the revised system; neglect the reverse reaction. Other available data include:

(Ans.: X = 0.68.) (California Professional Engineers Exam) [2nd Ed. P4-15]

CDP4-


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The irreversible liquid phase acid catalyzed reaction QB

is carried out in a semibatch reactor containing H2SO 4. A is fed at a constant rate of 10 mol/min. The volumetric flow rate of liquid entering the semibatch reactor is 5 dm 3/min. The initial volume of a 3 M solution of H2SO 4 catalyst in the reactor is 100 dm 3 (no A is present initially). The specific reaction rate is 0.05 min -1. The reaction is first order in A and zero-order in catalyst concentration. (a) Use POLYMATH or MATLAB to determine both the number of moles of A in the tank and the concentration of A and of H2SO 4 as a function of time. (b) Obtain an analytical solution for the number of moles of A, N A , and the concentration of A, C A , as a function of time. What is the concentration of A after 30 min? How many moles of A will there be in a tank )? (Ans.: N A = after long times (i.e., 200 mol) Explain why the number of moles remains virtually constant at long times. (c) Rework part (a) assuming the reaction is first order in H2SO 4with k = 0.02 dm 3/ mol min. (Hint: Do not try to use conversion in solving this problem!)



CRE -- Problem 5-B

Additional Homework Problems CDP5-B B When arterial blood enters a tissue capillary, it exchanges oxygen and carbon dioxide with its environment, as shown in this diagram.

The kinetics of this deoxygenation of hemoglobin in blood was studied with the aid of a tubular reactor by Nakamura and Staub [J. Physiol., 173, 161 (1967)]. Hb + O2

HbO2

Although this is a reversible reaction, measurements were made in the initial phases of the decomposition so that the reverse reaction could be neglected. Consider a system similar to the one used by Nakamura and Staub: the solution enters a tubular reactor (0.158 cm in diameter) that has oxygen electrodes placed at 5-cm intervals down the tube. The solution flow rate into the reactor is 19.6 cm3/s. Electrode Position

1

2

3

4

5

6

7

Percent Decomposition of HbO 2 0.00 1.93 3.82 5.68 7.48 9.25 11.00

Using the method of differential analysis of rate data, determine the reaction order and the forward specific reaction rate constant k for the deoxygenation of hemoglobin. Solution [Hint: Estimate the inlet concentration of HbO2 by using typical values of hemoglobin concentration in adult human blood, and by assuming that hemoglobin is saturated with oxygen]


CRE -- Problem 5-B



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Learning Resources Example CD5-1

Integral Method of Analysis of Pressure-Time Data

Use the integral method to determine the reaction order for the di-tert-butyl peroxide decomposition described in Example 5-1.

Solution Recalling Example 5-1, the combined mole balance and rate law for a constant-volume batch reactor can be expressed in the form

As a first guess we might try zero order,

Assuming a zero order reaction

Integrating gives us

= 0, for which equation (ES-1.5) becomes

(E51.5) (CDE51.1)

If this is the correct order, a plot of P T versus t should be linear. After using the data in Table CDE5-1.1 to obtain Figure CDE5-1.1, we see that P T is not a linear function of t. Consequently, we conclude that the reactionisnot zero-order.


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Figure CDE5-1-1

Next we try second order,

Integrating yields

Assuming a second order reaction

= 2:

If the reaction is second order, a plot of

versus t should be linear:

After forming Table CDE5-1.1, a plot of versus t was constructed and is shown in Figure CDE5-1.2. From the curvature of the plot, we conclude that the reaction is not second order.


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Figure CDE5-1-2

Finally, we try first order (i.e. = 1). If zero, first, or second order do not seems to describe the reaction rate equation, it is usually best to try some other method of determining the reaction order. for = 1,

Integrating with limits,

when t = 0 yields

If the reaction is first order, a plot of

versus t should be linear.

After completing Table CDE5-1.2 using the raw data, a plot of as a function of time is made using semilog paper as shown in Figure CDE5-1.3. From the

is indeed linear with time, and we therefore plot we see that conclude that the decomposition of di-tert-butyl peroxide follows first-order kinetics. From the slope of the plot in Figure CDE5-1.3, we can determine the specific reaction rate, k = 0.08 min -1


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Figure CDE5-1.3



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5.5.3 Weighted Least-Squares Analysis

Both the linear and nonlinear least-squares analyses presented assume that the variance is constant throughout the range of the measured variables. If this is not the case, a weighted least-squares analysis must be used to obtain better estimates of the rate law parameters. If the error in measurement is at a fixed level, the relative error in the dependent variable will increase as the independent variable increases (decreases). For example, in a first-order decay reaction ), if the error in concentration ( measurement is 0.01 C A0 , the relative error in the concentration measurement [0.01 C A0 /C A(t) ] will increase with time. When this error condition occurs the sum to be minimized for N measurements is

where is a weighting factor. For parameter estimation involving exponents, it has been shown that a weighted least-squares analysis is usually necessary.1 Two such cases that occur in the analysis of chemical reaction engineering data are concentration-time data for an irreversible first-order reaction,

and reaction rate-temperature data,2

In general, these equations are of the form

where


,

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respectively

Linearizing, we obtain

We want to find the values of A and B that minimize the weighted sum of squares. For a semilog arithmetic transformation, the weighting function is just the square of the independent variable itself3 (i.e., minimized is

). The function to be

Weighted least squares

There are also strategies available that suggest the experimental conditions to be used for each succeeding data point in order to converge most rapidly to the best values of the rate law parameters (Box et al.).4

Next



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Professional Reference Shelf 5.6 Experimental Planning Four to six weeks in the lab can save you an hour in the library--G.J. Quarderer, Dow Chemical Co.

So far, this chapter has presented various methods of analyzing rate data. It is just as important to know in which circumstances to use each method as it is to know the mechanics of these methods. In this section we discuss a heuristic to plan experiments to generate the data necessary for reactor design. However, only a thumbnail sketch will be presented; for a more thorough discussion the reader is referred to the books and articles by Box and Hunter. Text Figure 5-12 provides a road map to help plan an experimental program.

Figure 5-12 Flowchart for experimental projects

5.6.1 Do You Really Need the Experiments?


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When you are preparing to initiate an experimental program, be sure to question yourself and others to help guide your progress. The following questions will help you dig deeper into your project. Why perform the experiments? Can the information you are seeking be found elsewhere (such as literature journals, books, company reports, etc.)? Can you do some calculations instead? Have sufficient time and money been budgeted for the program? Are you restricted to specific materials or equipment? Will the safety of the investigators be endangered to such a degree that the program should not be carried out? These and other appropriate questions must be answered prior to beginning the experimental program so that the need for the experiments is clearly established. Text Figure 5-12 shows a flow sheet for experimental planning.

5.6.2 Define the Objectives of the Experiment Prepare a list of all the things you want to accomplish. Next try to prioritize your list, keeping in mind the following: What questions regarding your problem would you most like to answer? Are you sure you are not losing sight of the overall objectives and other possible alternative solutions ("can't see the forest for the trees" syndrome)? How comprehensive does the program need to be? Are you looking at an exhaustive study or a cursory examination of a narrow set of conditions? Specific answers to these questions will guide the rest of the project.

5.6.3 Choose the Responses You Want to Measure There are generally two different types of variables that are considered in an experimental program. The independent variables make things happen. Changes in the independent variables cause the system to respond. The responses are the dependent variables. Changing any one of the independent variables will change the system response (the dependent variable). As the experimental program is designed, the important dependent variables to be measured must be identified. What are the controlled or independent variables? What are the dependent variables? Are instruments or techniques available to make the measurements? Do they need to be calibrated? If so, have they been? Will the accuracy and precision of the expected results be sufficient to distinguish between different theories or possible outcomes?


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5.6.4 Identify the Important Variables In any experimental program there will always be many, many quantities you can measure. However, you must decide which independent variables have the greatest influence on the dependent variable. What are the really important measurements to make? What are the ranges or levels of these variables to be examined? Instead of changing each independent variable separately, can dimensionless ratios or groups be formed (i.e., Schmidt or Sherwood numbers) and varied so as to produce the same end results with fewer measurements?

5.6.5 Design the Experiment To obtain the maximum benefit from a series of experiments, they must be properly designed. How can the experimental program be designed to achieve the experimental objectives in the simplest manner with the minimum number of measurements and the least expense? A successfully designed experiment is a series of organized trials which enables one to obtain the most experimental information with the least amount of effort. Three important questions to consider when designing experiments are: What are the types of errors to avoid? What is the minimum number of experiments that must be performed? When should we consider repeating experiments?

5.6.6 Types of Errors There are two types of errors that should be avoided in experimental design. A type I error is one in which you declare that a variable has an effect on the experimental outcome when in fact it really doesn't. A type II error occurs when we fail to discover a real effect. A type II error results in lost information; a variable gets incorrectly classified as insignificant to the process or ignored and as a result, no further examination of it takes place. Type II errors can be avoided by researching fundamental principles related to the experiments, gathering sufficient information, and planning thoughtfully.

5.6.7 The Minimum Number of Experiments (or. . . "Getting the Most Bang for Your Buck") The minimum number of experiments that must be performed is related to the number of important independent variables that can affect the experiment and to how precisely we can measure the results of the experiment. One of the most important strategies to remember is to carry out first experiments at the


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extremes (maximum and minimum setting) of the range of the controlled variables. For example, if the range of pressures that can be used to determine the rate law of a gas-phase reaction is from 1 to 100 atm, it is somewhat best to determine the rate at 1 atm and then at 100 atm. If the independent variables have no effect on the dependent variables at the extremes, it is somewhat doubtful that there will be an effect in the intermediate range. Consequently, a lot of time, money, and energy would be lost if we progressed from a setting of 1 atm to 2 atm and found no effect, then to 10 atm and found no change, then to 50 atm and 80 atm with similar results. In designing the experiments, we will first choose two levels (i.e., settings) for each independent variable. Because these levels are usually at the extremes of the variable range, we refer to these settings as high and low (e.g., on/off, red/green, 100 psi/14.7 psi, 100°C/0°C, etc.). For example, consider an experimental program where the dependent variable is a function of three independent variables (A, B, and C), each of which can take on two possible values or levels.

If all possible variable combinations were to be tested, the number of experiments is equal to the number of levels, N, raised to the power of the number of independent variables, n. For the example for variables A, B, and C, the number of experiments necessary to test all combinations of experiments. These are detailed in independent variables is equal to Table A and Text Figure 5-13 [(+) indicates a high level, while (-) indicates a low level of a particular variable]. TABLE A. CONTROLLED VARIABLE SETTINGS Experiment No.

pH Temperature Concentration

1 2 3 4 5 6 7 8

+ + + +

+ + + +

+ + + + -

For example, let's design a series of experiments to determine the effect of pH file:///H:/html/05chap/html/05prof2.htm[05/12/2011 16:55:55]

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and temperature on the rate of the enzyme-catalyzed decomposition of urea.

Enzyme degradation is believed to occur at temperatures above 50°C and pH values above 9.5 and below 3.0. The rate of reaction is negligible at temperatures below 6°C. For a urea concentration below 0.001 M, the reaction will not proceed at a measurable rate and the rate appears to be

Figure 5-13 Placement (high/low) of controlled variables.

independent of concentration above 0.1 M. Consequently, the following high/low values of the parameters were chosen: A (-) pH 4 B (-) 10°C C (-) 0.005 M

(+) pH 8 (+) 40°C (+) 0.1 M

If there is no interaction among the variables (which may not be known beforehand), experiments 1-4 will yield all the necessary information (Table B). By no interaction it is meant that each of the variables affects the outcome of the experiment independently and there is no synergistic effect of a combined interaction. When there is no interaction, the effect of the pH variable changing from a high value to a low value is always the same, regardless of whether or not the values of temperature and urease concentration are high or low. Experiments 1-4 explore the effect of raising each variable, in turn, from its low level to its high level. In this type of situation, the minimum number of experiments that must be run is the number of independent variables plus one (3 + 1 = 4), and we can predict the results of the other experiments (5 through 8) by combinations of the appropriate responses. However, if there is interaction among the variables, it


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will not be detected from the tables. For example, if there is interaction between temperature and pH, the higher temperature may cause the pH to have a stronger effect on the reaction than at low temperatures. Then all eight experiments would be required to detect an interaction such as this one.

TABLE B: FIRST FOUR EXPERIMENTS Experiment No. 1 2 3

4

pH Temperature Concentration Comments Base case Reveals effect + of high pH Reveals effect + of high temperature Reveals effect + of high concentration

How good are the measurements? What modifications, if any, of the existing equipment are necessary to improv e the accuracy or precision of the measurements or to better achieve the overall experimental objectives? In the experiments, we might find that the urease reaction results are inconclusive with respect to pH in the high-temperature range, and additional runs will be necessary. Is there software available to perform least-squares analysis (see Appendix A.4), set confidence limits, or other statistical analyses? Is there any mathematical model or theory available that suggests how the data might be plotted or correlated? What generalizations can be made from the data? Should other experiments be run to extend the data into different regions? Has an error analysis been performed, and sources of error listed and discussed in relation to how they affect the final result (i.e., by what magnitude and in what direction?). Finally, have all experimental objectives been satisfied?

5.6.10 Report Communicate the results of your work with other members of your team. This is usually done by means of a technical report. Guides for writing such a report can be found in many books. One good source is Designing Technical Reports by J. C. Mathes and D. W. Stephenson (Indianapolis, Ind.: BobbMerrill, 1976), which has particularly useful examples. Typically, a report will file:///H:/html/05chap/html/05prof2.htm[05/12/2011 16:55:55]

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include the following sections: 1. Abstract. This one-page summary of the report is usually written last. It defines the problem, tells how you approached the problem, and states the important results that were found. 2. Introduction. The introduction section defines the problem, tells why it is an important problem worthy of being studied, gives background information, describes the fundamental issues, and discusses and analyzes how they relate to published work in the area. 3. Materials and methods. This section describes the equipment used to carry out the experiments, as well as instruments used to analyze the data. The purity of the raw materials is specified, as are the brand names of each piece of equipment. The accuracy of each measurement taken is discussed. The stepby-step procedure as to how a typical run is carried out is presented, and all sources of error are discussed. (If you developed a new model or theory, a theory section would come after section 3. The theory section would develop the governing equations that mathematically describe your phenomena and justify all assumptions in the development.) 4. Results. This section tells what you found. Make sure that figures and tables all have titles and the units of each variable are displayed. Discuss all sources of error and describe how they would affect your results. Put an error bar on your data where appropriate. 5. Discussion of results. This section tells why the results look the way they do. Discuss whether they are consistent with theory, either one you developed or that of others. You should describe where theory and experiment are in good agreement as well as those conditions where the theory would not apply. 6. Conclusion. The conclusion section lists all important information you learned from this work in numerical order; for example:(a) The reaction is insignificant below 0°C. (b) The results can be described by the BuckleyLeverette Theory. 7. References. List all resource material you referred to in this work in the proper bibliographical format. In addition to the sections of the technical report described above, many companies require an executive summary, which is usually an expanded version of the abstract that includes the conclusion and recommendations. However, at least of equal importance to the sections that are in the report are the style and clarity with which the report is written. A top ten list of things to look for in an effective report is shown below. TOP TEN LIST FOR EFFECTIVE WRITTEN REPORTS


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1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

No Disagreements with Strunk and White20 Logically Organized (Introduction..., Conclusions) Logical Flow of Ideas Within Each Section Concisely Written Interestingly Written, Using Wide Variety of Words Ideas Supported by Examples, Data, Evidence Appropriate Use and Placement, Figures, and Tables Passive Voice Clear Purpose Put in a Clear Plastic Binder (per Calvin and Hobbes)

In addition to written reports, you will also be expected to give oral reports throughout your career and you should refer to some of the references at the end of the chapter to prepare your presentation. The top ten list below identifies some of the key points to consider. TOP TEN LIST FOR EFFECTIVE PRESENTATIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

Well Organized (Introduction, Body, Closure) Logical Flow of Ideas Ideas Presented Concisely Ideas Supported by Examples, Data, etc. Clear Explanations Good Visual Aids Clearly Spoken at a Reasonable Speed Good Preparation and Thorough Run-Through Appropriate Dress Conclusions Supported by Evidence Presentation Matched to Audience Confident Appearance Good Diction and Grammar, Avoiding Slang Four of the Above Disregarded to Make a Top Ten Back Next

A full factorial design (all eight experiments in this case) is also useful for developing a model to predict the outcome of experiments whose independent variables can change continuously (i.e., they can assume a continuous range of values and not just two discrete values). Two levels (at least) of each of the


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variables are examined and the results can be interpreted in the form of a model to predict the outcome of future experiments. Deming5 discusses this method of statistically designing experiments.

5.6.8 Performing the Experiment: How Many Times?

If there is some error associated with measuring the outcome of an experiment, we must consider repeating some of the trials to be sure we have accurate information. But how much data is enough? The answer to this question depends on how precise (reproducible) the experiments are and on how small a change in the outcome or result of an experiment we wish to detect. Obviously, the less precise the measurements (i.e., the more error that is present) and the smaller the change we are interested in, the more data we must collect and average to be confident in our result. Averaging several runs under the same conditions is the best way to deal with such a situation to ensure reliable results. The required number of times that each run should be repeated prior to averaging can easily be calculated using a statistical procedure discussed by Hendrix.6 © 1999 Prentice-Hall PTR Prentice Hall, Inc. ISBN 0-13-531708-8 Legal Statement


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Professional Reference Shelf 5.7 Evaluation of Laboratory Reactors

The successful design of industrial reactors lies primarily with the reliability of the experimentally determined parameters used in the scale-up. Consequently, it is imperative to design equipment and experiments that will generate accurate and meaningful data. Unfortunately, there is usually no single comprehensive 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 shall closely follow the excellent strategy presented in the article by V. W. Weekman of Mobil Oil.7 The criteria used to evaluate various types of laboratory reactors are listed in Table 5-3. TABLE 5-3. CRITERIA USED TO EVALUATE LABORATORY REACTORS 1. Ease of sampling and product analysis 2. Degree of isothermality 3. Effectiveness of contact between catalyst and reactant 4. Handling of catalyst decay 5. Reactor cost and ease of construction

5.7.1 Integral (Fixed-Bed) Reactor

One advantage of the integral reactor is its ease of construction (see Text Figure 5-14). On the other hand, while the channeling or bypassing of some of the catalyst by the reactant stream may not be as fatal to data interpretation in the case of this reactor as in that of the differential reactor, it may still be a problem. There is more contact between the reactant and catalyst in the integral reactor than in the differential reactor, owing to its greater length. Consequently, more product will be formed, and the problems encountered in the differential reactor in analyzing small or trace amounts of product in the effluent stream are eliminated. However, if a reaction is highly


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endothermic or exothermic, significant axial and radial temperature gradients can result, and this reactor will receive a poor-to-fair rating on its degree of isothermality. If a reaction follows different reaction paths with different activation energies, different products will be formed at different temperatures. This makes it difficult to unscramble the data to evaluate the various reaction rate constants because the reaction mechanism changes with changing temperature along the length of the reactor.

Figure 5-14 Integral reactor.

If the catalyst decays significantly during the time an experiment is carried out, the reaction rates will be significantly different at the end of the experiment than at the start of the experiment (see Section 10.7). In addition, the reaction may follow different reaction paths as the catalyst decays, so that the selectivity to a particular product will vary during the course of the experiment. Consequently, it will be difficult to sort out the various rate law parameters for the different reactions and, as a result, this reactor receives a poor rating in the catalyst decay category. However, this type of reactor is relatively easy and inexpensive to construct, so it receives a high rating in the construction category.

5.7.2 Stirred-Batch Reactor

I n a stirred-batch reactor the catalyst is dispersed as a

slurry, as shown in Text Figure 5-15. Although this reactor has better contact between the catalyst and fluid than either the differential or integral reactors, it has a sampling problem. Samples of fluid are usually passed through cyclones or withdrawn through filters or screens to separate the catalyst and fluid, thereby stopping the reaction. However, slow quenching of the reaction in the cyclone or plugging of the filter sampling system by the catalyst particles is a constant concern, thus making the rating in


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the sampling category only fair. Since the system is well mixed, its isothermality is good. There is good contact between the catalysts and reactants, and the contact time is known since the catalyst and reactants are fed at the same time. However, if the catalyst decays, the activity and selectivity will vary during the course of data collection.

Figure 5-15 Stirred-batch reactor. [From V. Weekman, AIChE J., 20, 833 (1974). With permission of the AIChE. Copyright © 1974 AIChE. All rights reserved.]

5.7.3 Stirred Contained-Solids Reactor (SCSR)

Although there are a number of designs for containedsolids reactors, all are essentially equivalent in terms of performance. A typical design is shown in Text Figure 516. In this reactor, catalyst particles are contained in paddles that rotate at sufficiently high speeds to minimize external mass transfer effects (see Chapter 11) and, at the same time, keep the fluid contents well mixed. With this type of operation, isothermal conditions can be maintained and there is good contact between the catalyst and fluid. If the catalyst particle size is small, difficulties could be encountered containing the particles in the paddle screens. Consequently, it receives only a fair rating in the ease of construction and cost category. Although this type of reactor receives a good rating for ease of sampling and


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analysis of product composition, like the three previous reactors, it suffers from being unable to generate useful data when the catalyst being studied decays. As a result, it receives a poor rating in the catalyst decay category.

Minimizes external mass transfer resistance

Figure 5-16 Stirred contained-solids reactor. [From V. Weekman, AIChE J., 20, 833 (1974). With permission of the AIChE. Copyright © 1974 AIChE. All rights reserved.]

5.7.4 Continuous-Stirred Tank Reactor (CSTR)

In this reactor (Figure 5-17), fresh catalyst is fed to the reactor along with the fluid feed and the catalyst leaves the reactor in the product stream at the same rate that it is fed. As a result, the catalyst in the reactor is at the same level of catalytic activity at all times. Thus we are not faced with the problem encountered in the four previous reactors, in which the kinetic parameters evaluated at the beginning of the experiment will be different than those at the end. However, since there will be a distribution of time that the catalyst particles have been in the reactor, there will be a distribution of catalytic activities of the particles in the bed. The mean activity of a catalyst with first-order decay is derived in Example 14-3. If the mean residence time is large, selectivity disguise could be a problem. Since the reactor is well-mixed, isothermality and fluid--solid contact categories are rated as good. However, difficulties can arise


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in feeding the slurry accurately and, as with the stirredbatch reactor, it is difficult to quench the reaction products. Consequently, it receives only fair ratings in the first (sampling) and fifth (construction) categories.

One of the best reactors for isothermal operation

Figure 5-17 [From V. Weekman, AIChE J., 20, 833 (1974). With permission of the AIChE. Copyright © 1974 AIChE. All rights reserved.]

5.7.5 Straight-Through Transport Reactor

Commercially, the transport reactor (Text Figure 5-18) is used widely in the production of gasoline from heavier petroleum fractions (see Text Figure 1-13). In addition, it has found use in grain drying operations. In this reactor, either an inert gas or the reactant itself transports the catalyst through the reactor. With this reactor, any possibility of catalyst decay/selectivity disguise is virtually eliminated because the catalyst and reactants are fed continuously. For highly endothermic or exothermic reactions, isothermal operation will be difficult to achieve and it receives a poor-to-fair rating in this category. At moderate or low gas velocities there may be slip between the catalyst particles and the gas so that the gas-catalyst contact time will not be known very accurately. Consequently, this reactor receives only a fair-to-good rating in the gas-catalyst contacting category. This reactor is somewhat easier to construct than the CSTR, but salt or sand baths may be required to try to maintain isothermal operation, and it therefore receives a fair-to-good rating in the construction category. Difficulty in separating the


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catalyst and reactant gas or in thermally quenching the reaction results in a fair rating in the sampling category.

Best for catalyst decay

Figure 5-18 Straight-through transport reactor. [From V. Weekman, AIChE J., 20, 833 (1974). With permission of the AIChE. Copyright © 1974 AIChE. All rights reserved.]

5.7.6 Recirculating Transport Reactor

By recirculating the gas and catalyst through the transport reactor (Text Figure 5-19 ), a well-mixed condition can be achieved provided that the recirculation rate is large with respect to the feed rate. Consequently, isothermal operation is achieved. Since the reactor is operated at steady state, the kinetic parameters measured at the start of the experiment will be the same as those measured at the end. However, since fresh catalyst is mixed with decayed catalyst from the recycle, the product distribution and the kinetic parameters might not be the same as those measured in a straightthrough transport reactor where the gas "sees" only fresh catalyst. The incorporation of a recirculation system adds a degree of complexity to the construction, which gives it a lower rating in this category as well.


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5.7.7 Summary of Reactor Ratings

The ratings of the various reactors are summarized in Table 5-4 in the text. From this table note that the CSTR and recirculating transport reactor appear to be the best choices because they are satisfactory in every category except for construction.8 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 stirred-contained solids is the best choice. If the catalyst is nondecaying and heat effects are negligible, the fixed-bed (integral) reactor would be the top choice, owing to its ease of construction and operation. However, in practice, usually more than one reactor type is used in determining the reaction rate law parameters.

Figure 5-19 Recirculating transport reactor. [From V. Weekman, AIChE J., 20, 833 (1974). With permission of the AIChE. Copyright © 1974 AIChE. All rights reserved.]

Back


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Additional Homework Problems CDP5-AB

Penicillin G reacts with hydroxylamine (NO2OH) to form hydroxamic acid, which produces a colored complex with iron(III). To determine the overall reaction order, equal concentrations of penicillin and NH2OH were mixed together in a 250-mL flask [J. Chem. Educ., 49, 539 (1972)]. Samples were withdrawn every 10 minutes and added to a solution containing iron(III) chloride. With the aid of a colorimeter, the concentration of the colored complex, and hence the concentration of hydroxamic acid, was obtained as a function of time. The absorbance shown in the table below is directly proportional to the hydroxamic acid concentration.

(a) Use the nonlinear least squares (i.e., regression) and one other technique to determine the reaction order, , and specific reaction rate, k. (b) What experimental conditions would you suggest if you were to obtain more data?

[1st Ed. P5-10]

CDP5-BB

Although this is a reversible reaction, measurements were made in the initial phases of the decomposition so that the reverse reaction could be neglected. Consider a system similar to the one used by Nakamura and Staub: the solution enters a tubular reactor (0.158 cm in diameter) that has oxygen electrodes placed at 5-cm intervals down the tube. The solution flow rate into the reactor is 19.6 cm3/s.

[1st Ed. P5-3]

When arterial blood enters a tissue capillary, it exchanges oxygen and carbon dioxide with its environment. The kinetics of this deoxygenation of hemoglobin in blood was studied with the aid of a tubular reactor by Nakamura and Staub [J. Physiol., 173, 161 (1967)].


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CDP5-CC

Triaminoguanidine (TAGN) is an important propellent ingredient that is used as an oxidizer. It can be produced from guanidine (GN) and hydrazine in an aqueous solution [Ind. Eng. Chem. Res., 28, 431 (1989)]. The reaction is

The following data give the cumulative ammonia evolved as a function of time. The initial concentrations of GN and hydrazine were 0.5 and 2.575 M, respectively, in a 200-mL batch reactor. Neglect any volume change in the liquid with reaction for this aqueous reaction.

Determine as many rate law parameters as possible. [2nd Ed. P5-6]

CDP5-DB

Nitrogen oxide is one of the pollutants in automobile exhaust and can react with oxygen to form nitrogen dioxide according to

At 298 K the specific reaction rate is

or in parts per million,

(a) What is the half-life of 3000 ppm NO (a typical precontrol auto exhaust value) in air? (b) What is the half-life of 1 ppm NO (a typical polluted atmosphere value)? [1st Ed. P5-11] file:///H:/html/05chap/html/05prob.htm[05/12/2011 16:55:57]

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CDP5-EB

For the irreversible gas-phase dissociation of the dimer A2

determine the CSTR volume necessary to achieve 80% conversion and produce 1000 g mol of A per hour. The feed stream consists of 60% A2 and 40% inerts at a pressure of 10 atm and a temperature of 40°C. The following data were obtained in the laboratory in a wellmixed constant-pressure batch reactor, which had an initial charge consisting of 85% A2 and 15% inerts. Process your data in terms of the measured variables (i.e., time and volume).

[1st Ed. P5-6]



CRE -- Problem 5-B Solution

Additional Homework Problems CDP5-B B Solution The rate law for this reaction will be of the form: -rA = kC An Where subscript "A" refers to the reactant HbO2. Also, n = the order of the reaction and k = the specific reaction rate constant. These are the two parameters we need to find using the data given. Since the values of percent decomposition of HbO2 has been given, we can evaluate the conversion XA of the reaction (XA = % decomposition/100). Therefore, we need to express the rate law in terms of conversion as follows.

Where F A0 and C A0 are the the inlet molar flow rate and the feed concentration, respectively. Now dV = Acdz, where Ac is the tube cross-sectional area (= 0.0196 cm2). Further simplifying the rate equation leads to

where a is a lumped parameter containing k, n, F A0 , and C A0 . To evaluate the two parameters n and k, we first need to plot ln(dXA/dz) vs. ln(1-XA). We can estimate the values of dXA/dz by first calculating the X/ z, and then intrerpolating the dX/dz values graphically, as shown below. Electrode Position 1 2 3 4 5 6 Position (cm) 0 5 10 15 20 25 z (cm) 5 5 5 5 5 5 Conversion of HbO 2 (XA ) 0.0000 0.0193 0.0382 0.0568 0.0748 0.0925 (1-XA )

7 30 5 0.110

1.0000 0.9807 0.9618 0.9432 0.9252 0.9075 0.8900

X X/ z

file:///H:/html/05chap/html/ahs05-b.htm[05/12/2011 16:55:58]

-

0.0193 0.0189 0.0186 0.018 0.0177 0.0175 0.00386 0.00378 0.00372 0.00360 0.00354 0.00350


A histogram plot of X/ z vs. z is then produced. The values of dX/dz are evaluated using equal-area graphical differentiation as described in the appendix section:

The values of dXA/dz at electrode points can be estimated from the equal-area line: Electrode Position 1 2 3 4 5 6 7 dX/dz 0.00391 0.00382 0.00373 0.00365 0.00358 0.003514 0.00346

Using the values obtained above, a plot of ln(dXA/dz) vs. ln(1-XA) is produced:

As can be seen from this plot, the data points are nearly linear. The best-fit line is drawn in the plot, and the slope is 1.06. This slope corresponds to the order of the rate law (n), which can be approximated to be 1. Therefore, the rate law is now: file:///H:/html/05chap/html/ahs05-b.htm[05/12/2011 16:55:58]


-rA = kC A To calculate k, the specific reaction rate constant, we must employ the following relationship, using the data gathered at an electrode point (e.g., at p = 2):

All the values in the relationship are known, except k, F A0 , and C A0 . The values for the latter two can be estimated using the hint immediately following the problem statement. According to the hint, the inlet molar concentration of HbO2 can be estimated by using the typical values listed in the literature for hemoglobin concentration in the human blood (150 g of hemoglobin per liter of blood; molecular weight of hemoglobin = 64500)1,2. Therefore, C A0 = 2.33 x 10-6 moles/cm3, and F A0 = 45.67 x 10-6 moles/s. Substituting these values in the above expression leads to:

Therefore, the specific reaction rate is 3.8953 s -1, and the final reaction rate law is: -rA = 3.9CA mol/cm3s Back to problem 5-b 1

D. Voet and J.G. Voet, "Biochemistry", John Wiley & Sons, New York (1990). 2 A.L. Lehninger, "Principles of Biochemistry", Worth Publishers, New York (1982).



Lectures 11 and 12

Lectures 11 and 12

Multiple Reactions (Chapter 6) * Use molar flow rates and concentrations; DO NOT use conversions!

Types of Multiple Reactions 1. Series Reactions

2. Parallel Reactions

3. Combined: Series and Parallel

4. Independent

Selectivity and Yield (Section 6.1) Example:

Instantaneous Selectivity

Yield file:///H:/html/course/lectures/eleven/index.htm[05/12/2011 16:56:01]

Overall

Lectures 11 and 12

Series Reactions (Chapter 6) Example:

This series reaction could also be written as Reaction (1) Reaction (2)

Species A:

Species B:

Using the integrating factor, i.f.:

at t = 0, C B = 0

file:///H:/html/course/lectures/eleven/index.htm[05/12/2011 16:56:01]

Lectures 11 and 12

When should you stop the reaction to obtain the maximum amount of B? Let's see.

Then

And

Mole Balances (Chapter 6) Reactor Type

Gas Phase

Liquid Phase

Batch Semibatch

CSTR

PFR PBR

NOTE: The reaction rates in the above mole balances are net rates.

Net Rate of Reaction for Species A (Chapter 6) file:///H:/html/course/lectures/eleven/index.htm[05/12/2011 16:56:01]

Lectures 11 and 12

For N reactions, the net rate of formation of species A is:

Example: Liquid Phase Reaction (1)

NOTE: The specific reaction rate k1A is defined with respect to species A.

(2)

NOTE: The specific reaction rate k2C is defined with respect to species C.

CASE 1: PFR Mole Balances

A: B: C: D:

Rate Laws

(Rxn i)

Species A For reaction (2):


Lectures 11 and 12

Species B Species C Species D Stoichiometry

Combine Species A

Species B

Species C

Species D

Evaluate Using Polymath


Lectures 11 and 12

CASE 2: CSTR Species A

Species B

Species C

Species D

We will specify V, C Ao , C bo along with the specific reaction rates kij . This formulation leaves us with four equations and four unknowns (CA, C B , C C , and C D). Use Polymath's nonlinear equation solver.

CASE 3: Semibatch


Lectures 11 and 12

Species A Species B Species C Species D

Evaluate Using Polymath Parameters

Initial Conditions

*


Jump to Lecture(s): 1 & 2 | 3 & 4 | 5 & 6 | 7 & 8 | 9 & 10 11 & 12 | 13 & 14 | 15 & 16 | 17 & 18 | 19 & 20 21 & 22 | 23 & 24 | 25 & 26 | 27 & 28 | 29 & 30 31 & 32 | 33 & 34 | 35A | 35B | 36 & 37 | 38 & 39 file:///H:/html/course/lectures/eleven/index.htm[05/12/2011 16:56:01]

Lectures 11 and 12



Lectures 13 and 14

Lectures 13 and 14

Mulitple Reactions -- Gas Phase (Chapter 6) * Example: Gas Phase Reaction in a PFR (1) (2)

Mole Balances

A: B: C: D:

Rate Laws

(Rxn i)

Species A For reaction (2):

Species B file:///H:/html/course/lectures/thirteen/index.htm[05/12/2011 16:56:04]

Lectures 13 and 14

Species C Species D Stoichiometry

Combine

Evaluate Using Polymath Parameters

Initial Conditions Use the Polymath ordinary differential equation solver.

For a CSTR:

file:///H:/html/course/lectures/thirteen/index.htm[05/12/2011 16:56:04]

Lectures 13 and 14

A:

B:

C:

D:

Total: F T = F A + F B + F C + F D

Five Equations and Five Unknowns

Use the Polymath nonlinear equation solver.

Energy Balances (Chapter 8) ,

Need to relate X and T. Use the Energy Balance to achieve this:

Energy Balance:

General Energy Balance:


Lectures 13 and 14

For steady state operation:

We need to put the above equation into a form that we can easily use to relate X and T in order to size reactors. To achieve this goal, we write the molar flow rates in terms of conversion and the enthalpies as a function of temperature. We now will "dissect" both Fi and Hi.

Flow Rates, F i For the generalized reaction:

In general,

Enthalpies, Hi Assuming no phase change:

Mean heat capacities:


Lectures 13 and 14

Energy Balance with "dissected" enthalpies:

For constant or mean heat capacities:

Adiabatic Energy Balance: Adiabatic Energy Balance for variable heat capacities:

CSTR Algorithm (Section 8.3) 1.)

2.)

3.)

Given X Find T and V Solution: linear progression of calc T Given T Find X and V Solution: linear progression: calc k Given V Find X


cal k

cal KC

calc KC

calc X

calc -rA

calc -rA

calc V

calc V

Lectures 13 and 14

Solution: plot XEB vs. T and XMB vs. T on the same graph:

*





Introduction to Pharmacokinetics of Cobra Bites

Welcome to the Cobra Problem! Most people die within 30 minutes of being bitten by a cobra--unless they receive an injection of antivenom, that is. With that cheery thought in mind, let's take a look at how the human body is affected by cobra venom, how antivenom works, and how one might use chemical engineering principles to model these interactions. Background Information This module is a learning resource for Chapter Six of Elements of Chemical Reaction Engineering. Introduction | Background | Venom Effects Reactions | Problems | References file:///H:/html/web_mod/cobra/intro.htm[05/12/2011 16:56:05]

Introduction to Pharmacokinetics of Cobra Bites

Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss

file:///H:/html/web_mod/cobra/intro.htm[05/12/2011 16:56:05]

CRE -- Problem 6-B

Additional Homework Problems CDP6-B B The production of maleic anhydride by the air oxidation of benzene was recently studied using a vanadium pentoxide catalyst. [Chem. Eng. Sci., 43, 1051 (1988).] The reactions that occur are: Reaction 1: C6 H6 + (9/2)O2 Reaction 2: C4 H2 O3 + 3O2 Reaction 3: C6 H6 + (15/2)O2

C4 H2 O3 + 2CO 2 + 2H2 O 4CO 2 + H2 O 4CO 2 + 3H2 O

Because these reactions were carried out in excess air, volume change with reaction can be neglected, and the reactions can be written symbolically as this reaction sequence, where Reaction 1 is second-order and all others are first-order:

A = Benzene, B = Maleic anhydride, C = Products (H2O, CO 2), and D = Products (CO2, H2O) The specific reaction rates at the operating temperature are: k1 = 0.2 (dm6/mol/kg cat/s) k2 = 4*10-4 (dm6/kg cag/s) k3 = 6*10-4 (dm6/kg cat/s) These reactions are carried out isothermally in both a CSTR and a PBR. Benzene enters the reactor at a concentration of 0.01 mol/dm 3, and the total volumetric flow rate is 2.5 dm 3/s. (a) For a catalyst weight of 5000 kg, determine the exit concentrations from a "fluidized" CSTR. (b) What is the selectivity of B to C and of B to D in the CSTR? (c) Plot the concentrations of all species as a function of PBR catalyst weight in a PBR. Determine the catalyst weight for which the concentration of B is a maximum. (d) What is the effect of the entering concentration of benzene on your results. How would your product distribution change if CAo = 0.5 or 0.001 mol/dm3 ?

Solution file:///H:/html/06chap/html/ahp06-b.htm[05/12/2011 16:56:06]

CRE -- Problem 6-B


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Learning Resources Clarification: PFR with feed streams along the length of the reactor To keep C Ahigh throughout a plug-flow reactor, we might, at first glance, believe that we should feed A into the reactor at various points along the reactor, as shown in Figure CD6-1. However, further examination of the situation shows that although the addition of pure A (stream 2) to stream 1 increases the concentration of A in stream 1, stream 2 of pure A is diluted by this addition. Consequently, instead of diluting stream 2 (pure A) by feeding it into the side of the reactor, it is preferable to feed it to the entrance of either another reactor or of the reactor in Figure CD6-1, together with stream 1. The scheme shown in Figure CD6-1 will maintain C A at a low value.

Figure CD6-1 PFR with feed streams along the length of the reactor.



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Additional Homework Problems CDP6AB

The following parallel reactions occur when A is mixed with a catalyst HCl.

The HCl concentration is 0.1 M and the rates of reactions are independent of the HCl concentration above this concentration. The concentration of pure A is 2 mol/dm 3 but can be diluted with the HCl mixture to any concentration desired.

(a) What type(s) of flow reactor(s) and entering concentration of A would you use to maximize the yield and/or selectively to C. (b) Calculate the size of the reactors involved to achieve a 90% conversion of A for a feed rate of 1 mol of pure A per minute. The reactor is operated at 4 atm and 440°F. (Hint: Consider a scheme of series reactors. What should be done when SXY = SBY ?)

CDP6-BB

[2nd Ed. P9-5]

Reconsider Problem 6-14(b) for the case where Reaction 1 is second order, and the specific reaction rates at a different temperature are:

The entering concentration of benzene is 0.01 (mol/ dm 3) at a volumetric flow rate of 2.5 dm 3/s.

(a) For a catalyst weight of 5000 kg, determine the exit concentrations from a "fluidized" CSTR. (b) What is the selectivity of B to C and of B to D in the CSTR? (c) Plot the concentrations of all species as a function of PBR catalyst weight in a PBR. Determine the catalyst weight for which the


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concentration of B is a maximum. (d) What is the effect of the entering concentration of benzene on your results. How would your product distribution change if C A0 = 0.5 mol /dm 3 or if C A0 = 0.001 mol/dm 3? CDP6-CB

[2nd Ed. P9-8]

The reaction

is carried out in a batch reactor in which there is pure A initially.

(a) Derive an equation for the concentration of A as a function of time. If k1 = 0.001 s -1, what is the ratio C A/C A0 after 1.5 min? (b) Derive an equation that gives the concentration of B as a function of time. If k2 = 0.003 s -1, k3 = 0.002 s -1, and C A0 = 0.2 g mol/dm 3, what is the concentration of B after 2 min? (c) What is the concentration of C after 1 min? 2 min? as (d) Sketch the concentrations and functions of time. At what time is the concentration of B at a maximum? (e) If the series reaction is carried out in a CSTR, determine the reactor volume that will maximize the production of B for a volumetric flow rate of 20 dm 3 / min.

CDP6-DB

[2nd Ed. P9-12]

Isobutylene is catalytically oxidized to methacrolein, CO 2, and CO. Data, in terms of three pseudo-first-order, parallel, competing reactions, follow:


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where k 1 = Ai exp (-Ei /RT), i = 1,2,3

A1 = 2989 s -1 A2 = 9466 s -1 A3 = 11,127 s -1

E1 = 10.87 kcal/g mol E2 = 15.11 kcal/g mol E3 = 15.06 kcal/g mol R =0.001987 kcal/g mol K

The reactor is a fluidized bed of catalyst particles. Use a CSTR model to calculate the expected conversions of i -butylene for all reactions under the following conditions:

CDP6-EB

(O. M. Fuller, McGill University) [1st Ed. P9-16]

A vessel contains pure gas A initially at atmospheric pressure and temperature. The following isothermal gasphase reaction takes place. The reaction between B and D is instantaneous with an equilibrium constant of 0.5.

CDP6-FB

The velocity constant for reactions 1 and 2 are k1 = 0.002 min -1 and k2 = 0.001 min -1. Estimate the composition of the vessel 6.5 h after the start of the operation. (H. S. Shankar, IIT Bombay, 1984) [1st Ed. P9-11]

(a) The chlorination of benzene is carried out


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in a CSTR under the following conditions: 55°C, liquid saturated with chlorine gas at a pressure of 1 atm, and a benzene conversion of 30%. The reaction rates are given by

The rate of benzene disappearance by reaction (1) is

k1C b lb mol/ gal h

and the rate of monochlorobenzene disappearance by reaction (2) is k2C m lb mol/ gal h

where k1 = 0.412 h-1, k2 = 0.055 h-1, and C b and C m are the concentrations of benzene and monochlorobenzene in lb mol/gal. Assume that the feed concentration of benzene (C b0) is 0.10 lb mol/ gal and that liquid density is independent of composition. Do not assume that the formation of dichlorobenzene is negligible. What reactor volume is required to produce 10 lb mol of monochlorobenzene per hour? (O. M. Fuller, McGill University, Montreal) (b) Compose and solve (numerical answer required) a problem dealing with the design and operation of a CSTR for the production of monochlorobenzene. Give the principal parts of the problem and include your simplifying assumptions under constraints. Your problem must use the rate data for both reactions. [1st Ed. P9-14]


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Go to next problems




Additional Homework Problems CDP6-B B Solution Part A To solve this problem, we must follow the usual algorithm for multiple reactions with catalysis. Mole Balance Since there are four components, there are four mole balances for this problem. They are: IN - OUT + ACCUM = 0 voC Ao - voC A + rAW = 0 -voC B + rB W = 0 -voC C + rC W = 0 -voC D + rDW = 0 Remember, from the problem statement, that the volume change with reaction can be neglected since the reactions are carried out in excess air. Therefore, v = vo. Rate Laws Next, the rate laws must be formed, using the reaction sequence diagram and noting that Reaction 1 is second-order while Reactions 2 and 3 are first-order.

The rate laws become: rA = -k1C A2 - k3C A rB = rC = rD =

k1C A2 - k2C B k2C B k3C A

Polymath Solution



This set of equations, along with the reaction parameters in the problem statement, may be entered into Polymath to find the exiting concentrations. The equations entered in polymath are as follows:

Polymath then solves the equations and gives us an output file:



The final concentration values are listed in this results table. Therefore, the exit concentrations from a CSTR in this problem are: CA = 2.96 x 10-3 mol/dm 3 CB = 1.94 x 10-3 mol/dm 3 CC = 1.55 x 10-3 mol/dm 3 CD = 3.55 x 10-3 mol/dm 3 Part B Selectivity is a ratio of reaction rates. Therefore, the selectivity of B to C is the ratio of rB and rC , and the selectivity of B to D is the ratio of rB and rD.



Part C To solve this problem, we must set the up equations needed for a PBR with multiple reactions. The mole balances are:

These, combined with the same rate laws as in Part A, give the following Polymath equations:

These equations result in the following plot of concentrations as a function of catalyst weight in the PBR.



From this plot, we can see that the concentration of B is a maximum at a catalyst weight of approximately 2000 kg. To get the exact answer, we display C B as a function of W in a table of values:



This table shows that the concentration of B is a maximum at a catalyst weight of 1937.5 kg. Part D To model these changes in C Ao for the PFR, we just replace the original value of C A in the Polymath equations above. The new product distribution, with C Ao as a larger value (0.5 mol/dm 3) is:



Also, if we change the entering concentration of C A to a very low value, such as 0.001 mol/dm 3, we get this product distribution:





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Additional Homework Problems CDP6GC

Set up the stoichiometric table for the following multiple reaction systems and then write the plug- flow design equations.

CDP6HB

How many independent reactions are there? Compare the stoichiometric coefficient matrix of this problem with that of Example 9-6. What do you conclude? (H. S. Shankar, IIT Bombay, 1984) [1st Ed. P9-17]

The oxidation of formaldehyde to formic acid was studied over a vanadium-titanium oxide catalyst [Ind. Eng. Chem. Res. 28, 387 (1989).] Initially, we consider the occurrence of only the following two reactions:

These reactions occur simultaneously in the gas phase. As a first approximation, assume each reaction to be elementary. The following specific reaction rates are with respect to HCHO

(a) For an entering volumetric flow rate of 100 dm 3/ min at 5 atm and 140°C, plot the molar flow rates as a function of PFR volume. The feed is 66.7% HCHO file:///H:/html/06chap/html/06prob2.htm[05/12/2011 16:56:09]

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The space time can be varied by varying the flow rate. (c) For a CSTR space time of 10 minutes, vary the ratio of O2 to HCHO in the feed. Plot the yield of HCOOH and the selectivity of HCOOH to HCOOCH 2 as a function of (d) Assume that the relative values of k1 and k2 can be adjusted by varying temperature and catalyst composition. Vary the rations k1/k2 and simultaneously to learn their combined effects on selectivity and HCCOH yield. Write a paragraph describing your findings.

CDP6-I B

and 33.3% O2. (b) Repeat part (a) for a CSTR using space time as your variable rather than reactor volume.

[2nd ed. P9-13]

Next let's expand the number of reactions in Problem 7 to include the additional two elementary reactions:

(P6-18.1) (P6-18.2)

k3 = 0.07 min -1 and k4 = 0.09 min -1 as estimated from NCAA emissions spectra. For the entering conditions described in Problem CDP6-H,

(a) Plot the mloar flow rates of each species as a function of PFR volume. (b) Plot the existing molar flow rates of each species as a function of CSTR space time. (c) Assume that you could independently

file:///H:/html/06chap/html/06prob2.htm[05/12/2011 16:56:09]

06prob2.htm

vary the specific reaction rates and carry out a number of runs to learn the effect k1/k3, k2/k4, and so of the ratios of on, on selectivities and yields. Write a paragraph describing the trends that you would find. CDP6-JB

[2nd Ed. P9-14]

Let's further expand the number of reactions in Problem CDP6-I and CDP6-H to include the following:

where k5= 20.0 (dm3/mol)/min as estimated from dorosymmetry analysis.

(P6-19.1)

(a) Plot the molar flow rates of all species as a function of PFR volume for the case when all five reactions (Problems CDP6-H, CDP6-I, and CDP6-J) occur simultaneously. (b) Vary all the specific reaction rates

CDP6KC

for a CSTR with = 10 along with min. Describe your results. Look for any unusual trends (e.g. what happens if the ration of k4/k2 and k5/k2, are increased while the ratio of k1/k2 is decreased?) [2nd Ed. P9-15]

The alkylation of benzene (B) with propylene (P) was carried out over a zeolite catalyst FX-01 to produce isopropylbenzene (I) as the desired product and di-isopropylbenzene (D), which is an undesired product [Proc. 2nd Joint China/ USA Chem. Eng. Conf., III, 51 (1997)].

where


06prob2.htm

(a) Design an isothermal system and reactor that will process 1000 kg mol/day of benzene and keep disopropylbenzene at a minimum. Specify. (b) Try to see if ASPEN can be used to redo (a) for adiabatic operation.

CDP6MA

Propylene and chlorine react to give allyl chloride,

and 1,2,-dichloropropane,

The rate laws for reactions (1) and (2) are

where pp is the partial pressure of propylene and pc is the partial pressure of chlorine. Consider two types of reactors, a CSTR and an ideal tubular reactor, and two temperatures for constant temperature operation, 350°C and 450°C. Which combination of temperature and reactor type will give the largest selectivity ratio for making allyl chloride? (O. M. Fuller, McGill University)

CDP6NC

The production of p -xylene is to be carried out by the reaction of toluene and methanol over HZSM-8 zeolite catalyst [Ind. Eng. Chem. Res. , 28, 890 (1989)]. The reaction sequence is


06prob2.htm

For the first approximation, we are going to neglect the reverse reactions. The rate laws are:

where

The subscripts used above are: T = toluene, M = methanol, PX = p -xylene, MX = m -xylene, OX = o -xylene, TMB = mesitylene.

The feed conditions are: PM = 5 atm, PT = 5 atm, T = 450°C, and the total molar feed rate is 10 mol/ min. (a) Which of the reactions can be neglected to further simplify the kinetics? (b) Using the simplification in part (a)


06prob2.htm

plot the molar feed rates and selectivities as a function of catalyst weight in a (1) Packed-bed reactor. (2) "Fluidized" CSTR.

(c) How would you go about maximizing (e.g., feed concentrations, ratio of toluene to methanol fed, etc.) the production of

(1) p -xylene. (2)m -xylene. (d) The activation energies for the six reactions are (in kJ/mol): E1 = 60.5, E2 = 35.4, E3 = 45.9, E4 = 35.8, E5 = 32.8, and E6 = 29.8. How would changes in temperature affect your results in part (c)? (e) Look up the equilibrium constants for the xylene equilibrium reactions and comment on the validity of neglecting the reverse reactions in light of your results in part (b). CDP6OB

Reconsider Problem 9-7B for the case where Reaction 1 is second-order, and the specific reaction rates at a different temperature are: k1 = 0.2 (dm6/mol kg cat s), k2 = 4 x 10-4 (dm3/kg catás) and k3 = 6 x 10-4 (dm3/kg cat s) The entering concentration of benzene is 0.01 (mol/dm3) at a volumetric flow rate of 2.5 dm 3/s. (a) For a catalyst weight of 5000 kg, determine the exit concentrations from a "fluidized" CSTR.

(b) What is the selectivity of B to C and of B to D in the CSTR? (c) Plot the concentrations of all species as a function of PBR catalyst weight in a PBR. Determine the catalyst weight for which the concentration of B is a maximum. (d) What is the effect of the entering concentration


06prob2.htm

of benzene on your results. How would your product distribution change if C A0 = 0.5 or = 0.001 mol/dm 3?

The oxidation of propylene (P) to acrolein (A) is carried out in a fluidized-bed reactor. The reactions taking place are [Chem. Eng. Sci., 51, 2189 (1996)]:

Air and acrolein are fed to the reactor. The rate laws for the reaction (i= 1, 2, 3) are

CDP6-PC

ki (appropriate units)

ka

k1

k2

Frequency factor Activation energy (kJ/mol)

4 x 104

688

3.5 x 10-4 32.8

88,000

77.500 1312

The reaction is carried out isothermally. Plot the selectivity of acrolein as a function of temperature and feed composition.

Go to previous problems



k3

72,630

06prob2.htm



Lectures 25 and 26

Lectures 25 and 26

Active Intermediates / Free Radicals (Chapter 7) *

Mechanism: (1) (2) (3)

Pseudo Steady State Hypothesis (PSSH) The PSSH assumes that the net rate of species A* (in this case, NO3*) is zero.

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Lectures 25 and 26

This is why the rate decreases as temperature increases.

Enzymes (Chapter 7) Michaelis-Menten Kinetics (Section 7.4.2)

Bioreactors (Chapter 7)


Lectures 25 and 26

Rate Laws

Stoichiometry A.) Yield Coefficients

B.) Maintenance

A Word of Caution on A.) Growth Phase B.) Stationary Phase file:///H:/html/course/lectures/twenfive/index.htm[05/12/2011 16:56:11]

Lectures 25 and 26

Mass Balances Cell:

Also,

for most systems.

Substrate:

Polymath Setup 1.)

d(Cc)/d(t) = - D*Cc + (rg - rd)

2.)

d(Cs)/d(t) = D*(Cso - Cs) - Ysc*rg - m*Cc

3.)

d(Cp)/d(t) = - D*Cp + Ypc*rg

4.)

rg = (((1 - (Cp/Cpstar))**0.52) * mumax*(Cs/(Ks + Cs))*Cc

5.)

D = 0.2

6.)

kd = 0.01

7.)

rd = kd*C C

8.)

Cso = 250

9.)

Ypc = 5.6

10.)

m = 0.3

11.)

mumax = 0.33

12.)

Ysc = 12.5

13.)

Ks = 1.7


Lectures 25 and 26

Polymath Screen Shots Polymath Equations Summary Table Cc and Cp vs. Time Cs vs. Time

Wash Out:

1.) Neglect Death Rate and Cell Maintenance 2.) Steady State

*



Lectures 25 and 26

Back to the top of Lectures 25 and 26.




Lectures 36 and 37

Lectures 36 and 37 **

Polymerization (Chapter 7) * I. INTRODUCTION Polymers are macromolecules built up by the linking together of large numbers of much smaller molecules. The smaller molecules are called monomers and they repeat many times. A polymer is a molecule made up of repeating structural (monomer) units.

Examples of Polymers

Poly (vinyl chloride) Natural Polymers Proteins DNA/RNA Cellulose Fats Starch Synthetic Polymers

Poly (vinyl chloride)

Polyethylene

Pipes

High density: Plastic cups Low density:

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Lectures 36 and 37

Sandwich bags

Polystyrene

Coffee Cups

Poly (vinyl acetate)

Poly (tetra fluoro ethylene)


Plexiglas

Poly (2-hydroxyethyl methacrylate)

Stealth molecule

Poly (methyl methacrylite)

Shampoo/Thickener

Poly (ethylene glycol)

Chewing gum

Poly (vinyl alcohol)

Superglue

Poly (cyano acrylate)

Superglue (Dow)

Poly (acrylic acid)

Contact Lenses

Teflon

Lectures 36 and 37

Poly (ethylene teraphthalate)

Coke bottles Spinable fibers Structural Repeating Unit (mer)

Names/Nomenclature Polymers that are synthesized from a single monomer are named by adding the prefix poly such as polyethylene. However, a parenthesis is placed after the prefix poly when the monomer has a substituted parent name or multiword name such as poly (acrylic acid) or poly (vinyl alcohol). Homopolymers consist of a single repeating unit. All of the above are examples of homopolymers. POLYMERIZATION Two types: Chain Polymerization and Step Polymerization A. CHAIN POLYMERIZATION 1. Free Radical Example: Polyethylene

Linear addition

Back biting


Lectures 36 and 37

Branched Polyethylene resulting low density (0.92) 2. Cationic Polymerization

3. Anionic Polymerization

4. Ziegler-Natta Polymerization Ziegler-Natta Catalyst

Steps in Polymer Chain Growth

(4) Desorption from active site


Lectures 36 and 37

+

to produce linear polymer: Eq. High Density Polyethylene (0.98) (HDPE) B. Polymer structure 1. Linear Linear HDPE (70-90% crystalline) 2. Stereoregularity Can Crystallize. a. Botactic = isotatic = same side b. Syndiotatic = alternating

c. Atactic = random

Head to head (1,2 addition)

Head to tail (1,3 addition)

3. Branched Type A: Long Branches Off Backbone


Lectures 36 and 37

Branched Type B: Short Branches Off the Backbone

Branched Type C: Branches on Branches Off the Backbone

4. Cross linked

C. Copolymers More than one repeating unit.

For example, copolymers used to make records. PVC - hard - irrigation pipes, hard to engrave PVAc - easy to engrave PVC + PVAc copolymer phonograph records (these are a thing of the past) FIVE TYPES OF COPOLYMERS Alternating QSQSQS

Poly (vinyl acetate alt vinylchloride)

Block

QQQSSS

Poly (vinyl acetate-b-vinyl chloride)

Graph

QQQQQ ..... | . . . . . SSSS

Poly (vinyl acetate-g-vinyl chloride

Random

QSSQQQSQSSS

Poly (VA c-co-VC)

Statistical

QSSQSQQSS


Lectures 36 and 37

D. What affects polymer properties • Chemistry

• Molecular Weight (

) and Molecular Weight Distribution

Weight Average Molecular Weight

Molecular Weight Distribution • Crystalinity Amorphous Phase (Non-crystalline Phase) no order or orientation

Tg - characteristic of amorphous state Rubbery

glassy

Below glass transition temperature, Tg, there is a cessation of virtually all molecular motion (vibration , rotation). Crystalline Phase gives an order to the structure.


Lectures 36 and 37

Order means crystallinity Crystalline

liquid

Above the crystalline melting temperature, Tm, thre is no order. Fraction of total polymer that is in the crystalline state is the degree of crystallinity • Cross linking • Branching • Tacticity • Head to head attachment vs. head to tail attachment E. Molecular Weight (MW) 1. Measurement Membrane osmometry

Gel permeation chromatography

Viscosity

Light scattering


Lectures 36 and 37

2. Calculation Number average molecular weight

Weight average molecular weight

Example Calculate the mean molecular weight of a system that contains 95 wt % of polymer A with a molecular weight of 10,000 and 5% of polymer B with a molecular weight of 100. Solution: 1. Weight average molecular weight

Weight average molecular weight, 2. Number average molecular weight

Hence

gives a truer picture of the average molecular weight.


Lectures 36 and 37

3. Polydispersity TWO TYPES OF HOMOGENEOUS POLYMERIZATION: STEP AND CHAIN Step Polymerization. Monomer must be bifunctional. Polymerization proceeds by the reaction of two different functional groups. Monomer disappears rapidly, but molecular weight builds up slowly.

All species are treated as polymers. Mostly used to produce polyesters and polyamides. Chain Polymerization. Requires an initiator. Molecular weight builds up rapidly. Growing chains require 0.0001 to 1 to 10 seconds to terminate. Have high molecular weight polymers right at the start.

II. STEP POLYMERIZATION A. Functional Groups

1. Different functional groups on each end of monomer.

Structural Unit


Lectures 36 and 37

Here the structural unit is the repeating unit. 2. Same functional groups on each end. Example: diamines and diols Two structural units

and

Repeating unit = B. Polymerization Mechanism Monomer dimer ----> trimer ----> tetrameter ----> Pentamer ---->

C. Structural Units The number of structural units equals the number of bifunctional monomers present. 1. Monomers with different functional groups - one structural unit.

Here the repeating unit is the structural unit.


Lectures 36 and 37

Let p = fraction of functional groups of either A or B that have reacted. Let M = concentration of either A or B functional groups at time t. Let M 0 be the concentration of either A or B functional groups initially

Let N = total number (concentration) of polymer molecules present at time t. Let N0 = total number of polymer molecules initially Let M A = number of functional groups of A at time t. Let M A0 = number of functional groups of A initially. = number average degree of polymerization. It is the average number of structural units per chain.

therefore

the number average molecular weight.

Where is the mean molecular weight of the structural units and end group. Example : Calculate Xn for Monomers with Different End Groups 10 molecules of ARB initially MW R = 100 , M eg ~ 18, we neglect MW eg wrt MW R . At time t we have 5 polymer molecules with the following distribution By file:///H:/html/course/lectures/thirsix/index.htm[05/12/2011 16:56:18]

is the molecular weight of the

Lectures 36 and 37

MW* Species 100 100

product

200

ARB ARB AR 2 B + AB

300

AR 3 B + 2AB

300

AR 3 B + 2AB

Conversion of monomer where: NM = Concentration of monomer ARB (2ARB). NM0 = Initial concentration of monomer (10 ARB).

*Neglecting M eg p = fraction of functional groups of A that have reacted. There are 5 molecules of polymer that remain: 2 of ARB, 1 of AR 2 B, and 2 of AR 3 B. We are giving to calculate

by three different methods, all of which yield the same result.

Go back to beginning of example D. Monomers with Same End Group


Lectures 36 and 37

For a stoichiometric feed the number of A and B functional groups the same.

Example : Calculate Xn for Monomers with the Same End Groups

Initially we have 10 polymer molecules of ARA and 10 polymer molecules of BR¢ B, for a total of 20 structural units (10 R and 10R¢ ), a total of 20 A functional groups and 20 B functional groups. At time t there are seven polymer molecules and the distribution of molecules at time t is shown in the table below. Calculate the fractional conversion of functional groups, p, and the number average chain length . Molecules at time t ARA BR¢ B ARR¢ B ARR¢ RR¢ B ARR¢ RR¢ RR¢ B ARR¢ RA BR¢ RR¢ B

Structural Units(SU) 1 1 2 4 6 3 3 20 SU

Solution: M A0 = Number of functional groups of A initially = 20 M A = Number of functional groups of A remaining = 7 N0 = Number of polymer molecules initially. N = Number of polymer molecules remaining at time t file:///H:/html/course/lectures/thirsix/index.htm[05/12/2011 16:56:18]

Lectures 36 and 37

or

Go back to beginning of example E. Stoichiometry Imbalance in the Feed 1. Stoichiometry Imbalance Type 1: Monomers with the same end group and r not equal to 1

The maximum number average chain length is greatly reduced if the initial feed is not exactly stoichiometric 1. If p = 1 then Example : Calculating Xn for Stoichiometric Inbalance Case A: r = 0.99 , p = 1

Case B: r = 1 , p = 0.99


Lectures 36 and 37

Case C: p = 0.99 , r = 0.99

Case D: r = 1 , p = 0.999

Case E: r = 0.99 , p = 0.999

Summary r

p

Xn

1 1 0.99 0.99 0.99

0.99 0.999 1 0.99 0.999

100 1000 199 66.8 166.1

Only a 1% imbalance reduces the number average chain length by almost an order of magnitude Go back to beginning of example 1. 2. Stoichiometry Imbalance Type 2: Monomers with different end groups. Monofunctional Monomer Present

3. Stoichiometry Imbalance Type 3: Monomers with different end groups. Monofunctional Monomer Present

REACTION BETWEEN A DIOL (HOROH) AND A DIBASIC ACID


Lectures 36 and 37

(HOOCR1COOH)

Let

Then

Overall Reaction: The Mechanism Rate Law:

(1)

(2)

(3)


Lectures 36 and 37

Let

-- = ~=

The rate limiting step is Reaction (2)

Assume Reaction (1) is essentially in equilibrium

, Case 1. The acid itself acts as a strong acid catalyst: [HA] º [&emdash;COOH] and Stoichiometric Feed.

As the reaction proceeds and more ester is produced, the solution becomes less polar. As a result the uncatalyzed carboxylic acid becomes the major catalyst for the reaction, and the overall reaction order at high conversion is well described by a third order reaction (Case 1). The high conversion region is of primary importance because this region is where the high molecular weight polymers are formed.

+

At low conversions the solution is more polar and the proton, H is the more effective catalyst (Case 2) than the unionized carboxylic acid. Under these conditions, the reaction is self catalyzed and the reaction is 5/2 order. Case 2. Self catalyzed but acid acts as a weak acid catalyst, not completely dissociated


Lectures 36 and 37

[HA] = [-COOH]

3. External Acid Catalyzed H

+ is constant

F. Kinetics of Step Polymerization (1)

k is defined wrt the reactants


Lectures 36 and 37

Why 2k? Because there are two ways A and B can react (thus, 2k)

(2)

(3)

For all reactions of P1

In general for j „ 2

For j = 2

Mole balance on polymer of length j, in terms of the concentration Pj in a batch system

then


Lectures 36 and 37

At t = 0, P2 = 0

If we proceed further it can be shown that

Total number of polymer molecules (i.e. functional groups of either A or B) = Mole fraction

This is the Flory Distribution for the mole fraction of molecules with chain length j. The weight fraction is just

W = total weight =

G. Flory Distribution &emdash; Probability Approach Rule: The probability of several events occurring successively in a particular way equals the product of the probabilities that each event happens that way. P = probability that an A group will has reacted (1 &emdash; P) = probability group has not reacted. A &emdash; R &emdash; B HO &emdash; R C OO &emdash; H


Lectures 36 and 37

Probability of 1st link is P

Probability of Probability of A 2nd link is P or B unreacted = (1 &emdash; P)

M o = number of functional groups initially (no. of molecules) M = number of functional groups remaining

Number distribution function.

Weight distribution function


Lectures 36 and 37

On a number average basis there will always be more monomer than polymer. *


**

H. Scott Fogler would like to thank Dr. Kristi Anseth for her assistance in preparing the Summary Notes for Lectures 36 & 37. Back to the top of Lectures 36 and 37.




Lecture Notes 38 and 39

Lectures 38 and 39 Chain Polymerization * 1. Free Radical R· 2. Anionic/Cationic R- / R+ Chain polymerizations require an initiator.

Free Radical Polymerization 1. The Reaction INITIATION

This reaction produces the formation of the Primary Radical

PROPAGATION

TERMINATION Transfer To solvent

To monomer

To chain transfer agent

To initiator

Addition

Disproportionation

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MORE ABOUT INITIATION Types of initiators, homiletic, photo

Typical Initiators for Homolytic Dissociation

(1)

Temperature Range

Initiator

50-70°C

Azobisisobutyranitride (ABIN)

(2)

70-90°C

Acetyl Peroxide

(3)

80-95°C

Benzyl Peroxide

Kinetics



Initiator Efficiency "f" f = fraction of radicals produced in the homolysis reaction that initiate polymer chains. It is a measure of waste of initiator. Example How to determine f experimentally 1. In AIBN measure N2 evolution compare number of radicals produced with number of polymer molecules obtained. 2. Tag initiator 14C or 35S 3. Use scavengers - to stop growth.

2. Free Radical Polymerization Kinetics INITIATION PSSH applied to initiator

(1)

(2)

PROPAGATION



For all radicals the total rate of propagation

= total concentration of radicals (R1 + R2 + R3 . . . Rn ) Monomer balance

Long Chain Approximation (LCA)

The rate of disappearance of monomer, -r m ,

TERMINATION 1. Chain transfer A.

To monomer

Total rate of transfer for all radicals

B.

To solvent

C.

Transfer to a chain transfer agent

D.

Transfer to initiator



2. Dispropriation Termination

Disproportionation Define kd wrt reactants i.e.

Net rate of termination of all radicals by dispropriation. For every dead polymer molecule that is formed, one live polymer radical is lost.

3. Addition Termination

where ka is defined wrt to the reactant.



The net rate of termination of j radicals will all the Rk radical (k = 1, 2, …)

The net rate of termination of all radicals is

PSSH Applied to All Free Radicals

Recall

Example Termination by the Initiator Primary Radicals, I

Independent of Initiator Concentration



Monomer Concentration as a function of time

Dead ended polymerization occurs when the initiator concentration decreases to such a low value, the half life of the polymer chains approximates half life of initiator



Polymerization of Isoprene initiated by azobisisolulyronitride. KINETIC CHAIN LENGTH The kinetic chain length v is the average number of monomer molecules consumed (polymerized) per radial that initiates a polymer chain.

Chain Transfer

In Principles of Polymerization 3/e by George Odian, the definition of Rt, the factor of 2 is incorrect but does not matter because it cancels out since Rt is dived by 2 in the denominator in later equations. Also note Eqn. (3-118a) in Odians is altogether incorrect. Let the transfer coefficient be defined by

(1) (2) (3) (4) Term (1) file:///H:/html/course/lectures/thireigh/index.htm[05/12/2011 16:56:27]


Canceling terms

Term (4)

Solving for I 2

The Mayo-Walling Equation

Let’s neglect the term

For styrene In benzyl peroxide file:///H:/html/course/lectures/thireigh/index.htm[05/12/2011 16:56:27]


CM = 0.00006 CI = 0.055 In benzene CS = 2.3 ´ 10–6 In Butyl mercaptan CS = 21.1 Increasing the initiator concentration decreases Increasing the monomer concentration increases Further determination of Rate Constant A. Batch - Single Experiment

1. Dilatometry

Volume charge

2. Spectroscopically measure I 2 (t) and M(t)



B. Batch - Method of Initial Rates

Many experiments

Steady State Measurement

CSTR



No Chain Transfer

Determining Chain transfer Constants A. Only transfer to chain transfer agent S

Hold M and I constant, vary S

B. Transfer to monomer, initiation and a chain transfer agent

Hold I and S constant, vary M



Change I and repeat, the intercept will be the same but slope will be different, S2 . Change S and repeat S3 . Three equations and 3 unknowns. Could also use regression. Transfer Constants A. To Monomer 0.00005 < CM < .0015 CM is generally small, however chain transfer to monomer for vinyl chloride is sufficiently high to limit the molecular weight so that the maximum molecular weight of PVC is 50,000 to 500,000. B. To Initiator CI is a function of both the initiator and the reaction 0.0008 < CI < 0.3 Peroxides are usually the strongest chain transfer agents. C. To Chain Transfer Agent

For styrene 0.000002 < CS < 21 (Benzene) < CS < (n Butyl mercaptan)



Chain Transfer to Polymer, CP Chain transfer to polymer produces branched polymers. It is not important in determining CI , CM and CS because they are determined at low conversion. CP involves the determination of the number of branches produced relative to the number of polymer molecules polymerized. CP is the order of 10–4 The branching density, r , is the number of branches per monomer molecule polymerized.

For CP = 10–4 and 80% conversion, there will be 1.0 branches per 104 monomer units polymerized. There is one branch for every 4,000 to 10,000 monomer units and for a polymer molecular weight of 105 – 106 this corresponds to 1 polymer chain in 10 containing a branch. Polyethylene 1. Short branches (less than 7 carbon atoms) Formed by backbiting. Short branches out number the long branches by a fact of 20-50. They affect crystallinity giving maximum crystallinity of 60-70%. 2. Long branches – formed by normal chain transfer to polymer. Energetics (Free Radical)



Therefore chain transfer becomes more significant as temperature increases!!! CHAIN POLYMERIZATION Ionic Polymerization Cationic Used for monomers with electron releasing substituents

(a) (b) (c) (d) e.g. alkoxy, 1,1–dealky (a) covalent species, (b) tight ion pair, (c) loose ion pair, (d) free and highly solvated ion Anionic Used with monomers possessing electron withdrawing groups, e.g. nitride, carboxyl.

Anionic High molecular weight. No chain-chain termination. Initiation Alkyllithium used because soluble in hydrocarbon solvents.



Potassium amide

Propagation No effective termination - complete consumption of monomer to form living polymers.

Termination by: a. Impurities Moisture

b. Deliberate addition of chain transfer agent c. Spontaneous file:///H:/html/course/lectures/thireigh/index.htm[05/12/2011 16:56:27]


Hydride elimination, i.e.

Comparison with free radical polymerization Free Radical: Concentration of radicals is 10–9 to 10–7 mol/dm3 Anionic: Concentration of propagation anions is 10–4 to 10–2 mol/dm3 In hydrocarbon solvents In either solvents

is 10-100 times smaller than kP

is 10-100 times larger than kP Benzene

Tetrahydrofuran

1,2-Dimethoxyethane

2

550

3,800

Other values are given in Odian p.412 Why do the rates of polymerization vary by several orders of magnitude in different solvents? Kinetics of Ion/Ion Pair Initiation/Polymerization Initiation

Summing over all radicals



where

is the concentration of

concentration of

and all radicals initiated with

and

and radicals initiated with

We assume the ion and the ion pair are in equilibrium with the "salt."

Let

be the total concentration of all types of anionic living propagating centers.

where I o is the total amount of initiator added.

For small degrees of dissociation


is the


If K ~ 10–8

and

then

If K ~ 10–6 then

Polymerization of Styrene 160

80

22

2.2

1.5

0.02

6.5

6.5

6.5

Data of Bhattacharya et al., J. Phys Chem. 69, p.612 (1965) So we see that different solvents bring about different degrees of dissociation of the initiator resulting in different specific reaction rates. Anionic Polymerization



1. Determining the living polymer concentration as a function of time

For complete dissociation of the iniator

Assumptions Initiation is instantaneous, R10 = I o 2. No termination

Case 1 ko >> kp Immediate rate formulation of primary radical



Propagation with No termination

For the live polymer with the largest chain length n



Summing all these equations

Constant live polymer concentration

Let dq = kP M dt

t = 0, q = 0, R1 = I o



Convert back to real time from scaled time

Very small t (i.e., small I o kP t)

Very large t (i.e., large I o kP t)

Distribution of molecular weights of living polymers



Next consider a different set of initiation conditions Case 2 ko = kp



Anionic Polymerization in a CSTR

Monomer Balance



Balance on R1

Balance on Rj

Psuedosteady State Hypothesis (PSSH) Case 1 ko is essentially (i.e., ko >> kp ) infinite. I o is reacted immediately upon mixing with monomer to form R10

There is no initator, I, in the reactor



where

Let (j–1) = , \ j =

Substituting for


+1


Case 2 ko is finite

j = 1

*







07lr.htm

Learning Resources

CD7.1 Development of the Hydrogen Bromide Reaction

One of the most studied chain reaction is that between hydrogen and bromine:

H2 + Br2

2HBr

(CD7-1)

In studying this reaction we first deduce the rate law from reaction rate-concentration data and then formulate a mechanism and rate equation that are consistent with experimental observations. The relative rates of the reaction of bromine, hydrogen, and hydrogen bromide are (CD7-2)

Synthesis of the Rate Law from Experimental Data. An analysis of the experimental rate datashows the following observations. The reaction is:

1. First order with respect to the H2

Experimental observations from which to deduce the rate law

concentration. 2. Nearly independent of the HBr concentration at low HBr concentrations, and the decreases with increasing HBr concentrations at high HBr concentrations. 3. Three-halves order in Br2 at low Br2 concentrations and one-half-order in Br2 at high Br concentrations.

Example CD7-1 Deducing the Rate Law

CD7.2 The Mechanism

We first postulate that the active intermediates are bromine and hydrogen free radicals Br and H , and . Referring to the denominator of the rate


07lr.htm

equation and to part 1 in Text Table 7-1, we see that the active intermediates Br and H collide with HBr and Br2 o intermediates times two species) delineating these collisions or interactions are

Propagation steps

The last reaction has no effect on the overall reaction because one Br2 and one Br are produced for every Br2 and Br that are consumed. However, it is logical to assume that Br2 and H2 probably react instead of Br and Br2:

Note that our proposed scheme of reactions [Equations (CD7-3) through (CD7-6)] produces one of the active intermediates in each step. Now we only need a reaction yielding the initial formation of Br and H . Experience and, possibly, a good guess tell us that the fractional reaction order of Br2 in the numerator suggests that the primary reactive intermediate is Br? formed from Br2 and terminated by the reaction of two bromine free radicals:

Initiation and termination steps

As a first approximation we shall assume that the mechanism consists of reactions (CD7-3), (CD7-4), and (CD7-6) through (CD7-8). If this mechanism does not produce a rate equation that is consistent with the experimental observations in Equation (CDE7-1.6), we could propose a different mechanism that might include Equation (CD7-5) 2H and similar reactions. together with H2 The mechanism we first propose is


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The mechanism

The specific reaction rates k1 and k5 are defined

with respect to Br2.

Example CD7-2 Deriving the Rate Law from the Reaction Mechanism



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Learning Resources Example CD7-1 Deducing the Rate Law

Develop a rate law for the HBr reaction that is consistent with the experimental observations given above.

Solution It is customary in discussions of reactions of this sort to let the species formula enclosed in parentheses represent its concentration . In the azomethane reaction, we saw

could account for the change in the apparent reaction order when going from low to high concentration.

(CDE7-1.1)

where [s 1] represents a reaction constant or the concentration of one of the species that will be determined later

that a rate law of the form

For the HBr reaction

and

file:///H:/html/07chap/html/excd7-1.htm[05/12/2011 16:56:30]

(CDE7-1.2)

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We see that the concentration dependence of bromine given in Equation (CDE7-1.1) is consistent with experimental data. Next consider the HBr dependence.

One rate expression in which the concentration dependence of HBr is consistent with experimental observation is

(CDE7-1.3) Going to the extremes of high and low concentration

Consequently, the concentration dependence of HBr denoted by Equation (CDE7-1.31) is consistent with experimental observation.

This concentration dependence of the species H2, Br2, and HBr in the rate expression are, respectively,

Comparing Equations (CDE7-1.1) and (CDE7-1.3) file:///H:/html/07chap/html/excd7-1.htm[05/12/2011 16:56:30]

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we see that [s 1] in Equation (CDE7-1.1) could be (HBr) and that [s 2] in Equation (CDE7-1.3) could Therefore, Equations (CDE7-1.1) and be (CDE7-1.3) are combined to give

If we now incorporate dependence of the reaction rate on H2 [Equation (CDE7-1.4)] into Equation (CDE7-1.5), we arrive at the rate expression

(CDE7-1.5)

(CDE7-1.6)

or, identically,

(CDE7-1.7)

which is consistent with all experimental observations.



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Learning Resources Example CD7-2 Deriving the Rate Law From the Reaction Mechanism

Using the PSSH, develop a rate law for the rate of formation of HBr for the mechanism given by Equations (CD7-3), (CD7-4), and (CD7-6) through (CD7-8).

Solution Each reaction is to be considered elementary. First we write down the net rate of formation of the product resulting from all the reaction steps:

(CDE7-2.1)

(CDE7-2.2)

The intermediate species Br and H are free radicals, which are present in low concentrations and highly reactive. Consequently, the use of the PSSH is justified. We next write the net rate of formation of the intermediate species and set these rates equal to = 0). zero (r H = 0, r Br

Rate of formation of HBr

(CDE7-2.3)

(Br ) as a function of (Br 2 )

Adding Equations (CDE7-2.3) and (CDE7-2.4) gives


(CDE7-2.4) (CDE7-2.5)

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Subtracting Equation (CDE7-2.3) from (CDE7-2.2) gives

To eliminate (H ) from this equation, rewrite Equation (CDE7-2.3) as

r HBr= 2k3 (H

)(Br2 )

(CDE7-2.6)

(CDE7-2.7)

and solve for (H

):

(CDE7-2.8)

To eliminate (Br ), we use Equation (CDE7-2.5) to obtain

(H ) as a function of reactants and products

Substituting Equation (CDE7-2.9) into Equation (CDE7-2.6), we see that

By comparing Equations (CDE7-2.6) and (CDE72.11) we again see that the mechanism produces a rate law consistent with experimental observation!

(CDE7-2.10)

We can rewrite Equation (CDE7-2.10) in the form

(CDE7-2.11)


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Professional Reference Shelf CD7.3 Enzyme Inhibition 1

The maximum velocity, Vmax , in addition to being a function of the total enzyme concentration, may also be a function of other variables, such as ionic strength, pH, temperature, and inhibitor concentrations. For a number of enzymes Vmax exhibits an optimum in pH. (At a low pH the Vmax increases with increasing pH, while at a high pH it decreases with increasing pH, with the maximum value lying in the intermediate pH range.) One mathematical model that has been used to express this dependency on hydrogen ion concentration is.2

(CD79)

where H is hydrogen ion concentration and K a and K b are the ionization constants of the enzymesubstrate complex in the acidic, and basic solutions, respectively. The denominator of Equation (CD7-9) is referred to as the Michaelis pH function. The Michaelis constant, K m has been observed to exhibit a similar dependence on pH.

In addition to pH, another factor that greatly influences the rates of enzyme-catalyzed reactions is the presence of an inhibitor. The most dramatic consequences of enzyme inhibition are found in living organisms, where the inhibition of any particular enzyme involved in a primary metabolic sequence will render the entire sequence inoperative, resulting in either serious damage or death of the organism. For example, the inhibition of a single enzyme, cytochrome oxidase, by cyanide will cause the aerobic oxidation process to stop; death occurs in a very few minutes. There are also beneficial inhibitors such as the ones used in the treatment of leukemia and other neoplastic diseases. The three most common types of reversible inhibition occurring in enzymatic reactions are


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competitive, uncompetitive, and noncompetitive. The enzyme molecule is analogous to the heterogeneous catalytic surface in that it contains active sites. When competitive inhibition occurs, the substrate and inhibitor are usually similar molecules that compete for the same site on the enzyme. Uncompetitive inhibition occurs when the inhibitor deactivates the enzyme-substrate complex, usually by attaching itself to both the substrate and enzyme molecules of the complex. Noncompetitive inhibition occurs with enzymes containing at least two different types of sites. The inhibitor attaches only to one type of site and the substrate only to the other.

CD7.3A Competitive Inhibition

Competitive inhibition is of particular importance in pharmacokinetics (drug therapy). If a patient were administered two or more drugs simultaneously which react within the body with a common enzyme, cofactor, or active species, this could lead to competitive inhibition of the formation of the respective metabolites and produce serious consequences. In this type of inhibition another substance, I, competes with the substrate for the enzyme molecules to form an inhibitor-enzyme complex, (E I). Typical steps in an enzymatic reaction of this type of inhibition are

(CD710)

Example CD7-3 Derive a Rate Law For Competitive Inhibition

CD7.3B Uncompetitive Inhibition

Here, the inhibitor does not compete with the substrate for the enzyme; instead, it ties up the enzyme-substrate complex by forming an inhibitorenzyme-substrate complex, thereby restricting the breakdown of the (E S) complex to produce the


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desired product. Typical steps in an uncompetitive enzymatic reaction are

The rate of formation of the product is

Application of the pseudo-steady-state hypothesis to E S and IES yields the respective equations

(CD712) (CD713) (CD714)

To obtain E S, we solve Equations (CD7-12) and (CD7-14) together with the mole balance for the total enzyme initially present,

and substitute the result into Equation (7-49) to obtain the rate law for the mechanism above, involving the uncompetitive inhibition of an enzymatic reaction:

(CD715) (CD716)

A Lineweaver-Burk plot of Equation (CD7-16) for different inhibitor concentrations will result in a family of parallel lines all with a slope of K m /Vmax.

CD7.3C Noncompetitive Inhibition

In noncompetitive inhibition, the substrate and inhibitor molecules react with different types of sites on the enzyme molecule, and consequently, the deactivating complex, IES, can be formed by


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two reversible reaction paths: 1. After a substrate molecule attaches to the enzyme molecule at the substrate site, the inhibitor molecule attaches to the enzyme at the inhibitor site. 2. After an inhibitor molecule attaches to the enzyme molecule at the inhibitor site, the substrate molecule attaches to the enzyme at the substrate site.

These paths, along with the formation of the product, P. are as follows:

This reaction sequence can also be written as Reaction 1 E + S E S (CD7-17) Reaction 2 E + I I E (CD7-18) Reaction 3 I+E S IES (CD7-19) Reaction 4 I E+S IES (CD7-20) Reaction 5 E S P+E (CD7-21)

An alternative method of finding the rate law for the sequence presented in Equations (CD7-17) through (CD7-21) is to assume that the last reaction, (CD7-21), is rate limiting and that each of the other reactions, (CD7-17) through (CD720), is essentially in equilibrium. This method is analogous to that used to derive the rate law in heterogeneous catalysis, where one of the steps (e.g., adsorption) is rate controlling. Assuming that the rate-limiting step (reaction 5) is irreversible, the rate of formation of the product is r p = kp(E S) (CD7-22)

Using this equilibrium technique, the rate law for the first reaction, Equation (CD7-17),


(CD7-23)

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is rearranged and the ratio of the rate of reaction to the specific reaction rate is set equal to zero.

The resulting equation is solved for the concentration of the enzyme-substrate complex.

(E S) =K s (S)(E)

Example CD7-4 Derive a Rate Law For Non-Competitive Inhibition

Since the substrate and inhibitor attach at different sites, a reasonable assumption is that the equilibrium between the enzyme and the substrate is the same whether or not an inhibitor is attached to the enzyme (i.e., K s = ); then (EIS) = Ks (E I)(S) and (E S) = Ks (E)(S) By making a similar assumption for the equilibrium between the enzyme and the , we can rearrange Equation inhibitor, K I = (CDE7-4.6) so that

(CD7-25)

where

(CD7-24)

Equation (CD7-25) is in the form of the rate law that is usually given for an enzymatic reaction exhibiting noncompetitive inhibition. Heavy metal ions such as Pb 2+, Ag+ , Hg2+, and others, as well as inhibitors that react with the enzyme to file:///H:/html/07chap/html/07prof.htm[05/12/2011 16:56:32]

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form chemical derivatives, are typical examples of noncompetitive inhibitors. The various types of inhibition are compared with a reaction in which no inhibitors are present (dark line) on the Lineweaver-Burk plot shown in Figure CD7-1. We observe the following relationships: 1. In competitive inhibition the slope increases with increasing inhibitor concentration while the intercept remains fixed. 2. In uncompetitive inhibition the yintercept increases with increasing inhibitor concentration while the slope remains fixed. 3. In noncompetitive inhibition both the intercept and slope will increase with increasing inhibitor concentration.

Figure CD7-1

In addition to the Lineweaver-Burk plot, the Eadie plot is also used to present data of enzymatic reactions. The Eadie plot magnifies departures from linearity that might not be observed in the Lineweaver-Burk plot. In Figure CD7-2a, b, and c, the Eadie plot is used to present the data for three enzymatic reactions: one with the competitive type


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of inhibition, one with the uncompetitive type of inhibition, and the third with the noncompetitive type of inhibition. Each figure contains two lines corresponding to two different inhibitor concentrations; line a represents the higher inhibitor concentration and line b the lower inhibitor concentration.

Figure CD7-2

Example CD7-5 Match Eadie Plots to the Different Types of Inhibition

Next



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Example CD7-3 Derive a Rate Law For Competitive Inhibition

Show that the rate law for the mechanism in Equation (CD7-10) involving competitive inhibition, is

where I is the inhibitor concentration and r p is the rate of formation of product P. [Hint: Apply the pseudo-steady-state hypothesis to (E S) and (I E).]

Solution

The rate of formation of product P corresponding to the last step in this reaction sequence is

The uncomplexed or free enzyme concentration is

(CD7-11)

(CDE7-3.1)


cd7-3.htm

(CDE7-3.2)

Using the pseudo-steady-state approximation for the enzyme-substrate complex yields

(CDE7-3.3)

For the enzyme-inhibitor complex,

(CDE7-3.4)

Dividing Equation by k1(S) and Equation (CDE73.4) by k3(I) and rearranging each, we obtain

(CDE7-3.5)

(CDE7-3.6)

Subtracting Equation (CDE7-3.5) from (CDE7-3.6) and solving for E I, we find that

After substituting Equation (CDE7-3.7) into (CDE7-3.5) to solve for E S,

we substitute for E S in Equation (CDE7-3.1) to obtain the rate law for the inhibition of the competitive type:

(CDE7-3.7)

(CDE7-3.8)


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(CD7-11)

Letting = K m(1 + I/K i ), we see that the effect of a competitive inhibitor is to increase the . apparent Michaelis constant,

WHERE



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Example CD7-4 Derive a Rate Law For NonCompetitive Inhibition

Solve for the concentrations of the complexes (I E) and (IES) in terms of the free enzyme concentration, (E), and substrate concentration, (S), in a manner similar to that for (E S); substitute these expressions in the total enzyme balance and then obtain the rate law for noncompetitive , and represent inhibition. (Let Ks , KI, the equilibrium constants for reactions 1 through 4, respectively.)

Solution

For reaction 2,

We can use either Equation (CD7-19),

or (CD7-20),

to express IES in terms of I E, and S. The mole balance on the total amount of enzyme, bound and free (E), is

Substituting Equations (CD7-24), (CDE74.1), and (CDE7-4.2) into (CDE7-4.4) and rearranging gives us

(E I) =K I (I)(E)

(CDE7-4.1)

(CDE7-4.2)

(CDE7-4.3)

Et = E + E S + I E + IES


(CDE7-4.4)

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The rate of formation of the product for a noncompetitive inhibition of an enzymatic reaction is

(CDE7-4.5)

(CDE7-4.6)



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EXAMPLE CD7-5 Match Eadie Plots to the Different Types of Inhibition

Determine the figure that corresponds to each of the three types of inhibition: competitive, uncompetitive, and noncompetitive.

SOLUTION

For competitive inhibition:

Multiplying by and rearranging


, we obtain

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We see that a plot of r p vs. r p /S will be a straight line with a slope , where

A change in the inhibitor concentration will change the slope of the Eadie plot but not the intercept. The greater the inhibitor concentration, the greater (i.e., steeper) the slope.

For uncompetitive inhibition:

After rearranging (CD7-16) so that

we see that increasing the inhibitor concentration will decrease the slope and the intercept of an Eadie plot.

For noncompetitive inhibition:

Again,multiplying by (S + K m) and rearranging, we obtain

Ans.: Figure CD7-3b

Ans.:Figure CD7-3c.


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In the case of noncompetitive inhibition, increasing the inhibitor concentration decreases the intercept but does not affect the slope of an Eadie plot.

Ans.: Figure CD7-3a .



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CD7.4 Multiple Enzyme and Substrate Systems

In Section CD7.2 we discussed how the addition of a second substrate, I, to enzyme-catalyzed 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 other multiple-enzyme and multiple-substrate systems in which cyclic regeneration of the activated enzyme occurs.

CD7.4A Enzyme Regeneration

As a first example we shall consider the oxidation of glucose (S r ) with the aid of the enzyme glucose oxidase [represented as either G.O. or (Eo)] to give gluconolactone (P):

In this reaction, the reduced form of glucose oxidase (G.O.H2), which will be represented by Er , cannot catalyze further reactions until it is oxidized back to Eo. This oxidation is usually carried out by adding molecular oxygen to the system so that glucose oxidase, Eo, is regenerated. Hydrogen peroxide is also produced in this oxidation regeneration step.

In biochemistry texts, reactions of this type involving regeneration are usually written in the form

The reaction is believed to proceed by the following sequence of elementary reactions:

Overall, the reaction is written


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We shall assume that reaction involving the dissociation between reduced glucose oxidase and -lactone is rate limiting. The rate of formation of -lactone (P 1) is given by the equation

After applying the pseudo-steady-state hypothesis to the rates of formation of (Eo S r ), (Er S o), and (Er ),

we can solve for the following concentrations of the active intermediates in terms of the concentrations of glucose, oxygen, and unbound oxidized enzyme.

(CD726)

(CD727)

(CD728)

(CD729)

After substituting Equation (CD7-28) into Equation (CD7 26), the rate law is written as

The total enzyme initially present is given by the sum


(CD730)

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After using Equations (CD7-27), (CD7-28), and (CD7-29) to substitute for the active intermediate in Equation (CD731), one can then solve for the unbound oxidized enzyme concentration, Eo, and substitute it into Equation (CD7-30) to obtain the form of the rate law

The reaction above illustrates how an enzyme can be regenerated through the addition of another substrate, in this case O2.

CD7.4B Enzyme Cofactors

In many enzymatic reactions, and in particular biological reactions, a second substrate (i.e., species) must be introduced to activate the enzyme. This substrate, which is referred to as a cofactor 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 enzyme-cofactor complex is referred to as a holoenzyme. 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). After the enzyme is activated by combination with the cofactor in its reduced state, NADH,

the holoenzyme (ADH NADH) reacts with acetaldehyde in acid solution to produce ethanol and the oxidized form of the enzyme-cofactor coupling (ADH NAD+ ):

(Et ) = (Eo ) + (Eo S r ) + (Er

S o ) + (Er )

(CD731)

(CD732)

Example CD7-6 Construct a Lineweaver-Burk Plot for Different Oxygen Concentrations


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The inactive form of the enzyme-cofactor complex for a specific reaction and reaction direction is called an apoenzyme. This reaction is followed by dissociation of the apoenzyme (ADH NAD+ ), which is usually relatively slow.

3

The values of the specific rates are :

Typical initial concentrations for a small laboratory batch reactor experiment might be [acetaldehyde]0-1mol/L, [ADH] -7 -4 g mol/L. The overall 0 = 10 g mol/L, and [NADH] 0 = 10 reaction is often written in the form alcohol dehydrogenase

The ADH enzyme molecule produced by the dissociation of (ADH NAD+ ) can participate in subsequent reactions involving the formation of ethanol, while the nicotinamide adenine dinucleotide from the dissociation cannot participate until it is reduced back to NADH. Since the initial concentration of NADH is usually several orders of magnitude greater than the initial concentration of enzyme, the consumption of NADH will not limit the overall rate of formation of ethanol nearly so much as the slow dissociation of the (ADH NAD+ ) complex. This apoenzyme essentially ties up the enzyme, preventing it from becoming free (unbound) to combine with NADH to form the holoenzyme, which reacts with acetaldehyde to produce ethanol. We note that the reaction rate might be increased considerably if we had a way of going directly from (ADH NAD+ ) to (ADH NADH); that is,

rather than having the enzyme ADH go through the steps of dissociating from


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and then combining with NADH: Example CD7-7 Derive an Initial Rate Law for Alcohol Dehydrogenates

Equation (CD7-33) is of a form that is often used in the interpretation of initial rate data for enzymatic reactions involving two substrates. The parameters K 12, K 1, K 2, and Vmax in Equation (CD7-33), which was first developed by Dalziel,4 may be evaluated through a series of Lineweaver Burk plots. After substituting the numerical values for K 1, K 2, K 12 and recalling the initial concentrations specified (S 1,0 = 0.1 g mol/L, S 2,0 = 10-4 g mol/L), we see that we can neglect K 12 and K 2(S 2) with respect to the other terms in the denominator, in which case Equation (G25-7) becomes

10-6

The initial rate is r p0 = 3.7 x g mol/L s. In the next section we compare the rate above with one in which a third substrate is added to the system.

CD7.4C Multiple-Substrate Systems

In the preceding section we stated that the rate of formation of ethanol might be increased if the (ADH NAD1) complex could be converted by some means directly to the enzyme cofactor complex (ADH NADH) without having the enzyme ADH go through a series of reactions. This can be achieved by the addition of a third substrate, S 3, (e.g., propanediol), which during reaction (to form DLlactaldehyde) will also regenerate the cofactor (NADH). The overall reaction sequence for this case is

As a first approximation, this sequence of reactions could be file:///H:/html/07chap/html/07prof2.htm[05/12/2011 16:56:37]

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represented by the following elementary steps:

Example CD7-8 Derive a Rate Law for a Multiple Substrate System

This same reaction has been carried in the reverse direction by Gupta and Robinson.5 They measured the initial rate of conversion of DL-lactaldehyde to propanediol in the presence of NAD 1 and ADH. The rate of dissociation of the enzyme-cofactor complex (ADH NADH) is believed to be rate limiting. This is confirmed by the fact that when ethanol was added to the system, the reaction rate increased 100-fold by having the ethanol convert the (ADH NADH) directly back to (ADH NAD+ ).

Example CD7-9 Calculate the Initial Rate of Formation of Ethanol in the Presence of Propanediol

In analyzing multiple reactions in this manner, one should always question the validity of the application of the PSSH to the various active intermediates. Since the nicotinamide adenine dinucleotide is continually regenerated and the total concentration of the cofactor (in its oxidized, reduced, bound, and unbound forms) remains constant throughout the course of the reaction, it might be desirable to replace S 2 in the rate law in terms of the total cofactor concentration, S t . Neglecting any unbound, the total (initial) cofactor concentration is

The total concentration of enzyme is


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Subtracting Equation (CD7-34) from Equation (CD7-35), one obtains

Equation (G25-7) can be rewritten in the form

where

After adding Equations (CD7-36) and (CD7-37) and rearranging, we obtain

which is solved for the unbound enzyme concentration in terms of S 1, S 3, P 2, Et , and S t .

We can substitute Equation (CD7-39) into (CD7-34) and rearrange to determine the concentration of the unbound cofactor in its reduced form, that is,

The rate law was given by


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One could substitute Equations (CD7-39) and (CD7-40) for S 2 and E into Equation (CD7-41) to arrive at a reasonably complicated rate law involving S 1, S t , S 3, P 2, and Et . However, a computer solution would be used in most reaction sequences that are this involved algebraically, in which case further substitution would not be necessary and one could use Equations (CD7-39), (CD7-40), and (CD7-41) directly.

CD7.4D Multiple Enzymes Systems We shall again consider the production of ethanol from acetaldehyde which uses the cofactor NADH. However, the regeneration of NAD+ to NADH is brought about in a reaction catalyzed by acetaldehyde dehydrogenease (E2), which produces acetic acid from acetaldehyde:

One could apply the PSSH to E1 S 2, E1 E2, and E2 S and represent the total enzyme concentrations as

This sequence can be written in abstract notation as

in deriving the rate law for this system. We shall not carry through the necessary algebraic manipulation to obtain the rate law here, as all the principles for determining the rate file:///H:/html/07chap/html/07prof2.htm[05/12/2011 16:56:37]

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law have been presented and there would be little more to be accomplished by doing this.

The sections above on enzymatic reactions were meant to serve as a brief, yet somewhat encompassing discussion of enzyme kinetics. Further discussion can be found in the Supplementary Reading for Chapter 7.

Back Next



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Professional Reference Shelf EXAMPLE CD7-6 Construct a Lineweaver-Burk Plot for Different Oxygen Concentrations

Construct a Lineweaver-Burk plot for this system in which each line corresponds to a different oxygen concentration. Also discuss the effect of total gas pressure of fixed oxygen concentration on the rate of this liquid-phase reaction,r p. (Hint: Recall Henry's law.)

Solution We first invert Equation (CD7-32) to find

(CD7-6.1)

From Equation (CD7-6.1) we see that changes in the oxygen concentration will only affect the intercept of a LineweaverBurk plot. As the oxygen concentration increases, the intercept decreases.

Assuming equilibrium between oxygen in the gas phase and oxygen in the liquid phase, we obtain the following relationship (Henry's law):


cd7-6.htm

(CD7-6.2)

At high pressures,

Figure CDE7-6.1 shows the a LineweaverBurk plot for various oxygen concentrations, and Figure CDE7-6.2 shows the reaction rate as a function of total pressure for a fixed mole fraction of oxygen.

At low pressures,

Figure CDE7-6-1


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Figure CDE7-6-2



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Professional Reference Shelf EXAMPLE CD7-7 Derive an Initial Rate Law for Alcohol Dehydrogenates

Before considering the possibility of going directly from the apoenzyme to the holoenzyme, assume that the rate of dissociation of the complex (ADH NAD+ ),

is irreversible, and show that the initial rate law for ethanol in the enzyme cofactor reaction sequence discussed earlier is of the form

(CD7-33)

Solution

Let E = ADH, S 1= CH 3CHO, S 2 = = NAD+ , and P 1 = NADH, CH 3CH 2OH. Then

By adding the rate law for the rate of formation of ethanol (P 1),

to the equation for


excd7-7.htm

(below), we see that the rate law for ethanol can be written as

(CDE7-7.1)

Application of the PSSH to the holoenzyme (E S 2) and the apoenzyme ) (E

allows one to solve for the concentrations of the cofactor-enzyme complexes in terms of S , S 2, , and E.

(CDE7-7.2)

(CDE7-7.3)

The total concentration of bound and unbound enzyme is

(CDE7-7.4)

Substituting Equations (CDE7-7.2) and (CDE7-7.3) into Equation (CDE7-7.4), the unbound enzyme concentration is

(CDE7-7.5)

Setting (P 1) = 0, we obtain the initial rate law by combining Equations (CDE7-7.1), (CDE7-7.3), and (CDE77.5):

(CDE7-7.6)


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Professional Reference Shelf EXAMPLE CD7-8 Derive a Rate Law for a Multiple Substrate System

Derive the rate law for this sequence assuming that the rate of reaction between the (ADH NADH) complex and acetaldehyde is rate limiting.

Solution Assuming that the reaction between the cofactor-enzyme complex and acetaldehyde is rate limiting, the rate law is [Note: k 2 =k 2 (h+ ).] After applying the

PSSH to (E S 2) and (E

) to obtain the equations

we add Equations (CDE7-8.2) and (CDE7-8.3) and rearrange to obtain

The total enzyme concentration is again constant at (Et ):


excd7-8.htm

After solving Equation (CDE7-8.7), for E, we combine Equations (CDE7-8.1) and (CDE7-8.4) to obtain the rate law:

Setting P 2 = 0 and rearranging, we obtain the initial rate law:

where



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Professional Reference Shelf EXAMPLE CD7-9 Calculate the Initial Rate of Formation of Ethanol in the Presence of Porpanediol

Assume that the specific reaction rates for the conversion of propanediol to DL -lactaldehyde are the same order of magnitude as the corresponding specific reaction rates for the conversion of ethanol to acetaldehyde. Calculate the initial rate of formation of ethanol in the presence of propanediol and compare this rate with the initial rate when propanediol is absent from the system. Solution Assuming the corresponding specific reaction rates to be equal,

We can then obtain numerical values for Vmax , K3, and K23 in Equation (CDE7-8.9):

Vmax = (k 4)(E t) = (3.5 X 103)(10-7) = 3.5 x 10-4S-1

initially


excd7-9.htm

If

we can neglect the first two terms in the denominator with respect to the last term,

When no propanediol was present we found that

If the propanediol and acetaldehyde are present in the same initial concentration,

then

and we see that the addition of the third substrate increases the rate of reaction by a factor of 10!



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Professional Reference Shelf CD7.5 Oxygen-Limited Fermentation Oxygen is necessary for all aerobic fermentations (by definition) [cf. Equation (7-98)]. Maintaining the appropriate concentration of dissolved oxygen in fermentation is important for the efficient operation of a fermentor. For oxygen-limited systems, it is necessary to design a fermentor to maximize the oxygen transfer between the injected air bubble and the cell. Typically, a fermentor contains a gas sparger, heat transfer surfaces, and an impeller, such as the one shown in Text Figure 7-13 for a batch reactor. A chemostat has a similar configuration, with the addition of inlet and outlet streams.

Transport Steps. The overall transport mechanism of oxygen to a cell is very similar to that described for reactant gas in a slurry reactor and as shown in Figure CD7-3. The corresponding oxygen transport and reaction steps and equations are analogous to the slurry reactor transport steps discussed in Section 13.1, that is: (CD743) (CD742)

Generally, the diffusional resistance from the bulk liquid to the cell surface is ignored. However, depending on the cell size or cell floc size, the transfer step from the bulk liquid to the cell surface may be rate-limiting; one such example is the oxygenation of a culture of Streptomyces niveus. The transport of oxygen into a cell occurs by different mechanisms for yeast and bacteria. In the case of yeast, an oxygen molecule diffuses across an inert cell membrane before being consumed by the cell. The corresponding rate equations are:

Analogous to slurry reactor


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steps

Figure CD7-3 Oxygen transport to microorganisms.

(CD744)

(CD745)

where:

ac = cell surface area per mass of cell, m 2 /g De effective diffusivity across the cell, m 2 /s L = thickness of cell membrane, m

Cc = concentration of cells, g/ m 3 (analogous

to m in a slurry reactor, Text Chapter 12) Ci, Cb, C0, Cc = saturation, bulk, external surface, and internal cellconcentrations of oxygen, respectively

Combining Equations (CD7-42) through (CD7-45) and rearranging in a manner similar to arriving at Equation (12-xx), we obtain

Yeast

(CD746)

For many yeast cells, diffusion across the cell membrane can be neglected.


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In the case of bacteria, the oxygen begins to be consumed as soon as it diffuses into the cell membrane. In fact, bacteria consume oxygen primarily at the cell membrane, where most of the respiratory enzymes are located. As is the case of the catalyst pellet in the slurry reactor, the rate of oxygen consumption can be given by the product of the effectiveness factor and the rate of reaction that would occur if the entire interior of the cell were exposed to the concentration at the external surface, C0

The rate law for oxygen consumption (uptake) generally follows either Michealis-Menton or firstorder kinetics. In many systems it depends on the particular growth phase of the bacteria cell. Typical respiration rates for single-cell yeast and bacteria are on the order of 100 to 600 mg O2/g cell h. For firstorder kinetics we have

(CD747)

where kr is the specific reaction rate for oxygen

uptake, s -1, and h is the effectiveness factor for diffusion and reaction of oxygen inside the cell. Combining equations (CD7-42), (CD7-43), and (CD7-47) gives

Bacteria

(CD748)

We can observe from Equations (CD7-46) and (CD748) that at low cell concentrations, transport steps C, D, and E (mass transfer of oxygen to and within the cell) become rate limiting. The main variables affecting kL ab are the impeller diameter, Di , and speed, N (rpm); volumetric flow rate of the gas, Q; tank diameter, DT , and height L T ;

and power input to impeller, . Many fermentations produce products that cause the broth to exhibit non- Newtonian fluid behavior. Consequently, a


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characteristic relaxation time, , is included in the correlation for the mass transfer coefficient between the gas bubble and the bulk liquid, kL ab . Representative correlations for kL ab are given in Table CD7-1. The power input to the fermentor without gas bubbles present ( ) is a function of system variables:6,7

where the Reynolds number for this system is defined as

(CD75.5)

When gas is present, the power input, Pg is reduced for a given propeller speed 8 and is a function of gas flow rate, impeller speed and diameter, and the Reynolds number. The ration of the power inmput with gas present, Pg, to that without gas present ( ) is

(CD756)

Table CD7-1. Mass Transfer Coefficients In Fermentor 1. Low-viscosity broths Van't Reit (1): Range 2-2600 (CD749) dm 3

Pure Water Electrolytic (Solutions) where

= power input per unit volume of

vessel and is in units of W/m3 Q = volumetric flow rate, m 3/s DT = tank diameter, m Vs = superficial gas velocity


2-4400 (CD750) dm 3

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2. Non-Newtonian correlations

Perez and Sandall (2):

(CD751)

Yagi and Yoshida (3):

Ranade and Ulbrecht (4):

[Comment: These correlations were obtained in tanks having a volume of 12

dm 3

(CD752)

(CD753)

or less (5).]

Other parameters in the correlations are:

g = gravitational acceleration, m 2/s

e = effective viscosity, g/m s

d = viscosity of dispersed phase, g/m s

= fluid viscosity, g/m s

w = viscosity of water, g/m s = surface tension, N/m

N = impeller rotation speed, s-1 DAB = diffusivity, m 2/s

= density, g/m 3

3. Effect of solids (6):


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(1) K. Van't Reit, Fund. Eng. Chem. Proc. Des. Dev., 18, 357 (1979) (2) J.F. Perez and O.C. Sandall, AIChE J., 20, 770 (1974) (3) H. Yagi and F. Yoshida, Ind. Eng. CHem. Proc. Des. Dev., 14, 488 (1975) (4) V.R. Ranade and J.J. Ulbrecht, AIChE J., 24, 796 (1978) (5) D.W. Hubbard, L.R. Harris, and M.K. Wierenga, Chem.Eng. Prog., 84(8), 55 (1988) (6) D.B. Mills, R. Bar, and D.J. Kirwan, AIChE J., 33, 1542 (1987)

The functions F 1 and F 2 are generally given graphically for different types of fluids and different geometric configurations. 9,10

Back Next


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Professional Reference Shelf CD7.6 Fermentation Scale-up Scale-up 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 scaling from a pilot-plant bioreactor to a commercial plant reactor are shown in Table CD7-2. One key to a scale-up is to have the speed of the end (tip) of the impeller equal the velocity in both the laboratory pilot reactor and the full-scale plant reactor. If the impeller speed is too rapid, it can lyse the bacteria; if the speed is too slow, the reactor contents will not be well mixed. Typical tip speeds range from 5 to 7 m/s. This scale-up procedure has been applied11 to the data of Rogovin et al.12 to produce 205,000 kg of Phosphomannan per year using the yeast Hansenula holstii. The pilot reactor was 2.3 m 3 (600 gal) and the plant reactor was 50 m 3(13,200 gal). The relative sizes of the scaling goups for this plant are shown in Table CD7-3.

Perspective. In the limited space available for this topic we have presented this greatly simplified version of bioreactor design. Here we have tried to give an overview of some of the basic ideas and vocabulary that will serve as a springboard to a deeper study of bioreaction engineering. For example, we have considered only a single nutrient source and have not discussed the interrelated enzymatic reaction pathways that exist between all the species necessary for cell growth. Wang et al.13 and Bailey and Ollis 14 discuss the finer and more intricate details associated with the use of microorganisms to produce chemicals, antibiotics, and food products. In addition to the fundamentals already known about bioreactors, many challenging research areas exist. For example, animal cells are fragile and very susceptible to being lysed (killed)


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by even moderately large shear stress. Consequently, scale-up and thorough mixing of cells, nutrients, and oxygen become extremely difficult problems. In addition, cells can aggregate, which poses the problems of maintaining a supply of nutrients and removal of wastes. Fundamental studies of flocculation and surface interactions of microorganisms will aid in the solution to the aggregation problem as well as explore other frontiers of bioreactors, thus providing stimulating research in this area for many years to come.

TABLE CD7-2

1. Choose fermenter volume required based on desired capacity. Algorithm for fermentor

2. Choose impeller diameter, Di . 3. Calculate reactor dimension (e.g., DT = tank diameter) based on geometric similarity, with the impeller diameter being the characteristic length, for example,

Alternatively, we could have chosen the tank diameter, DT, in step 2 and then used Equation (A) to calculate the impeller diameter, Di . 4. Calculate impeller speed, N. ( D i N)PLANT

= ( Di N)LAB

Then

5. Choose mass transfer correlation for kbab. 6. Calculate gas flow rate using the correlation and setting


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(kbab) PLANT = (kbab)LAB For example, from Equation (CD7-51) the mass transfer coefficient kL ab depends on the following equipment parameters

Then

7. Calculate power requirement. 7A. An alternate (2) procedure for determining N and Q is to set either Q/ND = constant or VS = constant and then determine N from power or mass transfer correlations. These and other techniques, such as keeping the power per unit volume a constant, are discussed in Baily and Ollis.*

* T. J. Bailey and D. Ollis, Biochemical Engineering, 2nd ed., McGraw-Hill, New York, 1987. TABLE CD7-3 PHOSPHOMANNAN FERMENTATION SCALEUP *


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Back



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Additional Homework Problems CDP7-AA

The following observations have been made about the reaction

1. The reaction rate is independent of hydrogen concentration. 2. The reaction rate decreases with increasing concentrations of G=CH 2. 3. The initial rate appears to be firstorder with respect to GCH 3. 4. When there is an extremely high concentration of G=CH 2 and a low concentration of GCH 3,the reaction appears to be second-order with respect to GCH 3. (a) Deduce a form of the rate law from the information above. (b) Suggest a mechanism. [2nd Ed. P7-6]


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cdp7-ba

Additional Homework Problems

The reaction of hypochloride and iodide ions in aqueous solution,

occurs rapidly. The rate law for this reaction is reported to be

CDP7-BA

Suggest a mechanism. (Hint: It is believed that either HOCl or HOI or both may be active intermediates.) [2nd Ed. P7-8]


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CDP7-CA.htm

Additional Homework Problems CDP7CA

At high substrate concentrations, the substrate itself may tie up the enzyme-substrate complex by forming the nonreactive complex S E S:

Derive the rate law for the reaction

in which substrate inhibition is also takng place. [2nd Ed. P7-16]


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(POLYMATH) The enzymatic isomerization of glucose, G, having the formula C 6H12O6, to fructose, F, occurs in the following manner due to enzyme E.

A suitable isothermal reaction rate expression is given by

CDP7-DB

(a) Calculate the equilibrium conversion of glucose for an original water solution containing no fructose. (b) Determine the time in hours for a batch reactor containing only G0 = 2 g mol/ dm 3 and E = 15 units/mL to yield 60% of the equilibrium conversion. (c) A particular batch of glucose feed solution has impurities that degrade the activity of the enzyme by the function E = E0e-0.2t , where t is hours in the batch reactor. Determine the new reaction time for the conditions of part (b). (Note: E0 = 15 units/mL.) (d) There is uncertainty regarding the values of k3,k2, K G, and KF . Use a software package such as POLYMATH to find the sensitivity of the solution to different values of the rate and equilibrium constant. Recommended values:

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CDP7-CA.htm

[2nd Ed. P7-19]



CDP7-EB.htm

Additional Homework Problems CDP7-EB

Redo Problem P7-17 to include (a) Chain transfer to the solvent with ks = 5 x 10-3 dm 3/mol s. (b) Chain transfer to the monomer with km = 5 x 10-2 dm 3/mol s. (c) Vary the ratios of k0/kp, k0/k s , kp/km, and so on, and describe your results [2nd Ed. P7-23]


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Streptomyces, a mycelial organism, is grown commercially in a batch fermentor using glucose as the primary nutrient to produce various antibiotics. Air is bubbled through the liquid containing these mycelial cells such that the fermentor is analogous to a slurry reactor. The growth rate of the cells is a function of oxygen concentration. The cells grow in clumps, and consequently, oxygen must diffuse through the clumps to reach the inner core of the clump. As it diffuses it is consumed by the outer core, analogous to diffusion and reaction in a catalyst pellet. The growth rate of the cells is

where

CDP7-FB

Because the nutrient is in excess,

(a) Develop equations for the percentage resistance of each of the oxygen transport steps for (1) slow oxygen consumption rate by the cell. (2) fast oxygen consumption rate.

(b) What parameters should be changed to increase the growth rate? (c) Using any appropriate analogies with slurry reactors, derive an equation for cell concentration as a function of time. Assume that the cell clumps remain all the same size (i.e., only generation of new clumps occurs). (d) Repeat part (c) assuming that no new clumps are produced, so that clumps only grow in size. (e) Obtain numerical values for your model parameters in part (c) and plot

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CDP7-FB.htm

cell concentration as a function of time (see S. J. Pirt, Principles of Microbe and Cell Cultivation, Halsted Press, New York, 1975). Carry out a parameter sensitivity study on your model. [2nd Ed. P12-12]



CDP7-GB.htm


The growth of a bacterium B is carried out in a CSTR. The rate law for the growth of bacteria is

CDP7-G B

The substrate S is in great excess. Unfortunately, some amoeba inadvertently found their way into the reactor and now prey on the bacteria. The amoeba's, A, growth law is

The death law for amoeba is

(a) Write balances on both the bacteria and amoeba. Next assume steady state to arrive at an equation for the dilution rate. Assuming that the dilution rate is only variable under control, describe exactly what you would do (give exact number where possible) to rectify the situation (i.e., get rid of the amoeba). Assume that steady-state conditions are reached immediately after any changes have been made. [In part (b) this case will not apply.] (b) Use POLYMATH or some other ODE solver to plot the concentrations or the bacteria and amoeba as functions of time. and plot (c) Vary the parameters the concentrations as functions of time. Look for any unusual oscillations and/or chaotic behavior.


At the time the amoeba entered the tank,

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There are no bacteria or amoeba in the feedstream. [2nd Ed. P12-15]



CDP7-HC.htm


It is desired to scale up our current fermentation to produce 20 kg/h of a product in a continuous fermentor. A laboratory-scale fermentor is 1 dm in diameter and 2 dm high with an impeller that is 0.6 dm in diameter which rotates at 3 rpm. The oxygen flow rate to the laboratory fermentor is 2 dm 3/min. The fermentation is oxygen-limited. Scale up the laboratory fermentor to full-scale operation.

CDP7-H C


[2nd Ed. P12-16]


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CDP7-IB.htm


Experiments showed [Biotech. Bioeng., 17, 1137 (1975)] that the denitrifying ability of the bacterium Micrococcus denitrificans was retained even after these bacteria were encapsulated in a liquid surfactant membrane. These immobilized cells created an emulsion when suspended in water; each drop was viewed as a "bag of enzymes" that followed simple Michaelis-Menten kinetics. The following laboratory batch reactor data were obtained on the denitrification of various aqueous solutions of nitrite ion, all containing the same volume fraction of cell emulsion (concentration of cells in emulsion was 50 mg of wet cells/mL of emulsion);

CDP7-I B

The aqueous effluent from Acme Chemical Co. contains 0.25 M nitrite ion. Management wishes to fill an available 500-ft3 concrete basin (modeled as a well-mixed batch reactor) with this effluent, and then subject it to denitrification with a similar volume fraction of emulsion containing a similar charge of cells as was done in the lab. If the cells remain active for only 68 h, will it be possible to reduce the nitrite level to 500 ppm? If not, what parameter could be changed to achieve this reduction? (J. Pochodylo, University of Michigan, 1983) [2nd Ed. P12-18]


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Additional Homework Problems CDP7-JA

The data below (i.e., the reciprocal of the growth rate as a function of the reciprocal of the substrate concentration) were obtained in a continuous stirred tank fermenter (CSTF/ CSTR) for an uninhibited bioreaction. A feed of pure substrate at a concentration of 100 g/ dm 3 enters the fermenter at a volumetric flow rate of 10 dm 3/min. (a) What are the Monod rate law parameters? (b) What fermenter volume is required to produce an effluent product flow rate of 20 g/min with a cell concentration of 20 g/ dm 3?


[2nd Ed. P12-19]


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Lectures 15 and 16

Lectures 15 and 16

Evaluating the Heat Exchange Term, Q (Chapter 8) *

Energy transferred between the reactor and the coolant:

Assuming the temperature inside the CSTR, T, is spatially uniform:

At high coolant flowrates the exponential term will be small, so we can expand the exponential term as a Taylor Series, where the terms of second order or greater are neglected, then:

Since the coolant flowrate is high, Ta1

Ta2

Ta:

Reversible Reactions (Chapter 8, Appendix C)

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Lectures 15 and 16

For the special case of

:

Algorithm:

1. Choose X

calc T

calc k

(if gas, calc To/T)

calc KC

calc -rA

2. Increment X and then repeat calculations. 3. When finished, plot

vs. X or use some numerical technique to find V.

Levenspiel Plot for an exothermal, adiabatic reaction. The area under the curve is the PFR volume.

Steady State PBR with Heat Exchange (Chapter 8) file:///H:/html/course/lectures/fifteen/index.htm[05/12/2011 16:56:50]

Lectures 15 and 16

NOTE: The PFR formulas are very similar.

Heat exchange for a PFR: a = heat exchange area per unit volume of reactor; for a tubular reactor, a = 4/D

Catalyst weight is related to reactor volume by:

Heat exchange for a PBR:

Steady State Energy Balance (with no work):

Final Form of the Differential Equations:


Lectures 15 and 16

Solve simultaneously using an ODE solver.

Example: Exothermic, Reversible Reaction Rate Law:


Lectures 15 and 16

Reactor Inlet Temperature and Interstage Cooling (Chapter 8) Optimum Inlet Temperature:

Interstage Cooling:

*




Lectures 15 and 16



Lectures 17 and 18

Lectures 17 and 18

Multiple Steady States (Chapter 8) *

where

Now we need to find X. We do this by combining the mole balance, rate law, Arrhenius Equation, and stoichiometry.

For the first-order, irreversible reaction A --> B, we have:

where

At steady state:

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Lectures 17 and 18

Unsteady State CSTR, Semibatch, and Batch Systems (Chapter 9) Balance on a system volume that is well-mixed:

Adiabatic batch reactor with no work:

Make a table to generate a Levenspiel Plot:

or use one of the integration formulas, e.g.: reaction time, t.


, to find the

Lectures 17 and 18

Polymath The following reaction occurs in a batch reactor:

Multiple Reactions (Chapter 8) Multiple Reactions in a PFR


Lectures 17 and 18

CSTR Energy Balance q reactions and m species

Independent Reactions

Series/Parallel Reactions


Lectures 17 and 18

Combined Mole Balance, Rate Law, and Stoichiometry:

where F T = F A + F B + F C + F D, P = P 0, and

Energy Balance:

Control of Chemical Reactors (Chapter 9)


Lectures 17 and 18

Integral Control

For the reaction

in a CSTR:


Lectures 17 and 18

Proportional Control

Proportional-Integral Control

*



Lectures 17 and 18




Lecture 35, Part A

Lecture 35, Part A Runaway Reactions (Chapter 8) *

Exothermic Reaction

Tubular reactor with heat exchange

Figure 1. Temperature profile for different entering concentrations.

Phase Plane Plots We transform the above temperature and concentration profiles into a phase plane of temperature T and concentration CA in the following manner.

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Lecture 35, Part A

Figure 2. C A and T profile

Figure 3. Phase Plane.

For example, at volume V2 the temperature is T2 and the concentration is CA2

Figure 4. Temperature concentration phase plane plot for different entering concentrations. The governing equations for a PFR/PBR:

Mole Balance: Energy Balance: Solutions to these equations give the following typical profiles (in the absence of runaway):


Lecture 35, Part A

Figure 5. Temperature and concentration profiles. At the maximum temperature in the reactor, Tm

then

For a first order rxn:

, then

and we can solve for CAm to obtain

where Tm is the maximum temperature in the reactor and CAm is the corresponding concentration of A at that maximum temperature. For this reactor . The following figure shows a plot of CAm vs. Tm for different ambient temperatures (T aº Tw)

Figure 6. Maximum curves for a few wall temperatures. Simplified curve.


Lecture 35, Part A

We note that the concentration CAm goes through a maximum as the temperature Tm is varied. Different values of this maximum result for different values of Ta (T w). The line PS is the locus through these maximum values. We want to see how the concentration of A at the maximum temperature CAm varies with the maximum temperature Tm. Specifically we want to find the maximum value of CAm and the corresponding temperature Tm.

Criteria 1. The trajectory going through the maximum of the "maxima curve" is considered as critical and therefore is the locus of the critical inlet conditions for CA and T corresponding to a given wall temperature.

Figure 7. Critical trajectory on the CA -T phase plane plot. The critical trajectory goes through Tm. What is the inlet concentration that is related to this critical trajectory? The safe inlet concentration can be found by using adiabatic conditions.


Lecture 35, Part A

CRITERIA BASED ON INFLECTION POINTS

Figure 8. Temperature profile showing inflection points. Let

At the inflection point

and from the mole balance


Lecture 35, Part A

At

If we have a first order reaction Then we can solve this equation to find CAi as a function of Ti to find the locus of the inflection points. Dividing by

multiplying by

where

For a first order reaction


and rearranging we obtain

Lecture 35, Part A

Criteria 2. Runaway will occur when the trajectory starting at CA0 and To intersects the locus of the inflection points

Figure 9. CA - T Phase plane showing critical trajectory inersecting the locus of the inflection points. Criteria 2A. (Conservative Criteria) Runaway will occur if the trajectory starting at CA0 and To intersects the locus of the maxima of CAm and Tm for different values of To (i.e. Tw because To = Ta = Tw)

Figure 10. CA -T Phase plane plot showing locus of maxima. Runaway reaction figure from Froment-Bischoff (FB) "Chemical Reactor Analysis and Design"

For Comparing the term multiplying

in Froment-Bischoff (Eqn 11.5.2-2) with Fogler (Eqn 8-56) for DCP


Lecture 35, Part A

=0

therefore:

For no pressure drop CT = CTo . Note: The equation for (N/S) in Froment and Bischoff is not correct. If you check the units you will find (N/S) is not dimensionless. where:

rg

= Gas density (kg/m3)

Mm

= Mean molecular weight (kg/kmol) dt = Tube diameter (m) C T = Total concentration (kmol/m3) C A0 = Entering concentration of A (mol/m3) = Overall heat transfer coefficient (J/m 2 • U s • K) C Pio = Heat capacity J/kg•K) R = Gas constant (J/mol•K)


Lecture 35, Part A

E -rA0

= Activation energy (J/mol) = Rate of reaction (mol/s• m 3) at To,

DHRx = Heat of Reaction (J/mol) To

= Temperature (K) (To = Tw = Ta)

Figure 11. Runaway diagram ( Adapted from G. F. Froment and K. B. Bischoff, Chemical Reactor Design and Analysis, 2nd ed. New York: John Wiley and Sons, 1990). *

All chapter references are for the 3rd Edition of the text Elements of Chemical Reaction Engineering. Back to the top of Lecture 35A.



Lecture 35, Part A



excd8-1.htm

Learning Resources Example CD8-1 Heat Capacities Expressed as Quadratic Functions of Temperature

N2 + 3H 2

2NH3

The heat capacities are found in Perry's Handbook. The units are cal/mol K for C p and K for T.

(CDE8-1.1)

(CDE8-1.2)

(CDE8-1.3)

Then, substitute to obtain

© 1999 Prentice-Hall PTR


excd8-1.htm

Prentice Hall, Inc. ISBN 0-13-531708-8 Legal Statement


excd8-2.htm

Learning Resources

Example CD8-2 Second Order Reaction Carried Out Adiabatically in a CSTR

The acid-catalyzed irreversible liquid-phase reaction

is carried out adiabatically in a CSTR.

Chapter 8, Figure CDE8-2.1

The reaction is second order in A. The feed, which is equimolar in a solvent (which contains the catalyst) and A, enters the reactor at a total volumetric flowrate of 10 dm 3/min with the concentration of A being 4M. The entering temperature is 300 K. (a) What CSTR reactor volume is necessary to achieve 80% conversion? (b) What conversion can be achieved in a 1000 dm 3 CSTR? What is the new exit temperature?


excd8-2.htm

(c) How would your answers to part (b) change, if the entering temperature of the feed were 280 K? Additional Information:

Solution


ch8_lr.frm

Learning Resources

PFR/PBR Solution Procedure for a Reversible Gas-Phase Reaction

(CDT81.1)

(CDT81.2)

(CDT81.5)

(CDT81.6)

(CDT81.7)

(CDT81.8)

(CDT81.9)



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Steady-State Bifurcation Analysis

In reactor dynamics it is particularly important to

find out if multiple stationary points exist or if sustained oscillations can arise. Bifurcation analysis is aimed at locating the set of parameter values for which multiple steady states will occur.1 We apply bifurcation analysis to learn whether or not multiple steady states are possible. A bifurcation point is a point at which two branches of a curve coalesce as a parameter is , in which x varied. Consider the function is a scalar variable and is a parameter. Figure CD8-1 shows curves (AB, BC, and BD) for which

(CD81)

is satisfied. We see that as we start to increase TIME GIF along AB, there is only one value of x for a given that will satisfy Equation (CD8-1). However, as we continue to increase along AB, we reach a bifurcation point * beyond which there are two values of x that satisfy Equation (CD8-1) for a given value of . Consequently, we analyze our system of equations to learn if a bifurcation point exists that denotes multiple solutions. There is another condition that is necessary for a bifurcation point to exist. If we were to move an away from the bifurcation incremental amount point but still remain on BC or BD, we would have


(CD82)

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Figure CD8-1 Bifurcation point.

Expanding Equation (CD8-2) in a Taylor series, it can be shown that at the bifurcation point

If Equations (CD8-1) and (CD8-3) are satisfied, there will be a set of parameter values for which we will have multiple steady states (MSS).

(CD83)

We shall continue the discussion of the first-order

reaction taking place in a CSTR to illustrate bifurcation analysis. A slight rearrangement of a combination of Equations (8-68) and (8-69) from the energy balance gives

which is of the form

where

Similarly, with some minor manipulation, the mole balance on an isothermal CSTR can be put in similar forms,

and

are positive constants.


(CD84) (CD85)

(CD86)

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or

(CD87) We observe that both the CSTR energy and mole

balances are of the form

We can express the mole and energy balances by equations that are of similar forms

(CD88)

If F(y) is a monotonically increasing function as shown in Figure CD8-2, the derivative of the function with respect to y will never be zero, that is,

(CD8 Upon differentiating Equation (CD8-8), we have 9)

and we see that dF/dy can never be zero if the maximum value of the derivative of G is less than . Thus the sufficient condition for uniqueness is

Uniqueness

When Equation (CD8-10) is satisfied there will be no multiple steady states. However, if max , we do not know at this point whether or not multiple solutions exist and we must carry the analysis further.


(CD810)

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Figure CD8-2 MSS with a parabolic function.

Figure CD8-3 F(y ) curve common to chemically reacting systems.

We now apply the condition for a bifurcation point, Equations (CD8-1) and (CD8-3), to the CSTR balances, that is, Equation (CD8-8). If multiple steady states exist, there will be a bifurcation point, y*, at which the following conditions are satisfied:

Conditions for multiple steady states


(CD811) (CD8-

08prof.htm

12) Figure CD8-2 shows that a parabola satisfies both

conditions (1) and (2), that is, Equations (CD811) and (CD8-12), respectively. Consequently, we see that there will be a set of parameter values for which multiple solutions will exist, as demonstrated by the dashed-line parabola. Figure CD8-3 shows the shape of a typical curve in reaction systems with multiple steady states.



excd8-3.htm

Professional Reference Shelf Example CD8-3 Determining the Parameters That Give MSS 2

The oxidation of carbon monoxide is carried out in excess oxygen in a "fluidized" CSTR containing catalyst particles impregnated with platinum:

Combining a mole balance with the rate law gives

At the bifurcation point,

(CDE83.3)

The rate law for the disappearance of CO, (A), is (CDE83.1)


CDE83.2)

(CDE8-3.4)

and Equation (CD8-12) requires (CDE8-3.5)

Combining Equations (CDE8-3.4) and (CDE8-3.5) yields

(CDE8-3.6)


excd8-3.htm

Solving Equation (CDE8-1.6) for CA gives us

(CDE8-3.7)

For , no real roots exist and there are no possible steady states. A rearrangement of Equation (CDE8-1.4) gives

(CDE8-3.8)

Note that the right-hand side of Equation (CDE8-3.8) goes through a maximum as is increased from 1. To find this maximum we set the derivative of the right-hand side of Equation (CDE8-3.8) with respect to CA equal to zero, and solve to find that the maximum occurs at

Substituting Equation (CDE8-3.9) into the right-hand side of Equation (CDE8-3.8) gives the maximum as

CDE8-3.9)

(CDE8-3.10)

The maximum value of the right-hand side is 1/27; consequently, if is smaller than 27. Equation CDE8-3.10 can never be satisfied and there will be no MSS. Figure CDE8-1.1 shows a mapping of those regions where no multiple steady states will exist.

Figure CDE8-3.1 Mapping the regions of no multiple steady states.

To learn the regions where MSS exist, we need to

carry the analysis further. If we let


,

excd8-3.htm

Equation (CDE8-3.6) can be written as

(CDE8-3.11)

and Equation (CDE8-3.5) as

(CDE8-3.12)

Equations (CDE8-3.11) and (CDE8-3.12) are used to

form Table CDE8-3.1. Figure CDE8-3.2 shows a plot . The shaded area shows the of as a function of combinations of these variables that will produce multiple steady states.

Figure CDE8-3.2 Region of possible multiple steady states.


excd8-3.htm © 1999 Prentice-Hall PTR Prentice Hall, Inc. ISBN 0-13-531708-8 Legal Statement


08prob.htm

Additional Homework Problems CDP8 AB

The elementary gas-phase reaction

is to be carried out in the various reactors discussed below. The feed, which is at a temperature of 27°C, consists of 80% of A and the remainder inerts. The volumetric flow rate entering the reactor at this temperature is 100 dm 3/min. The concentration of A in the feed at 27°C is 0.5 mol/dm 3. For 80% of the adiabatic equilibrium conversion: (a) Calculate the volume of a plug-flow reactor when the reaction is carried out adiabatically. (b) Plot conversion and temperature as a function of reactor length if the reactor diameter is 5 cm. (c) Calculate the reactor volume for a CSTR that is operated adiabatically. (d) Calculate the temperature and conversion profiles down a tubular reactor 0.05 m in diameter that is not insulated. The ambient temperature is 27°C and the overall heat-transfer coefficient is 10 W/ m 2 K. (e) Repeat part (d) for northern winter and summer operation.

The heat of reaction is a function of temperature, and its value at 300 K is ó75,000 J per mole of A. At 300 K,


k1 varies with temperature as follows:

[2nd Ed. P8-16]

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08prob.htm



cdp8-bb.htm

Additional Homework Problems CDP8-BB

The catalytic reaction

is carried out isothermally on a solid platinum wire in which the flow of the reactant is normal to the wire. The rate law on the solid platinum surface is

where C A is the concentration at the wire surface in the gas phase. Use bifurcation analysis to answer the following:

(a) What is the lowest feed concentration, C A0 , one can have and still have the possibility of multiple steady states? (b) Derive an equation relating the wire diameter D to the velocity U, which divides a plot of U vs. D into two regions, one in which a multiple steady state (MSS) is possible and one in which a MSS is not possible.


(c) The smallest platinum wire available has a diameter of 0.025 cm. For this wire, what is the largest gas velocity one can have and still have a possibility of multiple steady states? For flow normal to a wire, Sh = 0.4 Re0.4 Sc1/3 . (See Chapter 11.) [2nd Ed. P8-26]


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CDP8-CB.htm


A gas-phase isomerization is carried out isothermally in a fluidized CSTR for a number of different catalysts. Determine which of the following catalysts might produce multiple steady states:

If any exhibit multiple steady states, map out the regions.

CDP8-CB

[2nd Ed. P8-27]


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cdp8-db.htm

Additional Homework Problems CDP8-DB

Nutrient A is fed to a CSTR in which microorganism B is produced:

The rate law for this biochemical reaction is

Map out the regions where multiple steady states are not possible and those where they are. [2nd Ed. P8-28]



cdp8-ec.htm

Additional Homework Problems CDP8-EC

(Sulfur dioxide oxidation in a tubular reactor with heat losses) Using the data, feed temperature, flow rate, and composition given in Example 8-10, reconsider the oxidation. Determine the effect of changing various ) on parameter values (e.g., the conversion and temperature profiles. Critically review all property data associated with this problem. [2nd Ed. P8-33]


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cdp8-fc.htm


(Interstage cooling) In the reactor system shown in Figure CDP8-F, reactor 1 is operated at feed conditions identical to those given in Example 810. The stream leaving reactor 1 is cooled to T 2 in a double-pass tube-and-shell heat exchange before entering reactor 2, which is operated adiabatically. Assuming that reactor 2 contains the same weight of catalyst as reactor 1, determine the optimum feed temperature to reactor 2 together with the conversion associated with this feed temperature.

Figure CDP8-F [2nd Ed. P8-34(a)]

CDP8-FC


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cdp8-gb

Additional Homework Problems CDP8GB

(a) Determine the minimum cost of the combination of the heat exchanger and reactor 2 to achieve an overall conversion of 95% (based on the feed to reactor 1). Air is to be used as the coolant and is available at 77°F and a rate of 100 mol /h. The cost of the reactor is

This cost includes both the catalyst and reaction container costs. The cost of the heat exchanger is ( b) Rather than using air as a coolant

in the heat exchanger, the feed stream to reactor 1 is used. The feed stream enters the exchanger at 600°F, is preheated to temperature T 0, and is fed directly to reactor 1. The stream then leaves reactor 1 at temperature T, again enters the air heat exchanger (this time as the hot steam), exits the exchange at T 2, and enters reactor 2. The weight of catalyst in reactor 1 is equal to the weight of catalyst in reactor 2. Determine the minimum cost of this reactor system (heat exchanger + reactor 1 + reactor 2) necessary to achieve an overall conversion of 92%. The cost functions are given in part (a). Reasonable first estimates of the catalyst weight and T 0 in each reactor are 1000 lb and 910°F, respectively. [2nd Ed. P8-34(b)&(c)]


cdp8-gb



cdp8-hb.htm

Additional Homework Problems CDP8-H B

You wish to use an existing CSTR for the liquidphase exothermic first-order reaction

Unfortunately, there is also an undesired side reaction

which is liquid phase, endothermic, and first order.

You have available as a heating medium saturated steam at 212°F. The feed is available at 130°F. What area of heating coil (if any) will you specify to maximize the production of B? (Ans.: A = 106 ft 2 .)


Reasonable assumptions: The density and specific heat of fluid are unchanged by reaction or temperature change. The effect of the coil on the effective reactor volume is negligible.

[1st Ed. P9-21]

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cdp8-hb.htm © 1999 Prentice-Hall PTR Prentice Hall, Inc. ISBN 0-13-531708-8 Legal Statement


cdp8-ib.htm

Additional Homework Problems CDP8-I B

The heat exchanger in Problem 8-20 is to be replaced with a larger unit that can adjust the heat exchange area. Assuming that everything else in the problem statement remains the same, what is the value of UA needed to obtain the maximum possible conversion in the reactor? What is the corresponding maximum conversion? [Old Exam]



cdp8-jb.htm


Phthalic anhydride is an important intermediate in the fiber industry. It can be produced by controlled oxidation of napthalene over vanadium pentoxide catalyst. Many reactions occur during the oxidation, all of which are extremely exothermic. A fluidizedbed reactor is preferred so that heat removal from the system can be achieved more easily. You are required to simulate the behavior of this reactor and recommend suitable conditions of operations.

CDP8-JB

Additional information

In such multiple reactions the behavior is always expressed in terms of any two of the following quantities:

You are required to simulate the behavior of the reactor so as to maximize the yield of intermediate product P.

(a) Express p, p, and p in terms of rate constants and residence time . fixed, it is desired to (b) Keeping choose the optimum operating temperature such that is maximized. Show that the temperature that maximizes p is given by

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cdp8-jb.htm

(c) A thermic fluid at 275°C is available for cooling purposes. The temperature of this fluid can be assumed to change negligibly as it flows through the heattransfer tubes immersed in the fluid bed. Estimate the surface area required per unit reactor volume for the following optimum temperatures of operation: T = 331, 365, and 423°C. Tabulate your results. (d) Discuss critically your results in part (c): in particular, the stability of steady state and the merits and demerits of decreasing the operating reactor temperature. (e) Discuss critically the implications of adiabatic operation of the system. (H. S. Shankar, IIT Bombay) [1st Ed. P9-23]



cdp8-kb.htm


The production of a drug intermediate in a batch reactor (Figure CDP8-K) was studied by the Sanofi Chimie Company [Chem. Eng. Sci., 51, 2243 (1996)]. The coded sequence for the liquid-phase reactions is represented by the following symbols:

CDP8-K B

Figure CDP8-K

The symbols correspond to the following compounds:

Radical groups R1, R2, and R3 have had to be kept secret. The desired product is C.

(a) Plot the concentration and yield as a function of time in both the pilot plant and the industrial plant (see Table CDP8-K).

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cdp8-kb.htm

(b) How should the reaction be carried out to maximize the desired product (e.g., temperature, reactor type, heating/cooling schedule, feed conditions)?


The reactions are considered elementary with



cdp8-lb.htm


The concentration-temperature curves for steady energy and mole balances are shown in Figure CDP8-L for a particular reaction. Sketch the various approaches to steady state starting at the different initial conditions shown for points 1 through 6. (Hint: See Figure 7.18, R. Aris, Elementary Reactor Analysis (Upper Saddle River, N.J.: Prentice Hall, 1969).]

CDP8-LB

Figure CDP8-L

Next consider a first-order irreversible exothermic carried out in an liquid-phase reaction adiabatic CSTR. The following data are given:

The following questions concern variation in one parameter with the others fixed at the values given.

(a) At what space-times do ignition and extinction (blowout) occur? (b) At what feed concentrations do ignition and extinction occur? (c) At what feed temperatures do ignition and extinction occur? For the parameter values given: (d) Plot the curves F = 0 and G =0 in

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cdp8-lb.htm

the phase space (CA , T ). (e) Identify stable and unstable operating points, and corresponding values of C A and T. (f) Calculate the location of the separatrix. (g) Plot the phase and time trajectories for the following initial conditions:

(R. L. Curl, University of Michigan) [2nd Ed. P8-22]



cdp8-nb.htm

Additional Homework Problems CDP8-NB

The reversible second-order liquid-phase reaction

is carried out adiabatically in a CSTR. Species A is soluble in a number of different inert solvents, each with a different heat capacity that can be used as a feed stream. Because the solvents are miscible, the heat capacity of the feed stream can be varied continuously by mixing different solvents. The molar feed ratio of inert solvent to reactant A is 4:1. We will use a mixture of two solvents, P and Q. Solvents P and Q may be mixed in any ratio to give the desired heat capacity. The heat capacity of solvent P is 20 cal /g mol °C all that of solvent Q is 40 cal/g mol °C and a 50:50 mixture would give a heat capacity of 30 cal /g mol °C. The feed temperature of all species is 380 K.


(a) What is the conversion if only solvent P is used? (b) What is the conversion if only solvent Q is used? (c) What should be the ratio of solvent P to solvent Q to obtain the maximum conversion? Assume a linear mixing rule to determine the heat capacity of the solvent mixture. What is the maximum conversion? (d) At the maximum conversion in part (c), are there other steady states, and if so, what are the corresponding conversion and reactor temperature? (See Table CDP8-N and Figure CDP8N.) (Note: The volumetric flow rate as well as the molar flow rate of A remains unchanged as the solvent ratio is varied.)

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cdp8-nb.htm

Figure CDP8-N

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cdp8-oc.htm

Additional Homework Problems CDP8OC

Consider the gas-phase reactions

which take place in a PFR. The molar feed rate of A is 100 mol/m.

(a) At what volume in the reactor is the temperature a maximum? (b) At what volume is the concentration of B a maximum? Additional information:


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cdp8-pb.htm

Additional Homework Problems CDP8-PB

The exothermic reversible gas-phase reaction follows an elementary rate law:

and is carried out in a packed-bed catalytic reactor. The reaction takes place adiabatically and with no pressure drop. At 600 K the equilibrium constant is 100 and the heat of reaction is (-210,000 cal /g mol of A). The specific reaction rate is 30.0 dm 3/g mol kg cat. min at 273 K, and the activation energy is 4000 cal/g mol A. Pure A enters the reactor at a temperature of 50°C, a flow rate of 100 g mol/min, and a concentration of pure A of 0.1 g mol/dm 3. A and B both have heat capacities of 8 cal/g mol K

(a) Summarize the simplified governing equations to determine the PBR weight required to achieve the desired conversion, substituting all the values from the data given above. (Temperature-heat of reaction relationship, Arrhenius equation, equilibrium constant as a function of conversion, concentrations, and reaction rate) (b) Calculate the adiabatic equilibrium conversion. Use any sketches or figures that you feel are appropriate. (c) Use an ODE solver such as POLYMATH or MATLAB, or construct a table similar to the one presented in Table E8-6.2 and determine the PBR weight needed to achieve 98% of the equilibrium conversion.


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Lecture 35, Part B

Lecture 35, Part B

Linearized Stability Theory (Chapter 9) * Energy Balance (Applied to a CSTR) 1.

2.

3.

4.

CSTR Mole Balance 5.

6.

Manipulating the Energy and Mole Balances Let

and

signify steady state values. Then at steady state:

7.

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Lecture 35, Part B

Adding equations (6) and (7):

8.

Linearizing

by expanding it in a Taylor Series:

To obtain: 9. 10.

Let

11.

12. 13.

Using these substitutions, we can arrive at the following equations that describe the behavior of temperature and concentration, when the steady state conditions are perturbed in a CSTR:


Lecture 35, Part B

14.

15.

16.

17. 18. 19.

20.

21.

22.

Predicting the Behavior of a CSTR using LST At time t = 0, y1 = y10, where:

Making use of Equation 20


Lecture 35, Part B

we notice that for the case of b2 = 4c: if b < 0 , then the amplitude (i.e., T - TS ) will increase if b > 0 , then the amplitude (i.e., T - TS ) will decrease and that: if b2 > 4c , then if b2 < 4c , then

is real (i.e., non-oscillatory behavior) is imaginary (i.e., oscillatory behavior)

1. Critically damped: b is positive and 2. Unstable growth: b is negative and

is real is real

3. Oscillatory and damped: b is positive and 4. Oscillatory: b is zero and

is imaginary

5. Unstable growth oscillation: b is negative and


is imaginary

is imaginary

Lecture 35, Part B

Reference: R. Aris, Elementary Chemical Reactor Analysis, Prentice Hall, New Jersey, (1969). *

All chapter references are for the 3rd Edition of the text Elements of Chemical Reaction Engineering. Back to the top of Lecture 35B.




excd9-1.htm

Learning Resources Example CD9-1 Startup of a CSTR

Again we consider the production of propylene glycol (C) in a CSTR with a heat exchanger. Initially there is only water at 75°F and 0.1 wt % H2SO 4 in the 1420-gallon reactor. The feed stream consists of 400 lb mol/h of propylene oxide (A), 5000 lb mol/h of water (B) containing 0.1 wt % H2SO 4 ,and 20 lb mol /h of methanol (M). Plot the temperature and concentration of propylene oxide as a function of time, and a concentration vs. temperature graph for different entering temperatures.

Solution

(CDE91.1)

(CDE91.2)

(CDE91.3)

(CDE91.4)

(CDE9-


excd9-1.htm

1.5) (CDE91.6)

(CDE91.7)

(CDE91.7)

(CDE91.9)

From Equation (E8-4.10):

(CDE91.10)

Neglecting because it changes the heat of reaction insignificantly over the temperature range of the reaction, the heat of reaction is assumed constant at

The POLYMATH program is shown in Table CDE9-1.1. Figures CDE9-1.1 and


excd9-1.htm

CDE9-1.2 show the reactor temperature and concentration of propylene oxide as a function of time. One observes that both the concentration of A and the temperature oscillate around their steady-state values before coming to rest at these values. The corresponding phase-plane plot of concentration and temperature shows a spiral approach to the steady state (Figure CDE9-1.3).

Table CDE9-1.1


excd9-1.htm

Figure CDE9-1.1

Figure CDE9-1.2


excd9-1.htm

Figure CDE9-1.3



excd9-2.htm

Learning Resources

Example CD9-2 Falling Off the Steady State

Propylene glycol is produced in a CSTR in which precautions have been taken to prevent product loss. Propylene oxide (A) is fed to a 413-gal reactor at a rate of 300 lb mol/h at a temperature of 75°F. Water (B) is fed at a rate of 2000 lb mol/h and methanol at a rate of 20 lb mol/h. Determine what happens to a CSTR operating at 172°F and a conversion of 86.4% when there is a drop in feed temperature from 75°F to 68°F.

The property values are the same as those given in Example 8-4.

We will first look at the steady-state conditions before the upset occurred. Recall Text Equation (E8-4.5), obtained from the mole balance,

and Equation (E8-4.6) obtained from the energy balance,

Solution

(E84.5)

(E84.6)

Neglecting and substituting the appropriate values, we have


excd9-2.htm

(CDE92.1)

(CDE92.2)

If we were to plot XEB and XMB as a function of temperature, we would we see that there are three steady-state conditions for this set of parameter values (Table CDE9 2.1). Before the drop in feed temperature occurred we are operating at the upper steady state (T = 172°F, X = 0.864). The ignition temperature of the feed is 78°F, and the extinction temperature is 61°F (review Figure 8-24).

We will now consider what happens when an upset occurs while operating at the upper steady state. Say that the inlet temperature drops from 75°F to 68°F. Figure CDE9-2.1 shows the corresponding XEB and XMB curves after this drop. There are still three steady states (Table CDE9-2.2) for this new inlet temperature, indicating that we have not dropped below the extinction temperature of 61°F.


excd9-2.htm

Figure CDE9-2.1 XEB and XMB after T O perturbation.

Based on a steady-state analysis, it would appear that after the drop in the inlet tmperature we should remain at the upper steady state with T = 164.5°F and X = 82.3%. However, using a dynamic simulation (see the development, equations E9-9.3 through E9-9.12, below Figure CDE9-2.2) with POLYMATH (Table CDE9-2.3), we will see that with this perturbation in the inlet temperature, the conversion and temperature drop to the lower steady-state values, as shown in Figure CDE9-2.2.

T ABLE CDE9-2.3


excd9-2.htm

Figure CDE9-2.2


excd9-2.htm

Unsteady-State Equations After Temperature Perturbation

Although there are slight differences in the densities of the liquid species, we shall assume constant density.

For constant volume, constant heat of reaction, and no work done by the system, Equation (9-9) when applied to this example becomes


excd9-2.htm

The number of moles N i in the denominator is just N i = C i V.

(E99.12)

The other parameter values remain the same as those in Example 9-3.

We see that while the entering temperature did not drop below the extinction temperature, the perturbation in the entering temperature was large enough to drive the temperature and conversion drop to their lower steady-state values of 89.5°F and 11.6%, respectively. To prevent this problem of moving to an undesired steady state, a control system is usually incorporated into the reactor system.


excd9-3.htm

Learning Resources Example CD9-3 Proportional-Integral (PI) Control

Repeat Example 9-6 using a proportional-integral controller instead of only an integral controller. Solution

We note that while the output temperature returns to 138°F, the temperature oscillates over a 20°F range, which may be unacceptable in many situations. To overcome this we will add proportional to integral control. The equations here to be added to those in Table CDE9-3.1 to represent the PI controller are

(CDE93.1)

(CDE93.2)

(Note that all equations are the same as before other than the perturbation variable, and the variable being controlled.) We will set the kC 8.5 and I = av = 0.165 hr. We see that the output outlet temperature does not oscillate as much as with just an integral controller. Try different controllers (differential, proportional, integral) and combinations thereof, along with different parameter values to study the behavior of CSTRs which have upsets to their system (Figures CDE9-3.1 and CDE9-3.2).

T ABLE CDE9-3.1


excd9-3.htm

Figure CDE9-3.1


excd9-3.htm

Figure CDE9-3.2



09prof.htm


9.1 Adiabatic Operation of Batch Reactors

For adiabatic operation and when the work done by the stirrer can be neglected Equation (9-12) can be written as

(913)

where An analytical solution relating X and T can be found for the case of constant or mean heat capacity. Combining Equations (2-6) and (9-13) yields

(915)

multiplying by dt, separating variables and integrating between the initial conditions, X = 0 and T =T 0, and later conditions X =X and T =T, yields

Taking the antilog, we have

solving for X yields

Then


09prof.htm

This form is the same as the equation relating X and T derived for flow reactors. Rearranging yields

(916)

Temperature conversion relationship for an adiabatic batch reactor

(9-17)



09prof2.htm

Professional Reference Shelf Approach to Steady State in a CSTR: Unsteady Mole Balance

A typical plot of C A vs. T for steady-state operation (i.e., MB = 0) is shown in Figure CD9-1. For example, if the reaction is firstorder and irreversible, the equation for the "MB = 0 curve" would be

The unsteady-state mole balance is given by Equation 4-44

where we use shorthand notation:

Using the mole balance (M B = 0) to divide the phase plane Figure CD9-1

This curve (i.e., MB = 0) divides the phase plane into two regions: where concentration increases with time, and where concentration decreases with time.

We now consider the relative magnitude of two positive terms,

When is positive and concentration increases with time. When (i.e. ), the derivative is negative and the concentration decreases with time.


09prof2.htm

To continue our analysis of the approach to steady state we need to consider the energy balance next. For operation of a CSTR with , the unsteady energy balance, Equation (9-11), can be rearranged in the form

Unsteady mole balance

where

Figure CD9-2 shows a typical plot of C A vs.T for EB = 0. For an irreversible first-order reaction, the equation for the "EB = 0 curve" would be

This curve divides the phase plane into two regions:

1. 2.

where temperature increases with time, and where temperature decreases with time.

Using the energy balance (EB = 0) to divide the phase plane

Figure CD9-2

Temperature can either decrease or increase with time depending and on the relative magnitudes of .

Next we combine Figures CD9-1 and CD9-2 to yield Figure CD93, which shows the curve for MB = 0 and EB = 0 for a particular set ). of parameter values (e.g.,


09prof2.htm

(Note that other parameter values will give a different set of intersections of MB = 0 and EB = 0). The steady-state value is, of course, the intersection of the MBand EBcurves. We can divide the phase plane into four quadrants:

Figure CD9-3

Figure CD9-4 Quadrants of the phase plane.

The dashed square around the steady state in Figure CD9-3 is expanded and shown in Figure CD9-4 identifying the various quadrants. Temperature is increasing in quadrants I and II and decreasing in quadrants III and IV, while the concentration of species A is increasing in quadrants I and IV and decreasing in II and III. With this information we may make qualitative sketches of the temperature-concentration pathways or trajectories. Figure CD9-4 shows a sketch of two different approaches to the steady state that result from starting the reactor up at the same


09prof2.htm

initial temperature, T i but at two different initial concentrations. Starting at T i and C A = 0 in quadrant IV, we know that C A , will increase and T will decrease until we cross the and EB = 0 curve into quadrant I, where both C A and T increase until we cross the MB = 0 curve into quadrant II, where C A decreases and T increases. The C A vs.T phase-plot trajectory continues its spiral approach to the steady state ( C As ,Ts ) as it crosses into quadrant III, where both C A and T decreases and then it reenters quadrant IV. This trajectory is shown in Figure CD9-5. A similar type of reasoning can be used to trace out the trajectory starting at T i and C Ai .

Figure CD9-5 Approach to the steady state.

Example CD9-1 Startup of a CSTR Example CD9-2 Falling Off the Steady State

Example CD9-3 Proportional-Integral (PI) Control


09prof2.htm © 1999 Prentice-Hall PTR Prentice Hall, Inc. ISBN 0-13-531708-8 Legal Statement


09prof3.htm

Professional Reference Shelf 9.6 Unsteady Operation of Plug-Flow Reactors We plan to reduce the energy balance into a more usable form than that given in Equation (8-60). To achieve this form for plug-flow reactors, we begin by applying the balance to a small differential volume, , in which there are no spatial variations (Figure CD9is 6). The number of moles of species i in

Figure CD9-6 PFR with heat gain or loss.

Substituting for N i in Equation (9-3) and dividing

Taking the limit as

Rearranging, we have

and noting that gives

by gives

(CD9A)

Comparing Equations (CD9-A) and (9-33)


(CD9-

09prof3.htm

33)

we note that the term in parentheses is just r i . The rate of reaction of species i is related to the rate of disappearance of species A through the stoichiometric coefficient

Then

Finally,

To include radial variations in temperature, see Problem P8-29

where a is the heat exchange area per unit volume. Differentiating yields

Recalling that

we substitute these equations into Equation (9-19) to obtain

Neglecting shaft work and changes in pressure with respect to time, we obtain

Transient energy balance on a PFR

This equation must be coupled with the mole balances,


(CD932) (CD933)

09prof3.htm

Numerical solution required for these three coupled equations

and the rate law,

(9-34)

and solved numerically. A variety of numerical techniques for solving equations of this type can be found in the book Applied Numerical Methods.1

We note, however, one can approximate the partial differential equation (PDE) of the unsteady PFR Energy Balance by replacing the PFR with a number of CSTRs in series. One then simply solves the coupled ODEs using POLYMATH or MatLab. For steady-state operation in which no work is done by the system, Equation (9-32) reduces to

(8-60)

Substitution for the molar flow rates Fi in terms of conversion gives Equation (8-56).

Use this equation for steady-state energy balance on a PFR with heat transfer

(8-56)

As stated previously, this equation is solved simultaneously with the mole balance.


09prof3.htm © 1999 Prentice-Hall PTR Prentice Hall, Inc. ISBN 0-13-531708-8 Legal Statement


cdp9-ab.htm

Additional Homework Problems Propylene glycol produced by the hyrolysis of propylene oxide as discussed in Example 8-4 is to be carried out in a semibatch reactor (Figure CDP9-A). A mixture of 33.3% methanol and 66.7% propylene oxide is fed to an 80-ft3 semibatch reactor containing 30 ft3 of water at 120°F with 0.1 wt % H2SO 4. The total molar feed rate is 2 lb mol/min at a

CDP9AB

Figure CDP9-A

temperature of 100°F. The cooling water flow rate is adjusted as a function of time to maintains a constant temperature in the reactor of 100°F. The inlet cooling water temperature is 50°F. (a) What is the cooling water flow rate initiatlly? (b) What is the cooling water rate after 6 min? You may neglect the heat of mixing. The product of the overall heat-transfer coefficient U and the heat-exhange area, A, is UA = 6 x 104 Btu/h °F. (c) Plot the conversion and temperature as a function of time for the case of a constant cooling water flow rate of 3000 lb/h. All other conditions remain the same.


cdp9-ab.htm



cdp9-bc.htm


A heat exchanger and a controller are to be added to the ammonium nitrate reactor discussed in Problem P8-3. The reactor temperature is to be controlled by manipulating the coolant flow rate, in the heat exchanger. The ambient temperature at the time the controller is turned on is 970°R. The product of the heat-transfer coefficient and heatexchanger area is UA = 10,000 Btu/h °F.

The controller is to be a PI controller with an integral time, of 1 h and a gain of -4.5°. A sudden upset causes the reactor temperature to rise to 1000°R, which is 25°R above the set point.

For each of the following three cases, plot the reactor temperature and mass in the reactor, M, as a function of time. [Hint: First show that the energy balance can be written as

CDP9-BC

(a) Consider Mconstant at 300 lb. (b) Calculate M from a mass balance, that is,

The set-point of the mass in the tank is 325 lb.

(c) Consider no control by setting the controller gain equal to zero (i.e., kC = 0.) (d)Vary kC and describe what you find. (e) Vary UA, C , , and so on, and describe what you find.

file:///H:/html/09chap/html/cdp9-bc.htm[05/12/2011 16:57:18]

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file:///H:/html/09chap/html/cdp9-bc.htm[05/12/2011 16:57:18]

Lectures 19 and 20

Lectures 19 and 20

Octane Rating (Chapter 10) * A Typical Engine Piston

1. a uniform burning front 2. spontaneous combustion producing detonation waves and knock

Determine the compression ratio, CR, to achieve the standard knock intensity. The more compact molecules are (for a given number of carbon atoms), the greater the octane number they will have.

Rationale:

n-pentane: Octane No. = 62 i-pentane: Octane No. = 90 The difference in octane ratings provides an economic incentive for carrying out this reaction.

Steps in this reaction:

file:///H:/html/course/lectures/nineteen/index.htm[05/12/2011 16:57:21]

Lectures 19 and 20

Focusing on the second reaction:

Mechanism Adsorption

Surface Reaction

Desorption

If the surface reaction is limiting:

Site balance: Substituting for C N-S , C I-S , and C V into C T = C V (1 + KN P N + KI P I) :

where KP is the thermodynamic equilibrium constant for the reactor. Linearizing the Initial Rate:


Lectures 19 and 20

Steps in a Catalytic Reaction (Chapter 10)

Adsorption on Surface Surface Reaction

Single Site Dual Site

Desorption from Surface

Adsorption on Surface

Surface Reaction

Dual Site

Eley-Rideal Desorption from Surface


Lectures 19 and 20

Strictly Speaking I Rate Limiting Step

For steady state operation we have:

where:

NOTE: We really cannot compare the magnitudes of kA , kS , and kD directly, because kA has different units than kS and kD . Consequently, we must compare the product (k A PC) with kS and kD to determine which reaction step may be limiting. If the surface reaction is limiting, we say kA PC and kD are very large with respect to kS :

Strictly Speaking II We only take ratios of quantities that have the same units.

If and are the fraction of free sites and the fraction of covered sites, respectively, and phase mole fraction of species A, then:


is the gas

Lectures 19 and 20

and

For surface reaction control:

Multiply both sides by Ct:

which is identical to the expression derived in the text, assuming that:

Strictly Speaking III NOTE: This is the same rate law we would get by comparing the k's directly.

Assume that the surface reaction is limiting, then:


Lectures 19 and 20

Regulations for Automotive Exhaust Emissions

Principle Reactions:

Surface reaction limiting:


Lectures 19 and 20

Let's see what fraction of sites are covered by CO at the optimum:

Multiplying by C V:

(A) (B)

*




Lectures 19 and 20



Lectures 21 and 22

Lectures 21 and 22

Molecular Adsorption (Chapter 10) *

At equilibrium:

Langmuir Isotherms (Chapter 10)

Dissociative Adsorption (Chapter 10)

At equilibrium:

file:///H:/html/course/lectures/twenone/index.htm[05/12/2011 16:57:22]

Lectures 21 and 22

Chemical Vapor Deposition, CVD (Chapter 10) Manufacture of a Silicon Layer

Site / Surface Area Balance:

For the homogeneous reaction:

then

where:


Lectures 21 and 22

*





Lectures 23 and 24

Lectures 23 and 24

Catalyst Deactivation (Chapter 10) * Separable Kinetics:

Types of Decay 1.) Sintering

2.) Coking

3.) Poisoning 4.) Slow Decay

Temperature-Time Trajectories

5.) Moderate Decay

Moving Bed

6.) Rapid Decay

STTR

Temperature-Time Trajectories The catalyst decay rate is a function of temperature, so you can vary the temperature with time to keep the rate of decay as constant as possible.

Then:

file:///H:/html/course/lectures/twenthre/index.htm[05/12/2011 16:57:25]

Lectures 23 and 24

or solving for

Catalyst Decay Example The gas-phase, irreversible reaction

is elementary with first order decay. The reaction is carried out at

constant temperature and pressure.

Batch Reactor

Moving Bed Reactor

Straight Through Transport Reactor

Mole Balance: Rate Law: Decay Law:

Stoichiometry:

gas phase, but


, T = T0 and P = P 0

Lectures 23 and 24

Combine:

Another Catalyst Decay Example The second-order, irreversible reaction

is carried out in a moving bed reactor. The catalyst loading rate is 1

kg/s to a reactor containg 10 kg of catalyst. The rate of decay is second order in activity and first order in concentration for the product, B, which poisons the catalyst. Plot the conversion and activity as a function of catalyst weight down the reactor. Additional information:

Solution:

Polymath Mole Balance: Rate Law:


Lectures 23 and 24

Decay Law:

Stoichiometry:

Combine:

Conversion vs. Catalyst Weight


Lectures 23 and 24

Catalyst Activity vs. Catalyst Weight


Lectures 23 and 24

Example 10-7: Strictly Speaking When there is a change in the velocity due to a change in the number of moles up through the STTR, one cannot directly substitute t = z/U in the coking activity equation:

(1)

Instead, one must add another equation to the Polymath program. We know that at any location, the gas velocity up the column is:

(2)

Then:


Lectures 23 and 24

(3)

where t = 0 at z = 0. You can use either Polymath or MatLab to solve this equation and substitute it for t in the activity equation:

Along with:

etc.

(same as the program in Table E10-7.1) *


Jump to Lecture(s): 1 & 2 | 3 & 4 | 5 & 6 | 7 & 8 | 9 & 10 11 & 12 | 13 & 14 | 15 & 16 | 17 & 18 | 19 & 20


Lectures 23 and 24

21 & 22 | 23 & 24 | 25 & 26 | 27 & 28 | 29 & 30 31 & 32 | 33 & 34 | 35A | 35B | 36 & 37 | 38 & 39



excd10-1.htm

Learning Resources

Example CD10-1 Analysis of a Heterogenous Reaction

are shown below. The limiting step in the reaction is known to be irreversible, so that the overall reaction is irreversible. The reaction was carried out in a differential reactor to which A, B, and C were all fed.

[Class Problem, Winter 1997]

Experimental data for the gas-phase catalytic reaction

(a) Sketch

as a function of P A

, as a

function of P B , and as a function of P C . (b) From your observations in part (a), which species would appear in the numerator of the rate expression? Which species would appear in the denominator of the rate expression? To what power is the denominator raised?

(c) From your conclusions from part (b), suggest a rate law consistent with the experimental data. (d) Evaluate the rate law parameters. (e) From your rate expression, which species can you conclude are adsorbed on the surface? file:///H:/html/10chap/html/excd10-1.htm[05/12/2011 16:57:26]

excd10-1.htm

(f) From your conclusions in part (e), suggest a mechanism and rate-limiting step for this reaction. (g) For an entering partial pressure of A of 2 atm in a PBR, what is the ratio of sites of A to C sites at 80% conversion of A? (h) At what conversion are the number of A and C sites equal?

Solution (a) Compare runs 1, 3, 6, and 7. PB and PC are fixed. As PA increases the rate first increases then levels off. Therefore, PA must be in both the numerator and denominator.

Compare runs 4 and 5. PA and PB are fixed. Because the reaction is irreversible, PC must be in the demoninator.

Compare runs 2 and 7. P and P are file:///H:/html/10chap/html/excd10-1.htm[05/12/2011 16:57:26]

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A

C

fixed. increases by a factor of 10 when PB increases by a factor of 10. PB is only in the numerator.

(b) A and B in numerator, C and A in denominator.

(c)

(d) For fixed PC and PB ,


excd10-1.htm

For fixed PA and PB


excd10-1.htm

We note that the non-linear regression techniques described in Chapter 5 could have easily been used. Find the Mechanism and Rate Limiting Step

(f) A and C are on the surface


excd10-1.htm



excd10-2.htm

Learning Resources

Example CD10-2 Least Squares Analysis to Determine the Rate Law Parameters k, KT , and KB (Example 6-2 in 2nd ed.)

Use the data in Table 10-5 together with Equation (10-81) to evaluate the rate law parameters.

Solution The calculations are shown in Table CDE10-2.1. A versus PT at constant PB will be plot of linear with slope 1/k (see Figure CDE10-2.1).


excd10-2.htm

Figure CDE10-2.1

Substituting these values into Equation (CDE10-2.1), we find that

The constant K B can be evaluated from the slope of versus PB for constant PT. From the plot of Figure CDE10-2.2,

Rearranging Equation (10-81) for PB = 0, we have

(CDE102.1)

From Figure CDE10-2.1,

(CDE102.2)


excd10-2.htm

Figure CDE10-2.2

In addition to the graphical determination, we can use linear regression to determine the rate law parameters. We will use Equations (5-31) through (533) along with Table CDE10-2.2 to determine k, K T, and K B. in Equations (10-81) and (5-22). Recall Equation (10-81),

where Y= , a0 = 1/kKT, a1 = K B /kKT, a2= 1/k, X1 =PB , and X2 = PT. Next, recall the linear regression equations and apply them for the 16 runs of this example.

(10-81)

and Equation (5-30),

(5-30)


(CDE102.3)

(CDE102.4)

excd10-2.htm

(CDE102.5)

If we were to use a software package such as POLYMATH, we would simply enter Y, X1 X2, and for each run and the parameters a 0 , a1, and a2 would be displayed in a few seconds. Alternatively, we can form Table CDE10-2.2 and carry out the numerical operations ourselves to determine the rate law parameters. We can either form the table using a calculator or by using a spreadsheet such as Excel or Lotus 123.

(CDE102.6) (CDE102.7)

Equation (CDE10-2.3) becomes 6.62 X 109 = 16a0 + 12a1 + 61.5a2

Equation (CDE10-2.4) becomes 5.67 X 109 = 12a0 = 44a = 12a1


excd10-2.htm

Equation (CDE10-2.5) becomes

Solving these three equations simultaneously, we obtain a0 = 7.12 x 107, a1 = 9.0 x 107, and a 2= 7.16 x 107. The corresponding rate law parameters are

6.0 X 1010 = 61.5a0 + 12a1 + 761.25a2

(CDE102.8)

Using POLYMATH, we obtain

with, of course, the parameter values being the same as those obtained from Table CDE10-2.2 and Equation (CDE10-2.5). After substituting the numerical values of k, K B, and K T into Equation (10-80), the rate law at 600°C for hydrodemethylation of toluene,

whereP i is in atm.

is given by the equation

(CDE102.9)

Once we have the adsorption constant K T and KB , we could calculate the ratio of sites. For example, the ratio of toluene sites to benzene sites at 40% conversion is

We see that at 40% conversion there are approximately 20% more sites occupied by toluene


excd10-2.htm

than by benzene.



excd10-3.htm

Learning Resources

Example CD10-3 Continued Hydrodemethylation of Tolune: No Pressure Drop

If we were to neglect pressure drop, we can solve for the catalyst weight necessary to achieve 65% conversion with the aid of Simpson's five-point formula.

Substituting for k, K T, K B ,PB , PH2, and PT in Equation (E10-3.2) yields

We can now proceed to solve for the catalyst weight necessary to obtain this conversion by using Equation (E10-3.6).

Hand calculations: The calculations for X versus (1/ ) are displayed in Table E10-3.2.

Using numerical integration (Appendix A.5), after dividing the area under the curve into two parts: X = 0 to X = 0.52 with h1 = 0.13 and X = 0.52 to X = 0.65 with h2 = 0.13, we have

(E10-3.6)


excd10-3.htm

Numerical evaluation

Since the reciprocal of the rate of reaction increases sharply as the conversion approaches unity, the numerical integration probably should have included a greater number of integrals than the six used in the calculation.



excd10-4.htm

Learning Resources

Example CD10-4 Decay in a Straight-Through Reactor

The gas-phase cracking of a Salina light gas-oil reaction

A typical cost of the catalyst is $1 million

gas-oil (g)

products (g) + coke (s) A B

is to be carried out in a straight-through transport reactor containing a catalyst that decays as a result of coking. The reaction is carried out at 750°F.

3

The entering concentration of A is 0.2 kmol/m . The catalyst particles are assumed to move with the mean gas velocity (Ug =Us = 7.5 m/s). The rate law is

with K

3 A=3m

3

/kmol and KB = 0.01 m /kmol. The

maximum value of the term K B C B is small (0.002), so it can be neglected with respect to the other terms (e.g., 1). Using this fact and the bulk density, b, the rate law becomes

with k = 8 s -1. At 750°F, the catalyst activity for light gas-oil over a synthetic catalyst for short contact times (i.e., less than 100 s) is

with A ´= 7.6 s

-1/2

. As a first approximation,


excd10-4.htm

neglect volume change with reaction, pressure drop, and temperature variations.

Determine the conversion as a function of distance down the reactor. The reactor length is 6 m.

Solution

For a catalyst particle traveling with a velocity of U, the time that the catalyst particle has been in the reactor when it reaches a height is just and where Stoichiometry. Gas-phase reaction with

(CDE104.4) (CDE104.5) = 0, T = T 0, and P = P0.

Combining yields


excd10-4.htm

(CDE104.6)

POLYMATH Solution. In using POLYMATH to obtain a solution, it is usually easiest to write the mole balance, rate law, and stoichiometry separately rather than combining them into a single equation. Therefore, starting with the mole balance

(CDE104.7)

we substitute for -r A and a using Equations (CDE10-4.2) and (CDE10-4.5). The POLYMATH program and solution are shown in Table CDE10-4.1 and Figure CDE10-4.1 respectively.

Table CDE10-4.1 POLYMATH Program Coking in a Straight-Through Transport Reactor

Figure CDE10-4.1 file:///H:/html/10chap/html/excd10-4.htm[05/12/2011 16:57:30]

excd10-4.htm

Conversion and Activity Profiles



10prof.htm


CD10.1 Hydrogen Adsorption

Consider the adsorption of a nonreacting gas onto the surface of a catalyst. Adsorption data are frequently reported in the form of adsorption isotherms. Isotherms portray the amount of a gas adsorbed on a solid at different pressures but at one temperature. A typical adsorption isotherm, shown in Figure CD10-1, is taken from the classic study of Ward, 1 who adsorbed hydrogen on powdered copper at 25°C (see Table CD10-1). The data appear to be quite precise. Only one point is slightly off a smooth curve, and there is no hysteresis, because the points taken while the pressure was being gradually increased lie on the same curve as those taken while the pressure was decreased.

Figure CD10-1 Adsorption of hydrogen on copper


10prof.htm

This isotherm can be used to gain insight into the adsorption process. An equation for the curve in Figure CD10-1 will be derived, and the derivation will reveal significant properties of the hydrogencopper system. First, a model system is proposed and then the isotherm obtained from the model is compared with the experimental data shown on the curve. If the curve predicted by the model agrees Postulate models, with the experimental curve, the model may then see which one(s) reasonably describe what is occurring physically in fit(s) the data the real system. If the predicted curve does not agree with that obtained experimentally, the model fails to match the physical situation in at least one important characteristic, and perhaps more. To describe Ward's data, two models will be postulated--one in which hydrogen is adsorbed as molecules, H2, on copper powder, and the other in which hydrogen is adsorbed as atoms, H, instead of molecules.

The former is called molecular or nondissociated (e.g., H2 ) adsorption and the latter is called Two models: dissociative adsorption. Whether a molecule 1. Adsorption as H 2 adsorbs nondissociatively or dissociatively depends 2. Adsorption as H on the metal (M) surface. For example, CO undergoes dissociative adsorption on iron and molecular adsorption on nickel.


10prof.htm

A. Molecular Adsorption

The adsorption of hydrogen molecules will be

considered first. Since the hydrogen does not react further after being adsorbed, we need only to consider the adsorption process:

(CD101)

In obtaining a rate law for the rate of adsorption, the reaction in Equation (CD10-1) can be treated as an elementary reaction. The rate of attachment of the hydrogen molecules to the surface is proportional to the number of collisions that these molecules make with the surface per second. In other words, a specific fraction of the molecules that strike the surface become adsorbed. The collision rate is, in turn, directly proportional to . Since the hydrogen partial pressure, hydrogen molecules can adsorb only on vacant sites and not on sites already occupied by other hydrogen molecules, the rate of attachment is also directly proportional to the concentration of vacant sites, C . Combining these two facts means that the rate of attachment of hydrogen molecules to the

surface is directly proportional to the product of file:///H:/html/10chap/html/10prof.htm[05/12/2011 16:57:32]

10prof.htm

the partial pressure of H and the concentration of vacant sites:

rate of attachment

The rate of detachment of molecules from the surface can be a first-order process; that is, the detachment of hydrogen molecules from the surface is usually directly proportional to the concentration of sites occupied by the molecules,

The net rate of adsorption is equal to the rate of molecular attachment to the surface minus the rate of detachment from the surface. If kA and k-A are the constants of proportionality for the attachment and detachment processes, then

rate of detachment

(CD102)

( The ratio K A = kA / k- A is the adsorption equilibrium constant. Using it to rearrange Equation (CD10-2) gives

(CD103)

The parameters kA, k-A, and K A are all functions of temperature, exhibiting an exponential temperature dependence. The forward and reverse specific reaction rates, kA and k-A increase with increasing temperature, while the adsorption equilibrium constant, K A, decreases with increasing temperature. At a single temperature, in this case 25°C, they are, of course, constant in the absence of any catalyst deactivation.

Since hydrogen is the only material adsorbed on the catalyst, the site balance gives

(CD104)

The points plotted in Figure CD10-1 were taken at file:///H:/html/10chap/html/10prof.htm[05/12/2011 16:57:32]

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equilibrium conditions. The experimental details present in the original work support this fact, and the absence of hysteresis confirms it. At equilibrium, the net rate of adsorption equals zero. Setting the right-hand side of Equation (CD10-3) equal to zero and solving for the concentration of H2 adsorbed on the surface, we get

(CD105)

Using Equation (CD10-4) to give C in terms of and the total number of sites C t we can solve in terms of constants and the pressure of for hydrogen:

Rearranging gives us

(CD106)

This equation thus gives as a function of the partial pressure of hydrogen, and so is an equation for the adsorption isotherm. This particular type of isotherm equation is called a Langmuir isotherm. 2 Langmuir isotherm A quick look shows that this equation shares many for adsorption of properties with the curve in Figure CD10-1. As molecular hydrogen

was shown in Chapter 5, one method of checking whether a model predicts the behavior of some experimental data is to linearize the model's equation and then plot the indicated variables against one another. For example, Equation (CD10-6) may be arranged in the form

(CD107)

and the linearity of a plot of function of

as a

will determine if the data conform to a Langmuir single-site isotherm. The data in


10prof.htm

Figure CD10-1 are replotted in Figure CD10-2 in

the form suggested by Equation (CD10-7). The

data indicate a slight but definite curvature. Thus there is a significant question as to whether these data really conform to a model of hydrogen adsorbing as molecules

B. Dissociative Adsorption

Next, the isotherm for hydrogen adsorbing as atoms is derived:

When the hydrogen molecule dissociates upon adsorption, it is referred to as the dissociative adsorption of hydrogen. As in the case of molecular adsorption, the rate of adsorption here is proportional to the pressure of hydrogen in the system, because this rate governs the number of gaseous collisions with the surface. For a molecule to dissociate as it adsorbs, however, two adjacent vacant active sites are required rather than the single site needed when a substance adsorbs in its molecular form. The probability of two vacant sites occurring adjacent to one another is proportional to the square of the concentration of vacant sites. These two observations mean that the rate of adsorption is proportional to the product of the hydrogen partial pressure and the square of the . vacant site concentration,

For desorption to occur, two occupied sites must be adjacent, meaning that the rate of desorption is proportional to the square of the occupied-site 2 ) . The net rate of adsorption concentration, ( can then be expressed as


(CD108)

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On this particular catalyst the data show that hydrogen does not adsorb as molecules (H2 )

Figure CD10-2 Adsorption of molecular hydrogen

Factoring out kA, the equation for dissociative adsorption is

Rate of dissociative adsorption

where

At equilibrium, r AD = 0, and

or

From Equation (CD10-1),

(CD109)

This value may be substituted into Equation (CD10-9) to give an expression that can be solved . The resulting isotherm equation is for


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Langmuir isotherm for adsorption as atomic hydrogen

(CD1010)

then multiplying through by

Taking the inverse of both sides of the equation, , yields

(CD1011)

This equation is the linearized Langmuir isotherm for dissociative adsorption. It says that if the hydrogen is dissociatively adsorbed on the copper, is a straight line should result when plotted as a function of .

The data in Figure CD10-1 are replotted in Figure CD10-3 in the form suggested by Equation (CD10-11). An excellent straight line is obtained, giving support to the postulate of hydrogen being dissociatively adsorbed on the copper powder. A comparison of the results of the models indicates that hydrogen is adsorbed on the copper as atoms rather than as molecules. Hydrogen-deuterium tracer studies have confirmed this interpretation. Some comments seem deserved here. The data presented seemed very precise. If they had not been so, and there had been significant scatter, it would have been impossible to distinguish between the two models. The curvature in Figure CD10-2 is slight, so the difference between the two plots is somewhat subtle. A discerning eye is necessary to distinguish between the two mechanisms in this situation. This subtle difference is one reason why this type of agreement between a model and the behavior of experimental data usually requires supporting spectroscopic and tracer experiments.

The dissociative adsorption of H 2 model fits the experimental data for this catalyst


10prof.htm

Figure CD10-3 Dissociative adsorption of hydrogen.

Next



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CD10.2 Catalyst Poisoning in a ConstantVolume Batch Reactor

For the case of a constant-volume batch reactor, a mole balance of the poison in the gas phase combined with the rate law for poisoning [Equation (10-109)] gives

(CD10-12)

where kA =kCT0 is the initial total concentration of unpoisoned site, C P•S is the concentration of poisoned sites, and ƒ is the fraction of poisoned sites. Taking the ratio of Equation (10-111) to Equation (CD10-12)

Constantvolume batch

and then integrating with limits a = 1 and C P =C P0, (CD10-13)

or

(CD10-14)

The decay can now be written in the form

(CD10-15)

which is of the form:

where file:///H:/html/10chap/html/10prof2.htm[05/12/2011 16:57:33]

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Back Next



10prof3.htm


CD10.3 Differential Method of Analysis to Determine the Decay Law The differential method of analysis can also be used to determine the order of decay for experiments carried out in either a CSTR or a plug-flow reactor. To illustrate this method, we consider a stirred contained solids reactor (SCSR), in which we measure the concentrations as a function of time. We assume the decay rate law to be of the form

The design equation for the SCSR is the same as that for a CSTR. Consequently, for the n th-order irreversible reaction

Taking the log and differentiating with respect to time, we get

Substituting Equation (10-141) for a and taking the ln of both sides gives us

(CD10-16)

the activity at any time t is given by

(CD10-17)

(CD10-18)

Combining Equations (10-136) and (10-137) yields (CD10-19)


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We see that the reaction order can be found from a plot on log-log paper similar to that shown in Figure CD10-4. The data to construct this plot could be arranged as shown in (CD10Table CD10-2. By plotting the last column as a function of 20) the reciprocal of the second column on log-log paper the decay order q can be obtained.

Figure CD10-4 Differential method for determining order of decay

Back Next

© 1999 Prentice-Hall PTR


10prof3.htm

Prentice Hall, Inc. ISBN 0-13-531708-8 Legal Statement




CD10.4 Etching

We have seen in Figure 10-34 that etching (i.e., the dissolution or physical or chemical removal of material) is also an important step in the fabrication process. Etching takes on a priority role in microelectronics manufacturing because of the need to create well-defined structures from an essentially homogeneous material. In integrated circuits, etching is necessary to remove unwanted material that could provide alternative pathways for electrons and thus hinder operation of the circuit. Etching is also of vital importance in the fabrication of micromechanical and optoelectronic devices. By selectively etching semiconductor surfaces it is possible to fabricate motors and valves, ultrasmall diaphragms that can sense differences in pressure, or cantilever beams that can sense acceleration. In each of these applications proper etching is crucial to remove material that would either short out a circuit or hinder movement of the micromechanical device. CD10.4.A Dry Etching

Dry etching involves gas-phase reactions (usually in plasmas) which form highly reactive species that impinge on the surface to react with the surface, to erode the surface, or both. In the fabrication of optoelectronic devices, dry etching is used almost exclusively. Indium phosphide is a material that is used in the fabrication of optoelectronic devices, but it is an extremely difficult material to etch. One technique that has been used to etch InP successfully is reactive ion etching (RIE). RIE involves the formation of highly reactive species, usually in a plasma, which impinge on the surface to react with and to erode the surface. To illustrate the principles of RIE, we will study the etching of InP in an argon plasma. Typical reactions that occur in the



gas-phase plasma (which generates high-energy electrons) are

A schematic of the plasma is shown in Figure CD10 5.

Figure CD10-5 Schematic of ion bombardment taking place on the wafer (Courtesy of Semiconductor Int., 15 (6) 72 (1992)

Typical operating conditions are shown in Table CD10-3.



Figure CD10-6 shows a schematic of the various processes occurring in the plasma directly above the surface to be etched as well as those occurring in the substrate.3 As shown in Figure CD10-6, the incident ion may either be reflected and neutralized, causing the surface to eject an electron or atom (sputtering) or be implanted in the surface. Any or all of these processes can result in structural rearrangement of the surface, making the surface move reactive to other gas molecules (e.g., Cl) present in the plasma.

Figure CD10-6 Plasma interactions with a surface

We now look specifically at the reactive ion etching



of InP. The removal of InP from the surface can occur by two mechanisms. In the first mechanism, InP is removed as a result of bombardment from a highenergy argon ion which knocks InP off the surface. This type of etch, called sputtering or sputter etching, is simply erosion of the surface; no chemical reactions are involved (Figure CD10-7). In the second mechanism a reaction takes place between InP and chlorine to form InCl.

Figure CD10-7 Etching by sputtering

3 The

more volatile InCl 3 is removed from the surface either by argon or by vaporization ion bombardment (Figure CD10-8). Because a reaction takes place before removal of the material it is called reactive ion etching. We shall now consider the RIE of InP in more detail. The chlorine atoms that are generated react with the surface according to the following sequence:4 This last step is a sputtering reaction.

Figure CD10-8 RIE etching



We now proceed to prepare a law for the rate of etching in terms of the concentrations of argon and chlorine. We assume that each step in the etching process is elementary. The net rate of removal of indium from the surface is the sum of that removed by both Reactions (4) and (5): (CD1021)

where (Ar) is the concentration of argon above the is the fraction of the surface surface and covered by InCl 3.

Net Rate of Formation of Reactive Intermediates. The chemical species InCl and InCl 2 are unstable reactive intermediates. Because the etching takes place at steady state, there is no net rate of formation of these species. That is, they react virtually as fast as they are formed. For Reactions (1) and (2), which are elementary, the net rate is (CD1022)

Solving for the fraction of the surface covered by InCl gives (CD1023)

Similarly, the net rate of formation of InCl 2 is zero, and Reactions (2) and (3) give


(CD1024)


Solving for the fraction of surface covered by InCl 2 yields

(CD1025) (CD1026)

Finally, the net rate of formation of InCl 3 is also zero at steady state. Reactions (3), (4), and (5) give

Solving for the fractional coverage of InCl 3, we obtain (CD1027)

Next we carry out a fractional surface-area balance: (CD1028)

Substituting for InCl, InCl 2, and InCl 3 yields (CD1029)

Combining Equations (CD10-21) and (CD10-27), we obtain (CD1030)

Substituting for° InP in Equation (CD10-30), and

multiplying numerator and denominator by and dividing by k1, the etch rate law for RIE of indium phosphide becomes


(CD1031)


where

Figure CD10-9

A nonlinear least squares analysis (discussed in Chapter 5) gives kt = 400, k4/A= 0.001, and kt/A = 0.095. Figure CD10-9 shows the experimental etch rate as a function of the predicted etch rate shown on the right-hand side of Equation (CD10-31). A photocorrelation spectros copy analysis of the surface being etched reveals that the surface is primarily InP and InCl 3, with virtually no InCl and InCl 2:

Consequently, for all practical purposes the etch rate can be written as (CD10-



32)

CD10.4.B Wet Etching

Wet etching is used primarily in micromachining, where acids are used to dissolve the solid substrate to form intricate channels and slopes, but it has also been used in the fabrication of computer chips. Figure CD10-10 shows a schematic of the etching process of material A. When a chemical etch is to be carried out, the portion of the material that is not to be etched is covered by a polymer coating called a photoresist, which prevents it from contacting the acid. Next, the area of material A that is unprotected by the photoresist is etched away until the acid reaches material B and completes the channel formation. Finally, the photoresist is removed by immersing the chip in the appropriate solvent. In many instances we want to stop the etching process as soon as we reach the A/B interface. To achieve this etch stop as precisely as possible, we need to know the rate of etching. Consequently, in this section we

Figure CD10-10 Wet etching

focus on the rate of etching (i.e., rate of dissolution of solid materials) to develop mechanisms and rate laws.

As an example of etching, consider the dissolution of a silicate in hydrofluoric acid. First consider the dissolution in which no liquid-phase catalysts are used. The acid, A (e.g., HF), adsorbs reversibly on a site S on the substrate surface (Figure CD10-11).



Symbolically, we have

Figure CD10-11 Adsorption of HF

The adsorbed acid breaks the Si-O bonds on the substrate surface, thereby dissolving it and generating products that go into the liquid phase. These products (Si, O) undergo a series of rapid homogeneous reactions which produce SiF 4 and H2O. When these products go into solution, they uncover the next molecular layer of substrate to be dissolved, as shown in Figure CD10-12.

Figure CD10-12 Surface reaction

We can write the surface reaction symbolically as

The symbol S in this reaction represents the new substrate surface that has been exposed to acid once the layer above it has been removed. As with catalytic reactions, the surface reaction is most often rate-limiting, in which case we have


(CD1033)


where f A S is the fraction of the surface that has A adsorbed. For the adsorption step, the rate of adsorption (CD1034)

where aA is the activity of the acid in solution. For ideal solutions we can replace a by the concentration C A . Because surface reaction is rate-limiting, CD1035)

The fractional surface area balance is (CD1036)

Substituting for f A

S

and solving for f v gives us (CD1037)

Combing Equations (CD10-33), (CD10-35), and (CD10-37), we obtain the etch rate as (CD1038)

For activity coefficients of approximately unity, the rate expression becomes (CD1039)

Figure CD10-13 shows the dissolution of the aluminosilicate montmorillonite as a function of HF concentration. 5 The corresponding rate law is



(CD1040)

Figure CD10-13 Rate of etching as a function of HF concentration

CD10.4.C Dissolution Catalysis

Dissolution catalysis is the process in which a species that has no dissolving capacity is added to a system to accelerate the rate of dissolution. To illustrate this process, we shall consider the aluminosilicate system just discussed, except that this time we will add a dissolving catalyst B (e.g., H+ ). In this system, B can only adsorb on specific sites on the surface, S ´ . These sites are different from those sites upon which A can adsorb. The adsorption process shown schematically in Figure CD10-14 can be written symbolically as (CD1041)

Next we have the adsorption of A (HF) on S-type sites (Figure CD10-15): (CD10-



42)

Figure CD10-14 Adsorption of H+ (i.e. Species B)

Figure CD10-15 Adsorption of HF (i.e. Species A)

Finally, the surface reaction takes place in which silicon and oxygen are removed from the surface exposing the next layer of silicon and oxygen (Figure CD10-16). The surface reaction is (CD1043)

and the corresponding rate law for the catalytic dissolution is


(CD1044)


Figure CD10-16 Surface reaction

Because the adsorption of B is not rate-limiting,

Solving for the fraction of S ´ sites occupied by the catalyst B gives us (CD1045)

from a site balance of only those sites on which B can adsorb: (CD1046)

Combining Equations (CD10-39), (CD10-44), (CD1045), and (CD10-46), we obtain (CD1047)

(CD1048)

The total rate of dissolution is the sum of the



uncatalyzed rate [Equation (CD10-39)] and the catalyzed rate [Equation (CD10-48)]:

(CD1049)

(CD1050)

The degree of catalysis, DAC ,is defined as the catalyzed rate minus the uncatalyzed rate, divided by the uncatalyzed rate: (CD1051)

Substituting for the uncatalyzed rate equation (CD1039) and the catalyzed rate equation (CD10-48) gives (CD1052)

Figure CD10-17 shows a plot of the degree of catalysis as a function of hydrogen ion activity for the dissolution of the aluminosilicate kaolinite. Taking the reciprocal of Equation (CD10-52), we obtain


(CD1053)


Figure CD10-17 Acceleration of kaolinite dissolution rates by HCI

Figure CD10-18 shows the etch rate as a function of catalyst activity.6 We see that a plot of 1/DAC versus 1/a B should be a straight line.

Figure CD10-18 Linearization of model for the catalyzed dissolution of kaolinite

Back


cdp10-aa.htm

Additional Homework Problems CDP10AA

The catalytic oxidation of ethanol has been carried out over a tantalum oxide catalyst [J. Catal., 25, 194 (1972)]. The overall reaction

is believed to proceed by the following mechanism, in which ethanol is adsorbed on one type of site, S, and oxygen is adsorbed on a different type of site, S´ . Ethanol undergoes dissociative adsorption on two type S sites:

Oxygen also undergoes dissociative adsorption on two type S´ sites (presumably metal ions):

(a) Assuming that the rate-limiting step is

and that the adsorbed hydroxyl may be converted into water by

show that the initial rate law is

(b) What generalizations can you make after studying this problem? [2nd Ed. P6-17]


cdp10-aa.htm



cdp10-bb.htm

Additional Homework Problems CDP10- The vapor-phase esterification of acetic acid and ethanol BB to form ethyl acetate and water was carried out over a

resin catalyst at 118°C [Ind. Chem. Eng., 26, 198 (1987)].

What can you learn from these data?



cdp10-cb.htm

Additional Homework Problems CDP10CB

Silicon dioxide is formed by chemical vapor deposition from dichlorosilane (DCS) and nitrous oxide [Proceedings of the Third World Congress of Chemical Engineering, Tokyo, p. 290 (1986)].

The deposition rate is independent of HCl and nitrogen. At 900°C

At 900°C the rate of deposition is

(a) Can the reaction order with respect N2O and DCS be explained by some other means than powers that the author used? If so, formulate a new rate law and evaluate the parameters. (b) Is the following mechanism consistent with the rate law?

In this mechanism we assume that the last step is extremely rapid in that as soon as SiO S is formed on the surface it reacts instantaneously with N2O S to form SiO 2 (i.e., ). file:///H:/html/10chap/html/cdp10-cb.htm[05/12/2011 16:57:40]

cdp10-cb.htm

(c) The wafers are 0.125 m in diameter and set upright along the length of the reactor. Silicon dioxide is deposited uniformly on both sides of the wafers. The reactor diameter is 250 mm and the wafers are spaced 20 mm apart. This arrangement corresponds to a deposition surface area of 250 m 2 per cubic meter of reactor volume. Assume that the gas phase behaves as a plug-flow reactor at steady state. Develop the equations for the axial deposition profile. Specifically, determine the thickness of the deposited film at 900°C on the 1st, 50th, and 110th wafers in the reactor after 30 min. Dichlorosilane is fed at a partial pressure of 157 mT and a rate of 0.00368 mol/ min, while nitrous oxide is fed at a partial pressure of 447 mT and a rate of 0.013 mol/ min. [2nd Ed. P6-13]



cdp10-db.htm

Additional Homework Problems CDP10 DB

The autocatalytic reaction

is carried out isothermally in a moving-bed reactor. The reaction is elementary, follows separable catalyst decay kinetics, and the catalyst decay law is

(a) Neglect pressure drop and determine the activity and concentration profiles as a function of catalyst weight along the reactor. (b) Repeat part (a) by accounting for pressure drop for the case where the gas density varies down the reactor [i.e. . Both reactants and fresh catalyst enter at the same end.



cdp10-eb.htm

Additional Homework Problems CDP10EB

The dissolution of the aluminosilicate feldspar is shown in Figure CDP10-E. Determine the reaction mechanism and the rate law parameters.

Figure CDP10-E



cdp10-fb.htm


Titanium nitride films are used in decorative coatings as well as in wear-resistant tools. There is increasing interest in TiN because of its thermal stability, good diffusion barrier properties, and its low electrical resistivity [J. Electrochem. Soc., 138, 500 (1991)]. Titanium nitride films were formed by CVD from a mixture of TiCl 4, NH3, H2 and Ar. The following observations can be made from the article:

CDP10-FB

The rate of deposition is independent of Ar and H2. At low partial pressures of both TiCl 4 and NH3 the deposition rate appears to be first-order in TiCl 4 and second-order in NH3 . At high partial pressures of NH3 the rate varies inversely with TiCl 4 The following mechanism has been suggested for the reaction:

It is believed that the gas-phase reaction to form the complex TiCl 4 (NH3)2 is in equilibrium.

(a) Determine the rate expression for the suggested mechanism. Does it agree with experimental observations? (b) Determine the reaction rate parameters. (c) Do the data selected from the article agree with the data given below?


cdp10-fb.htm



cdp10-gc.htm

Additional Homework Problems When the decomposition of cumene was carried out over a LaY zeolite catalyst, deactivation of the catalyst was found to occur by coking. The reaction is first-order and was carried out in a differential reactor containing 1 g of catalyst at 430°C [Ind. Eng. Chem. Process Des. Dev., 22, 609 (1983)]. The exit molar flow rate of propylene was monitored as a function of time.

CDP10-G C

(a) From the data below, determine the decay functionality, with the amount of coke on the catalyst together with the apparent cumene kinetics. (b) Determine the functionality of the decay function with respect to time and partial pressure of cumene. (Ans.:

(c) Calculate the reactor height necessary to achieve 60% conversion in a STTR. The catalyst particle velocity is 1 m/s. Cumene is to enter the reactor at a rate of 7 mol/ s. The total molar feed rate to the reactor is 10 mol/ s. The bulk density of the catalyst is 0.5 kg/m 3 and the reactor cross-sectional area is in dm 3.

Run 1: Entering partial pressure of cumene: 1 atm Entering flow rate of cumene 0.32 mol/s

Run 2: Entering partial pressure of cumene: 0.4 atm Entering flow rate of cumene 0.01 mol/s

[2nd Ed. P6-27]

file:///H:/html/10chap/html/cdp10-gc.htm[05/12/2011 16:57:42]

cdp10-gc.htm


file:///H:/html/10chap/html/cdp10-gc.htm[05/12/2011 16:57:42]

cdp10-hb.htm


The dehydrogenation of ethylbenzene is one of the most important methods for the manufacture of styrene. The reaction has been studied using a Shell 105 catalyst (93 wt % Fe2O3, 5% Cr2O3, 2% KOH) in a differential reactor [Ind. Eng. Chem. Process Des. Dev. 4, 281 (1965)]. It was observed that the rate of reaction was decreased when styrene was added to the feedstream. Initial rate data showed that as the partial pressure of ethylbenzene was increased to moderate values, the rate of reaction became independent of the partial pressure of ethylbenzene. However, the rate of reaction was also decreased when styrene and hydrogen were formed. The equilibrium constant K p was taken to be 0.415 atm at 630°C. Suggest a mechanism that is consistent with the experimental observations and derive the corresponding rate law. Using the information from the following table, taken at T = 630°C

CDP10-H B

evaluate all constants in your model and the determine the total cost of catalyst necessary to produce 2000 kg of styrene per day in (a) A CSTR fluidized-bed reactor.

(b) A fixed-bed reactor. (c) Also calculate the cost of the PFR and CSTR needed to hold the catalyst. Steam is used as a diluent in the feedstream, with an H2O/EtB molar ratio equal to 1/5. In each case, the conversion of ethylbenzene is to be 45%.


cdp10-hb.htm

[2nd Ed. P6-20]



cdp10-ib.htm

Additional Homework Problems CDP10-I B


is carried out in a moving-bed reactor in which there is a significant pressure drop and significant catalyst decay. The decay is first-order. Pure A enters the reactor at a temperature of 400 K, a concentration of 2 mol/dm 3, and a molar feed rate of 4 mol/s at a pressure of 50 atm. The pressure at the exit of the reactor is 5 atm. The catalyst loading is such that the catalyst activity at the exit of the reactor is 0.1. There is 100 kg of catalyst in the reactor. (a) What is the conversion at the exit of the reactor?

(b) Calculate the conversion at two or three catalyst weights, and then use these points to sketch the conversion of A down the length of the reactor. The specific reaction rate at 400 K is 0.09 dm 3/ mol s. (c) How would your answers change if the catalyst and reactants were fed to opposite ends of the reactor?


cdp10-ib.htm



cdp10-jb.htm

Additional Homework Problems CDP10-JB

The elementary isothermal gas-phase reaction

is carried out in a moving-bed reactor. The catalyst decays by sintering. The moving bed contains 100 kg of catalyst and the catalyst flow rate through the bed is adjusted so that the exiting catalytic activity is one-fourth of the entering activity. The specific reaction rate for fresh catalyst is 10-4 dm 3/g cat. s. Pure A enters at a concentration of 0.005 mol/dm 3 and a volumetric flow rate of 3.75 dm 3/ s. The residence time of a catalyst particle in the reactor is 5 min. What conversion can be expected? [Final Exam, Winter 1994]



cdp10-kb.htm

Additional Homework Problems CDP10-K B

The cracking of normal paraffins (P n) can be modeled as

with surface reaction as the rate-limiting step. The rate of reaction has been found to increase with increasing temperature up to a ) [J. Wei, carbon number of 15 (i.e., Chem. Eng. Sci., 51, 2995 (1996)]. The rate was found to decrease with increasing temperature for a carbon number greater than 16. The reactions are carried out at sufficiently high temperatures that

Plot the heat of adsorption (i.e., ) and the apparent activation energy E * as a function of carbon number. Suggest an explanation for the unusual temperature dependence discussed above.

The value En appears to be independent of carbon number.



cdp10-lb.htm

Additional Homework Problems CDP10-LB

The use of a differential reactor to study the formation of methane from hydrogen and carbon monoxide over a nickel catalyst gave the following:

Suggest a mechanism and rate-limiting step that is consistent with the experimental observation.

It is desired to produce 20 tons/day of CH 4. Calculate the catalyst weights necessary to achieve 80% conversion in:

(a) A fixed bed. (b) A fluidized bed. The feed consists of 75% H2 and 25% CO at a temperature of 500°F and a pressure of 10 atm. Assume both molecular and atomic hydrogen are adsorbed on the surface.



cdp10-mb.htm

Additional Homework Problems CDP10-M B

The elementary irreversible gas phase catalytic reaction

is to be carried out in a moving-bed reactor at constant temperature. The reactor contains 5 kg of catalyst. The feed is stoichiometric in A and B. The entering concentration of A is 0.2 mol/dm 3. The catalyst decay law is zero-order with kD = 0.2 s +1.

(a) At what catalyst loading rate (kg/s) will the catalyst activity be exactly zero at the exit of the reactor? (b) Sketch the catalyst activity as a function of catalyst weight (i.e., distance) down the length of the reactor (from 0 to 5 kg) for a catalyst feed rate of 0.5 kg/s. What does an activity of zero mean? Can catalyst activity be less than zero? (c) What conversion will be achieved for a catalyst feed rate of 0.5 kg/s? (d) What catalyst loading rate is necessary to achieve 40% conversion? (e) What is the maximum conversion that could be achieved (i.e., at infinite catalyst loading rate)?


file:///H:/html/10chap/html/cdp10-mb.htm[05/12/2011 16:57:46]

cdp10-na.htm

Additional Homework Problems CDP10-NA

Experimental data for the gas-phase catalytic reaction

is shown below. The limiting step in the reaction is known to be irreversible, so that the overall reaction is irreversible. The reaction was carried out in a differential reactor to which A, B, and C were all fed.

(a) Sketch

as a function of PA, as a function of PB , and as a function of PC . (b) From your observations in part (a), which species would appear in the numerator of the rate expression? Which species would appear in the denominator of the rate expression? To what power is the denominator raised? (c) From your conclusions from part (b), suggest a rate law consistent with the experimental data. (d) Evaluate the rate law parameters. (e) From your rate expression, which species can you conclude are adsorbed on the surface? (f) From your conclusions in part (e), suggest a mechanism and rate-limiting step for this reaction. (g) For an entering partial pressure of A of 2 atm in a PBR, what is the ratio of sites of A to C sites at 80% conversion of A? At what conversion are the number of A and C sites equal?

file:///H:/html/10chap/html/cdp10-na.htm[05/12/2011 16:57:46]

cdp10-na.htm © 1999 Prentice-Hall PTR Prentice Hall, Inc. ISBN 0-13-531708-8 Legal Statement


cdp10-ob.htm

Additional Homework Problems The irreversible, gas-phase, solid-catalyzed reaction

CDP10OB

is carried out in a differential reactor. (a) Determine a mechanism and ratelimiting step consistent with the following data:

(b) Evaluate all rate law parameters. (c) Where would you take additional data points? (d) Discuss what you learned from this problem. [2nd Ed. P6-18]


file:///H:/html/10chap/html/cdp10-ob.htm[05/12/2011 16:57:47]

cdp10-pb.htm

Additional Homework Problems CDP10PB

The vapor-phase dehydration of ethanol to give diethyl ether and water was carried out over a solid catalyst at 120°C, with the following results:

Suggest a rate law and mechanism consistent with the experimental data. [2nd Ed. P6-21]



cdp10-qa.htm

Additional Homework Problems CDP10QA

The conversion of buckeyelene (B) to wulfrene (W) and carbon dioxide,

follows an elementary rate law over fresh catalyst. This gas-phase reaction is carried out in a 1-dm 3 batch reactor filled with 1 kg of fresh catalyst. The catalyst follows a zero-order decay law with a decay constant kd equal to 0.05 min -1. If the reaction rate constant is 0.01 dm 6/(kg cat. mol min) and 10 mol of B is introduced into the batch reactor, what is the conversion after 40 min?


file:///H:/html/10chap/html/cdp10-qa.htm[05/12/2011 16:57:47]

cdp10-qb.htm

Additional Homework Problems CDP10 RB

Reforming catalysts are used to transform normal paraffins into branched paraffins. These reactions can be written symbolically as

This isomerization is carried out in a moving-bed reactor containing 100 kg of a catalyst that follows first-order decay. The pressure-drop parameter at 400 K is = 9.8 x 10-3 kg cat-1. The entering temperature and feed rate are 400 K and 1 mol/s of pure A. Describe how you would operate (PA0 , Us , T) the bed if

(a) B were the desired product. (b) C were the desired product. (c) What temperature-time trajectory would you recommend for a constant? a CSTR to keep (d) If the reactions were carried out adiabatically with what changes would you recommend in part (a)?


file:///H:/html/10chap/html/cdp10-rb.htm[05/12/2011 16:57:48]

Lectures 27 and 28

Lectures 27 and 28

External Diffusion Across a Stagnant Film (Chapter 11) *

1.

Mole Balance on Species A

Integrating:

2.

Rate Law / Constitutive Equation

3.

Boundary Conditions

file:///H:/html/course/lectures/twensev/index.htm[05/12/2011 16:57:51]

Lectures 27 and 28

kC is the mass transfer transfer coefficient. It can be found from a correlation for the Sherwood number:

which in turn is a function of the Reynolds Number

and the Schmidt Number

For packed beds:

The flux to the surface is equal to the rate of reaction on the surface:

when kr >>> kC : , rapid reaction on the surface


Lectures 27 and 28

kC increases with increasing velocity, while kr is independent of velocity:

Packed Bed of Catalyst Particles

Mole Balance

Rate Law / Constitutive Equation


Lectures 27 and 28

for single pellets

for packed beds

If kr >>> kC Then

We want to know how the mass transfer coefficient varies with the physical properties (e.g., DAB) and the system operating variables.

Isothermal


Lectures 27 and 28

This equation tells us how the product of our mass transfer coefficient and surface area would change, if we were to change our operating conditions. In other words, it will help us answer "What if..." questions about our system. For example: What effect does pressure drop have here?

For

:

Substituting:

If

Then

Diffusion and Reaction in Catalyst Pellets (Chapter 12) We shall carry a mole balance on species A as it diffuses and reacts in a catalyst pellet.

1. Mole Balance


Lectures 27 and 28

2. Rate Law / Constitutive Equation

3. Boundary Conditions at r = R, C A = C As and at r = 0, C A is finite

4. Put the Equation in Dimensionless Form

5. Concentration Profile (see text)


Lectures 27 and 28

6. Effectiveness Factor

Summary on the Swimming of Small Creatures (P12-7)

The rate of swimming is a function of the rate of diffusion into the tail. Therefore, consider only diffusion in the tail. 1. Mole Balance

2. Constitutive Equation and Rate Law file:///H:/html/course/lectures/twensev/index.htm[05/12/2011 16:57:51]

Lectures 27 and 28

3. Boundary Conditions at z = 0, CA = CA0 at z = L, dCA / dz = 0 4. Put the Equation in Dimensionless Form

When When We shall assume the reaction is zero order:

where:

Solving:


= 0, then = 1, then

= 1. = 0.

Lectures 27 and 28

effectiveness factor = 1.0 (for zero order reactions) Dependence on Catalyst Particle Size:

*





Lectures 27 and 28


excd11-1.htm

Learning Resources Example CD11-1

Calculating Steady-State Fluxes

Determine the molar flux of ammonia, WA, and the mass flux of ammonia, mA,through the stationary frame at steady state for the system shown in Figure CDE11-1.1. Ammonia is diffusing at steady state from point 1 to point 2. There are 12.04 x 1023 molecules of ammonia that pass through an area bounded by a stationary frame with a dimension of 1 m x 2 m in the direction of the lower concentration and over a time interval of 10 s.

Solution

The flux is the flow rate per unit area normal to the

direction of flow. The number of moles of ammoniacrossing the boundary in the 10-s interval is

Figure CDE11-1.1

Two moles pass through an area of 2 m 2 in 10 s.

Therefore, the number of moles passing through 1 m per second is


2

(CDE111.1)

excd11-1.htm

The molar flow rate is

(CDE111.2)

The mass flux of A is the molar flux of A times the molecular weight of A, MA:

(CDE111.3)



excd11-2.htm

Learning Resources

Example CD11-2 Relating the Fluxes WA, B A , and

JA

Derive Equation (11-6) starting from the fact that the molecular diffusional flux is the flux relative to the molar average velocity.

A is moving relative to the average velocity of all the species. The diffusive flux JA of A relative to the molar average velocity is

Solution The difference in velocities (VA -V) tells us the velocity at which species

JA = C A (VA -V)

Velocity of A relative to the average velocity of all the species

CDE11-2.1) (CDE112.2)

JA = C AV A -C AV Using Equation (11-3) yield

(CDE112.3) (CDE112.4)

The second term on the right-hand side can be expanded to

Factoring the reciprocal of the total concentration, 1/C T , from the term in parentheses, we have

JA = WA -C AV

Rearranging gives us

WA = JA +C AV

C AV = C A(yAVA + yB VB )

(CDE112.5)

(CDE11-


excd11-2.htm

2.6)

= yA(WA + WB )

Then

(11-6)

Recalling that B A = CAV

= YA (WA + WB



excd11-3.htm

Learning Resources

Example CD11-3 Diffusion Through a Stagnant Gas

Calculate the steady-state concentration profile and diffusion rate for the diffusion of gas A through a stagnant gas B. The geometry (Figure E11-2.1) and boundary conditions are identical to those given in Example 11-2. The pressure and temperature, and hence the total concentration, are constant throughout the system.

Solution Steps 1 and 2: The procedure for obtaining the differential of the molar flux with respect to distance is identical with that for the diffusion of species A through a liquid. Consequently, we can start with Equation (E11-1.3):

Mole balance

(E111.3)

Step 3: We now need to relate the flux to concentration. Recalling the discussion in Section 11.2.1 and Equation (11-19), for diffusion through a stagnant film we have

Evaluating bulk flow term

(CDE113.1)

Combining Equations (CDE11-1.1) and (E11-1.3) yields

or

Differential equation

Integrating, we obtain


(CDE113.2)

excd11-3.htm

(CDE113.3) (CDE113.4)

Step 4: The constants of integration can be evaluated using the following boundary conditions:

(CDE113.5)

Boundary conditions

(CDE113.6)

Step 5: Applying the first boundary condition, we obtain

(CDE113.7)

(CDE113.8)

Using the second boundary condition yields

Combining Equations (CDE11-3.4), (CDE11-3.7), and (CDE11-3.8) gives us

(CDE113.9)

Rearranging yields

Concentration profile


(CDE113.10)

excd11-3.htm

The concentration profile is shown in Figure CDE11-3.1. The concentration profiles for the case of dilute gas or EMCD are compared with diffusion through a stagnant film in Figure CDE11-3.1.

Figure CDE11-3.1 Concentration profiles.

Step 6: To obtain the molar flux, we differentiate Equation (CDE11-3.9) with respect to and multiply by cD AB . That is, we combine Equations (CDE11-3.9) and (E11-8.1) to obtain

(CDE113.11)



11prof.htm


1. Mass Transfer-Limited Reactions on Metallic Surfaces

In this section we develop the design equations and give the mass transfer correlations for two common types of catalytic reactors: the wire screen or catalyst gauze reactor and the monolith reactor.

A. Catalyst Monolith. The previous discussion in this chapter focused primarily on chemical reactions taking place in packed-bed reactors. However, when a gaseous feedstream contains significant amounts of particulate matter, dust tends to clog the catalyst bed. To process feedstreams of this type, parallel-plate reactors (monoliths) are commonly used. Figure CD11- 1 shows a schematic diagram of a monolith reactor. The reacting gas mixture flows between the parallel plates, and the reaction takes place on the surface of the plates. In deriving the design equation we carry out a balance on a differential section of the reactor (Figure CD11-2).

Figure CD11-1 Catalyst monolith.


11prof.htm

Figure CD11-2 Top view of monolith.

Mole balance for a monolith catalyst

(CD11-1)

where am is the catalytic surface area per unit volume of reactor and A cis the crosssectional area normal to the direction of gas flow. The rate of surface reaction is equal to mass flux to the surface. Taking the surface concentration equal to zero for mass transfer-limited reactions gives

Substituting Equation (11-71) into (11-70) 0 yields and taking the limit as

(CD112)

(CD113)

In terms of volume


(CD114)

11prof.htm

The surface area per unit volume, a, for n plates is

(CD11-5)

Typical spacings between the plates are usually between 0.005 and 0.01 m. The length ranges between 0.05 and 0.5 m and gas velocities between 5 and 20 m/s are not uncommon. The mass transfer coefficient can be calculated from the correlation

Mass transfer correlation for a monolith catalyst

(CD11-6)

The approximate error in the correlation is 20%. Other limitations of the correlation can be found in the article just cited by Arashi et al.1

For no volume change with reaction, Equation (CD114) can be integrated to give

(CD117)

Ford and Chrysler use monolith catalytic afterburners

A variation of the monolith reactor has the gas flowing through square (or other shape) channels as shown in Figure CD11-3. This reactor is also known as a honeycomb reactor. Monolith reactors are used as catalytic afterburners on automobiles and are manufactured by Chrysler and Ford.2


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(b) Figure CD11-3 (a) Honeycomb reactor; (b) catalytic afterburner. (Photo courtesy of Engelhard Corporation)

B. Wire Gauzes Wire gauzes are commonly used in the oxidation of ammonia and hydrocarbons. A gauze is a series of wire screens, stacked one on top of another (Figure CD11-4). The wire is typically made out of platinum or a platinum-rhodium alloy. The wire diameter ranges between 0.004 and 0.01 cm.


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Figure CD11-4 Wire gauzes.

As a first approximation, one can assume plug flow through the gauze, in which case the design equation is similar to that for monolith reactors,

Differential form of the wire gauze design equation

(CD118)

where ag = total screen surface area per total volume of one screen, m 2/m 3 or in 2/in 3 n = number of screens in series V = n (volume per screen)

The values of ag can be calculated from the equations3

where d = wire diameter, in. N = mesh size, number of wires per linear inch

In calculating the volume of the screen, the thickness is taken as twice the wire diameter (i.e., 2d). The porosity can be calculated from the equation


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Mass transfer correlation for wire gauzes

The mass transfer coefficient can be obtained from the correlation for one to three screens, (CD11-9) (CD1110)

For one to five screens, the correlation is

(CD1111)

where is the minimum fractional opening of a single screen:

In the commercial process for the oxidation of ammonia, typical parameter values are

(CD1112)

conversion of ammonia. When more than one or two screens are necessary, some backmixing takes place. Shimizu et al. 4 account for this backmixing by introducing dispersion in the axial direction:

(CD1113)

Equation (CD11-13) is then combined with Equation (CD11-8) and solved. When dispersion is significant it was shown that, Too few screens? depending on the flow conditions, 33 to 300% more screens were required than predicted by the plug-flow model.


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cdp11-ab.htm


The reaction

is carried out on a metallic surface and follows Langmuir-Hinshelwood kinetics at room temperature with the corresponding rate law:

However, at higher temperatures the reaction is mass transfer-limited. The reaction currently takes place on a monolith catalyst where 45% conversion is achieved. It is proposed to double the number of plates for the same width and to halve the length of the reactor. What conversion can be expected for

(a) Monoliths operated at low Reynolds numbers? (b) Monoliths operated at high Reynolds numbers? [2nd Ed. P10-11]



cdp11-bb.htm


Currently, 35% conversion is achieved in a 20mesh gauze screen with a 0.51-mm wire diameter.

CDP11-BB

(a)What mesh and corresponding wire diameter would be required to achieve 80% conversion? (b)For the same mesh size, how many screens would be necessary to achieve 80% conversion? (c) Repeat part (b) for 99.9% conversion. (d) If the mesh size and wire diameter could be varied independently, which would bring about a greater increase in conversion, doubling the mesh size or halving the wire diameter? [2nd Ed. P10-12]



cdp11-cb.htm

Additional Homework Problems CDP11-CB

What reduction in monolith reactor length could be achieved if the absolute temperature were doubled while everything else remained the same? [2nd Ed. P10-13]



cdp11-db.htm


The elementary second-order reaction

takes place inside a catalyst monolith at 400 K and 101.3 kPa.

(a) Calculate the conversion in a monolith that is 0.3 m high, 0.3 m wide, and 1.0 m in length. The spacing between the 0.1-cm-thick plates is 0.5 cm. The molar feed rate is pure A at 30.5 mol/s. (b) How would your answer change if the reaction

were carried out under identical conditions?


[2nd Ed. P10-14]



cdp11-ec.htm

Additional Homework Problems (Fracture acidizing) One technique used to increase the productivity of oil wells is to inject acid under high pressure until it fractures the porous formation. At the point of fracture, acid flows out from the well bore along the fracture slit dissolving the face of the slit as it flows. When the pressure on the well is released, the faces of the slit try to come together but a small channel usually remains where the faces have been dissolved. In addition to reacting with the slit, some of the acid "leaks off" through the side of the slit and is consumed in the porous rock. See Figure CDP11-E (Fracture acidizing).

CDP11EC

Fracture acidizing

(a) Assuming steady-state flow, derive a differential equation describing the flow and reaction of the acid in the channel assuming that the reaction is mass


cdp11-ec.htm

transport-limited to the surface for flow in a slit between two surfaces separated by a distance D,

As a first approximation, assume that Sh is independent of the leak-off rate.

(b) For the case of a linear leak-off rate solve for the concentration profile, C A. (c) Solve for channel thickness as a function of channel length drawing on your understanding of shrinking core analysis. (d) Repeat part (a) accounting for the effect of leak-off on Re and Sh. [2nd Ed. P10-15]



cdp11-fc.htm

Additional Homework Problems CDP11- The oxidation of silicon to form SiO 2 is an important step FC in the fabrication of microelectronic devices. The

oxidation process can be modeled by three events: (1) the diffusion of oxidant from the bulk gas to the surface of the oxide layer, (2) the diffusion of oxidant through the growing oxide layer to the Si-SiO 2 interface, and (3) the reaction of oxidant with Si at the Si-SiO 2 interface [J. Appl. Phys. 36, 3770 (1965)]. Refer to the sketch below for additional details.

The reaction at the Si surface is assumed to be first-order in oxidant and zero-order in Si. Derive an expression for the oxide layer thickness as a function of time, and show that it can be written in the following form:

(P. E. Savage, The University of Michigan)

State explicitly any further assumptions you make. Assume that the oxide layer has some initial thickness T 0 at time t = 0. [2nd Ed. P10-17]


cdp11-fc.htm



cdp11-gb.htm


Lead titanate, PbTiO 3, is a material having remarkable ferroelectric, pyro- electric, and piezoelectric properties [J. Elec. Chem. Soc., 135, 3137 (1988)]. A thin film of PbTiO 3 was deposited in a CVD reactor. The deposition rate is given below as a function of a temperature and flow rate over the film.

CDP11-G B

What can you learn from these data? [2nd Ed. P10-18]



cdp11-hb.htm


Read the section in Perry's Handbook on multicomponent diffusivities. Using the appropriate equations together with the Fuller correlation [E. N. Fuller, P. D. Schettler, and J. C. Giddings, Ind. Eng. Chem., 48(5), 19 (1966)], calculate the diffusivities of SO 2, O2, N2, and SO 3 when the reaction

CDP11-H B

is taking place on a V2O5-coated pellet. The reaction mixture consists of 20% SO 2 and 80% air. [2nd Ed. P10-11]



cdp11-tb.htm

Additional Homework Problems The following data were obtained from the oxidation of FeS 2 rock samples that produced acid mine drainage

CDP11-I B


file:///H:/html/11chap/html/cdp11-tb.htm[05/12/2011 16:58:00]

Lectures 29 and 30

Lectures 29 and 30 **

Fluidization Engineering (Chapter 12) * A. Fluid Mechanics Ergun Equation

Nomenclature Note: In the text and lecture = porosity, while in the CD-ROM section on Fluidization = porosity. (Our apologies for the inconsistancy.)

At fluidization:

1. Minimum Fluidization Velocity

valid for Red < 10 Need 2. Minor Break Through to Find

file:///H:/html/course/lectures/twenine/index.htm[05/12/2011 16:58:02]

Lectures 29 and 30

3. Terminal Velocity,

4. Slugging Velocity,

First calculate

then

then

Operate the reactor (column) at a flow velocity between

and

or

Assumptions of the Bubbling Bed Model

1. The emulsion exits at minimum fluidization conditions,

and

2. All bubbles are the same size. 3. The concentration of solids in the wake is the same as the emulsion. file:///H:/html/course/lectures/twenine/index.htm[05/12/2011 16:58:02]

.

.

Lectures 29 and 30

4. The solids in the emulsion phase flow downward in plug flow with velocity,

.

5. Velocity of a Single Bubble Rise

For two or more bubbles, bubble rise is:

6. Factors That Determine Bubble Diameter a. bed diameter, b. type of distributor plate c. height of bubble above plate d. gas velocity e. internals (i.e., baffles)

One usually evaluates db at h/2, but this is an area for future research. = Fraction of the total bed occupied by bubbles (not including wakes)


Lectures 29 and 30

7. Balance on Solid Solids flowing down = Solids flowing up in wakes

B. Mass Transport Between Bubble, Cloud, and Emulsion

1. Transport Between Bubble and Cloud


Lectures 29 and 30

2. Transport Between Cloud and Emulsion

Mole Balances 1. Balance on Bubble for Species A

Divide by Vb , then:

2. Balance on Cloud for Species A


Lectures 29 and 30

or

3. Balance on Emulsion for Species A


Lectures 29 and 30

For first order reaction:

where:

*


**

This material was developed from notes by Dr. Lee F. Brown and H. Scott Fogler. Back to the top of Lectures 29 and 30.



Lectures 29 and 30



12prof.htm

Professional Reference Shelf CD12.1 Trickle Bed Reactors

In a trickle bed reactor the gas and liquid flow (trickle) concurrently downward over and uses of a a packed bed of catalyst particles. trickle bed reactor Industrial trickle beds are typically 3 to 6 m deep and up to 3 m in diameter and are filled with catalyst particles ranging from to in. in diameter. The pores of the catalyst are filled with liquid. In petroleum refining, pressures of 34 to 100 atm and temperatures of 350° to 425°C are not uncommon. A pilot-plant trickle bed reactor might be about 1 m deep and 4 cm in diameter. Trickle beds are used in such processes as the hydrodesulfurization of heavy oil stocks, the hydrotreating of lubricating oils, and reactions such as the production of butynediol from acetylene and aqueous formaldehyde over a copper acetylide catalyst. It is on this latter type of reaction,

Characteristics

that we focus in this section. In a few cases, such as the Fischer-Tropsch synthesis, the liquid is inert and acts as a heat transfer medium.

CD12.1.A Fundamentals

(CD121)

The basic reaction and transport steps in trickle bed reactors are similar to slurry reactors. The main differences are the correlations used to determine the mass transfer coefficients. In addition, if there is more than one component in the gas phase (e.g., liquid has a high vapor pressure or one of the entering gases is inert), there is one additional transport step in the gas phase. Figure CD12-1 shows the various


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transport steps in trickle bed reactors. Following our analysis for slurry reactors we develop the equations for the rate of transport of each step.

1. Transport from the bulk gas phase to the gas-liquid interface. The rate of transport per mass of catalyst is

Transport from bulk gas to gasliquid interface to bulk liquid to solidliquid interface Diffusion and reaction in

catalyst pellet


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Figure CD12-1 (a) Trickle bed reactor; (b) reactant concentration profile.

2. Equilibrium at gas-liquid interface:

(CD123)

C Ai = concentration of A in liquid at the interface H = Henry's constant

3. Transport from interface to bulk liquid:


(CD124)

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where k1 = liquid-phase mass transfer coefficient, m/s C Ai = concentration of A in liquid at the interface, kmol/m 3 C Ab= bulk liquid concentration of A, kmol/m 3

4. Transport from bulk liquid to external catalyst surface:

(CD125)

5. Diffusion and reaction in the pellet. If we assume a first-order reaction in dissolved gas A and in liquid B, we have

(CD126)

Combining Equations (CD12-2) through (CD12-6) and rearranging in a manner identical to that leading to the development of Equation (12-82) for slurry reactors, we have

The overall rate equation for A

(CD127)

that is,


(CD128)

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where is the overall transfer coefficient for the gas into the pellet (m3 of gas/g cat. s). A mole balance on species A gives

We next consider the transport and reaction of species B, which does not leave the liquid phase.

6. Transport of B from bulk liquid to solid catalyst interface:

where CB and C Bs are the concentrations of B in the bulk fluid and at the solid interface, respectively.

7. Diffusion and reaction of B inside the catalyst pellet:

Combining Equations (12-33) and (12-34) and rearranging, we have

The overall rate equation for B

(CD129)

(CD1210)

(CD1211)

(CD1213)

A mole balance on species B gives

Mole balance on B

(CD1214)


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One notes that the surface concentrations of A and B, C As and C B s , appear in the denominator of the overall and . transport coefficients

Consequently, Equations (CD12-7), (CD12-9), (CD1212), and (CD12-14) must be solved simultaneously. In some cases analytical solutions are available, but for complex rate laws, one resorts to numerical solutions.1 However, we shall consider some limiting situations.

CD12.1.B Limiting Situations

Mass Transfer of the Gaseous Reactant Limiting. For this situation we assume that either the first three terms in the denominator of Equation (CD12-7) are dominant or that the liquid-phase concentration of species B does not vary significantly through the trickle bed. For these conditions

is constant and we can integrate the mole balance. For negligible volume change = 0, then

Catalyst weight necessary to achieve a conversion XA of gasphase reactant

Mass Transfer and Reaction of Liquid Species Limiting. Here we assume that the liquid phase is entirely saturated with gas throughout the column. As a result, C Asa constant and therefore so is . Consequently, we can integrate the combined mole balance and rate law to give .

Catalyst weight necessary to achieve a conversion XB of gasphase reactant

(CD1215)

(CD1216)

CD12.1.C Evaluating the Transport Coefficients file:///H:/html/12chap/html/12prof.htm[05/12/2011 16:58:04]

12prof.htm

The mass transfer coefficients, kg , kl , and k c depend on a number of variables, such as type of packing, flow rates, wetting of particle, and geometry of the column, and as a result the correlations vary significantly from system to system. Consequently, we will not give all the correlations here but instead will give correlations for particular systems and refer the reader to four specific references where other correlations for trickle bed reactors may be found.2 Typical correlations are given in Table CD12-1. Note that the correlation for organic particles tends to underpredict the transport coefficient.

The representative correlations given in Table CD12-1 assume complete wetting of the catalyst particles. Corrections for incomplete wetting as flow regimes, pressure-drop equations, and other mass transfer correlations can be found in the reviews by Shah, Smith, and Satterfield. 3

Criterion for assuming that plug flow is valid

The plug-flow design equation may be applied successfully provided that the ratio of reactor length L to particle diameter dp satisfies the criterion (Satterfield, 1975)

CD1217)

where Pe = Péclet number = dp U l /DAX DAX = axial dispersion coefficient n = reaction order


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aAlso

see N. Midoux, B. I. Morsi, M. Purwasasmita, A. Laurent, and J. C. Charpentier, Chem. Eng. Sci., 39, 781 (1984), for a comprehensive list of correlations. b In some cases this gives a low estimate of k a ; see M. l i Herskowitz and J. M. Smith, AIChE J., 29, 1983); F. Turek and R. Lange, Chem. Eng. Sci., 36 569 (1981)

The CSTR design equations apply to the trickle bed when4

(CD12-18)


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Techniques for determining the single-phase axial dispersion coefficient are given in Chapter 14.

Example CD12-1 Trickle Bed Reactor

The material presented in this example is meant to serve as an introduction to trickle bed reactors. Other worked-out trickle bed example problems can be found in an article by Ramachandran and Chaudhari6. In addition, the hyrodesulfurication of a hydrocarbon in a trickle bed reactor is given in detail by Tarhan7. Next



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Professional Reference Shelf CD12.2 Fluidized-Bed Reactors * CD12.2.3 Mass Transfer in Fluidized Beds CD12.2.6 Limiting Situations

CD12.2 Fluidized-Bed Reactors8 When a man blames others for his failures, it's a good idea to credit others with his successes. -Howard W. Newton

The fluidized-bed reactor has the ability to process large volumes of fluid. For the catalytic cracking of petroleum naphthas to form gasoline blends, as an example, the virtues of the fluidized-bed reactor drove its competitors from the market.

Fluidization occurs when small solid particles are suspended in an upward-flowing stream of fluid, as shown in Figure CD12-2. The fluid velocity is sufficient to suspend the particles, but not large enough to carry them out of the vessel. The solid particles swirl around the bed rapidly, creating excellent mixing among them. The material fluidized is almost always a solid and the fluidizing medium is either a liquid or a gas. The characteristics and behavior of a fluidized bed are strongly dependent on both the solid and liquid or gas properties. Nearly all of the significant commercial applications of fluidized-bed technology concern gas-solid systems, so these are treated in this section. The material that follows is based on what is seemingly the best model of the fluidized-bed reactor developed thus far--the bubbling-bed model of Kunii and Levenspiel.

file:///H:/html/12chap/html/12prof2a.htm[05/12/2011 16:58:07]

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Figure CD12-2 From D. Kunii and O. Levenspiel, Fluidization Engineering (Melbourne, Fla.: Robert E. Krieger Publishing Co., 1969). Reprinted with permission of the publishers.

CD12.2.1 Overview

We are going to use the Kunii-Levenspiel bubbling-bed model to describe reactions in fluidized beds. In this model the reactant gas enters the bottom of the bed and flows up the reactor in the form of bubbles. As the bubbles rise, mass transfer of the reactant gases takes place as they flow (diffuse) in and out of the bubble to contact the solid particles, where the reaction product is formed. The product then flows back into a bubble and finally exits the bed when the bubble reaches the top of the bed. The rate at which the reactants and products transfer in and out of the bubble affects the conversion, as does the time it takes for the bubble to pass through the bed. Consequently, we need to describe the velocity at which the bubbles move through the column and the rate of transport of gases in and out of the bubbles. To calculate these parameters we need to determine a number of fluid-mechanics parameters associated with the fluidization


12prof2.htm

process. Specifically, to determine the velocity of the bubble through the bed we first need to calculate:

1. Porosity at minimum fluidization,

The algorithm

2. Minimum fluidization velocity, 3. Bubble size, db

To calculate the mass transport coefficient we first must calculate:

1. Porosity at minimum fluidization, 2. Minimum fluidization velocity, 3. Velocity of bubble rise, Ub 4. Bubble size, db

To determine the reaction-rate parameters in the bed we first need to calculate:

1. Fraction of the total bed occupied by bubbles, 2. Fraction of the bed consisting of wakes, 3. Volume of catalyst in the bubbles, clouds, and

emulsion,

b

,

c

and

e

,

It is evident that before we begin to study fluidizedbed reactors we must obtain an understanding of the fluid mechanics of fluidization. In Section CD12.2.2 equations are developed to calculate all the fluid ,) necessary to mechanic parameters (e.g., db

obtain the mass transfer and reaction parameters.In

Section CD12.2.3, equations for the mass transfer parameters are developed. In Section CD12.2.4 the reaction-rate parameters are presented, and the mole balance equations are applied to the bed to predict conversion in Section CD12.2.5.

CD12.2.2 Mechanics of Fluidized Beds In this section we describe the regions of fluidization and calculate the minimum and maximum fluidization velocities. Next, the Kuniifile:///H:/html/12chap/html/12prof2a.htm[05/12/2011 16:58:07]

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CD12.2.2A Description of the Phenomena

We consider a vertical bed of solid particles supported by a porous or perforated distributor plate, as in Figure CD12-3a. The direction of gas flow is upward through this bed. There is a drag exerted on the solid particles by the flowing gas, and at low gas velocities the pressure drop resulting from this drag will follow the Ergun equation, Equation (4-22), just as for any other type of packed bed. When the gas velocity is increased to a certain value however, the total drag on the particles will equal the weight of the bed, and the particles will begin to lift and barely fluidize. If c is the density of the solid catalyst particles, Ac the crosssectional area, hs the height of the bed settled before the particles start to lift, h the height of the bed at any time, and s and the corresponding porosities of the settled and expanded bed, respectively, then the mass of solids in the bed, Ws , is

Levenspiel bubbling-bed model is described in detail.9 Finally, equations to calculate the fraction of the bed comprising bubbles, the bubble size, the velocity of bubble rise, and the fractional volume of bubbles, clouds, and wakes are derived.


(CD1219)

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Figure CD12-3 Various kinds of contacting of a batch of solids by fluid. Adapted from D. Kunii and O. Levenspiel, Fluidization Engineering (Melbourne, Fla.: Robert E. Krieger Publishing Co., 1977).

(Note nomenclature change: In the text and lecture, = porosity, while in this section, = porosity.) This relationship is a consequence of the fact that the mass of the bed occupied solely by the solid particles is the same no matter what the porosity of the bed. When the drag force exceeds the gravitational force, the particles begin to lift and the bed expands (i.e., the height increases), thus increasing the bed porosity, as described by Equation (CD12-19). This increase in bed porosity decreases the overall drag until it is again balanced by the total gravitational force exerted file:///H:/html/12chap/html/12prof2a.htm[05/12/2011 16:58:07]

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on the solid particles (Figure CD12-3b).

If the gas velocity is increased still further, expansion of the bed will continue to occur; the solid particles will become somewhat separated from each other and begin to jostle each other and move around in a restless manner. Increasing the velocity just a slight amount causes further instabilities and some of the gas starts bypassing the rest of the bed in the form of bubbles (Figure CD12-3c). These bubbles grow in size as they rise up the column. Coincidentally with this, the solids in the bed begin moving upward, downward, and around in a highly agitated fashion, appearing as a boiling frothing mixture. With part of the gas bubbling through the bed and the solids being moved around as though they were part of the fluid, the bed of particles is said to be fluidized. It is in a state of aggregative, nonparticulate, or bubbling fluidization. A further increase in gas velocity will result in slug flow (Figure CD12-3d) and unstable chaotic operation of the bed. Finally, at extremely high velocities the particles are blown or transported out of the bed (Figure CD12-3e). The range of velocities over which the Ergun equation applies can be fairly large. On the other hand, the difference between the velocity at which the bedstarts to expand and the velocity at which the bubbles start to appear can be extremely small and sometimes nonexistent. This observation means that if one steadily increases the gas flow rate, the first evidence of bed expansion may be the appearance of gas bubbles in the bed and the movement of solids. At low gas velocities in the range of fluidization, the rising bubbles contain very few solid particles. The remainder of the bed has a much higher concentration of solids in it and is known as the emulsion phase of the fluidized bed. The bubbles are shown as the bubble phase. The cloud phase is an intermediate phase between the bubble and emulsion phases.

Once the drag exerted on the particles equals the net gravitational force exerted on the particles, that is,


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(CD1220)

the pressure drop will not increase with an increase in velocity beyond this point (see Figure CD12-4). From the point at which the bubbles begin to appear in the bed, the gas velocity can be increased steadily over quite an appreciable range without changing the pressure drop across the bed or flowing the particles out of the bed. The bubbles become more frequent and the bed more highly agitated as the gas velocity is increased (Figure CD12-3c), but the particles remain in the bed. This region is bubbling fluidization. Depending on the physical characteristics of the gas, the solid particles, the distributor plate, and internals (such as heat exchanger tubes) within the bed, the region of bubbling fluidization can extend over more than an order of magnitude of gas velocities (e.g., 4 to 50 cm/s in Figure CD12-4). In other situations, gas velocities in the region of bubbling fluidization may be limited; the point at which the solids begin to be carried out of the bed by the rising gas may be a factor of only three or four times the velocity at incipient fluidization.

If the gas velocity is increased continuously, it will eventually become sufficiently rapid to carry the solid particles upward, out of the bed. When this begins to happen, the bubbling and agitation of the solids are still present, and this is known as the region of fast fluidization, and the bed is a fast-fluidized bed. At velocities beyond this region, the particles are well apart, and the particles are merely carried along with the gas stream. Under these conditions, the reactor is usually referred to as a straight-through transport reactor (STTR) (Figure CD12-3e).

The various regions described above display the behavior illustrated in Figure CD12-4. This figure presents the pressure drop across a bed of solid particles as a function of gas velocity. The region covered by the Ergun equation is the rising portion of the plot (section I: .The section of the figure where the pressure drop remains essentially constant over a wide range of velocities is the region cm/s). of bubbling fluidization (section II: Beyond this are the regions of fast fluidization and of


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purely entrained flow.

Figure CD12-4 From D. Kunii and O. Levenspiel, Fluidization Engineering (Melbourne, Fla.: Robert E. Krieger Publishing Co., 1977). Reprinted with permission of the publishers.

CD12.2.2B Minimum Fluidization Velocity

Fluidization will be considered to begin at the gas velocity at which the weight of the solids gravitational force exerted on the particles equals the drag on the particles from the rising gas. The gravitational force is given by Equation (CD12-19) and the drag force by the Ergun equation. All parameters at the point where these two forces are equal will be characterized by the subscript mf, to denote that this is the value of a particular term when the bed is just beginning to become fluidized. The combination g( c- g) occurs very frequently, as in Equation (CD12-19), and this grouping is termed :

(CD1221)

The Ergun equation, Equation (4-22), can be written in the form (CD1223)

where the shape factor of catalyst particle, sometimes called the sphericity.


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At the point of minimum fluidization the weight of the bed just equals the pressure drop across the bed:

(CD1223)

(CD1224)

For (Rep = gdpU/ ) we can solve Equation (CD12-24) for the minimum fluidization velocity, to give

Calculate

(CD12-25)

Reynolds numbers below 10 represent the usual situation, in which fine particles are fluidized by a gas. Sometimes, higher values of the Reynolds number do exist at the point of incipient fluidization, and then the quadratic equation (CD12-24) must be used.

Two dimensionless parameters in these two equations for deserve comment. The first is , the sphericity, which is a measure of a particle's nonideality in both shape and roughness. It is calculated by visualizing a sphere whose volume is equal to the particle's and dividing the surface area of this sphere by the actually measured surface area of the particle. Since the volume of a spherical particle in

(CD12-26)

and its surface area is

(CD12-27)

Calculate


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(CD12-28)

Measured values of this parameter range from 0.5 to 1, with 0.6 being a normal value for a typical granular solid.

The second parameter of special interest is the void . fraction at the point of minimum fluidization, It appears in many of the equations describing fluidized-bed characteristics. A correlation exists which apparently gives quite accurate predictions of (within 10%) when the measured values of particles in the fluidized bed are fairly small:10

Calculate

(CD12-29)

Another correlation commonly used is that of Wen and Yu 11 :

(CD12-30)

and/or

(CD12-31)

When the particles are large, the predicted can below 0.40 is be much too small. If a value of predicted, it should be considered suspect. Kunii and Levenspiel12 state that is an easily file:///H:/html/12chap/html/12prof2a.htm[05/12/2011 16:58:07]

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measurable value. However, if it is not convenient to do so, Equation (CD12-29) should suffice. around 0.5 are typical. If the Values of distribution of sizes of the particles covers too large a range, the equation will not apply because smaller particles can fill the interstices between larger particles. When a distribution of particle sizes exists, an equation for calculating the mean diameter is

where dpi.

CD12.2.2C Maximum Fluidization

Maximum velocity through the bed u t

(CD12-32)

is the fraction of particles with diameter

If the gas velocity is increased to a sufficiently high value, however, the drag on an individual particle will surpass the gravitational force on the particle, and the particle will be entrained in a gas and carried out of the bed. The point at which the drag on an individual particle is about to exceed the gravitational force exerted on it is called the maximum fluidization velocity. When the upward velocity of the gas exceeds the free-fall terminal velocity of the particle, ut , the particle will be carried upward with the gas stream. For fine particles, the Reynolds numbers will be small, and two relationships presented by Kunii and Levenspiel13 are:

(CD12-33)

We now have the maximum and minimum superficial velocities at which we may operate the bed. The entering superficial velocity, u0 , must be


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above the minimum fluidization velocity but below the slugging, ums and terminal, ut , velocities.

Both of these conditions must be satisfied for proper bed operation.

CD12.2.2D Descriptive Behavior of a Fluidized Bed: The Model of Kunii and Levenspiel

At gas flow rates above the point of minimum fluidization, a fluidized bed appears much like a vigorously boiling liquid; bubbles of gas rise rapidly and burst on the surface, and the emulsion phase is thoroughly agitated. The bubbles form very near the bottom of the bed, very close to the distributor plate, and as a result the design of the distributor plate has a significant effect on fluidized-bed characteristics.

Literally hundreds of investigators have contributed to what is now regarded as a fairly practical description of the behavior of a fluidized bed; chief among these is to be the work of Davidson.14 Early investigators saw that the fluidized bed had to be treated as a two-phase system: an emulsion phase and a bubble phase (often called the dense and lean phases). The bubbles contain very small amounts of solids. They are not spherical; rather, they have an approximately hemispherical top and a pushed-in bottom. Each bubble of gas has a wake which contains a significant amount of solids. These characteristics are illustrated in Figure CD12-5, which were obtained from x-rays of the wake and emulsion, the darkened portion being the bubble phase.

As the bubble rises, it pulls up the wake with its solids behind it. The net flow of the solids in the emulsion phase must therefore be downward. The gas within a particular bubble remains largely within that bubble, penetrating only a short distance into the surrounding emulsion phase. The region penetrated by gas from a rising bubble is called the file:///H:/html/12chap/html/12prof2a.htm[05/12/2011 16:58:07]

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cloud.

Davidson found that he could relate the velocity of bubble rise and the cloud thickness to the size of bubble. Kunii and Levenspiel15 combined these observations with some simplifying assumptions to provide a practical, usable model of fluidized-bed behavior. Their assumptions are presented in Table CD12-2. Several of these assumptions had been used by earlier investigators, particularly Davidson and Harrison.16 With the possible exception of assumption 3, all of these assumptions are of questionable validity, and rather obvious deviations from them are observed routinely. Nevertheless, the deviations apparently do not affect the mechanical or reaction behavior of fluidized beds sufficiently to diminish their usefulness.

Figure CD12-5 Schematic of bubble, cloud, and wake.


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Single bubble

CD12.2.2E Bubble Velocity and Cloud Size From experiments with single bubbles, Davidson found that the velocity of rise of a single bubble could be related to the bubble size by

(CD12-35)

When many bubbles are present, this velocity would be affected by other factors. The more bubbles that are present, the less drag there would be on an individual bubble; the bubbles would carry each other up through the bed. The greater number of bubbles would result from larger amounts of gas passing through the bed (i.e., a larger value of u0). Therefore, the larger the value of u0, the faster should be the velocity of a gas bubble as it rises through the bed. Other factors that should affect this term are the


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viscosity of the gas and the size and density of the solid particles that make up the bed. Both of these terms also affect the minimum fluidization velocity, so this term might well appear in any relationship for the velocity of bubble rise; the higher the minimum fluidization velocity, the lower the velocity of the rising bubble.

Adopting an expression used in gas-liquid systems, Davidson proposed that the rate of bubble rise in a fluidized bed could be represented simply by adding and subtracting these terms:

Velocity of bubble rise u b

(CD12-36)

Bubble Size. The equations for the velocity of bubble rise, Equations (CD12-35) and (CD12-36), are functions of the bubble diameter, an elusive value to obtain. As might be expected, it has been found to depend on such factors as bed diameter, height above the distributor plate, gas velocity, and the components that affect the fluidization characteristics of the particles. Unfortunately, for predictability, the bubble diameter also depends significantly on the type and number of baffles, heat exchangers tubes, and so on, within the fluidized bed (sometimes called "internals"). The design of the distributor plate, which disperses the inlet gas over the bottom of the bed, can also has a pronounced effect on the bubble diameter. Studies of bubble diameter carried out thus far have concentrated on fluidized beds with no internals and have involved rather small beds. Under these conditions the bubbles grow as they rise through the bed. The best relationship between bubble diameter and height in the column at this writing seems to be that of Mori and Wen,17 who correlated the data of studies covering bed diameters of 7 to 130 cm, minimum fluidization velocities of 0.5 to 20 cm /s, and solid particle sizes of 0.006 to 0.045 cm. Their principal equation was


12prof2.htm

db

(CD12-37)

In this equation, db is the bubble diameter in a bed of diameter Dt observed at a height h above the distributor plate, db0 is the diameter of the bubble formed initially just above the distributor plate, and dbm is the maximum bubble diameter attained if all the bubbles in any horizontal plane coalesce to form a single bubble (as they will do if the bed is high enough). The maximum bubble diameter, dbm , has been observed to follow the relationship

d maximum

(CD12-38)

for all beds, while the initial bubble diameter depends on the type of distributor plate. For porous plates the relationship

is observed, and for perforated plates the relationship

d minimum

(CD12-39)

(CD12-40)

appears to be valid, in which nd is the number of perforations. For beds with diameters between 30 and 130 cm, these relations appear to predict bubble diameters with an accuracy of about 50%; for beds with diameters between 7 and 30 cm, the accuracy of prediction appears to be approximately +100%, -60% of the observed values. Werther developed the following correlation based on a statistical coalescence model:18


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(CD12-41)

The bubble size predicted by this model is close to that predicted by Mori and Wen19 for large-diameter beds (2 m) and smaller than that suggested by Mori and Wen for small-diameter beds (0.1 m).

CD12.2.2F Fraction of Bed in the Bubble Phase

Using the Kunii-Levenspiel model, the fraction of the bed occupied by the bubbles and wakes can be estimated by material balances on the solid particles and the gas flows. The parameter is the fraction of the total bed occupied by the part of the bubbles that does not include the wake, and is the volume of wake per volume of bubble. The bed fraction in the wakes is therefore . The bed fraction in the emulsion phase (which includes the clouds) is (1 - ). Letting Ac and c represent the cross-sectional area of the bed and the denity of the solid particles, respectively, a material balance on the solids (Figure CD12-6) gives

(CD12-42)


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Figure CD12-6 Wake angle w and wake fraction of threedimentional bubbles at ambient conditions; evaluted from x-ray photographs by Rowe and Partridge. Adapted from D. Kuknii and O. Levenspiel, Fluidization Engineering, 2nd. Ed., (Stoneham, Mass.; ButterworthHeinemann 1991).

Velocity of solids u s

or

(CD12-43)

A material balance on the gas flows gives

(CD12-44)

The velocity of rise of gas in the emulsion phase is

Velocity of gas in emulsion ue

(CD12-45)

(In the fluidization literature, us is almost always taken as positive in the downward direction.) Factoring the cross-sectional area from Equation (CD12-44) and then combining Equations (CD1244) and (CD12-45), we obtain an expression for the fraction of the bed occupied by bubbles:

Volume fraction bubbles

CD12-46)

The wake parameter, , is a function of the particle size in Figure CD12-5. The value of a has been observed experimentally to vary between 0.25 and 1.0, with typical values close to 0.4. Kunii and Levenspiel assume that Equation (CD12-46) can be


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simplified to

(CD12-47)

which is valid for (e.g.,

Example CD12-2 Maximum Solids Holdup

Back Next

* This material was developed from notes

by Dr. Lee F. Brown and H. Scott Fogler. Return to the top of the page.



excd12-3.htm


Example CD12-3 Catalytic Oxidation of Ammonia

25

Massimilla and Johnstone studied the catalytic oxidation of ammonia in a fluidized-bed reactor. Under their experimental conditions, the reaction was first order, dependent only on the ammonia concentration, and without a significant change in volumetric flow rate. In one of their runs, 4 kg of catalyst was used with a gas flow rate of 818 cm3/s at reaction conditions. A conversion of 22% of the entering ammonia was obtained. Predict this conversion using the KuniiLevenspiel model.

Other data:

Solution

A. Mechanical characteristics of bed

Step 1. Gravitation term, :


excd12-3.htm

Step 2. Porosity of bed at minimum fluidization:

(CD1229)

Step 3. Gas velocity at minimum fluidization:

(CD12-25)

Step 4. Entering gas velocity:

Step 5. Is u0 within a reasonable operating range? Check ut .

(CD1233)


excd12-3.htm

Are the N Re in the proper range for use of Equations (CD12-25) and (CD12-33)?

Thus u0 is 5.4 times

, and well below ut .

Step 6. Bubble sizes, db 0, dbm , and db:

(CD1238)

(CD1239)

Step 7. Bubble sizes, db 0, dbm , and db: The unexpanded bed height is 38.9 cm. The expanded bed height will probably be 40 to 50% greater, say about 60 cm. We will therefore assume that the average bubble size will be taken as the one calculated for h /2 = 30 cm. Step 8. Average bubble diameter:

Step 9. Rise velocity of single bubble:


excd12-3.htm

Step 10. Rise velocity of a bubble when many bubbles are present:

(CD12-35)

(CD12-36)

From Figure CD12-6 for glass spheres with dp = 0.105 mm, = 0.4.

Step 11: Fraction of bed in bubble phase:

(CD12-46)

Since the estimated bed height of 60 cm is sufficiently close to the calculated value of 63.2 cm, one can proceed further in the calculations without making a new estimate of h.

Good guess of h = 60 cm

Step 12. Bed height:

B. Mass transfer and reaction parameters

Step 1. Bubble-cloud mass transfer coefficient:


excd12-3.htm

Order of magnitude parameters

(CD1253)

Step 2. Cloud-emulsion mass-transfer coefficient.,

(CD12-55)

Step 3. Volume of catalysts in the bubble per volume of bubble:

= 0.01 (assumed) b

Step 4. Volume of catalyst in clouds and wakes/ cm3 of bubbles:

(CD12-63)

Step 5. Volume of catalyst in emulsion/cm 3 of bubbles:

(CD12-64)


excd12-3.htm

Step 6. Calculate K R and X from Equation (CD1227):

rearranging Equation (CD12-75).

where

(CD12-72)

This is close to the observed value of 22% conversion.

Solving for X gives


excd12-3.htm



excd12-4.htm


Example CD12-4 Calculation of Resistances

Calculate each of the resistances to reaction and transfer and the relationship between CAb, C Ac , and C Ae for the ammonia oxidation reaction described in Example CD12-2. Assume that .

To relate C Ae and C Ac we rearrange Equation (CD12 67) for a first-order reaction as

The analog electrical resistance for the system is shown in Figure CD12-11 along with the corresponding resistances for this reaction. As with its electrical analog, the reaction will pursue the path of least resistance, which in this case is along the right hand-side branch of Figure CD12-11. If the major resistance in this side, the resistance to reaction in the emulsion Rre , could be reduced, a greater conversion could be achieved for a specific catalyst weight. To reduce Rre one needs to look for ways of increasing e.

Electrical analog

Solution


excd12-4.htm

(CD1280)

Figure CD12-11 Electrical Analog

Examination of Equation (CD12-80) shows that decreasing the bubble fraction and the minimum fluidization velocity would increase e and hence the conversion. The minimum fluidization velocity could be decreased by decreasing the particle size. We will investigate how the various parameters will affect the conversion for different limiting situations.



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Professional Reference Shelf Example CD12-5 Effect of Particle Size on Catalyst Weight-Slow Reaction

Suppose that you are operating at five times the What minimum fluidization velocity, uo = 5 would be the effect of doubling the particle diameter on the catalyst weight for the same throughput and conversion?

Solution Substitution for u0 into Equation (CD12-89) gives (CDE12-5.1)

(CDE12-5.2)

Since the temperature remains constant and there are no inter- and intraparticle resistances, kcat1 = kcat2, the throughput (u01 = u02) and conversion (X1 =X 2) are the same for cases 1 and 2. The ratio of Equations (CDE12-5.1) and (CDE12-5.2) yields

(CDE12-5.3)

Recalling Equation (CD12-25) gives

(CDE12-5.4)

and neglecting the dependence of on dp, we see that the only parameters that vary between case 1 (dp)

and case 2 (dp2= 2dp1) are


and W. Taking the

excd12-5.htm

ratio of 2 to 1 and substituting for d p2 in terms of dp1, we obtain

Thus in the situation we have postulated, with a firstorder reaction and reaction limiting the bed behavior, doubling the particle size will reduce the catalyst by approximately 75% and still maintain the same conversion.

(CDE12-5.5)

and therefore

(CDE12-5.6)



excd12-6.htm

Professional Reference Shelf Example CD12-6 Effect of Catalyst Weight on Particle Size- -Rapid Reaction

We again consider the effect of doubling particle size while keeping all other variables the same. Case 1,dp1 = dp1; case 2, dpI2 = 2dp1.

Solution

In this case we see that doubling the particle diameter decreases the catalyst weight by 89% while maintaining the same conversion. However, for a fast reaction, a significant decrease in effectiveness factor could offset this advantage.

Using Equation (CD12-77) and taking the ratio of case 1 to case 2 gives

(CDE12-6.1)

Recalling from previous examples that

then

(CDE12-6.2)

or

(CDE12-6.3)


excd12-6.htm



12prof3.htm

Professional Reference Shelf CD12.3.1 Chemical Vapor Deposition Reactors As discussed in Section 6.6, CVD is a very important process in the microelectronics industry. The fabrication of microelectronic devices may include as few as 30 or as many as 200 individual steps to produce chips with up to 106 transducers per chip. An abbreviated schematic of the steps involved in producing a typical computer chip is shown in Figure CD12-12. Although we will focus on CVD, it is worthwhile to give an overview of microelectronic fabrication. Starting from the upper left we see that single crystal silicon ingots are grown in a Czochralski crystalizer, then sliced into wafers, and chemically and physically polished. These polished wafers serve as a starting material for a variety of microelectronic devices. A typical fabrication sequence is shown for processing the wafer beginning with the formation of an SiO 2 layer on top of the silicon. The SiO 2 layer may be formed either by oxidizing a silicon layer or by laying down a SiO 2 vapor deposition (CVD). Next the wafer is masked with a polymer photoresist (PR), a layer by chemical template with the pattern to be etched onto the SiO 2 layer is placed over the PR, and the wafer is exposed to ultraviolet irradiation. If the mask is a positive PR , the light will cause scission in the polymer so that the exposed areas will dissolve when the wafer is placed in the developer. On the other hand, when a negative PR mask is exposed to ultraviolet irradiation, cross-linking of the polymer chains occurs and the unexposed areas dissolve in the developer. The undeveloped portion of the PR (in either case) will protect the covered areas from etching.


12prof3.htm

Figure CD12-12 Microelectric fabrication steps

After the exposed areas of SiO 2 are etched to form trenches (either by wet etching (see Problem P5-12) or plasma etching), the remaining PR is removed. Next the wafer is placed in a furnace containing gas molecules of the desired dopant, which then diffuse into the exposed silicon. After diffusion of dopant to the desired depth in the wafer, it is removed and then covered with SiO 2 by CVD. The sequence of masking, etching, CVD, and metallization continues until the desired device is formed. A schematic of a final chip is shown in the lower righthand corner of Figure CD12-12. One of the key steps in the chip-making process is the deposition of different semiconductors and metals on the surface of the chip. This step can be achieved by CVD. CVD mechanisms were discussed in Chapter 10. Consequently, this section will focus on CVD reactors. A number of CVD reactor types have been used, such as barrel reactors, boat reactors, and horizontal and vertical reactors. A description of these reactors and modeling equations is given by Jensen.26


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CD12.3.1A Boat Reactors

One of the more common CVD reactors is the horizontal low-pressure CVD (LPCVD) reactor. This reactor operates at pressures of approximately 100 Pa. The main advantage of the LPCVD is its capability of processing a large number of wafers without detrimental effects to film uniformity. Owing to the large increases in the diffusion coefficient at low pressures (recall Table 11-2), surface reactions are more likely to be controlling than mass transfer. A schematic of a LPCVD boat reactor is shown in

Figure CD12-13 LPCVD boat reactor

To illustrate LPCVD modeling we shall use a specific but simplified example, the deposition of silicon from a gas stream of SiH 2. The reaction mechanism is

CVD reaction sequence in silicon deposition

Here we have assumed that the equilibrium for the dissociation of SiH discussed in Problem P10-12 lies far to the right. The corresponding rate law is

Recalling that the adsorption constants K 1 and K 2 decrease with increasing temperatures, an excellent approximation at high temperature is


12prof3.htm

Consequently, the deposition rate can be modeled as firstorder in SiH 2:

where A

SiH 2.

Modeling Concepts. We shall model the axial flow in the annular region as being laminar. This assumption is reasonable because a typical Reynolds number for flow in a LPCVD reactor is less than 1. As the reactant gases flow through the annulus, the reactants diffuse from the annulus radially inward between the wafers to coat them.27 The reacting gas flows through the annulus between the outer edges of the cylindrical wafers and the tube wall (see Figure CD12-14). The corresponding cross-sectional area of the annulus is

Flow in the

annulus

where Rt and Rw are the radii of the tube and wafer, respectively. Because SiH 2 is being consumed by CVD, the mole fraction of SiH 2 (i.e., the reactant) in the annulus, yAA, decreases as the reactant flows down the length of the annulus. The reacting gases diffuse out of the annular region into the space between the wafers where the mole fraction is represented by yA. As molecules diffuse radially inward, some of them are adsorbed and deposited on the wafer surface. The reaction products then diffuse radially outward into the gas stream axially flowing in the annulus. This system can be analyzed in a manner analogous to flow through a packed catalyst bed where the reaction gases diffuse into the catalyst pellets. In this analysis we used an effectiveness factor to determine the overall rate of reaction per volume (or mass) of reactor bed. We can extend this idea to LPCVD reactors, where the reactants diffuse from the annular flow channel radially inward between the wafers.


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Figure CD12-14 LPCVD boat reactor with peripherals

CD12.3.2 Effectiveness Factor for a LPCVD Reactor

Silicon will deposit on the wafers, the reactor walls, and on the boat support. Deposition on the walls and support will take place at the reactant concentrations in the annulus. However, the concentration of A between the wafers is less than the concentration in the annulus. Consequently, the rate of deposition on the wafer will be less than the rate at conditions in the annulus. Fortunately, these two concentrations can be related by the effectiveness factor. We can determine the effeciveness factor once the concentration profile in the region between the wafers is obtained.

where l is the distance between wafers and file:///H:/html/12chap/html/12prof3.htm[05/12/2011 16:58:13]

is the rate of

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disappearance of A at the concentration of A in the annular

region, C AA. We now use to express the actual rate of reaction per unit surface area of wafer in terms of the rate of reaction at conditions in the annulus:

Letting a be the wafer surface area per unit volume of reactor, the rate of consumption of species A by the wafer per unit volume of reactor is

Example CD12-7 Diffusion Between Wafers

Deposition on the Peripherals. Silicon will desposit on the walls of a reactor and on the boat support in addition to the wafers. This rate of deposition on the walls and support is

Accounting for Si deposition on the walls and support

Owing to high temperature and low pressure, radiation is the dominant heat transfer mechanism; therefor, small temperature differences exist between the wafer and reactor wall. Consequently, there is no need to couple the mole and energy balances for these small temperature gradients.

Example CD12-8 CVD Boat Reactor



cdp12-ab.htm



is carried out in a packed-bed reactor. A stoichiometric mixture of A and B enters the reactor at a total molar flow rate of 10 g mol/min, a temperature of 400 K, and a pressure of 20 atm. Currently, 1000 kg of catalyst measuring 1.5 mm in diameter is used to achieve 0.05 conversion with an exiting pressure of 19 atm. In an effort to increase conversion, data were taken correlating the diameter and effectiveness factor of the catalyst pellet. Use the data below to determine the catalyst size that gives the highest conversion. Flow is completely turbulent throughout the bed.

Additional Information:


cdp12-bd.htm


It is proposed that the wafer thickness in a LPCVD reactor could be made more uniform if the temperature were to be increased along the length of the reactor. The feed composition is 15% A and 85% carrier gas and enters at 1000 K and 1 torr (1mmHg). Determine the temperature profile to achieve uniform thickness at the center of the wafer. Also determine the wafer shape as a function of distance down the reactor. (Note: The feed composition as well as flow rate can also be varied.)

CDP12BD


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cdp12-cb.htm

Additional Homework Problems (CVD boat reactor)

CDP12-CB

(a) With increasing temperature, the wafer thickness increases to some maximum value, then decreases. Explain! (b) How do the wafer thickness and curvature vary as the pressure is increased from a very low pressure to a high pressure? (c) What are the effects of wafer radius and wafer spacing on the curvature and thickness of the wafer? (d) What are the effects of volumetric flow rate and entering concentration on wafer thickness and curvature? Hint: You may want to program Equations (E12-5.11) and (E125.12) to carry out a series of computer simulations to help answer these questions. Parameter values are given in Problem CDP12-D.



cdp12-db.htm

Additional Homework Problems CDP12DB

(CVD boat reactor) The following table gives the range of typical parameter values for the boat reactor.

(a) Determine the effect of each of the parameter values in the cases A, B, and C on the thickness and curvature.

The following parameter values are the same for all three cases.

(b) Vary other parameters and describe your findings.


cdp12-db.htm



cdp12-ec.htm


The silicon compound described in Example CD12-3 is to be deposited on the large sheets separated by a distance d shown in Figure CDP12-E. The mass transfer correlation for flow across a flat plate varies along the length of the plate and is given by the expression

CDP12-EC

Figure CDP12-E

(a) Derive an expression for the deposition thickness as a function of position. (b) How would you suggest changing the parameters (e.g., d, U) or operating conditions to make the deposition thickness uniform? (c) Suppose that a wire screen were placed a small distance S from the sheets so that the gas in that region was virtually stagnant. How would this addition affect the deposition thickness?



cdp12-fc.htm


Rework the CVD boat reactor examples (a) CD12-2 and / or (b) CD12-3 taking into account the equilibrium reaction

CDP12-FC

occurring throughout. Note that we no longer have equimolar counterdiffusion! A mixture of 60% SiH 4 and 40% inert carrier enters at 1000 K. All other parameter values are the same as in Problem CDP12-D.



cdp12-gb.htm


The hydrogenation of an unsaturated organic is to be carried out in a trickle bed reactor packed with 0.4-cm-diameter spherical catalyst particles.

CDP12-G B

The reaction in the pellet is first-order in both hydrogen and the organic. Hydrogen and nitrogen are fed in equimolar portions at a total pressure of 2 atm. The reactor diameter is to be 1.0 m. It is proposed to operate at a superficial liquid mass velocity of 5 kg/m 2 s and a total gas flow rate of 36 mol/s. As a first approximation assume that the concentration of organic is constant and the pseudo-first-order specific reaction rate is 2.5 x 10-5 m 3/kg cat. s at 400 K.

(a) Calculate the catalyst weight necessary to achieve 55% conversion of the hydrogen. (b) If 55% conversion could not be obtained for the operating conditions specified, what should the operating conditions be to obtain this conversion? (c) For each transport step, determine its fraction of the total resistance to mass transport and reaction. (d) The bed is now to be operated under conditions of complete saturation of the hydrogen throughout the bed. Calculate the weight of catalyst necessary to achieve 95% conversion of the organic. To achieve this conversion the pressure is increased to 100 atm and the feed is pure H2. The particle size is reduced by a factor of 4. (e) Assuming an axial dispersion coefficient of 0.38 cm2/s, determine


cdp12-gb.htm

if the plug-flow model was a good assumption.


Liquid viscosity: 3.1 cP = 0.0031 kg/m s Liquid density: 700 kg/m 3 Hydrogen liquid diffusivity in oil: 7.0 x 10-9 m 2/s Organic diffusivity in organic product: 2.5 x 109 m 2 /s Molecular weight of organic: 256 daltons Hydrogen solubility: 0.004 kmol/m 3 atm Pellet porosity: 0.45 Pellet density: 1600 kg/m 3 Bed porosity: 0.4 Superficial gas velocity (kg/m2 s): 0.75 0.5 0.1 Pressure drop (kPa/m): 70 50 20 [From AIChE J., 29, 1 (1983)]



cdp12-hb.htm


The oxidation of ethanol is to be carried out in a trickle bed reactor. The reaction is first-order with respect to oxygen and zero-order with respect to ethanol. The reactor is 30 cm in diameter and 6 m long and the bed porosity is 40%. The Pd/Al 2O catalyst pellets are 0.5 cm in diameter and have a density of 1800 kg/m 3 . The diffusivity of oxygen in ethanol is 4 x 10-10 m 2/s and the equilibrium solubility is 4 x 10-2 kmol/m 3 at reaction conditions. The superficial gas velocity is 10 cm/s and the corresponding pressure gradient is 30 kPa/m. The superficial liquid-phase velocity is 0.25 cm/s. The specific reaction rate is 0.03 s -1. Tortuosity = 3.9, p = 0.6, = 0.8.

CDP12HB

(a) Calculate the percent of the total resistance for each transport step. (b) Determine the conversion of ethanol. (c) Assuming an axial dispersion coefficient of 0.02 cm2/s, determine if the plug-flow model was a good assumption.



cdp12-jb.htm

Additional Homework Problems The hydrogenation of aromatics in a napthenic lube oil distillate takes place in a trickle bed reactor. The bed is 1 m in diameter and 6 m deep. The solubility of hydrogen in the oil is 0.0068 mol/dm 3 atm. The concentration of aromatics in the feed is 0.07 mol/ dm 3. The superficial liquid flow rate is 8 kg/m 2 s, while the hydrogen flow rate is 0.2 kg/m 2 s. The reaction is first-order in hydrogen and in aromatics on the

CDP12-JB

-in. catalyst pellet. The true specific reaction rate is 2 x 10-5 m 6/kmol kg s. Nitrogen enters with the hydrogen at a rate of 0.05 kg/m 2 > s. The reaction occurs at 100°C and 50 atm.

(a) Make a plot of conversion of aromatics as a function of bed depth. (b) Calculate the percent resistance of each transport step. (c) How would the exit conversion change if the gas flow rate were reduced by a factor of 4? (d) How could the conversion be increased? Additional information:

Effective liquid diffusivity of H2 in oil: 8 x 10-9 m 2/s Effective liquid diffusivity of aromatics in oil: 6 x 10-11 m 2/s Viscosity of oil: 3.0 cP Bed porosity: 0.35 Pressure gradient for flow of 0.25 kg gas/m2 s: 30 kPa/m Density of oil: 900 kg/m 3 Pellet Density: 1500 kg/m 3 Tortuosity: 3.8 Porosity: 0.5 Construction: 0.8


cdp12-jb.htm



cdp12-kc.htm

Additional Homework Problems Read the article by C. Chavarie and J. R. Grace: Performance Analysis of a Fluidized Bed Reactor III Modification and Extension of Conventional Two Phase-Models. [Ind. Eng. Chem. Fundamentals 4 2 86-91 1975.]

CDP12-K C

(a) Compare the models discussed in the article with the KuniiLevenspiel model--e.g. list major difference, the advantages, disadvantages, and strong and weak points of each model. (b) Using the correlations suggested in this paper (e.g. Eqn (10)) recalculate the various conversion in Example 12-2. (c) What are the most sensitive parameters in each of the models?


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cdp12-lc.htm

Additional Homework Problems Reconsider example problem 12-2, using the correlations of C. Chavarie and J. R. Grace [Ind. Engrg Chem. Fundamentals, Vol. 14. No. 2 pp. 75-79 (1975)]

CDP12-LC

(a) Calculate the average bed porosity, bubble diameter, frequency of bubble rise, and velocity of bubble rise, (b) Conversion (c) Compare the results with those given in the example problem, Calculate the conversion using the Orcutt, DPPF and DPPM models. (d) What is the effect on conversion if the particle size is increased by a factor of 5 for each of the limit situations discussed in section 126?


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cdp12-ma.htm

Additional Homework Problems If the temperature were increased from 273°K to 546°K, the minimum fluidization velocity of the gas would approximately:

CDP12-M A

(a) Increase by a factor of a) 1.4, b) 2.8, c) 2.0.

(b) Decrease by a factor of a) 1.4, b) 2.8, c) 2.0. (c) Remain the same. (d) Cannot be calculated from the information given. (e) None of the above. Assuming other variables (e.g. pressure, mole fractions, , particle size, etc.) remain virtually unchanged and the Reynolds number is less than 10.


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cdp12-na.htm

Additional Homework Problems In Example 12-3 the specific reaction rate was given as

CDP12-NA

At the very top of the column, what is the rate of reaction in the emulsion phase in [gmole/(cm3 of bubble)(s)] as predicted by the Kunii-Levienspiel model?



cdp12-ob.htm

Additional Homework Problems When a fluidized column is operated at a superficial velocity of 10 cm/sec which is 5 times the minimum fluidization velocity, it was found the bubble size at the mid-point in the column was 10 cm and that bubbles occupied exactly 10.6% of the column.

CDP12-O B

(a)What is the size of the particles in the column? Do not neglect any terms in your calculations. (b)If the cross sectional area of the column is 314.16 cm2, what is the tower height? (c) If the particle density is 2 gm per cm3, the porosity at minimum fluidization conditions 0.5, and the specific reaction rate, kcat= 0.05 sec1 , and K = 5 for an irreversible first R order isomerization reaction, A B, what will be the conversion? If you were unable to find h in part B, take h= 20 cm


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cdp12-pb.htm

Additional Homework Problems CDP12-PB

The irreversible gas phase isomerization reaction

is being carried out in a fluidized bed reactor. Currently, 63.2% conversion is achieved when the bed is operated at 2 times the minimum fluidization velocity. It is proposed to increase production by doubling the volumetric flow rate through the bed while at the same time replacing the catalyst in the bed with one whose particle diameter is one third that currently in the bed. If the proposed change is carried out, what will the new conversion be if the bed height remains fixed if; (a) The reaction rate is limiting. (b) The reaction is transport limited with very few catalyst particles present in the bubble phase. remains virtually You may assume unchanged and DAB is small.



cdp12-qb.htm

Additional Homework Problems CDP12QB

When operated at twice the minimum fluidization velocity 86.5% conversion is realized in a fluidized of 0.5. It is proposed to bed reactor which has a increase the pressure from 1 atm. to 4.0 atm. and decrease the temperature from 427°C to 227°C. Assuming the height to diameter ratio is large, what new column height (relative to the old height) will be needed for a column of the same diameter to achieve the same conversion if the reaction is: (a) Mass Transport limited with very few particles in the bubble phase and very small gas diffusivity. (b) Reaction limited and the activation energy is 19,870 cal/mole. The volumetric flow rate, 0 (hence superficial velocity) is to be the same in both cases.


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cdp12-rb.htm

Additional Homework Problems CDP12-RB

The irreversible catalytic isomerization

is to be carried out in fluidized bed reactor. It is desired to process 8200 liters per hour of feed stock (at 100°C) containing 50% A and 50% inerts and to achieve 90% conversion. The reaction is first order in A. The feed is considered to have the same physical--properties as air.

(a) What is the minimum fluidization velocity? (b) What is the bed porosity at the minimum fluidization velocity um? (c) If you are to operate at 5 times um , what should your bed diameter be? (d) What is the entrainment velocity for this particle size? (e) What is the minimum bubble diameter? (f) What is the maximum bubble diameter? (g) What is the bubble velocity? (h) What is the cloud diameter? (i) What is the fraction of the bed occupied by the bubble? (j) What is the fraction of bed occupied by the wake? (k) What is the velocity of the solids flowing downward? (l) What is the velocity of rise of the gas in the emulsion phase? (m) Use the balances on emulsion and cloud phases to eliminate C AC and C Ae in order to show that the


cdp12-rb.htm

balance on the bubble phase can be written as

(n) Calculate K bc (o) Calculate K ce

(p) Calculate c and e (q) Calculate K R (r) Using the fact that h = tu b calculate the bed height to achieve 90% conversion of A. (s) Calculate the corresponding mass of catalyst. (t) Calculate the bubble diameter at the half way up the reactor and compare this value with your estimate of db. Do you need to redo your calculations? (i) Calculate the mass of catalyst necessary to achieve 90% conversion in 1. a CSTR 2. a packed bed reactor ( =

.20)



cdp12-sb.htm

Additional Homework Problems CDP12-SB

The production of methane from carbon monoxide and hydrogen over a 1/8´ nickel catalyst is carried out at 500°F. The total feed rate to the fluidized bed reactor is 500 lbmoles/hr and is stoichiometric in CO and H2. If the operating velocity is 6 times the minimum fluidization velocity calculate the catalyst weight, bed diameter and bed height necessary to achieve 55% conversion of carbon monoxide. Plot the conversion and species concentration as a function of bed height.


file:///H:/html/12chap/html/cdp12-sb.htm[05/12/2011 16:58:22]

cdp12-tb.htm

Additional Homework Problems CDP12-TB

Using tracer studies to characterize fluidized bed reactors. Originally, it was suggested that a reasonable model for a fluidized bed was a PER and a perfectly-mixed CSTR in series. We know now that such a model is seriously in error. The purpose of this problem is to see how such an interpretation could arise. It is desired to calculate the response of a fluidized bed to cutting off a steady input of one of the components of the fluidizing gas stream (i.e., a negative-step tracer experiment). Assume that you are testing the fluidized-bed reactor of Massimilla and Johnstone, described in Example 12-2. You will use a steady input of air, containing 0.1% helium. The gas flow rate, temperature, and catalyst loading are all exactly the same as in the example. At time t =0, the helium flow is cut off. Calculate and plot the concentration of helium in the outlet gas stream as a function of time. Does the resulting plot compare what would happen if the same test were interpreted as being a PFR-CSTR series system? If the plot you created were interpreted as being a PFRCSTR series system, could any parameters in the hypothetical series system be related to critical parameters in the fluidized-bed system? If so, how would they be related?


file:///H:/html/12chap/html/cdp12-tb.htm[05/12/2011 16:58:22]

P12-UB.htm

Additional Homework Problems The hydrogenation of aromatics in a napthenic lube oil distillate takes place in a trickle bed reactor. The bed is 1 m in diameter and 6 m deep. The solubility of hydrogen in the oil is 0.0068 mol/dm 3 atm. The concentration of aromatics in the feed is 0.07 mol/dm 3. The superficial liquid flow rate is 8 kg/m 2 s, while the hydrogen flow rate is 0.2 kg/m 2 s. The reaction is first-order in hydrogen and in aromatics on the - catalyst pellet. The true specific reaction rate is 2 x 10-5 m 6/kmol kg s. Nitrogen enters with the hydrogen at a rate of 0.05 kg/m 2 s. The reaction occurs at 100°C and 50 atm.

P12U B

(a) Make a plot of conversion of aromatics as a function of bed depth. (b) Calculate the percent resistance of each transport step. (c) How would the exit conversion change if the gas flow rate were reduced by a factor of 4? (d) How could the conversion be increased? Additional data: Effective liquid diffusivity of H2 in oil: 8 x 10-9 m 2/s Effective liquid diffusivity of aromatics in oil: 6 x 10-11 m 2/s Viscosity of oil: 3.0 cP Bed porosity: 0.35 Pressure gradient for flow of 0.25 kg gas/m2 s: 30 kPa/m Density of oil: 900 kg/m 3 Pellet Density: 1500 kg/m 3 Tortuosity: 3.8 Porosity: 0.5 Construction: 0.8

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P12-UB.htm


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P12C-1.htm

Additional Homework Problems P12C-1

Using values of the minimum fluidization velocity provided by Kurian and Raja Rao (Ind. J. of Tech., 8, 275 (1970)), in their table 2 as calculated using equations by Wen and Leva along with data listed in their table 1, calculate corresponding minimum fluidization velocity from the Kunii-Levenspiel model. Comment on accuracy of results.


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Lectures 31 and 32

Lectures 31 and 32

Residence Time Distribution [RTD] (Chapter 13) * Inject a tracer and measure exit concentration, CT(t).

A. RTD

= Fraction of molecules exiting the reactor that have spent a time between (t) and (t + dt) in the reactor. = Fraction of molecules that have spent a time t or less in the reactor. = Fraction of molecules that have spent a time t or greater in the reactor.

file:///H:/html/course/lectures/thirone/index.htm[05/12/2011 16:58:25]

Lectures 31 and 32

Mean Residence Time

Variance

Space Time

Internal Age Distribution, = Fraction of molecules inside the reactor that have been in the reactor between a time

and

.

Life Expectancy = Fraction of molecules inside the reactor with age reactor in a time

to

.

for Ideal Reactors


that are expected to leave the

Lectures 31 and 32

PFR

CSTR

Laminar

B. Models to Calculate the Exit Concentrations and Conversions Zero parameter models One parameter models Two parameter models 1. Segregation Model Models the real reactor as a number of small batch reactors, each spending a different time in the reactor. Mixing occurs at the last possible moment.


Lectures 31 and 32

Little batch reactors (globules) insides a CSTR.

Mixing occurs at the latest possible moment. Mean Conversion

Mean Concentration


Lectures 31 and 32

For multiple reactions:

1. Segregation Model Mixing occurs at the earliest possible moment.

Mixing occurs at the earliest possible moment.


Lectures 31 and 32

Modeling maximum mixedness as a plug flow reactor with side entrances.

To obtain solutions with an ODE solver, let is recorded.


where T is the largest time in which

Lectures 31 and 32

*





Introduction to Attainable Regions

Choosing the best reactor for a given system is not an easy thing to do. It is hard to work out which is the best type of reactor to use. The Attainable Region (AR) method is a technique for chemical reaction engineering that helps us to build and optimize the flowsheet for our reaction system. In order to have the optimum reactor system, we need to know what reactors to use, as well as the best way to connect these reactors together in a flowsheet. For more information on Attainable Regions, please visit the Attainable Regions Homepage at: http://sunsite.wits.ac.za/wits/fac/engineering/procmat/ARHomepage/frame.htm The Attainable Regions Homepage is a creation of Dianne Hildebrandt, David Glasser, Craig McGregor, and Brendon Hausberger from the University of the Witwatersrand in Johannesburg, South Africa.


file:///H:/html/arpages/intro/intro.htm[05/12/2011 16:58:26]

excd13-1.htm

Learning Resources Example CD13-1 RTD Calculations for a Series Reaction

Consider the reaction

occurring in two different reactors with the same mean residence time t m=1.26 min, but with the following residence-time distributions which are quite different: (a) Calculate the conversion predicted by an ideal

1. PFR. 2. CSTR.

(b) Fit a polynomial to each RTD (Figures CDE13-1.1 and CDE13-1.2). Note: These curves are identical to Figures E13-9.1 and E13-9.2 in the text.

Figure CDE13-1.1 E1 (t ): asymmetric distribution.


excd13-1.htm

Figure CDE13-1.2 E2 (t ): bimodal distribution.

(c) For the multiple reaction sequence given above, determine the product distribution in each reactor for 1. The segregation model. 2. The maximum mixedness model.

Solution PFR Combining the mole balance and rate laws for a PFR reactor, we have

(CDE131.1)

(CDE131.2)

(CDE131.3)

The initial conditions are V = 0, C A = 1, C B = C C = 0. In order to compare the performance of the different models and different RTDs, the mean residence time, , was set equal to 1.26 min.

The POLYMATH program used to solve this PFR system is shown in Table CDE13-1.1. The solution is C A = 0.284, C B = 0.357, C C = 0.359, X = 71.6%.

Table CDE13-1.1. Polymath Program for Reactions in a Series in a PFR


excd13-1.htm

CSTR

The mole balances on A, B, and C for a CSTR are

(CDE131.4) (CDE131.5)

(CDE131.6)

The equations can be solved with a nonlinear equation solver. Again, comparison reasons.

was set to 1.26 min for

The POLYMATH program for the CSTR model is shown in Table CDE13-1.2.The solution is C A = 0.443, C B = 0.247, C C = 0.311, X = 55.7%

Table CDE13-1.2. Polymath Program for Reactions in a Series In a CSTR


excd13-1.htm

Segregation Model

Combining the mole balance and rate laws for a constant-volume batch reactor, we have:

For the globules

(CDE131.7)

(CDE131.8)

For the exit concentrations

(CDE131.9) (CDE131.10)

(CDE131.11)

(CDE131.12)

The initial conditions are t = 0, C A = 1, C B = C C = 0 The POLYMATH program used to solve

these equations is shown in Table CDE13-1.3 forE (t) (1) and in Table CDE13-1.4 for E (t) 1 2

(2).The results are listed in Table CDE13-1.5.

Table CDE13-1.3. POLYMATH Program for Segregation Model with Asymmetric RTD


excd13-1.htm

Table CDE13-1.4. POLYMATH Program for Segregation Model with Bimodal Distribution


excd13-1.htm

Maximum Mixedness Model The equations describing the variations in concentrations with position (life expectancy) are

(CDE131.13)

(CDE131.14)

(CDE131.15)

The POLYMATH program is shown in Table CDE13-1.6 for E1(t) (1) and in Table CDE13-1.7 for (2). The results are listed in Table CDE13-1.8.

Table CDE13-1.6. POLYMATH Program for Maximum Mixedness Model with Asymmetric RTD


excd13-1.htm

Table CDE13-1.7. POLYMATH Program for Maximum Mixedness Model with Bimodal Distribution


excd13-1.htm

Summary: Example CD13-1

Additional Tables for Example 13-9 Table CDE13-9.5. POLYMATH Program for Segregation Model with Bimodal Distribution (Multiple Reactions)

Table CDE13-9.6 POLYMATH Program for Maximum Mixedness Model with Asymmetric RTD (Multiple Reactions)


excd13-1.htm



excd13-2.htm

Learning Resources

Example CD13-2 Effect of Variance

If the variance is changed for the asymmetric RTD in Example 13-1, how is the conversion affected?

Solution

Below are three similar RTDs, with the same mean residence time but with different variances. The variances are listed below. Note: E1.1(t) is the same as curve (1) in Example CDE13-2.1.

for the original RTD E1.1(t) is 0.157. for the second RTD E1.2(t) is 0.302.

for the third RTD E1.3(t) is 0.0834.

A comparison of the RTDs is shown in Figure CDE132.1.

Figure CDE13-2.1. E(t) for different variations

The only equations that need to be changed in the file:///H:/html/13chap/html/excd13-2.htm[05/12/2011 16:58:28]

excd13-2.htm

POLYMATH examples above are those for the E(t) curves. They are changed to:

Comparing the conversion and variance of the RTD curves, one can deduce that as the variance increases, the conversion decreases. This result can be explained by analyzing the RTD. With the higher variance curves, more reactants are able to leave the reactor at an earlier time, not allowing the reaction time to occur, thus decreasing the conversion.

The conversions are given below.



13prof.htm

Professional Reference Shelf CD13.1 Attainable Region Analysis

To illustrate attainable reagion analysis (ARA), we will examine a reaction

where the rate law is given by

Substituting for concentration in terms of conversion and for some parameter values for a, b, k, and C A0 , we obtain

We would like to find the highest possible conversion for a given reactor space-time. For this example ARA suggests that the highest conversion will be achieved in a combination of PFRs and CSTRs in series with bypass

If we wish to obtain conversion between points A and B on Figure 13-19 in the smallest space-time, we would use a PFR with feed C A0 For conversions between B and C we would start with a PFR with a feed C A0 and with output at point B. We would then feed some of this material from B into a CSTR that operates at point C and then bypasses the rest around the CSTR. Note that the values of the rates of disappearance of A (-r A) are the same at points B and C. Depending on the amount of bypass, we could achieve any conversion between points B and C. For conversions from point C onward we would first use a PFR with feed C A0 and then feed all the exit material (point B) into a CSTR which operates at point C A0 and finally use this as feed to a subsequent PFR. One could also plot the results above which represent the highest possible conversion for a given space-time (or mean residence time t m) on a graph (Figure 13-20) and label it as the attainable region, AR. At each point on this curve there is an associated RTD. For example the E(t ) between points A and B is just , between points B and C


13prof.htm

Figure 13-19(a) Levenspiel plot

Figure 13-19(b) Model reactor arrangement to achieve maximum (AR) conversion for a given space time

There are two questions one might ask at this stage: For each of these RTDs, what is the conversion that either a segregation model (SM) or maximum mixedness model (MM) with the same space-time will give? For a given space time, , we will have a given RTD. If the space time produces a . This conversion in the attainable region A-B, the RTD will be RTD will be used in the segregation model (SM AR RTD) and the maximum


13prof.htm

mixedness model (MM AR RTD) to calculate the conversion. To obtain the curves in Figure 13-20. After calculating the conversion for a given the space time is increased resulting in a new AR RTD (e.g. it may move into the BC region of Figure 13-19(b)) and the conversion again calculated. The processed is then repeated. The conversions versus space-time achieved by these reactors have been plotted on the same graph as SM (AR RTD) and MM (AR RTD), respectively. Notice that as cautioned above, the best achievable conversions for a given spacetime are not necessarily bounded between the SM and MM. However, we might ask even a broader question. What is the best conversion we can achieve for a given space-time for an SM and MM for all possible RTDs? Answering this, we obtain the curves labeled "Best SM" and "Best MM" on Figure 13-20. The two curves represent the bounds on the performance that can be obtained with SMs and MMs, respectively. Thus it can again be seen even for this less constrained example that the highest conversion for a given space-time does not lie between those of the SM and MM

Figure 13-20 Comparison of the conversion achieved for different models.

To obtain the best SM and MM we simply ask what E(t) will give the maximum conversion for a given space time,

where file:///H:/html/13chap/html/13prof.htm[05/12/2011 16:58:29]

13prof.htm

which is subject to the constraint that is satisfied for the chosen E(t).

For more details on how the curves in Figure 13-20 were obtained, the reader is referred to Glasser, Hildebrandt and Gadorr. We note that the problem above could be solved using the ARA method. In general it is a method that is able to find the "best" reactor for a system of reactions with given kinetics. No other method is able to solve such problems. However, the attainable region analysis (ARA) on the CD-ROM shows that for certain rate law functionalities (cf. Figure 13-18) there may be other reactor systems, with the same RTD, that can give rise to conversions that lie outside the conversion range calculated from the bounds (early vs. late) on mixing.

For more information, consult the article by Glasser and Hildebrandt, and please visit the Attainable Regions Homepage at: http://sunsite.wits.ac.za/wits/fac/engineering/procmat/ARHomepage/frame.htm The Attainable Regions Homepage is a creation of Dianne Hildebrandt, David Glasser, Craig McGregor, and Brendon Hausberger from the University of the Witwatersrand in Johannesburg, South Africa.



13prof2.htm


CD13.2 Comparison Between Segregation Model and Maximum Mixedness Model for Conversion for Reaction Orders Between 0 and 1 In these two models, one has to be able to perceive any macroelement of a fluid composed of numerous microelements. Macroelements have physical dimensions, but microelements don't have any physical dimension (like a point). When we consider mixing within a reactor with respect to the microelements present in the reactor, we can view two extreme levels of mixing. In one extreme, microelements are retained in a single macroelement. In other words, each macroelement is independent of another macroelement and only the microelements within a macroelement mix with each other. This level of micromixing is called complete segregation. The other extreme level of micromixing is when microelements can permeate through the boundary of a macroelement. In other words, microelements from one macroelement can mix with microelements from another macroelement. In this type of mixing all macroelements eventually loose their identity. This level is called maximum mixedness. Hence, we can see that in the segregation model, the different macroelements behave like batch reactors. In the maximum mixedness model, however, this is not the case because microelements of different macroelements can mix together. We have so considered two extreme levels of micromixing. As we discussed in the text, there are also two extreme levels of macromixing. At one extreme, we have an ideal PFR which has no macromixing; at the other extreme, we have a stirred tank which has a maximum degree of macromixing. For a better understanding of mixing in a reactor, let us consider

CD13.2.1 Segregation Model For the two microelements (say 1 and 2) flowing together in the reactor which may or may not be macroscopically well mixed, the segregation model can be pictured as shown in Figure CD13-2. Here, there is no exchange of reactants among macroelements, but there is complete mixing of microelements in each macroelement. Hence, each macroelement can be visualized as a file:///H:/html/13chap/html/13prof2.htm[05/12/2011 16:58:31]

13prof2.htm

well-mixed batch reactor.

Figure CD13-1 Mixing Extremes

Figure CD13-2 Segregation

CD13.2.2 Maximum Mixedness Model The maximum mixedness model can be visualized as shown in Figure CD13-3.


13prof2.htm

Figure CD13-3 Micromixing

There is complete mixing and exchange of microelements among all macroelements. Hence, the boundaries of all macroelements are not defined and we can not distinguish one macroelement from another. The two macroelements, 1 and 2, (as shown in Figure CD13-2 for the segregated model) are completely mixed here. In segregation model, these two macroelements have different ages (i.e. different arriving time to the reactor), hence they have different concentrations. C A1 , C A2 , V 1, and V 2, respectively.

CD13.2.3 First Order Kinetics

For the Segregation Model, we have

Net of Formation = Rate in Macroelement 1 + Rate in Macroelement 2


(CD131)

13prof2.htm

For Maximum Mixedness Model

(CD132)

From Equations (CD13-1) and (CD13-2), we get the same net reaction rate for both models.

CD13.2.4 Second Order Kinetics

For the segregation model, we have Net Rate of Formation = Rate in Macroelement 1 + Rate in Macroelement 2

(CD133)

(CD134)

Now,

From Equation (CD13-3) and (CD13-4), and the above inequality, file:///H:/html/13chap/html/13prof2.htm[05/12/2011 16:58:31]

13prof2.htm

we observe that the segregation model gives greater rate of formation than the maximum mixedness model.

CD13.2.5 Negative First Order Kinetics


(CD135)

For the maximum mixedness model, Total Volume = VT = 2V

(CD136)

Now,

From Equations (CD13-5) and (CD13-6), and the above inequality, we observe that the segregation model gives greater reaction rate than the maximum mixedness model.

CD13.2.6 Half Order Kinetics


13prof2.htm

For the segregation model, we have

Net Rate of Formation = Rate in Macroelement 1 + Rate in Macroelement 2

(CD137)

For the maximum mixedness model, Total Volume = VT = 2V

Now,

(CD138)

From Equations (CD13-7) and (CD13-8), and the above inequality, we observe that the maximum mixedness model gives greater reaction rate than the segregation model.

CD13.2.7 nth Order Kinetics


(CD139)

For the maximum mixedness model,


13prof2.htm

(CD1310)

Now, if the inequality

holds then from (CD13-9) and (CD13-10), we observe that the segregation model gives greater reaction rate than the maximum mixedness model.

CD13.2.8 Mathematical Analysis

Figure 13-4

Let us derive the second order differential of -r A with respect to C A at the point C A= C AT .


(CD1311)

13prof2.htm

Comparing (CD13-10) and (CD13-11), for the segregation model to give greater net reaction rate than the maximum mixedness model for inequality must hold

Similarly, for the maximum mixedness model to give greater net rate of formation

If both of the models give equal net reaction rate, then

CD13.2.9 Conclusions

Consider an nth order reaction

The maximum mixedness model will give greater net reaction rate (or conversion) as compared to segregation model if

as was shown in the analysis of half order reaction kinetics.

Similarly, the segregation model will give greater net rate of formation (or conversion) as compared to the maximum mixedness model if


13prof2.htm

as was shown in the analysis of second order and negative first order reaction kinetics.

as we see in the analysis of first order reaction kinetics. Zero order case is trivial

CD13.2.10A Simple Langmuir-Hinshelwood Kinetics

Hence if the average concentration

then the segregation model will give higher conversion than the maximum mixedness model, otherwise maximum mixedness will give higher conversion. [Prepared with Probjot Singh]

Back



cdp13-ac.htm


(a) Show that E(t) for two CSTRs in series having different volumes is

CDP13-AC

(b) Make a plot of E(t ) as a function of t / for different values of m (e.g., m = 0.1, 0.2, 0.5) for = 5 min. You could also use POLYMATH to solve the unsteady-state concentration balances and generate these curves. (c) Use the segregation model to calculate the conversion of the second-order gas-phase reaction

for which the combined constant, kCA0 = 10, for m = 0.1, 0.3, and 0.5. Pure A enters the reactor. (d) Repeat part (c) for the maximum mixedness model. [2nd Edition]


file:///H:/html/13chap/html/cdp13-ac.htm[05/12/2011 16:58:32]

cdp13-bb.htm

Additional Homework Problems CDP13BB

It is difficult to carry out a pulse tracer test on a particular commercial reactor because of trouble injecting the tracer all at once. The following data were recorded at the inlet and outlet of the reactor. The concentration is given as a relative concentration.

Tracer input:

Tracer output:

(a) Derive the relationship between C input,

C output , and E(t). (b) Compare your answer to part (a) with the equation presented by E. B. Nauman [Chem. Eng. Commun., 8, 60 (1981)]. Determine:

(c) The mean residence time. (d) The variance. (e) The conversion for the elementary gasphase reaction

where A and B are fed in stoichiometric proportions with

using both the segregation model and the maximum mixedness model.


cdp13-bb.htm



cdp13-cb.htm


The flow of non-Newtonian fluids that react chemically are of great industrial interest [Chem. Eng. Sci., 43, 1209 (1988)]. Consider one such fluid, a Bingham plastic in laminar flow in a tubular reactor. This fluid has the following properties:

where U is the velocity, B the viscosity, y the yield stress, and r stress at radius r. For laminar flow of a Bingham plastic in a cylindrical tube, the shear< there will be core region of radius r c near the axis where the axial velocity is uniform,

where y =r/R and yc = r c/R, which can be calculated from the equation

where

(a) Derive equations for E(t) and F(t) for a Bingham plastic flowing in a straight cylindrical reactor. (b) Derive an equation for the mean residence time.

CDP13CB

The average velocity, U, over the cross section is

, the Bingham number.


cdp13-cb.htm © 1999 Prentice-Hall PTR Prentice Hall, Inc. ISBN 0-13-531708-8 Legal Statement


cdp13-db.htm

Additional Homework Problems CDP13DB

The third-order gas-phase reaction

is to be carried out in a nonideal reactor whose macromixing can be characterized by the RTD in Figure 13-12. The space-times corresponding to the figure are CSTR = 50 s and PFR = 50 s. The specific reaction rate is k = 500 dm 6/mol 2 s and the entering concentration of A is 0.02 mol/dm 3. Calculate the conversion assuming that the micromixing is described by: (a) Figure E13-4.1. (b) Figure E13-4.2. (c) Three reactors in series: a PFR (< = 25 s) followed by a CSTR ( = 50 s) and then a PFR ( =25 s). (d) The complete segregation and maximum mixedness models.



cdp13-eb.htm

Additional Homework Problems Review the data by Murphree on the pilot-plant reactor.

CDP13EB

(Figure 13-3) . (a) How many tanks in series are necessary to model the RTD data? (b) What conversion will be predicted by the model for an elementary isomerization with k = 3.1 h-1? (c) What conversion will be predicted by this model for a second-order isomerization with kCA0 = 3.1 h-1? (d) What conversion is predicted by the segregation model for parts (b) and (c)?



cdp13-fa.htm


Gasoline shortages in the United States have produced long lines of motorists at service stations. The table below shows a distribution of the times required to obtain gasoline at 23 Centre County service stations.

(a) What is the average time required? (b) If you were to ask randomly among those people waiting in line, "How long have you been waiting?", what would be the average of their answers?

(c) Can you generalize your results to predict how long you would have to wait to enter a five-story parking garage that has a 4-hour time limit? (R. L. Kabel, Pennsylvania State University)

CDP13-FA


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cdp13-gb.htm


Review Examples 13-6 through 13-10 concerning segregation and maximum mixedness. Use POLYMATH or another software package to rework these problems for the following cases

Vary the parameters to learn their effects on conversion. What is the effect of varying the product kt 0on conversion? Write a paragraph describing you findings

(a) Start with a second-order reaction with kCA0 = 0.1 min -1,

CDP13GB

(b) Example 13-6: Vary t 0 around 8 min. (c) Example 13-7: Vary t 0 around 100 min. (d) Example 13-8: Vary t 0 around 50 min. (e) Example 13-9: Vary t 0around 1 min. (f) Example 13-10: Vary t 0 around 1 min (g) How would your answers to part ____ (to be assigned) change if the reaction were carried out adiabatically with the parameter values given by Equation [P13-2(h).1]? When varying A, B, and t 0, remember that the values chosen for A, B, and t 0must be such that


cdp13-gb.htm



cdp13-hb.htm

Additional Homework Problems CDP13-H B

The reactions described in Problem 6-16 are to be carried out in the reactor whose RTD is described in Example 13-7. However, the reactions are to be carried out at a lower temperature with k1 = 0.007 (kmol/m3)1/2 /s and C T0 = 0.5 kmol/m 3. Determine the exit selectivities (a)Using the segregation model. (b) Using the maximum mixedness model. (c) Compare the selectivities in parts (a) and (b) with those that would be found in an ideal PFR and ideal CSTR in which the space-time is equal to the mean residence time.



Lectures 33 and 34

Lectures 33 and 34

One Parameter Models (Chapter 14) * A. Tanks-In-Series A real reactor modeled as a number of tanks-in-series. The number of tanks necessary, n, is determined from the E(t) curve.

For a first order reaction

For reactions other than first order and for multiple reactions, the sequential equations must be solved:

B. Dispersion

file:///H:/html/course/lectures/thirthre/index.htm[05/12/2011 16:58:37]

Lectures 33 and 34

Determine Per from E(t). For a Closed-Closed Vessel

then t m and Per is found from the relationship:

For an Open-Open Vessel


Lectures 33 and 34

then:

For a First Order Reaction

Two Parameter Models Model the real reactor with combinations of ideal reactors.


Lectures 33 and 34

Two parameters,

and

:

1. Use the tracer data to find

and

.

2. Then use the mole balances and the rate law to solve for CA1 and CA2.

Reactor 1: (Eqn. A) Rate Law: for a first order reaction or

for a second order reaction

Reactor 2: (Eqn. B) Rate Law:

Solve Equations (A) and (B) to obtain C A1 as a function of

, , k,

, and C A0.

Overall Conversion

*


Jump to Lecture(s): 1 & 2 | 3 & 4 | 5 & 6 | 7 & 8 | 9 & 10 11 & 12 | 13 & 14 | 15 & 16 | 17 & 18 | 19 & 20 21 & 22 | 23 & 24 | 25 & 26 | 27 & 28 | 29 & 30


Lectures 33 and 34

31 & 32 | 33 & 34 | 35A | 35B | 36 & 37 | 38 & 39



excd14-1.htm

Learning Resources Example CD14-1 Two CSTRs with Interchange

is carried out in a nonideal CSTR with k = 0.03 min -1. The flow patterns seem to approximate two CSTRs with interchange (Figure CDE14-1.1). Species A enters the reactor at a rate of 25 dm 3/min and a concentration of 0.02 mol/dm 3. The total reactor volume is 1000 dm 3. The results of a pulse tracer test are shown in Table CDE14-1.1 Using the results of these tests, determine the conversion.

The elementary first-order liquid-phase reaction

Figure CDE14-1.1


excd14-1.htm

Solution

A tracer balance yields

(mass added at t = 0) = (mass out over all time)

(CDE14-1.1) (CDE14-1.2)

We could also have evaluated Equation (CDE141.2) by taking the area under the curve of a plot of C( ) versus (Figure CDE14-1.2).


excd14-1.htm

Figure CDE14-1.2 Dimensionless tracer concentration as a function of dimensionless time

We will now determine the decay constants, m1 andm2 , from which the fraction exchanged, , can be determined. The dimensionless concentration is obtained from Equation (CD14-27).

Plotting the ratio C(t) /C 10 as a function of on semilog coordinates, we get the graph shown in Figure CDE14-1.3. At long times, the first term containing m2 in the exponent is negligible with respect to the second term. Consequently, if we extrapolate the portion of the curve for long times back to = 0, we have

(CDE14-1.3)


excd14-1.htm

(CDE14-1.4)

Solving for , we obtain = 0.1. The two parameters for this model are then

Substituting for 18) yields

= 0.8 and = 0.1 = (40 min) (0.03 min -1) = 1.2

, , and

in Equation (CD14-

(CDE14-1.5)

So X = 0.51. For a single ideal CSTR,

(CDE14-1.6) (CDE14-1.7)

For a single ideal plug-flow reactor,

(Xmodel = 0.51)

(XCSTR = 0.55)


(XPFR )= 0.70

(CDE14-1.8)

excd14-1.htm

Figure CDE14-1.3 Dimensionless tracer concentration as a function of dimensionless time



14prof.htm

Professional Reference Shelf CD14.1 Derivation of Equation for Taylor-Aris Dispersion

In a laminar flow reactor we know that the axial velocity varies in the radial direction according to the Hagen-Poiseuille equation:

(13-47)

In arriving at this distribution E(t) it was assumed that there was no transfer of molecules in the radial direction between streamlines. Consequently, with the aid of Equation (13-44), we know that the molecules on the center streamline (r=0) exited the reactor at a time t = / 2, and molecules traveling on the streamline at r = 3R/4 exited the reactor at time

(13-44)

where U is the average velocity. For laminar flow we saw that the RTD function E(t) was given by

The question now arises: What would happen if some of the molecules traveling on the streamline at r = 3R/4 jumped (i.e., diffused) to the streamline at r = 0? The answer is that they would exit sooner than if they had stayed on the streamline at r = 3R/4. Analogously, if some of the molecules from the faster streamline at r = 0 jumped (i.e., diffused) to the streamline at r = 3R/4, they would take a longer time to exit (Figure CD-1). In addition to the molecules diffusing between streamlines, they can also move forward or backward relative to the average fluid velocity by molecular diffusion (Fick's law). With both axial and radial diffusion occurring, the question arises as to what will be the distribution of residence times when molecules are transported between and along streamlines by diffusion. To


14prof.htm

answer this question we will derive an equation for the axial dispersion coefficient, Da, that accounts for the axial and radial diffusion mechanisms. In deriving Da, which is referred to as the Aris-Taylor dispersion coefficient, we closely follow the development given by Brenner and Edwards.1

Figure CD14-1 Radial diffusion in laminar flow

The convective-diffusion equation for solute (e.g., tracer) transport in both the axial and radial direction is

(CD141)

(CD143)

Because we want to know the concentrations and conversions at the exit to the reactor, we are really only interested in the average axial concentration , which is given by

(CD144)

We are going to change the variable in the axial direction z to , which corresponds to an observer moving with the fluid (CD142)

A value of = 0 corresponds to an observer moving with the fluid on the center streamline. Using the chain rule, we obtain

Consequently, we are going to solve Equation (CD14-3) for the solution concentration as a function of r and then substitute the solution c(r, z, t) into Equation (CD14-4) to find (z, t).

To solve the equations above to determine the Aris-Taylor


14prof.htm

dispersion coefficient, we make the following four assumptions: We now apply the approximations above to Equation (CD14-4) to arrive at the following equation:

(CD145)

Because is independent of r, Equation (CD14-5) can be rearranged and integrated with respect to r:

We now apply the approximations above to Equation (CD14-4) to arrive at the following equation:

Symmetry conditions dictate that at r = 0,

, and

therefore the constant of integration, K 1 , is zero.

After dividing both sides by r, a second integration yields

(CD146)

The constant of integration K 2 is independent of r and can be evaluated using the equation for the average concentration:


14prof.htm

(CD147)

The final step to arrive at the celebrated Aris-Taylor dispersion coefficient,D*, is to multiply Equation (CD14-3) by the differential cross-sectional area 2 r dr and carry out the integration over the tubular reactor radius R:

By changing the order of differentiation and integration for the first and last terms and recalling Equation (CD14-4) gives (CD148)

Because there is no transport of solute through the tube wall,

Integrating the first term on the right-hand side of Equation (CD14-8) along with the boundary condition above gives

(CD149)

substitute it into the second term on the left-hand side of

Next we differentiate Equation (CD14-7) with respect to Equation (CD14-8) to yield file:///H:/html/14chap/html/14prof.htm[05/12/2011 16:58:40]

and

14prof.htm

(CD1410)

Carrying out the integration between 0 and 1 (r = R) gives

(CD1410)

Changing our variable

(CD1411)

where D* is the Aris-Taylor dispersion coefficient:

back to z gives

ArisTaylor dispersion coefficient

(CD1412)

That is, for laminar flow in a pipe

Figure 14-5 shows the dispersion coefficient D* in terms of the ratio as a function of the product of the Reynolds and Schmidt numbers. Next



14prof2.htm


CD14.2 Real Reactor Modeled as an Ideal CSTR with an Exchange Volume

In 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 CD14-2). There is considerable material transfer between the two regions. Both inlet and outlet flow channels connect to the highly agitated region. We shall model the highly agitated region as one CSTR, the quieter region as another CSTR, with material transfer between the two. The material balances describing the steady-state behavior of the two reactors are

Figure 14-2 (a) Real reaction systems; (b) model reaction system

(CD14-13) (CD14-14)

Mole balances

The two parameters in this model are the exchange flow rate and the volume of the highly agitated region, V1. Note that the measured volume V is just the sum of V1 and V2.

These two simultaneous algebraic equations can readily be solved for most file:///H:/html/14chap/html/14prof2.htm[05/12/2011 16:58:42]

14prof2.htm

rate laws. We shall first consider the case of a first-order reaction:

(CD14-15)

Let represent that fraction of the total flow that is exchanged between reactors 1 and 2:

(CD14-16)

and let represent that fraction of the total volume V occupied by the highly agitated region:

Two parameters: and

(CD14-17)

With these specifications the balance on reactor 2 becomes

Substituting this value for C A2 into the mole balance on reactor 1, Equation (CD14-13) yields

Then

The space-time is

(CD14-18)

(CD14-19)

Solving for C A2 gives us

(CD14-20)

(CD14-21)

Solving for C A1 we have


(CD14-22)

14prof2.htm

In terms of conversion,

Conversion for the two-CSTR model

(CD14-23)

For large values of the product

,

we observe that the exit concentration is a function of the relative sizes of the reactor volumes but not of the fluid exchanged between the two volumes. This statement will . For small also be true for the case where. values of the product ,

Limiting situations

(CD14-24)

(CD14-25)

which shows the consistency of our equation. If is very small, we should expect no significant amount of reaction. Let's examine the case where the Damköhler number, , has an intermediate value, say = 1; then

We shall soon show how the two parameters may be determined from tracer tests.

Determination of the Parameters

and

and

.

In more complex models, unfortunately, the determination of the parameters in the model is not straightforward, and they usually must be calculated by nonlinear regression using the predicted tracer response curve directly. Exceptions do exist to this general rule, though: Consider, for example, the two-CSTRswithinterchange model described earlier (shown in Figure CD14-3 in simplified form). A mole balance on a tracer


14prof2.htm

pulse injected at t = 0 for each of the tanks is

Unsteadystate balance of inert tracer

Model system

Figure CD14-3 Model system: two CSTRs with interchange

C T1and C T2 are the tracer concentrations in reactors 1 and 2, respectively, with C T10 =N T 0/V1 and C T20= 0. As before,

(CD14-16)

(CD14-17)

Substituting, we arrive at two coupled differential equations describing the unsteady behavior of the tracer that must be solved simultaneously.

(CD14-26)

(CD14-27)

To obtain a solution, we first differentiate Equation (CD14-26) with respect to u and then multiply through


14prof2.htm

by (1 - ) to get (CD14-28)

Substituting Equation (CD14-27) for the bracketed term on the right-hand side gives us

(CD14-29)

The term C T2 in Equation (CD14-29) is eliminated by solving Equation (CD14-26) for

Solution technique commonly encountered in reactor modeling

(CD14-30)

Combining Equations (CD14-29) and (CD14-30) and rearranging, we get

(CD14-31)


For the problem at hand, the initial conditions at = 0 are:

The corresponding solution for the tracer outlet concentration is

the solution to which is


14prof2.htm

(CD14-32)

Plot in CT1 as a function of time find and

When tank 1 is rather small in comparison with tank 2 (small ), and the rates of transfer between the two reactors are small (small ), then during the first portion of the response to a pulse input the second exponential term approximates to 1. During the second portion of the response, the first exponential term approximates to zero. If the logarithm of the tracer concentration is plotted as a function of time, the response curve will approach a straight line at the two ends of the curve, and the parameters may be obtained from the slopes and intercepts of these lines. This concept has been used in physiological systems.2

See Example CD14-1



cdp14-ac.htm

Additional Homework Problems CDP14- Figure CDP14-A shows a combination of ideal AC reactors used to model a real reactor. Each CSTR is

the volume of the PFR.

Figure CDP14-A

(a) Qualitatively sketch the RTD (i.e., E(t ), F (t)) that would result if two of the following combinations of ideal reactors were connected in series (i.e. n = 1). (b) How would your results change if two were connected in series (i.e. n = 2)? (c) How about n = 5 or 10?


file:///H:/html/14chap/html/cdp14-ac.htm[05/12/2011 16:58:43]

cdp14-bb.htm


You have a nonideal batch tank reactor, and your predecessor (who was promoted for his brilliance at modeling reactors) modeled it as two well-mixed batch reactors with transfer between them

CDP14BB

Figure CDP14-B.

You wish to check this model. Your predecessor stated that was 30 ft3 and was 10 ft 3s . The transfer rate was 0.1 ft3/s. At time t = 0, you put 0.1 lb mol of tracer into the region near the agitator (reactor I). If your predecessor was correct, what should be the concentration of tracer in reactor 100 s after you put in the tracer?



cdp14-cc.htm

Additional Homework Problems CDP14 CC

The following data were obtained from a step tracer input to a reactor:

(a) Develop a model that is consistent with the experimental data. (b) Evaluate all the model parameters. For the secondorder reaction

with kCA0 = 0.01 min -1, a reactor volume of 1 m 3, and a volumetric flow rate of 0.01 m 3/min, determine the conversion of A: (c) Using the model developed in part (a). (d) Using the segregation model. (e) Using the maximum mixedness model. Also: (f) What is the Peclet number? (g) How many tanks in series are necessary to model the data above? (h) For a first-order reaction with k = 0.07 min -1, what conversion is predicted by the dispersion model? (i) Calculate the conversion for the first-order reaction for the tanks-in- series model (k = 0.07 min -1).


file:///H:/html/14chap/html/cdp14-cc.htm[05/12/2011 16:58:44]

cdp14-db.htm


A packed-bed reactor is injected with a tracer. The pulse is injected into the bed about 4 particle diameters from the entrance, which is not perfect. The variance in the injection is = 15 s 2. The variance at the measuring point 3 m downstream is = 50 2 s . The superficial velocity is 0.01 m/s. (a) Calculate the dispersion coefficient (b) Calculate the conversion for a first-order reaction in which k = 0.02 s -1.



cdp14-eb.htm

Additional Homework Problems CDP14-EB

The Chemical Technology Division at Oak Ridge National Laboratory has developed the use of fluidized beds as bioreactors (ORNL/MIT 319, 1981). Fluidized beds provide an effective means of promoting contact and hence mass transfer between the fluid and solid phases. The advantages of these features have been demonstrated in biological systems for ethanol production and wastewater treatment. For example, Pseudomonas bacteria adhering to coal particles in a fluidized bed successfully convert wastewater nitrates to nitrogen gas at 20 to 100 times the rate achieved in a stirredtank bioreactor. Although global reaction rates and operating conditions of fluidized bioreactors have been studied thoroughly, an understanding of their flow patterns is needed to allow modeling and scale-up. From the exit concentration data of a typical run, E( ), given below, fit the data with: (a) The axial dispersion model. (b) The tanks-in-series model. (c) A model of your own (Level D).


cdp14-eb.htm



Polymath Short Course, Page 2

Polymath Short Course -- Page 2

2. Common Commands A. Main Menu The first screen you see when you start Polymath is the main menu, which looks like this:

As you can see, Polymath is equipped with four Solvers that you can use. Of these four solvers, only three of them are needed in order to "play around" with the Living Example Problems (LEPs). (The one that is not used, is the Linear Equation Solver.) For each of the remaining solvers, the LEPs are split into library files, one for each of those three solver types. When you choose a given solver, you will automatically have access to the library file that corresponds to that solver. file:///H:/html/toolbox/polymath/short2.htm[05/12/2011 16:58:45]


For example, if you select the Simultaneous Differential Equation Solver (option 1), you will have access to the library file that contains files that have differential equations to solve. You will not have access to files from any of the other solvers' libraries, so you cannot accidentally open up a Living Example Problem using the wrong solver. NOTE: From this point forward, I will demonstrate the commands you will commonly use by showing you how to take a look at a Living Example Problem. In this case, we'll examine Example 4-10 from Chapter 4. As I stated earlier, all of the Living Example Problems are stored in library files that are specific to a given solver. If you have a particular example in mind that you would like to view, then see what the Polymath code for it look like (either by looking at the example itself in the text, or by following the link to a given example file from its corresponding Chapter Outline page on the CD-ROM). For instance, Example 5-6 holds Polymath code for a non-linear regression. The code has data in a tabular format, since you enter regression data in a table. By comparison, Example 4-10 holds Polymath code for the Simultaneous Differential Equation Solver. The code presents differential equations and constants as a list of equations, since that is how the information would be entered for that solver. Since we want to look at Example 4-10 from Chapter 4, and we know we need to open the Simultaneous Differential Equation Solver to look at it, we'll select option 1 from the Polymath main menu by hitting the 1 key. IMPORTANT: If this stuff about looking at the code for an example problem to determine which solver to use has got you confused, then don't worry! The easiest way to figure out which solver you need for a given example is to open a solver and look at the contents of its library file. If you don't see your example listed in that library, then that's not the solver you want--back out to the main menu and start again. (Be careful, though, because some examples were solved using more than one solver, and some examples have multiple parts.) Page 1 | Page 2 | Page 3 | Page 4 | Page 5 Polymath Main | Using Polymath | Installing Polymath | Short Course




2. Common Commands B. Intermediate Screen The next screen looks like this:

Once you become more familiar with Polymath, you might choose to hit RETURN to begin entering a new problem. Since you're following this example, you will hit F9 for file options.



C. Task Menu This will take us to the TASK MENU screen:

Hit F9 again for library options.

D. Library Files This will take us to a listing of the contents of the library file for this solver:



As you can see, Example 4-10 is in this library file (near the bottom of the screen shot). You can use the arrow keys to move the on-screen pointer down to Example 4-10, and then hit L to load that example. When you are told that the example has loaded, press any key (e.g., the spacebar) to look at the example. Page 1 | Page 2 | Page 3 | Page 4 | Page 5 Polymath Main | Using Polymath | Installing Polymath | Short Course




3. "Playing Around" A. Initial Values Okay, let's assume you have successfully loaded Example 4-10. Now let's play around with it a little! Your screen should now look like this:

To solve the problem with its original conditions, you can hit SHIFT-F7 (the SHIFT key and the F7 key, simultaneously). When the solution process is complete, hit any key (e.g., the spacebar) to proceed. The next screen you see will be the Summary Table for the solution results: file:///H:/html/toolbox/polymath/short4.htm[05/12/2011 16:58:47]


Let's say you want to plot the molar flowrate of species A (Fa) as a function of reactor volume (V). Hit g to plot results in graphical form. At the new dialog box that comes up, type Fa/V:



NOTE: You have to type the variable names exactly as they appear in the Summary Table, or else Polymath will gripe at you. Here we have our plot of Fa versus V:



We can also plot the molar flowrates of all species versus V, since Polymath let's us plot up to three dependent variables versus one independent variable at a time:



Page 1 | Page 2 | Page 3 | Page 4 | Page 5 Polymath Main | Using Polymath | Installing Polymath | Short Course



Polymath Short Course -- Page 5 3. "Playing Around" B. What If I Do This . . . ? That's neat, but now what? Well, let's see what happens if we vary the transport coefficient (kc). Since Example 4-10 is an example about Membrane Reactors, we know that if we vary kc (not to be confused with Kc, the equilibrium constant for the example), it should affect the rate of transport of molecules through the membrane. Taking a closer look at Example 4-10 in the text, we see that species B is the only one that can pass through the membrane, so we should see a difference in the molar flowrates of all species leaving the reactor, but especially the molar flowrate of species B (Fb). Hit SHIFT-ENTER (the SHIFT key and the ENTER key, simultaneously) to make changes to the current problem. Let's change the transport coefficient from its initial value of 0.2 min-1 to a new value of 2 min1 . This should increase the amount of species B passing through the membrane. Move the on-screen pointer down to kc and hit ENTER to edit it:



Then hit SHIFT-F7 to solve the problem using this new value for kc. The Summary Table shows that the values of the molar flowrates for species A, B, and C (Fa, Fb, and Fc, respectively) have changed quite a bit (compared to the Summary Table for the initial values), but a plot of these flowrates versus reactor volume (V) makes it a little easier to notice the change:



compare with the earlier plot

That's the end of the Polymath Short Course. Now the rest is up to you. If you run into any problems, you can refer to the Polymath Manual for more information. Have fun! Page 1 | Page 2 | Page 3 | Page 4 | Page 5 Polymath Main | Using Polymath | Installing Polymath | Short Course


file:///H:/html/toolbox/polymath/pics/instpm01.gif[05/12/2011 16:58:48]




















Installing Polymath

Installing Polymath The following files have been included on your CD-ROM in the Polymat4\Files\poly402 directory:

These files will allow you to install Polymath 4.0.2 on your hard drive. Please read the Polymath 4.0.2 Manual (the mpoly402.pdf file) to learn how to use Polymath. The library.zip and the library.exe files are used to install the Living Example Problems, discussed below. In addition to the installation instructions that follow, please read the readme.txt file, since it also contains information about installing Polymath. The installation files for Polymath 4.1 are also on the CD (in the Polymat4\Files\poly41 directory).

Why Two Versions of Polymath? This is the second printing of the text and the CD. Polymath 4.1 has better printing features than Polymath 4.0.2, but it was not available for the first printing of the CD. We could have replaced Polymath 4.0.2 with Polymath 4.1, but we

file:///H:/html/toolbox/polymath/install2.htm[05/12/2011 16:58:48]

Installing Polymath

decided to make both versions available instead.

1. Installing Polymath This is the Polymath 4.0.2 installation procedure. The Polymath 4.1 installation procedure is similar, yet much simpler, so refer to the Polymath 4.1 readme.txt file, if you plan to install Polymath 4.1 instead. Since Polymath is a DOS-based program, directions have been written so that you may use either Windows 95/NT or a DOS Prompt to install Polymath. The installation instructions for DOS are given below. The Windows 95/NT installation instructions are located on a separate page.

A. DOS Procedure 1. If you are already in DOS (or in a DOS window), then move to the next step. Otherwise, open a DOS window by double-clicking on the DOS Prompt. In Windows 95/NT, the DOS Prompt should be available as an icon on your desktop or in your Start menu. 2. Switch to your CD-ROM drive and change to the directory containing the Polymath files. As mentioned above, they should be in the Polymat4\Files\poly402 directory. 3. Type install.exe and hit RETURN to run the Polymath installation program. 4. The installation program will ask you the following questions: a. Enter drive and directory for POLYMATH [C:\POLYMAT4] : Polymath is suggesting that you install to a default directory called Polymat4. Please select this directory by simply hitting the RETURN key. NOTE: It is important for you to install Polymath in the default directory, because you will also have to install the library files for the Living Example Problems in this directory. b. Is this a network installation? [N] Type N when you're asked this question, even if you're running your computer on a network. Trust me. c. Please select the type of output device you prefer [1]: Polymath wants to know if you have a (1) printer or (2) plotter. The default response is that you have a printer, which is what we assume you will select, too. d. Please select the type of printer you have: There should be a long list of printers here. If you don't see your printer in the list, we suggest you select the one that comes closest. NOTE: If you have trouble printing with with Polymath (especially if you print on a network), then contact the authors of Polymath directly. Their contact information should be in the Polymath Manual. e. Please select the mode you want output in [1]: You'll be offered a list of printing options to choose from; the options are similiar, yet different for each file:///H:/html/toolbox/polymath/install2.htm[05/12/2011 16:58:48]

Installing Polymath

printer. We suggest you choose a portrait (or full page) option for printing code or tables, and a landscape option for printing graphs. f. Please select the port your printer is connected to [1]: You'll be offered a list of printing ports to choose from; most people have printers on LPT1. 5. Polymath should now be able to run on your machine. Close the DOS window and open a new DOS window with the DOS Prompt. 6. Switch to the Polymat4 directory on your hard drive. 7. Type polymath at the command prompt and hit RETURN to run Polymath. IMPORTANT: You will need to follow the instructions for installing the library files before you can access the Polymath code for the Living Example Problems. 8. Now read the instructions on using Polymath or take the Polymath Short Course.

2. Installing the Library Files The Polymath code for the Living Example Problems is contained in a file called library.zip, which is located in the Polymat4\Files directory on your CD-ROM. For your convenience, a second file called library.exe is in that same directory. This program will automatically install the Polymath code for the Living Example Problems in the default Polymath directory on your hard drive (C:\Polymat4).

A. DOS Procedure 1. The self-extractor program (library.zip) is Windows-based, so you won't be able to use it in DOS. You will need to unzip library.zip with a DOS-based unzip utility (like PKUnzip). 2. If you use PKUnzip, then extract the contents of the Polymath directory on your hard drive. (Remember, the default Polymath directory location is C:\Polymat4.) Please refer to the documentation for your unzip utility for exact instructions on how to do this. Polymath Main | Using Polymath | Installing Polymath | Short Course

file:///H:/html/toolbox/polymath/install2.htm[05/12/2011 16:58:48]

POLYMATH VERSION 4.02

USER-FRIENDLY NUMERICAL ANALYSIS PROGRAMS - SIMULTANEOUS DIFFERENTIAL EQUATIONS - SIMULTANEOUS ALGEBRAIC EQUATIONS - SIMULTANEOUS LINEAR EQUATIONS - POLYNOMIAL, MULTIPLE LINEAR AND NONLINEAR REGRESSION for IBM and Compatible Personal Computers THIS MANUAL AND SOFTWARE ARE TO ACCOMPANY Elements of Chemical Reaction Engineering - 3rd Edition by H. Scott Fogler and published by Prentice Hall

POLYMATH Internet Site http://www.polymath-software.com Users are encouraged to obtain the latest general information on POLYMATH and its use from the above Internet site. This will include updates on this version and availability of future versions.


i -2 PREFACE

POLYMATH 4.0 PC

POLYMATH LICENSE AGREEMENT The authors of POLYMATH agree to license the POLYMATH materials contained within this disk and the accompanying manual.pdf file to the owner of the Prentice Hall textbook Elements of Chemical Reaction Engineering, Third Edition, by H. Scott Fogler. This license is for noncommercial and educational uses exclusively. Only one copy of this software is to be in use on only one computer or computer terminal at any one time. One copy of the manual may be reproduced in hard copy only for noncommercial educational use of the textbook owner. This individual-use license is for POLYMATH Version 4.02 and applies to the owner of the textbook. Permission to otherwise copy, distribute, modify or otherwise create derivative works of this software is prohibited. Internet distribution is not allowed under any circumstances. This software is provided AS IS, WITHOUT REPRESENTATION AS TO ITS FITNESS FOR AND PURPOSE, AND WITHOUT WARRANTY OF ANY KIND, EITHEREXPRESS OR IMPLIED, including with limitation the implied warranties of merchantability and fitness for a particular purpose. The authors of POLYMATH shall not be liable for any damages, including special, indirect, incidental, or consequential damages, with respect to any claim arising out of or in connection with the use of the software even if users have not been or are hereafter advised of the possibility of such damages. HARDWARE REQUIREMENTS POLYMATH runs on the IBM Personal Computer and most compatibles. A floating-point processor is required. Most graphics boards are automatically supported. The minimum desirable application memory is 450 Kb plus extended memory for large applications. POLYMATH works with PC and MS DOS 3.0 and above. It can also execute as a DOS application under Windows 3.1, Windows 95, and Windows NT. It is important to give POLYMATH as much of the basic 640 Kb memory as possible and up to 1024 Kb of extended memory during installation. A variety of drivers are provided for most printers and plotters. Additionally some standard graphics file formats are supported. POLYMATH 4.0 PC

PREFACE i-3


1 2 2 2 2 2 2 3 3 4 4 5 5 6 6 6 6


2 2 2 2 2

-

1 1 1 2 3


3 3 3 3 3

-

1 2 2 3 3


- 1 - 1 - 1 - 3 - 3 - 5 - 5 - 6 PC


4 - 7 4 - 7 4 - 8 4 - 8 4 - 8 4 - 9 4 - 9 4-10 4-10 4- 10 4-11 4-11


5 5 5 5 5 5 5 5 5 5 5 5 5 5

-

1 1 1 2 3 4 4 5 6 7 8 9 9 9


6 6 6 6 6 6 6 6 6

-

1 1 1 3 5 7 7 7 8

POLYMATH 4.0 PC

PREFACE i-5


7 7 7 7 7


8 8 8 8 8 8 8 8 8 8 8 8 8 8 8

APPENDIX INSTALLATION AND EXECUTION INSTRUCTIONS................... DOS INSTALLATION......................................................................... DOS EXECUTION............................................................................... WINDOWS 3.1 INSTALLATION....................................................... WINDOWS 3.1 EXECUTION.............................................................. WINDOWS 95 INSTALLATION........................................................ WINDOWS 95 EXECUTION............................................................... INSTALLATION QUESTIONS............................................................ "OUT OF ENVIRONMENT SPACE" MESSAGE............................. CHANGING PRINTER SELECTION.................................................. PRINTING FOR ADVANCED USERS................................................ PRINTING TO STANDARD GRAPHICS FILES...............................

9 9 9 9 9 9 9 9 9 9 9 9

i-6 PREFACE

-

1 1 1 3 4

- 1 - 1 - 1 - 1 - 2 - 4 - 5 - 6 - 7 - 7 - 8 -11 -13 -15 -15

-

1 1 1 2 2 3 3 3 7 8 8 9

POLYMATH 4.0 PC


INTRODUCTION 1-1


POLYMATH 4.0 PC







POLYMATH 4.0 PC

INTRODUCTION 1-3


1-4 INTRODUCTION

POLYMATH 4.0 PC



Symbol + * / x.x10a



LOGICAL EXPRESSIONS An "if" function is available during equation entry with the following syntax: if (condition) then (expression) else (expression). The parentheses are required, but spaces are optional. The condition may include the following operators: > greater than < less than >= greater than or equal <= less than or equal == equals <> does not equal | or & and The expressions may be any formula, including another "if" statement. For example: a=if(x>0) then(log(x)) else(0) b=if (TmaxT) then (maxT) else (T)) POLYMATH MESSAGES There are many POLYMATH messages which may provide assistance during problem solving. These messages will tell you what is incorrect and how to correct it. All user inputs, equations and data, are checked for format and syntax upon entry, and feedback is immediate. Correct input is required before proceeding to a problem solution. HARD COPY If there is a printer connected to the computer, hard copy of the problem statements, tabular and graphical results etc. can be made by pressing F3 key wherever this option is indicated on the screen. Complete screen copies can also be made if the DOS graphics command is used before entering POLYMATH and your printer accepts this graphics mode. Problem statements and results can be also printed by saving them on a file and printing this file after leaving POLYMATH. GRAPHICS POLYMATH gives convenient displays during problem entry, modification and solution. Your computer will always operate in a graphics mode while you are executing POLYMATH. 1-6 INTRODUCTION

POLYMATH 4.0 PC

GETTING STARTED This chapter provides information on the hardware requirements and discusses the installation of POLYMATH. HARDWARE REQUIREMENTS POLYMATH runs on IBM compatible personal computers from the original IBM PC XT to the latest models with Pentium chips. Most graphics boards are automatically supported. The minimum application memory requirement is 560Kb, and a new feature uses extended memory when it is available. POLYMATH works with PC and MS DOS 3.0 and above. POLYMATH can be a DOS application under Windows 3.1 and Windows 95. Calculations are very fast since the floating point processor is used when it is available. POLYMATH SOFTWARE The complete set of POLYMATH application programs with a general selection menu is available on a single 3-1/2 inch 1.44 Mb floppy in compressed form. It is recommended that a backup disk be made before attempting to install POLYMATH onto a hard disk. Installation is available via an install program which is executed from any drive. INSTALLATION TO INDIVIDUAL COMPUTERS AND NETWORKS POLYMATH executes best when the software is installed on a hard disk or a network. There is a utility on the POLYMATH distribution disk which is called "install". Detailed installation instructions are found in the Appendix of this manual. Experienced users need to simply put the disk in the floppy drive, typically A or B. Type "install" at the prompt of your floppy drive, and press return. Follow the instructions on the screen to install POLYMATH on the particular drive and directory that you desire. Note that the default drive is "C:" and the default directory is called "POLYMAT4". Network installation will require responses to additional questions during installation. Latest detailed information can be found on the INSTALL. TXT and README.TXT files found on the installation disk. The PRINTSET program which can be used to change the printer specifications without completely reinstalling POLYMATH is discussed in the Appendix. POLYMATH 4.0 PC

GETTING STARTED 2-1



POLYMATH 4.0 PC

EXITING POLYMATH PROGRAM The best way to exit POLYMATH is to follow the instructions on the program display. However, a Shift F10 keypress (⇑F10) will stop the execution of POLYMATH at any point in a program and will return the user to the Polymath Program Selection Menu. THIS ACTION WILL DELETE THE EXISTING PROBLEM so be sure to store your problem as an individual file or in the library before exiting the program in this manner. A query is made to determine if the user really wants to end the program in this manner while losing the current problem. This ⇑ F10 keypress is one of the few POLYMATH commands which is not always indicated in the various Display Menus. It is worth remembering.

POLYMATH 4.0 PC

GETTING STARTED 2-3



3-1 HELP

POLYMATH 4.0 PC




HELP 3-2



3-3 HELP

POLYMATH 4.0 PC




UTILITIES 4-1


4-2 UTILITIES

POLYMATH 4.0 PC

ASSIGNMENT FUNCTIONS The assignment function is a way of storing your results. You may specify a variable name in which to store the results of a computation by first typing in the variable name, then an equals sign, then the expression you wish to store. For example, if you wish to store the value of sin (4/3) 2 in variable 'a', you would enter: a = sin (4/3)**2 Variable names must start with a letter, and can contain letters and digits. There is no limit on the length of the variable names, or on the number of variables you can use. You can then use the variable 'a' in other calculations. These variables are stored only as long as you remain in the current POLYMATH program. Please note that all stored values are lost when the particular program is exited. Calculator information is not retained during problem storage. CALCULATOR EXAMPLES Example 1. In this example the vapor pressure of water at temperatures of 50, 60 and 70 o C has to be calculated using the equation: log10 P = 8.10765 – 1750.29 235.0 + T For T = 50 the following expression should be typed into the calculator: 10^(8.10675 - 1750.29 / (235+50)) CALCULATOR: Enter an expression and press to evaluate it. Press to leave or press F6 for information.


POLYMATH 4.0 PC

UTILITIES 4-3


Where

2 2 a = 27 R Tc Pc 64

b = RTc 8 Pc






4-4 UTILITIES

POLYMATH 4.0 PC







POLYMATH 4.0 PC



deka 10

UTILITIES 4-5



e. hp hr






e. hp hr


4-6 UTILITIES

To units: kilo-cal

POLYMATH 4.0 PC




UTILITIES 4-7




POLYMATH 4.0 PC




POLYMATH 4.0 PC

UTILITIES 4-9





POLYMATH 4.0 PC




A B C 1 0 0 0.74081822 0.19200658 0.067175195 0.54881164 0.24761742 0.20357094 0.40656966 0.24127077 0.35215957 0.30119421 0.21047626 0.48832953 0.22313016 0.17334309 0.60352675 0.16529889 0.13797517 0.69672595 0.12245643 0.10746085 0.77008272 0.090717953 0.082488206 0.82679384 0.067205513 0.062688932 0.87010556 0.049787068 0.047308316 0.90290462


UTILITIES 4-11

PROBLEM OUTPUT AS GRAPHICS FILES Advanced POLYMATH users can direct their printed output to many standard graphics files for direct use in word processing, desktop publishing, etc. This is accomplished though special polymath.bat files which direct any printer output to specified graphics files with user-defined file names. Details of this option are found in Chapter 9 of this manual. A typical example would be to create the following output in a TIFF file for inclusion in a written reporting using word processing or desktop publishing. The problem is again the for Quick Tour Problem 1 from the next chapter. Note that the figure below has utilized the title and axis definition options.

The modified bat file which lead to the above TIFF image was ECHO OFF SET BGIPATH=C:\POLYMAT4\BGI SET PM_WORKPATH=C:\POLYMAT4 SET PM_PRINTER=_TIF,0,FILE:E:\PMOUT+++.TIF POLYMENU CLS and the first tiff image was stored as PMOUT001.TIF in C:\POLYMAT4 directory. See Chapter 9 for more information. 4-12 UTILITIES

POLYMATH 4.0 PC


POLYMATH 4.0 PC




POLYMATH 4.0 PC

ENTERING THE EQUATIONS The equations are entered into POLYMATH by first pressing the "a" option from the Problem Options Menu. The following display gives the first equation as it should be entered at the arrow. (Use the Backspace key, to correct entry errors after using arrow keys to position cursor.) Press the ) to indicate that the equation is to be entered. Don't worry Return key ( if you have entered an incorrect equation, as there will soon be an opportunity to make any needed corrections. d(A)/d(t)=-ka*A_







POLYMATH 4.0 PC



POLYMATH 4.0 PC




A, B, C __



POLYMATH 4.0 PC


B/A






POLYMATH 4.0 PC

TROUBLE SHOOTING SPECIFIC ERROR MESSAGES Most error messages given by POLYMATH are self-explanatory, and they suggest the type of action which should be taken to correct the difficulty. NONSPECIFIC ERROR MESSAGES "Circular dependency detected." This message appears during the inputting of equations when the equations are not all explicit. For example, an attempt to define y=z/x when z has been previously defined will cause this error message to appear. This version of POLYMATH Differential Equations Solver can only solve variables which can be explicitly expressed as a function of other variables. "The expression ... is undefined at the starting point." This common problem can be solved by starting the integration from t=eps where eps is a very small number and t represents the independent problem variable. "Solution process halted due to a lack of memory." This message may result when the default Runge-Kutta-Fehlberg algorithm is used for a stiff system of differential equations, and thus very small step sizes are taken. Consequently, a large number of data points for possible plotting of the results. Use the F10 to stop the integration and switch to the stiff algorithm. If the message persists, then take the following steps to resolve the difficulty: (1) If you are running under Windows, make sure the PIF for POLYMATH specifies 640K of conventional memory and 1024K or more of XMS(but see item (3) below). (2) Remove other memory-resident programs from your computer. (3) Reduce the number of equations. This is most easily accomplished by introducing the numerical values of the constants into the equations, instead of defining them separately. Due to a limitation of some versions of HIMEM, POLYMATH may not be able to access most of the XMS in the number of equations exceeds approximately 24. (The exact number depends on the computer's configuration.) (4) Reduce the integration interval. POLYMATH 4.0 PC



.


POLYMATH 4.0 PC


POLYMATH 4.0 PC





POLYMATH 4.0 PC











POLYMATH 4.0 PC

Press enter (


A shift-return (⇑











POLYMATH 4.0 PC




POLYMATH 4.0 PC


POLYMATH 4.0 PC





POLYMATH 4.0 PC



POLYMATH 4.0 PC



) when the




POLYMATH 4.0 PC


POLYMATH 4.0 PC

REGRESSION 8 -1



8 -2 REGRESSION

POLYMATH 4.0 PC

POLYMATH 4.0 PC

REGRESSION 8 -3




8-4 REGRESSION

POLYMATH 4.0 PC



REGRESSION 8-5



Press integration.



8-6 REGRESSION

POLYMATH 4.0 PC


POLYMATH 4.0 PC

REGRESSION 8-7




POLYMATH 4.0 PC




POLYMATH 4.0 PC

REGRESSION 8-9




8-10 REGRESSION

POLYMATH 4.0 PC



POLYMATH 4.0 PC

REGRESSION 8-11




POLYMATH 4.0 PC





POLYMATH 4.0 PC

REGRESSION 8-13



POLYMATH 4.0 PC


REGRESSION 8-15

XT XA = XT Y where

Y=

y y1 .2 . . y

A=

N

ao a1 . . . an

x0 X=

1 x0 2 x0 N

x1

1 x1 2 x1 N

. . . . . . . . . . . .

xn

1 xn 2 xn N


X=

1

x1,1

x2,1 . . . x n,1

1

x1,2

x2,2 . . . x n,2

1

x1,N x2,N . . . x n,N



POLYMATH 4.0 PC

APPENDIX INSTALLATION FROM THE CRE-99 CD-ROM ACCOMPANYING "ELEMENTS OF CHEMICAL REACTION ENGINEERING" AND EXECUTION INSTRUCTIONS FOR VARIOUS OPERATING SYSTEMS This Appendix provides complete instructions for the installation and execution of POLYMATH for the DOS, Windows 3.1 and Windows 95 operating systems. Detailed information is provided for advanced options. Latest updates are on the README.TXT and INSTALL.TXT files. DOS INSTALLATION Place the CD-ROM in an appropriate drive. This is usually drive D. First set your current directory to that of the CD-ROM and indicate the location of the POLYMATH installation software. For example, if the drive to be used is D, then insert “D:” at the cursor and press Enter. Then enter “cd\html\toolbox\polymath\files”. At the cursor enter “dir” to verify that you have the following POLYMATH installation files in that directory: PMUNZIP.EXE BGI.EXE INSTALL.EXE README.TXT INSTALL.TXT POLYMATH installation is initiated by typing “install” at the cursor the drive and subdirectory containing the above files. For a CD-ROM as drive D, this would be: D:\html\toolbox\polymath\files>install The installation program will ask a number of questions. These are listed later on this file. Please refer to the details of these questions as needed. Follow the instructions on the screen. Note that the initial recommendation for the Printer Output setting is option 1. Half Page, Low Resolution. When the installation is complete, press Enter. DOS EXECUTION 1. Change your directory to the POLYMATH directory by entering "C:" and then "cd\polymat4" at the cursor. Substitute your specified location if you did not use the default directory location. 2. Enter "polymath" at the cursor. 3. Always end POLYMATH by exiting from the Program Selection Menu. POLYMATH 4.0 PC

APPENDIX 9-1

WINDOWS 3.1 INSTALLATION 1. Put the CD-ROM in the appropriate drive such as “D:”. 2. Double click on the Main icon in the Program Manager window. 3. Double click on the MS-DOS Prompt icon. 4. Change the directory to the CD-ROM drive to where POLYMATH is stored. For example, if the drive to be used is D, then insert “D:” at the cursor and press Enter. Then enter “cd\html\toolbox\polymath\files”. 5. At the D:\html\toolbox\polymath\files> prompt, enter “dir”. 6. The following five files should be listed: PMUNZIP.EXE, INSTALL.EXE, INSTALL.TXT, BGI.EXE, and README.TXT. 7. At the D:\html\toolbox\polymath\files> prompt, enter “install” 8. Follow the instructions on the screen. Please note that the initial recommended setting for Output is option 1. Half Page, Low Resolution. 9. At the DOS prompt enter “exit” to return to Windows CONTINUE STEPS BELOW TO CREATE A PIF FILE FOR POLYMATH ONLY IF POLYMATH DOES NOT EXECUTE PROPERLY 10. From the Program Manager Window click on Main. 11. From the Main Window double click on PIF Editor. 12. Please enter the following in the PIF Editor: Program Filename: POLYMATH.BAT Window Title: POLYMATH 4.0 Startup Directory: C:\POLYMAT4 Video Memory: Low Graphics Memory Requirements: -1 -1 EMS Memory: 0 1024 XMS Memory: 1024 1024 Display Usage: Full Screen 13. From File use Save As “POLYMAT4.PIF”. WINDOWS 3.1 EXECUTION* 1. From the File options in the Program Manager Window, select Run. 2. The Command Line for your POLYMATH location should be entered such as “c:\polymat4\polymath”. 3. Click on “OK” 4. If POLYMATH does not run properly, then create a PIF file starting with Step 10 as given above. 5. Always end POLYMATH by exiting from the Program Selection Menu. * A Program Group and a Program Item can be created under Windows to allow POLYMATH to be executed conveniently from the desktop. 9-2 APPENDIX

POLYMATH 4.0 PC

Windows 95 Installation 1. Put the installation disk in your floppy drive. 2. Click on the Start button. 3. Click on the My Computer icon. 4. Double click on the CD-ROM icon indicating cre-99. Then continue double clicking on html, toolbox, polymath, and files. 5. Double click on the Install.exe file 6. Follow the directions on the screen. Please note that the initial recommended setting for Output is option 1. Half Page, Low Resolution. 7. When installation is complete, press Enter. 8. Close the DOS window. Windows 95 Execution* 1. Click on the Start button. 2. Click on the Run icon. 3. Enter “c:\polymat4\polymath” 4. Always end POLYMATH by exiting from the Program Selection Menu. * A Program Group and a Program Item can be created under Windows 95 to allow POLYMATH to be executed conveniently from the desktop. INSTALLATION QUESTIONS 1. Enter drive and directory for POLYMATH [C:\POLYMAT4] : ==> The default response is indicated by the contents of the brackets [...] which is given by pressing Enter key. The full path (drive and directory) where you wish the POLYMATH program files to be stored must be provided here. If the directory does not exist, then the installation procedure will automatically create it. NOTE: Network clients will need read and execute permission for this directory and its subdirectories. This procedure does not provide the needed permissions. 2. Is this a network installation? [N] If you are installing POLYMATH on a stand-alone computer, take the default or enter "N" for no and GO TO 5. on the next page of this manual. If you are installing on any kind of network server, answer "Y" and continue with the installation. POLYMATH 4.0 PC

APPENDIX 9-3

3. What will network clients call [POLYdir]? ===> _ This question will only appear if you answered "Y" to question 2 to indicate a Network installation. Here "POLYdir" is what was provided in question 1. On some networks, the clients "see" server directories under a different name, or as a different disk, than the way the server sees them. This question enables POLYMATH to print by indicating where the printerdriver files are located. They are always placed in subdirectory BGI of the POLYMATH directory by the installation procedure. During run time, they must be accessed by the client machines, thus POLYMATH must know what the client's name is for the directory. 4. Enter drive and directory for temporary print files [C:\TMP]: ===> _ This question will only appear if you answered "Y" to question 2 to indicate a Network installation. Depending on the amount of extended memory available, and the type of printer is use, POLYMATH may need disk workspace in order to print. Since client machines are not normally permitted to write on the server disk, you are requested to enter a directory where files may be written. The temporary print files are automatically deleted when a print is completed or cancelled. 5. The POLYMATH installation program now copies all of the needed files according to your previous instructions. This may take some time as the needed files are compressed on the installation diskette. 6. Please select the type of output device you prefer [1]: _ 1. Printer 2. Plotter ===> _ Your answer here will take you to an appropriate selection menu. 7 A. POLYMATH INSTALLATION - printer selection Please select the type of printer you have: 1. Canon Laser Printer 2. Canon BJ 200 3. Canon BJC 600 4. HP LaserJet II 9-4 APPENDIX

POLYMATH 4.0 PC

5. HP LaserJet III 6. HP LaserJet IV 7. HP LaserJet III or IV (HPGL/2) 8. Epson 9-pin Dot Matrix Printer 9. Epson 24-pin Dot Matrix Printer 10. Epson 9-pin Dot Matrix Printer (color) 11. Epson 24-pin Dot Matrix Printer (color) 12. Epson Stylus 13. Epson Color Stylus 14. IBM Proprinter 15. IBM Proprinter X24 16. IBM Quietwriter 17. Kodak Diconix 18. OkiData Dot Matrix Printer (native mode) 19. PostScript printer 20. Toshiba 24-pin Dot Matrix Printer (native mode) 21. Xerox CP4045/50 Laser Printer (USA) 22. Xerox CP4045/50 Laser Printer (International) 23. HP DesignJet 24. HP DeskJet (Black & White) 25. HP DeskJet 500C (Color) 26. HP DeskJet 550C (Color) 27. HP DeskJet 1200C 28. HP PaintJet 29. HP PaintJet XL300 30. HP ThinkJet ===> _ Type in the number of your printer (or a type of printer that your printer can emulate) and press Enter. 7 B. POLYMATH INSTALLATION - plotter selection Please select the type of plotter you have: 1. HP 7090 Plotter 2. HP 7470 Plotter 3. HP 7475 Plotter 4. HP 7550 Plotter 5. HP 7585 Plotter 6. HP 7595 Plotter 7. Houston Instruments DMP/L Plotters ===> _ POLYMATH 4.0 PC

APPENDIX 9-5

Type in the number of your plotter (or a type of plotter that your plotter can emulate) and press Enter. Once you have selected either a printer or a plotter, you will be asked to select the mode for your output. 8. POLYMATH INSTALLATION - printer/ plotter mode selection Please select the mode you want output in [1]: 1. Half Page, Low Resolution 2. Half Page, Medium Resolution 3. Half Page, High Resolution 4. Landscape, Low Resolution 5. Landscape, Medium Resolution 6. Landscape, High Resolution 7. Portrait, Low Resolution 8. Portrait, Medium Resolution 9. Portrait, High Resolution ===> _ Generally, the fastest printing with adequate resolution is option (1), which is also the default. This choice is recommended. It is easy to change your printer/plotter selection and mode by using the PRINTSET program which is installed in the POLYMATH directory. 9. POLYMATH INSTALLATION - printer/ plotter port selection Please select the port your printer/plotter is connected to [1] : 1. LPT1 2. LPT2 3. COM1 4. COM2 5. LPT3 ===> _ The default port, LPT1, is the first parallel port on most personal computers and usually is connected to the printer. For network installations, these are usually "logical" printer names, but they work just as well. If your network uses other printer logical names, please read the Appendix section entitled "PRINTING FOR ADVANCED USERS" which starts on the next manual page.

9-6 APPENDIX

POLYMATH 4.0 PC

10. POLYMATH Execution The POLYMATH program can now be executed by first changing the directory of your personal computer to the one where the program has been installed. Then you should enter "polymath" at the cursor as shown below and press Enter: C:\POLYMAT4> polymath_ Windows users must use POLYMATH as a DOS program. Advanced users may wish to place the polymath program in the "path" statement of the autoexec.bat file so that POLYMATH is more easily available from any cursor. "OUT OF ENVIRONMENT SPACE" MESSAGE If you receive this message or if you are having difficulty in printing from POLYMATH, then follow ONE of the following instruction set for your particular operating system. Window 95, Windows 98, or Windows NT 4.0 1. Open a DOS prompt window (if yours opens full-screen, hit Alt+Enter to get a window). 2. Click on the "Properties" button at the top. 3. At the end of the Cmd line, add the text "/e:2048" (if you already have this, then change the existing number to a higher number in increments of 1024). Windows 3.x and Windows 95 1. Open a DOS prompt window. 2. At the C:\ prompt type "CD/WINDOWS, and press Enter. 3. Type "EDIT SYSTEM.INI" and press Enter. 4. Locate a line that reads "[NonWindows App]". 5. Make sure that this section contains the following entry: "CommandEnvSize=2048". 6. Save the modified file. 7. Reboot your computer. 8. Adjust the size upwards in increments of 1024 as necessary. DOS or Windows 3.1 1. Open a DOS prompt window. 2. Type "SET" at the prompt. POLYMATH 4.0 PC

APPENDIX 9-7

3. Look for the the line that displays the value of the COMSPEC environment variable. If COMSPEC is set to C:\COMMAND.COM then add the following line to the CONFIG.SYS file: SHELL=C:\COMMAND.COM C:\ /E:2048 /P If COMSPEC is set to C:\DOS\COMMAND.COM then add this line to the CONFIG.SYS file: SHELL=C:\DOS\COMMAND.COM C:\DOS /E:2048 /P 4. Reboot your computer. 5. Adjust the size of E: in Step 3 upwards in increments of 1024 as necessary. CHANGING PRINTER SELECTION The printer selection may be changed without a complete reinstallation of POLYMATH by using the PRINTSET utility program which is stored on the directory containing POLYMATH. This program is executed by entering "printset" while in the POLYMATH directory. If there are several printers to be used with POLYMATH, then different POLYMATH.BAT files can be created for each printer. The batch file for printers is discussed in the following section. PRINTING FOR ADVANCED USERS POLYMATH printing is accomplished using GRAF/DRIVE PLUS which is a trademark for copyrighted software from Fleming Software. The printer setup is controlled by three environmental variables which are set in the POLYMATH.BAT file. This file is created by the INSTALL procedure, and it can be subsequently be modified by the PRINTSET utility or with an editor. CAUTION: This file should only be altered with an editor by advanced users. A typical POLYMATH.BAT file is shown below: ECHO OFF SET BGIPATH=C:\POLYMAT4\BGI SET PM_WORKPATH=C:\POLYMAT4 SET PM_PRINTER=_LJ3R,0,LPT1 POLYMENU CLS 9-8 APPENDIX

POLYMATH 4.0 PC

In this file, the three environmental variables are: BGIPATH - This is the directory in which the printer driver "BGI" files for POLYMATH reside. It is usually the BGI subdirectory of the POLYMATH directory. This variable is not normally changed. PM_WORKPATH - This is the directory in which temporary print files are stored. PM_PRINTER - This variable has the following form: ,, This variable in the previous POLYMATH.BAT file is defined as: PM_PRINTER=_LJ3R,0,LPT1 where _LJ3R indicates the printer_type as HP LaserJet III, 0 (the number zero) indicates the page_format as half page with low density print, and LPT1 indicates the printer_port as LPT1. PM_PRINTER Options for and A detailed listing of the and options is given in Table 1 at the end of this Appendix. PM_PRINTER Options for The may be any of the following: LPT1, LPT2, LPT3, COM1, COM2 or other physical/logical device names. PRINTING TO STANDARD GRAPHICS FILES It is possible to have all output which is "printed" by POLYMATH to be saved as various graphics files for use in word processing, desktop publishing, etc. This involves specialized use of the PM_PRINTER variable which is not available during the INSTALL procedure or the PRINTSET program for printer modification. All printing to graphics files requires the creation of a special batch file for this purpose. POLYMATH 4.0 PC

APPENDIX 9-9

A typical file for this purpose which is arbitrarily called POLYGRAP.BAT is shown below: ECHO OFF SET BGIPATH=C:\POLYMAT4\BGI SET PM_WORKPATH=C:\POLYMAT4 SET PM_PRINTER=_TIF,0,FILE:C:\GRAPHS\PMOUT+++.TIF POLYMENU CLS In the above file, the PM_PRINTER variable _TIF indicates the printer_type as a Tagged Image Format (TIFF), 0 indicates low resolution with two colors, and FILE:C:\GRAPHS\PMOUT+++.TIF indicates the location and name of the resulting graphics file. PM_PRINTER Options for and for Graphics Files A detailed listing of the and options for graphics file output is given in Table 2 at the end of this Appendix. PM_PRINTER Options for for Graphics Files This variable can be defined as FILE: when the output is to be to a file and not a port. (Note that if is a logical device name, such as LPT3, the output will be printed.) may be a particular filename or a general template, including '+' signs where a sequence number is to be written. The above example will create a sequence of files named PMOUT001.TIF, PMOUT002.TIF, etc., stored in directory C:\GRAPHS .

POLYMATH 4.0 PC

APPENDIX 9-10




Epson-compatible 9-pin Dot Matrix, and IBM Proprinter _FX

0

HalfLo

5

LandHi

1

HalfMed

6

FullLo

2

HalfHi

7

FullMed

3

LandLo

8

FullHi

4

LandMed

Epson-compatible 24-pin Dot Matrix _LQ

0

HalfLo

5

LandHi

1

HalfMed

6

FullLo

2

HalfHi

7

FullMed

3

LandLo

8

FullHi

4

LandMed

Epson-compatible 9-pin Dot Matrix Printer (color) _CFX

0

HalfLoC

3

LandMedC

1

HalfMedC

4

FullLoC

2

LandLoC

5

FullMedC

Epson-compatible 24-pin Dot Matrix Printer (color) _CLQ

9-11 APPENDIX

0

HalfLoC

3

LandMedC

1

HalfMedC

4

FullLoC

2

LandLoC

5

FullMedC

POLYMATH 4.0 PC




0

HalfLo

5

LandHi

1

HalfMed

6

FullLo

2

HalfHi

7

FullMed

3

LandLo

8

FullHi

4

LandMed

0

HalfLo

5

LandHi

1

HalfMed

6

FullLo

2

HalfHi

7

FullMed

3

LandLo

8

FullHi

4

LandMed

2

Full

2

Full

IBM Proprinter X24 _PP24

IBM Quietwriter _IBMQ

OkiData Dot Matrix Printer (native mode) _OKI92

0

Half

1

Land

Toshiba 24-pin Dot Matrix Printer (native mode) _TSH

0

Half

1

Land

POLYMATH 4.0 PC

APPENDIX 9-12




LaserJet II, LaserJet III, LaserJet IV, DeskJet (black cartridge), Canon laser _LJ _LJ3R _LJ4 _DJ _Canon

0

HalfLo

5

LandHi

1

HalfMed

6

FullLo

2

HalfHi

7

FullMed

3

LandLo

8

FullHi

4

LandMed

DeskJet 500C (color cartridge), and DeskJet 550C _DJC _DJC550

0

HalfLoC8

5

LandHiC8

1

HalfMedC8

6

FullLoC8

2

HalfHiC8

7

FullMedC8

3

LandLoC8

8

FullHiC8

4

LandMedC8

0

HalfLoC2

8

FullLoC8

1

LandLoC2

9

HalfHiC8

2

FullLoC2

10

LandHiC8

3

HalfHiC2

11

FullHiC8

4

LandHiC2

12

HalfLoC16

5

FullHiC2

13

LandLoC16

6

HalfLoC8

14

FullLoC16

7

LandLoC8

PaintJet _PJ

9-13 APPENDIX

POLYMATH 4.0 PC




0

Half

6

HalfC16

1

Land

7

LandC16

2

Full

8

FullC16

3

HalfGR

9

HalfC256

4

LandGR

10

LandC256

5

FullGR

11

FullC256

0

DraftPL

2

DraftPLB

1

LQPL

3

LQPLB

_HP7470

0

DraftPL

1

LQPL

_HP7475 _HP7550

0

DraftPL

4

DraftPLr

1

LQPL

5

LQPLr

2

DraftPLB

6

DraftPLBr

3

LQPLB

7

LQPLBr

0

DraftPL

5

LQPLC

1

LQPL

6

DraftPLD

2

DraftPLB

7

LQPLD

3

LQPLB

8

DraftPLE

4

DraftPLC

9

LQPLE

PostScript Printers _PS

Hewlett-Packard Plotters _HP7090

_HP7585 _HP7595

POLYMATH 4.0 PC

APPENDIX 9-14




0

HalfLo

5

LandHi

1

HalfMed

6

FullLo

2

HalfHi

7

FullMed

3

LandLo

8

FullHi

4

LandMed

Canon BJ 200, and Epson Stylus _BJ200 _ESCP2

Canon BJC600, and Epson Color Stylus _BJC600 _ESCP2C

0

HalfLoC8

5

LandHiC8

1

HalfMedC8

6

FullLoC8

2

HalfHiC8

7

FullMedC8

3

LandLoC8

8

FullHiC8

4

LandMedC8

HPGL/2 (DeskJet 1200C, PaintJet XL300, and DesignJet) _HGL2

9-15 APPENDIX

0

HalfC2

5

FullC8

1

LandC2

6

HalfC256

2

FullC2

7

LandC256

3

HalfC8

8

FullC256

4

LandC8

POLYMATH 4.0 PC




0

LoResC16

3

LoResC2

1

MedResC16

4

MedResC2

2

HiResC16

5

HiResC2

0

ColorMode

1

MonoMode

0

LoResC2

2

HiResC2

1

MedResC2

Zsoft PCX _PCX

Windows BMP _BMP GEM IMG _IMG

Tagged Image Format (TIFF) compressed and uncompressed _TIF _UTIF

0

LoResC2

1

MedResC2

2

HiResC2

1

MonoMode

1

MonoMode

Computer Graphics Metafile (ANSI) _CGM

0

ColorMode

0

ColorMode

AutoCad DXF _DXF

Video Show (ANSI NAPLPS) _VSHO

0

ColorMode

Word Perfect Graphics _WPG

0

POLYMATH 4.0 PC

ColorMode

APPENDIX 9-16




0

ColorMode

1

MonoMode

ColorMode

1

MonoMode

ColorMode

1

MonoMode

Windows Metafile _WMF

Adobe Illustrator PostScript _AI

0

Color QuickDraw (PICT) _PCT

9-17 APPENDIX

0

POLYMATH 4.0 PC

Example 5-6 Nonlinear Regression 100 29 RATE Pe Pea PH2 1.04 1 1 1 3.13 1 1 3 5.21 1 1 5 3.82 3 1 3 4.19 5 1 3 2.391 0.5 1 3 3.867 0.5 0.5 5 2.199 0.5 3 3 0.75 0.5 5 1 * RATE=k*Pe**a*PH2**b k=1 a=1 b=1

file:///H:/html/toolbox/polymath/examples/ch5/e5-6[05/12/2011 16:58:51]

Example 4-10 d(Fa)/d(V)=ra d(Fb)/d(V)=-ra-kc*Cto*(Fb/Ft) d(Fc)/d(V)=-ra kc=0.2 Cto=0.2 Ft=Fa+Fb+Fc k=0.7 Kc=0.05 ra=-k*Cto*((Fa/Ft)-Cto/Kc*(Fb/Ft)*(Fc/Ft)) V(0)=0 Fa(0)=10 Fb(0)=0 Fc(0)=0 V(f)=500

file:///H:/html/toolbox/polymath/examples/ch4/e4-10[05/12/2011 16:58:52]

file:///H:/html/toolbox/polymath/pics/poly1.gif[05/12/2011 16:58:52]























Lectures 1 and 2 - More Examples

Lectures 1 and 2 Other Rate Laws

Back to Lectures 1 and 2

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Lectures 1 and 2 - Self Test

Lectures 1 and 2 Self Test At a particular time t, the rate of formation of B in the reaction, r B, is 10 mole/dm3 *min. Which of the following are true? a. The rate of disappearance of B is -10 moles/dm 3 *min. b. The rate of formation of A is -10 mole/dm3 *min. c. The rate of disappearance of A is 10 moles/dm 3 *min. d. r A = -10 moles/dm 3 *min e. -r A = 10 moles/dm 3 *min f. -r B = -10 moles/dm 3 *min g. All of the above h. None of the above Solution

Solution g. (all are true) Back to Lectures 1 and 2

file:///H:/html/course/lectures/one/test1.htm[05/12/2011 16:58:54]


Lectures 1 and 2 Self Test Calculate the time to reduce the number of moles by a factor of 10 in a batch reactor for the reaction with -r A = kCA , when k = 0.046 min-1 Solution

Solution



Therefore, t = 50 minutes Back to Lectures 1 and 2


Lectures 1 and 2

Lectures 1 and 2 Derivation of Batch Reactor Design Equations



Lectures 1 and 2

Lectures 1 and 2 Derivation of PFR Reactor Design Equations



Lectures 1 and 2

Lectures 1 and 2 Numerical Evaulation of Integrals FAo = 2 mol/s, fed to a plug flow reactor

To reach 80% conversion, your PFR must be 293.3 dm 3 . Back to Lectures 1 and 2


Lectures 1 and 2

Lectures 1 and 2 Reactors in Series



Thoughts on Problem Solving: Chemical Reaction Engineering

Please note that the use of these examples requires an understanding of the material from Chapter 4 of Elements of Chemical Reaction Engineering, 3rd Edition. PFR/CSTR Example Sample Registration Exam Problem - this problem was taken off a Professional Engineer's Exam

file:///H:/html/probsolv/closed/alg/cre-exam/cre.htm[05/12/2011 16:59:04]

Industrial Reactor Photos

Industrial Reactor Photos There are a wide variety of reactors in use in the chemical processing industry. The list of photos (below) is just a small sampling of these reactors, and we have received permission from the respective companies to show them to you here. Reactor System [Amoco] Spherical Ultraforming Unit [Amoco] Three Spherical Ultraforming Units [Amoco] Spherical Ultraforming Unit in Series [Amoco] Hydrotreating Unit [Amoco] CSTR (cutaway view) [Pfaudler, Inc.] Batch Reactor [Pfaudler, Inc.] Batch Reactor Stirring Apparatus [Pfaudler, Inc.] file:///H:/html/01chap/html/reactors/photos.htm[05/12/2011 16:59:05]


We have received a number of photos from SASOL, Ltd. and British Petroleum (BP), so we made separate pages for them. SASOL Reactors BP Reactors

Reactor System Used at Amoco

file:///H:/html/01chap/html/reactors/photos.htm[05/12/2011 16:59:05]


Spherical Reactor

Three Spherical Reactors



Spherical Reactors Connected in Series

Hydrotreating Unit file:///H:/html/01chap/html/reactors/photos.htm[05/12/2011 16:59:05]


Cutaway View of CSTR



Batch Reactor



Batch Reactor Stirring Apparatus




CRE -- Reactors Module, Page One

Reactors Module The Reactors Module covers a wide variety of reactor types, from CSTRS and PFRs (commonly encountered in homework problems), to Moving-Bed and Fixed Film reactors. You have the opportunity to navigate through descriptions of each of these reactor types:

For example, if you wanted to learn more about CSTRs, you would click on the button for CSTRs, and then you would be presented with the following options:

Advance to page two.

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CRE -- Reactors Module, Page One


file:///H:/html/01chap/html/reactors/module.htm[05/12/2011 16:59:06]

Aspen Plus - Introduction

Introduction to the Chemical Engineering Student Engineers are constantly being called upon to predict the behavior of systems. Chemical engineers in particular must be able to predict the actions of chemical species, often a very difficult task. As chemical engineering students, when confronted with a large chemical system, you might ask, "Where do I even begin? Mass balances? Energy balances? Thermodynamic properties? Reaction Kinetics?" Over the past few years as a student you've learned about each of these crucial topics separately, however, "real world" situations will require an engineer to incorporate all of these areas. This is where the idea of a process model is helpful. A process model can be defined as an engineering system's description or "blue print." The process model is a complete layout of the engineering system including the following: 1. Flowsheet. The process model flowsheet maps out the entire system. The flowsheet shows one or more inlet streams entering into the system's first unit operation (i.e., heat exchanger, compressor, reactor, distillation column, etc.) and continues through the process, illustrating all intermediate unit operations and the interconnecting streams. The flowsheet also indicates all product streams. Each stream and unit operation is labeled and identified. 2. Chemical Components. The process model specifies all chemical components of the system from the necessary reactants and products, to steam and cooling water. 3. Operating Conditions. All unit operations in the process model are kept under particular operating conditions (i.e., temperature, pressure, size). These are usually at the discretion of the engineer, for it is the operating conditions of the process that effect the outcome of the system. Sound confusing? Well can you imagine keeping track of all of this by hand, then solving all the mass and energy balances, determining thermodynamic behavior, and using reaction kinetics just to determine what size reactor to use, or how much product you'll achieve? ASPEN PLUS TM allows you to create your own process model, starting with the flowsheet, then specifying the chemical components and operating conditions. ASPEN PLUS TM will take all of your specifications and, with a click of the mouse button, simulate the model. The process simulation is the action that executes all necessary calculations needed to solve the outcome of the system, hence predicting its behavior. When the calculations are complete, ASPEN PLUS TM lists the results, stream by stream and unit by unit, so you can observe what happened to the chemical species of your process model. So continue on and check out how to use ASPEN PLUS TM as a tool in solving some reaction engineering problems. I. Introduction II. Accessing ASPEN PLUSTM

file:///H:/html/toolbox/aspen/intro.htm[05/12/2011 16:59:07]

Aspen Plus - Introduction

III. Creating a Reaction Engineering Process Model A. Building a Process Flowsheet B. Entering Process Conditions IV. Running the Process Model A. Interpreting the Results B. Changing Process Conditions and Rerunning the Model V. Example Problems A. 8-6: Adiabatic Production of Acetic Anhydride B. 8-7: Operation of a PFR with Heat Exchanger VI. Other Need-to-Knows A. Saving your Process Model B. Printing your Process Model C. Changing Names of Streams and Unit Operations D. Changing Units of Parameters E. Exiting ASPEN PLUSTM VII. Credits BACK to ASPEN PLUS TM MAIN PAGE aspen plus > introduction

file:///H:/html/toolbox/aspen/intro.htm[05/12/2011 16:59:07]

Aspen Plus - Accessing

Accessing ASPEN PLUSTM At the University of Michigan, Ann Arbor ASPEN PLUS TM is installed on an on-line server directory which can be accessed from a UNIX workstation. Log on to a UNIX Ultra Sparc® workstation, 3rd floor Media Union. See the CAEN office, 2nd floor Media Union, for other locations of these machines. At Other Universities Inquire at your computing sites about the ASPEN PLUS TM software package. It may be installed on an on-line directory, or on an individual computer's hard-drive. Once you are logged on to a UNIX Ultra Sparc®, follow these instructions to register for ASPEN PLUS TM : 1. At the % prompt, type: swselect. A new window will appear. 2. With the mouse pointer, highlight Aspen Plus from the left column of packages. Click the Select Package arrow. 3. A window will appear asking for the version. Double click on 9.2-1. Aspen Plus will now appear in the right column. 4. Log off the workstation or type source ~/.software at the % prompt. In steps 1-4, you gain access to ASPEN PLUS TM under your login id. After selecting the package, you no longer have to do steps 1 - 4 each time you log on to a UNIX workstation to work on ASPEN PLUS TM . Simply start at step 5 and continue through step 9. 5. Log back onto the workstation. At the % prompt, type: aplus 6. The ASPEN PLUS TM flowsheet window will appear. A series of little "start up" windows will appear, one after the other. 7. At the window asking about connecting to the on-line directory, click OK. 8. A window that reads Connection Established will appear. Click OK. 9. Select Create a New Run from the next window to appear. 10. Click OK on English Units when the next window appears (you can always change the units later). Lastly, click on Flowsheet Simulation for the run type. Once the start up windows are completed, you'll be presented with an empty flowsheet window titled "UNNAMED" like the one below. You're now ready to begin creating your process model.

file:///H:/html/toolbox/aspen/access.htm[05/12/2011 16:59:07]

Aspen Plus - Accessing

I. Introduction II. Accessing ASPEN PLUSTM III. Creating a Reaction Engineering Process Model A. Building a Process Flowsheet B. Entering Process Conditions IV. Running the Process Model A. Interpreting the Results B. Changing Process Conditions and Rerunning the Model V. Example Problems A. 8-6: Adiabatic Production of Acetic Anhydride B. 8-7: Operation of a PFR with Heat Exchanger VI. Other Need-to-Knows A. Saving your Process Model B. Printing your Process Model C. Changing Names of Streams and Unit Operations D. Changing Units of Parameters E. Exiting ASPEN PLUSTM VII. Credits BACK to ASPEN PLUS TM MAIN PAGE aspen plus > accessing

file:///H:/html/toolbox/aspen/access.htm[05/12/2011 16:59:07]

Aspen Plus - Creating

Creating a Reaction Engineering Process Model Now that you have gained access to ASPEN PLUS TM , you are ready to begin creating a process model. The following series of steps will create a process model for the tubular reactor (PFR) example problem 4-4 taken from the 3rd Edition of Elements of Chemical Reaction Engineering by H. Scott Fogler. Here is a summarized version of the problem: Example 4-4 Problem Statement Determine the plug-flow reactor volume necessary to produce 300 million pounds of ethylene a year from cracking a feed stream of pure ethane. The reaction is irreversible and elementary. We want to achieve 80% conversion of ethane, operating the reactor isothermally at 1100K at a pressure of 6 atm. C2 H6 (g)

C2 H4 (g) + H2 (g) A

B+C

Where A is gaseous ethane, B is gaseous ethylene, and C is gaseous hydrogen. Other information: Fao = 0.425 lbmol/s (calculated from 300 million pounds of ethylene at 80% conversion) k = 0.072s -1 at 1000K Activation Energy, E = 82 kcal/gmol Building the Process Flowsheet The first step in creating a process model is drawing the flowsheet in ASPEN PLUS TM , much like you would on paper. Note that while you're constructing the flowsheet, a red text prompt in the upper left corner under the main tool bar will state "Flowsheet Not Complete." This will change to a black text "Flowsheet Complete" prompt when the flowsheet is finished. Inlet (Feed) Streams To draw the inlet stream do the following steps 1-5. 1. First, you must draw the inlet stream (ethane in this case). On the right side of the flowsheet window is the drawing tool bar. Under Type highlight Feed/Prod with the left mouse button. 2. Under Model highlight Feed. 3. Under Icon highlight Material. 4. Move the mouse pointer (which now looks like cross-hairs) to the middle left of the empty flowsheet window. 5. Click the left mouse button to place the inlet stream on the flowsheet. The inlet stream is file:///H:/html/toolbox/aspen/creating.htm[05/12/2011 16:59:08]


represented by an outline of an arrow pointing to the right. You have finished adding the inlet arrow. Note this example only has one arrow (representing the ethane feed). More than one inlet stream can be drawn, each representing different reactants. Your screen should look like this:

Outlet (Product) Streams You are now ready to add the outlet stream (containing both ethylene and hydrogen) to your flowsheet. Repeat the steps described above, however under 'Model' highlight 'Prod'. Place the product arrow about 2-3 inches to the right of the inlet arrow on the flowsheet. Of course, for other examples, there could be more than one outlet stream. Your screen should now look like this:

file:///H:/html/toolbox/aspen/creating.htm[05/12/2011 16:59:08]


Unit Operations Now that inlet and outlet streams are placed on the flowsheet, you are ready to place the unit operation (the PFR) on the flowsheet. The steps are similar to placing the streams on the flowsheet. 1. Under Type highlight Reactor. 2. Under Model highlight RPLUG (reactor plugflow). 3. Under Icon highlight either Block, Icon1,or Icon2. The only difference is the way the PFR will look on the flowsheet. The display box above Type will give you a preview of the PFR icon before you place it on the flowsheet . 4. Move the mouse pointer (which now looks like cross-hairs) in between the inlet and outlet arrows. Click the left mouse button to place the PFR on the flowsheet. The PFR is represented by whichever icon you chose. It will arbitrarily be named B1 (you can change the name later). Your flowsheet should now look like this:



Connecting the Streams to the Unit Operation Once the inlet arrow, PFR, and outlet arrow are on the flowsheet, it is time to connect the flow streams to the PFR. Inlet to PFR 1. With the left mouse button, click on the inlet arrow so it is highlighted by four black dots. 2. Now with the left mouse button, double click on the highlighted inlet arrow. The feed port, represented by a small arrow, will appear on the inlet arrow. Notice that when you place the mouse pointer on the feed port, it automatically highlights the small arrow. 3. With the left mouse button, double click on the feed port arrow. Notice how port arrows appear on the PFR. 4. To connect the inlet arrow to the PFR, move the mouse pointer to the port arrow called FEED on the PFR and click the left mouse button. Note: When you move the mouse pointer over the PFR port arrows, their names will appear. A line show now connect the inlet arrow to the PFR. The line should be labeled with a 1 (you can change the name of the stream later). This line represents the ethane inlet stream. Your screen should look like this:



PFR to Outlet 1. With the left mouse button, click on the PFR so it is highlighted by four black dots. 2. Now with the left mouse button, double click on the highlighted PFR. Various ports represented by small arrows appear on the PFR. 3. With the left mouse button, double click on the product port arrow on the PFR. Notice how the outlet port arrow appears on the outlet arrow to the right of the PFR. 4. To connect the outlet arrow to the PFR, move the mouse pointer to the outlet port arrow called PROD on the outlet arrow and click the left mouse button. A line should now connect the outlet arrow to the PFR. The line should be labeled with a 2 (you can change the name of the stream later). This line represents the ethylene and hydrogen product stream. Your screen should look like this:



Entering Process Conditions Now that the process flowsheet is complete, it is time to enter the process conditions. ASPEN PLUS TM will guide you through the required input windows, simply click on the Next button on the upper right of the main toolbar. When each input window is complete, the prompt "# Complete" will appear in the upper left corner of the screen. 1. Click Next with the left mouse button. A window will prompt you that the flowsheet is complete and asks whether the next input form be displayed. Click OK. 2. The first input window will be called Setup.Main. With the left mouse button, click once on the Title box to highlight it, enter the title of your process model. 3. Check that the desired units are correct. Change the units by clicking with the right mouse button on the unit box to pull down a unit menu. Click on the desired units (you can always change the name later while entering the stream properties). 4. # Complete should be stated in the upper left corner. Click Next. The next input window is Components.Main. Here is where all of the chemical species in your process model are specified. 5. For this particular example, the components are: ethane, ethylene, and hydrogen. Start with the right most column called Component Name. Click on the first row in the column and type in: ethane. Hit Enter. ***Note: If you are unsure of how to spell the chemical name, or do not know whether it's in the ASPEN PLUS TM library, simply click (with the right mouse button) where you normally type in the chemical name to pull down a list of all the chemicals in the library. You can scroll through the list with the up/down arrow keys. To select the desired chemical, highlight it and then click on it or hit Enter.



6. Ethane is in the ASPEN PLUS TM chemical library. Notice how the molecular formula automatically appears after typing it in. Now click under Comp ID. Enter an id name (to show up in results) for ethane, perhaps ETHA. Hit Enter. 7. Repeat steps 5 and 6 for ethylene and hydrogen. In this example, the component id names used were ETHY and H2, respectively. 8. # Complete appears. Click Next. The next window to appear is Properties.Main. Here is where you specify the solving engine used to simulate your process model. 9. Highlight the Opsetname box and type: SYSOP3 Hit Enter. 10. A window will appear asking if you wish to use SYSOP3. Click OK. 11. # Complete appears. Click Next. A window stating Required Properties Input Complete will appear. Read it and click OK. The next set of input windows will ask for the information state in the problem statement such as: flowrates, rate law, stoichiometry, etc. The first window will be titled Streams 1(Stream.Main). This indicates the stream labeled 1 on your flowsheet: the ethane inlet stream. You may have already assigned a different id to the inlet stream, if so that name will appear as the subtitle. 12. Highlight the Description box. Enter a description, perhaps Ethane Inlet Stream. 13. Highlight Temp. Enter 1100. Highlight the unit box to the right of Temp. Click on it with the right mouse button to pull down a unit menu. Click on K for Kelvin. 14. Highlight Pres. Enter 6. Highlight the unit box to the right of Pres. Click on it with the right mouse button to pull down a unit menu. Click on ATM for atmospheres. ***Note. Since Temp. and Pres. are specified, the vapor fraction, Vfrac, is set for the ethane inlet stream (Gibbs Phase Rule). 15. Now look to the middle of the window for Enter stream composition and flow. The Composition Basis should read MOLE-FLOW. If not, pull down the menu by clicking on the box with the right mouse button. Select MOLE-FLOW. Change the unit box via the pull down menu method to read LBMOL/SEC. 16. Since ethane is the only component of the inlet stream, highlight the box under Value and enter 0.425 for the molar flowrate. Enter 0.425 again for Total MOLE-FLOW. Be sure the units are LBMOL/SEC. 17. # Complete appears. Click Next. The next input window is titled B1(Rplug.Main). This window is where you will specify the operating conditions of the PFR. This is the window you will come back to upon running the simulation to change any operating conditions as well. 18. Highlight the Description box. Enter a description for the PFR, perhaps Isothermal PFR. 19. Highlight the Type box. Click on the box with the right mouse button to pull down the menu.



Select T-Spec. This stands for a PFR with a temperature specification. You will be describing the temperature profile in the PFR. 20. Highlight the Length box. Enter a length in feet. 10 feet is a good starting point. Be sure the units are FT. Change the units accordingly via the pull down menu method if necessary. 21. Do the same for the Diam box. 3 feet is a good guess for the diameter. Be sure the units are in FT. ***Note. You are solving this problem by guessing a volume. When you run the simulation you will see what conversion is achieved with the guessed volume. You will keep changing the volume (increasing/decreasing the length while keeping the diameter constant) and rerunning the simulation until the desired conversion is achieved. Remember that a PFR is a cylinder with a volume of V = (P/4)D 2 L. 22. Look down the window for Temperature specifications. Here is where you'll enter the temperature profile of the PFR. Since the PFR is isothermal, it will be 1100K (as given in the problem statement) throughout the entire reactor. Enter the following profile:

Location Temp K 0

1100

0.5

1100

1

1100

Be sure Temp is in K. Change it via the pull down menu method if necessary. The Location is given by a fraction of the distance down the PFR. Hence 0 = the beginning, 0.5 = the midpoint, and 1 = the end of the PFR. 23. Since the PFR has no heat exchanger jacket, do not enter any values under External coolant. 24. Under Optional reactor specifications highlight Pres. Type 6 and change the units to ATM. 25. # Complete appears. Click Next. The next input window is called Blocks B1(Rplug.Reac). The stoichiometry of the reaction occurring in the PFR will be described here. 26. Highlight the box called Reaction Number. The cracking of ethane in this problem is the only reaction occurring in the PFR. Therefore, this reaction will be called reaction 1. Enter 1 in the box. 27. Under Reactants, highlight the box under Comp ID. Enter your id name for ethane, in this case ETHA. 28. Now look at the balanced reaction equation in the problem statement. The coefficient of ethane is 1, however since ethane is a reactant it's disappearing so the coefficient is really -1. Enter -1 under Coefficient. 29. Under Products, enter the id names for the products ethylene and hydrogen (ETHY and H2 here) under Comp ID. Their coefficients will both be 1. Enter 1 for both coefficients. # Complete appears. Click Next. The last input window is titled Blocks B1(Rplug.Kinetics). Here you'll describe the rate law of the reaction in the PFR. 30. Highlight the Reaction Number box. Enter 1. Thus, you will be describing reaction 1, the cracking file:///H:/html/toolbox/aspen/creating.htm[05/12/2011 16:59:08]


of ethane is this case. ***Note. This is where a calculator comes in handy. ASPEN PLUS TM asks for two parameters of the Arrhenius equation: the activation energy, E, and the pre-exponential factor, A. With these parameters and the temperature of the reaction, ASPEN PLUS TM solves the equation of the specific rate of reaction, k. In this example problem, you are given E = 82 kcal/mol and k = 0.072s -1 when T = 1000K. Using this information, you can back track and solve for the preexponential factor A: k = Aexp(-E/RT) A = k/[exp(-E/RT)] A = 0.072 s -1 /[exp((-82000 cal/mol)/((1.987 cal/mol-K)(1000 K))] A = 6.02x1016 s -1 31. Hence type 6.02E16 for Pre-Exp. Type 82 for Act-energy, with units of KCAL/MOL. TempExpon will automatically be 0. 32. Lastly, the rate law must be entered. This example has an elementary rate law, hence, -rA= kCA . Therefore, under the Reactants, enter the id name for ethane (ETHA). Type 1 for the Exponent. 33. Under Products, enter the id names for ethylene and hydrogen (ETHY and H2). Enter 0 for the Exponents. 34. # Complete. Click Next. You are finished entering all the required data for the process model! The next window is called Rplug Input Menu. Click Next. Required Input Complete will appear. This window asks if you wish to run the simulation. Click OK. I. Introduction II. Accessing ASPEN PLUSTM III. Creating a Reaction Engineering Process Model A. Building a Process Flowsheet B. Entering Process Conditions IV. Running the Process Model A. Interpreting the Results B. Changing Process Conditions and Rerunning the Model V. Example Problems A. 8-6: Adiabatic Production of Acetic Anhydride B. 8-7: Operation of a PFR with Heat Exchanger VI. Other Need-to-Knows A. Saving your Process Model B. Printing your Process Model C. Changing Names of Streams and Unit Operations D. Changing Units of Parameters E. Exiting ASPEN PLUSTM VII. Credits



BACK to ASPEN PLUS TM MAIN PAGE aspen plus > creating


Aspen Plus - Running

Running the Process Model Congratulations on completing the flowsheet and entering the process model conditions. Now you are ready to put your model to the test. Example 4-4, the cracking reaction, can now be simulated. Once you've clicked OK to run the process model, the Control Panel window will appear. This window gives you a look at the ASPEN PLUS TM "thinking process." Phrases indicating the PFR block is being executed will scroll across the sc reen. When the simulation is complete, the Control Panel will read: Simulation completed successfully. Interpreting the Results 1. Once the simulation is complete, click the Results button on the Control Panel. 2. A window titled Results-Status.Main will appear. Click on the >> button in the upper right corner to jump forward to the stream-by-stream results page. For the Example 4-4 simulation the results screen should look like this:

Note that, down the left side of the screen, are the different parameters: temperature, pressure, mole flow, etc. Along the top row are the stream id names, in this case 1 and 2 (inlet and product). This forms a grid of information that can be interpreted easily. In the upper left corner of the screen, you will see Display, Units, and Format. You can change what results are shown and their units by using the pull down menu method for each of these areas. In Example 4-4, the problem asks for the PFR volume that will achieve an 80% conversion. As you

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recall, when entering the process model conditions, you guessed a volume by entering an arbitrary length and diameter of the PFR. In order to complete the problem, you must see what conversion your process model obtained. Recall that conversion is defined as: X = (moles of limiting reagent reacted) (moles of limiting reagent fed) Where moles reacted = moles in - moles out To solve for conversion, do the following with the results: 1. Under Mole Flow for ethane, calculate the moles of ethane that reacted: stream 1 - stream 2 2. Divide this number by the molar flowrate of ethane into the PFR: stream 1. If you followed the example exactly, using a length of 10 feet and a diameter of 3 feet, you should get a conversion of 76%. X = stream 1 - stream 2 stream 1 X = 693.996 kmol/hr - 169.362 kmol/hr 693.996 kmol/hr -orX = 1530 lbmol/hr - 373.380 lbmol/hr 1530 lbmol/hr X = 0.76 = 76% The dimensions of the PFR did not achieve an 80% conversion. Therefore, you need to go back and adjust them. It is easiest to vary the length of the PFR while holding the diameter constant. In this example, since the conversion was too low, you must INCREASE the length of the PFR. Changing Process Conditions and Rerunning the Model Adjusting the input conditions is very straightforward, just follow these steps: 1. With the mouse pointer, click on the Flowsheet window to make it active. 2. With the left mouse button click on the inlet stream or unit operation whose conditions you wish to change. In this example, click on the PFR. 3. Click the right mouse button to pull down a large menu. Click onInput with the right mouse button. 4. Click the >> button to forward you ahead to the input window you need. In this example, you are looking for Rplug.Main. 5. Highlight the value you wish to alter and type in the new value. In this case change the Length to a value greater than 10 feet (Hint: Try 11.42 feet!). 6. Click the >> or << buttons to move you forward or backward through the input windows if you need to change more parameters. If not, keep clicking Next until you are prompted with "Run Simulation Now?"



7. Rerun the simulation, and check your results. In Example 4-4, you will find that a length of 11.42 feet and a diameter of 3 feet will achieve an 80% conversion. To finish the problem, the volume of a PFRwith these dimensions is V = 80.72 ft3 . I. Introduction II. Accessing ASPEN PLUSTM III. Creating a Reaction Engineering Process Model A. Building a Process Flowsheet B. Entering Process Conditions IV. Running the Process Model A. Interpreting the Results B. Changing Process Conditions and Rerunning the Model V. Example Problems A. 8-6: Adiabatic Production of Acetic Anhydride B. 8-7: Operation of a PFR with Heat Exchanger VI. Other Need-to-Knows A. Saving your Process Model B. Printing your Process Model C. Changing Names of Streams and Unit Operations D. Changing Units of Parameters E. Exiting ASPEN PLUSTM VII. Credits BACK to ASPEN PLUS TM MAIN PAGE aspen plus > running


Aspen Plus - Examples

Example Problems This section is devoted to example reaction problems. The problems were taken from the 2nd Edition of Elements of Chemical Reaction Engineering by H. Scott Fogler. Please note, it is assumed that the user knows how to create a flowsheet and enter process conditions, since these examples explain only the values to enter for each input window. Example 8-6 Adiabatic Production of Acetic Anhydride Jeffreys, in a treatment of the design of an acetic anhydride manufacturing facility, states that one of the key steps is the vapor-phase cracking of acetone to ketene and methane: CH 3 COCH 3

CH 2 CO + CH 4

He states further that this reaction is first-order with respect to acetone and that the specific reaction rate can be expressed by ln k = 34.34 - 34,222/T (E8-6.1) where k is in reciprocal seconds and T is in Kelvin. In this design, it is desired to feed 8000 kg of acetone per hour to a tubular reactor. If the reactor is adiabatic, the feed pure acetone, the inlet temperature 1035K, and the pressure 1 62 kPa (1.6 atm), a tubular reactor of what volume is required for 20% conversion? Creating the Flowsheet The flowsheet consists of one inlet stream, a PFR, and one product stream. It should look like this:

Create a flowsheet like this in ASPEN PLUS TM . If you do not know how, see Example 4-4. When the flowsheet is complete,"Flowsheet Complete" should appear in the upper left corner of the screen. Click the Next button. Click OK when prompted to Enter Required Data. Entering Process Conditions This section will explain what values to type in for each input window. If you do not know how to enter values, change units, or navigate through the input windows, see Example 4-4. 1st Window: Setup.Main 1. Title: Enter any title you wish. 2. Flow / Frac Options: MOLEFLOW and MASSFLOW 3. Click Next.

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2nd Window: Components.Main 1. Under Component Name type the following in a column: ACETONE, KETENE, METHANE 2. Under Comp ID type in any id names for the above components: A, K, C1 3. Click Next. 3rd Window: Properties. Main 1. 2. 3. 4.

Opsetname: type SYSOP3 Click YES when prompted about SYSOP3 Click Next. Click OK when prompted about continuing entering stream input.

4th Window: Stream. Main 1. 2. 3. 4. 5.

Description: Enter any description of stream 1. Temp: 1035 K (change units if necessary) Pres: 1.6 atm (change units if necessary) Composition Basis: Change to MASS-FLOW KG/HR For a mass flowrate of A (acetone) type 8000. Leave ketene and methane at zero (no mass flow in reactant stream). 6. Total: Change to MASS-FLOW and enter 8000 KG/HR 7. Click Next. 5th Window: Rplug.Main 1. 2. 3. 4. 5.

Description: Enter any description for PFR, perhaps Adiabatic PFR. Type: ADIABATIC Length: Need to guess a length, 3 METERS is a good starting point. Diam: Need to guess a diameter, 1 METER is a good starting point. Click Next.

6th Window: Rplug.Reac 1. 2. 3. 4.

Description: Should be stated already as that you used for Rplug.Main. Reaction Number: type 1 Under Reactants: type A for acetone. Coefficient (stoichiometric): -1 Under Products: type K for ketene. Coefficient (stoichiometric): 1. Type C1 for methane. Coefficient (stoichiometric): 1 5. Click Next. 7th Window: Rplug.Kinetics 1. Description: Should already be stated. 2. Reaction Number: type 1 3. Pre-Exp: Enter the pre-exponential factor, A, of the Arrhenius equation, 8.2e14. This number was solved for by putting equation E8-6.1 in the form of k = Aexp(-E/RT) Take the natural log of both sides of equation E8-6.1 to get k (s-1 ) = 8.2x1014exp(-32,444/T) 4. Act-Energy: Enter the activation energy E of the Arrhenius equation, 67999 CAL/MOL. Again this file:///H:/html/toolbox/aspen/example.htm[05/12/2011 16:59:10]


value was solved for using equation E8-6.1: k (s-1 ) = 8.2x1014exp(-32,444/T) (Note that R is missing in the denominator.) Activation Energy = E = (32,444)(R) E = (32,444)(1.987 cal/mol K) = 67999 cal/mol 5. Under Reactants: Type A for acetone. Exponent (in rate law): 1 6. Under Products: Type K for ketene. Exponent (in rate law): 0. C1 for methane. Exponent (in rate law): 0 7. Click Next. Running the Simulation and Interpreting the Results Click Next again until you are prompted to run the simulation. Click OK. When the simulation is complete, click on Results (Control Panel). If you do not know how to interpret the results window, see Example 4-4. Otherwise, check the conversion (X = moles reacted/moles fed). Does X = 20%? If X < 20%, you must increase the length of the PFR. If X > 20%, you must decrease the length of the PFR. In this case where length = 3 m, diam = 1m, the conversion was greater than 20%. Therefore, you need to go back to the PFR and input a smaller length. You must access the Rplug.Main window to do this. If you do not know how to reenter inputs, see Example 4-4. This time, try a length of 2.5 m while holding the diameter constant at 1 m. When you rerun the simulation, you will find that X = 20%! Finishing up the example, the volume of the PFR with these dimensions is V = 1.96 m 3 .

Reference: G. V. Jeffreys, A Problem in Chemical Engineering Design: The Manufacture of Acetic Anhydride, 2nd ed. (London: Institution of Chemical Engineers, 1964).

Example 8-7 Operation of a PFR with Heat Exchanger We again consider the vapor-phase cracking of acetone used in Example 8-6: CH 3 COCH 3

CH 2 CO + CH 4

The reactor is to be jacketed so that a high-temperature gas stream can supply the energy necessary for this endothermic reaction (see Figure E8-7.1). Pure acetone enters the reactor at a temperature of 1035K and the temperature of the external gas in the heat exchanger is constant at 1150K. The reactor consists of a bank of one thousand 1-in. schedule 40 tubes. The overall heat-transfer coefficient is 110 J/m 2 -s-K. Determine the temperature profile of the gas down the length of the reactor.



Figure E8-7.1 Creating the Flowsheet Use the same flowsheet in Example 8-6. Entering Process Conditions Repeat windows 1 - 4, and 6 as in Example 8-6. For window 5: Rplug.Main do the following: 5th Window: Rplug.Main 1. 2. 3. 4. 5. 6.

Description: Enter any description for PFR, perhaps Jacketed PFR. Type: TCOOL-SPEC Under External coolant, enter a Coolant-Temp of 1150 K with a U of 110 J/SEC-SQM-K. Length: Need to guess a length, 3 METERS is a good starting point. Diam: Need to guess a diameter, 1 METER is a good starting point. Click Next.

Run the simulation. Again, adjust the length until the conversion is X = 20%. In this example, the proper length was 1.9 m with a diameter of 1 m. Thus the volume was V = 1.49 m 3 . Temperature Profiles down the Length of the PFR To see the temperature profile down the length of the PFR, do the following: 1. Click with the left mouse button on the PFR to highlight it. 2. Click the right mouse button to pull down a large menu. Select Results. 3. A results menu window will appear. Double click with the right mouse button on the black text: Profiles.\ 4. You will see columns of values: lengths, pressures, temperatures, vapor fractions, etc. 5. Click on the Length column to highlight it then click on Plot (main tool bar). 6. Click on Independent Variable (this is the x-axis). You assigned the length as an independent variable. 7. Click on the Temp column to highlight it then click on Plot (main tool bar). 8. Click on Dependent Variable (this is the y-axis). You assigned the temperature as an independent variable. 9. Pull down the Plot menu again and click on Display Plot. You should see a plot of the temperature profile that looks like this:



I. Introduction II. Accessing ASPEN PLUSTM III. Creating a Reaction Engineering Process Model A. Building a Process Flowsheet B. Entering Process Conditions IV. Running the Process Model A. Interpreting the Results B. Changing Process Conditions and Rerunning the Model V. Example Problems A. 8-6: Adiabatic Production of Acetic Anhydride B. 8-7: Operation of a PFR with Heat Exchanger VI. Other Need-to-Knows A. Saving your Process Model B. Printing your Process Model C. Changing Names of Streams and Unit Operations D. Changing Units of Parameters E. Exiting ASPEN PLUSTM VII. Credits BACK to ASPEN PLUS TM MAIN PAGE aspen plus > examples


Aspen Plus - Other

Other Need-to-Knows The following important information applies to UNIX Ultra Sparc® workstations. Saving Your Process Model It is wise to save your process model periodically while working. To save the model: 1. Select the File pull down menu from the tool bar. Click on Save As. 2. Enter the name of your process model. Click OK. 3. The process model will be saved under your afs directory. 4. The name of your process model will now appear on the window, replacing model UNNAMED. From this point on, select Save under the File menu to save your process while you work. Printing Your Process Model The flowsheet and results are valuable documents verifying your work. ASPEN PLUS TM allows you to individually print a flowsheet, stream-by-stream result pages, and a history file. The Flowsheet. To print your process model's flowsheet from a UNIX workstation do the following while the flowsheet window is active: 1. Select the File pull down menu from the tool bar. Click on Print. 2. A small window will appear asking for the name of the flowsheet. Type in an arbitrary name of the flowsheet (typing in 'flowsheet' works fine). Click OK. 3. Moving the mouse pointer off the ASPEN PLUS TM window, use the middle mouse button to open a New Shell. 4. At the % prompt, type: lpr -P For example, to print a flowsheet entitled 'flowsheet' on the 3 South printer in the Media Union type: lpr -P mu3s flowsheet (hit enter) Stream-by-Stream Results. To print the stream-by-stream results, follow steps 1-4 above while the results window is active. Instead of naming the flowsheet, however, you will be assigning an arbitrary name to the results ('results' works fine). The History File. Printing the history file of your process model will allow you to step through all of the computer code used in solving the simulation. The history file keeps track of all previous ASPEN PLUS TM runs executed while you've been logged on. Depending on the complexity of the process model, the history file can be very lengthy (100 pages or more!). Therefore, evaluate whether a hard copy of this file is necessary before you print.

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Aspen Plus - Other

ASPEN PLUS TM creates the history file after completing one run of the process simulation. To print the history file do the following: 1. Open a New Shell (step 3 above). 2. At the % prompt, type in: lpr -P For example, if you saved your process model as CSTR, to print its history file on the 3 South printer in the Media Union type: lpr -P mu3s CSTR.his (hit enter) Changing Names of Streams and Unit Operations ASPEN PLUS TM arbitrarily assigns id names to all streams and unit operations on the flowsheet you create. If you wish to change the id name, do the following: 1. With the left mouse button, click on the id box of the stream or unit operation to highlight it. 2. Click the right mouse button to pull down a large menu. 3. Click on the Rename option. 4. When the Rename box appears, type in the desired id name and click OK. Changing Units of Parameters When entering the input conditions, you may need to change the units of parameters such as temperature, pressure, mole flow, etc. Do the following, commonly referred to as the pull-down menu method: 1. Click on the unit box with the right mouse button. A list of units should appear. 2. Scroll up/down the list with the cursor keys. Click on the appropriate unit. Exiting ASPEN PLUSTM When finished with your process model, do the following to exit the ASPEN PLUS TM program: 1. Pull down the File menu. 2. Click on Exit. 3. A window will appear asking if you wish to save your process model run. Click No if you do not wish to save the run, or Yes if you do. 4. The ASPEN PLUS TM window will disappear and you will be left with the % prompt. I. Introduction II. Accessing ASPEN PLUSTM III. Creating a Reaction Engineering Process Model A. Building a Process Flowsheet file:///H:/html/toolbox/aspen/other.htm[05/12/2011 16:59:10]

Aspen Plus - Other

B. Entering Process Conditions IV. Running the Process Model A. Interpreting the Results B. Changing Process Conditions and Rerunning the Model V. Example Problems A. 8-6: Adiabatic Production of Acetic Anhydride B. 8-7: Operation of a PFR with Heat Exchanger VI. Other Need-to-Knows A. Saving your Process Model B. Printing your Process Model C. Changing Names of Streams and Unit Operations D. Changing Units of Parameters E. Exiting ASPEN PLUSTM VII. Credits BACK to ASPEN PLUS TM MAIN PAGE aspen plus > other

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Credits The ASPEN PLUS TM web site was created by Ellyne E. Buckingham at the University of Michigan, Summer 1997. Special thanks to: Anuj Hasija and Professor H. Scott Fogler The University of Michigan at Ann Arbor, MI Dr. J. Mahalec ASPENTech at Cambridge, MA All ASPEN screen shots courtesy of ASPENTech.

Final editing of the ASPEN PLUS TM web site done by Dieter Andrew Schweiss, Fall 1997. I. Introduction II. Accessing ASPEN PLUSTM III. Creating a Reaction Engineering Process Model A. Building a Process Flowsheet B. Entering Process Conditions IV. Running the Process Model A. Interpreting the Results B. Changing Process Conditions and Rerunning the Model V. Example Problems A. 8-6: Adiabatic Production of Acetic Anhydride B. 8-7: Operation of a PFR with Heat Exchanger VI. Other Need-to-Knows A. Saving your Process Model B. Printing your Process Model C. Changing Names of Streams and Unit Operations D. Changing Units of Parameters E. Exiting ASPEN PLUSTM VII. Credits BACK to ASPEN PLUS TM MAIN PAGE aspen plus > credits

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Thoughts on Problem Solving: Bloom's Taxonomy

Problem solving is an activity whereby a best value is determined for an unknown, which is subject to a set of constraints. To determine how to attack a problem, three problem classifications can be used to direct the investigation: by type of unknown, difficulty, and open-endedness. The author’s preference in problem classification is that of Bloom, who classifies and identifies six problem-solving skills.

A. Level of Difficulty or Skill Level Each successive skill level calls for more advanced intellectual ability. 1. Knowledge: The remembering of previously learned material. Can the problem be solved simply by defining terms and by recalling specific facts, trends, criteria, sequences, or procedures? This level solves the type of problems such as recallin g the type of continuous flow reactor normally used for liquid-phase reactions. This is the lowest intellectual skill level. Examples of knowledge level questions are the following: Write the equations for a batch reactor and list its chara cteristics. Define . Which reactors operate at steady state? Other words used in posing knowledge questions: Who..., When..., Where..., Identify..., What formula .... 2. Comprehension: This is the first level of understanding. Given a familiar piece of information, such as a scientific principle, can the problem be solved by recalling the appropriate information and using it in conjunction with manipulation, tr anslation, interpretation, or extrapolation of the equation or scientific principle? For example, given the reactor volume [V=(v 0 /k)ln(CA0/CA )], can one manipulate the design equation formulas to find the effluent concentration to find the reactor volume if the inlet concentration were doubled? Compare and contrast the advantages and uses of a CSTR and a PFR. Construct a plot of NA as a function of t. Other comprehension words: ... Relate..., Show..., Distinguish..., Reconstruct..., Extrapolate... This is skill level 2. A. Drilling the concepts A1. Plug and chug

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B. Extrapolate What is the time at 400 K?

C. More difficult/Think Problems C1.Intermediate calculation Example: 80% conversion is achieved for a first order reaction in a 1000 gal PFR, what conversion could be achieved in a CSTR for the same conditions? C2.More than one intermediate calculation Example: 80% conversion is achieved for a first order reaction achieved 1000 gal PFR, what conversion could be achieved in a CSTR operated at a temperature 100°C higher. 3. Application: The next higher level of understanding is recognizing which set of principles ideas, rules, equations, or methods should be applied, given all the pertinent data. Once the principle is identified, the necessary knowledge i s recalled and the problem is solved as if it were a comprehension problem (skill level 2). An application level question might be: Make use of the mole balance to solve for the concentration exiting a PFR. Other words: ... Apply..., Demonstrate..., Determine..., Illustrate.... 4. Analysis: This is the process of breaking the problem into parts such that a hierarchy of subproblems or ideas is made clear and the relationships between these ideas is made explicit. In analysis, one identities missing, redundant, and contradictory information. Once the analysis of a problem is completed, the various subproblems are then reduced to problems requiring the use of skill level 3 (application). An example of an analysis question is: What conclusions did you come to after reviewing the experimental data? Other words: ... Organize..., Arrange..., What are the causes ..., What are the components.... 5. Synthesis: This is the putting together of parts to form a new whole. Synthesis enters problem solving in many ways. A synthesis problem would be one requiring the type, size, and arrangement of equipment necessary to make styrene from ethylben zene. Given a fuzzy situation, synthesis is the ability to formulate (synthesize) a problem statement and/or the ability to propose a method of testing hypotheses. Once the various parts are synthesized, each part (problem) now uses the intellectual skill described in level 4 (analysis) to continue toward the complete solution. Examples of synthesis level question are: Find a way to explain the unexpected results of your experiment. Propose a research program that will elucidate the reaction mechanism. Other words: ... Speculate..., Devise..., Design..., Develop..., What alternative..., Suppose..., Create..., What would it be like..., Imagine..., What might you see.... 6. Evaluation: Once the solution to the problem has been synthesized, the solution must be evaluated. Qualitative and quantitative judgments about the extent to which the materials and methods satisfy the external and internal criteria should be m ade. An example of an evaluation question is: Is the author justified in concluding that the reaction rate is the slowest step in the mechanism. Other Words: ... Was it wrong..., Will it work..., Does it solve the real problem..., Argue both sides..., Which do you like best..., Judge..., Rate.... B. Classification as Closed- or Open-Ended As an application of the strategy outlined in Figure PS1-1, we consider a thumbnail sketch of the design of a chemical plant. Specifically, we want to produce 200 million pounds of styrene per year from ethylbenzene. First we synthesize a sequence of processing operations as shown in the synthesis level (row 1) in Figure PS1-2. In synthesis level we develop the type and arrangement of operations and equipment to produce ethylbenzene. Next we evaluate this sequence to learn if additional operations are necessary, such as a heat-exchange system following the separation sys tem or a feed purification system before the first heat exchanger. Following this file:///H:/html/probsolv/open/blooms/index.htm[05/12/2011 16:59:12]


cursory evaluation of our sequence, we analyze (row 2) each system (i.e. break it down into a number of subproblems). For example, Figure PS1-2 shows an analysis of t he reaction system. Here, we determine the type of reactor and catalyst to be used, the best temperature at which to carry out the reaction, the type and arrangement of reactor peripherals (e.g., heating/cooling of the reactor), and the optimum feed condi tions. After breaking down the reaction system into a number of subsystems, we proceed to the application skill level (row 3) to decide which laws or principles are to be applied to each subsystem. For example, to calculate the catalyst weight, W, we make use of the design equation (Ch. 6) for a packed-bed reactor,

In using our comprehension (level 2) skills we recall or look up the equation that gives - as a function of concentration, express concentration as a function of conversion X (Ch. 2), carry out the integration, and finally determine the catalyst weight necessary to achieve a conversion X. C. Exercises A. Use a specific example to explain how one could work backward from Bloom’s levels 5 through 1 to solve ill-defined open-ended problems. B. For each of Bloom’s skill levels, construct an example that illustrates the skill used in that level (e.g., for level 1, what assumptions are used to derive the CSTR equation?).

Take a look at our cobra example or move on to Closed-Ended Problems.

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First Steps in Solving Open-Ended Problems From Strategies for Creative Problem Solving by H. Scott Fogler and Steven E. LeBlanc, 1995 1.

Write an initial problem statement. Include information on what you are to solve, and consider why you need to solve the problem.

2.

Make sure you are proceeding to solve the real problem as opposed to the perceived problem (chapter 3). Carry out one or more of the following: A. B. C. D. E.

Find out where the problem came from Explore the problem Apply the Duncker Diagram Use the statement-restatement technique Apply Problem Analysis

3. Generate solutions (chapter 4) A.

Understand what conceptual blocks can occur so that you will be aware of them when they surface. 1. 2. 3. 4. 5. 6.

Perceptual Emotional Cultural Environmental Intellectual Expressive

B.

Brainstorm

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1. Free association 2. Osborn’s Check List 3. Lateral Thinking a. Random Stimulation b. Other People's Views C.

Analogy 1. State the problem 2. Generate analogies 3. Solve the analogy 4. Transfer the analogy to the solution D.

Organize the ideas/solutions that have been generated. 1. Fishbone Diagram E.

Cross Fertilize 1. Draw analogies from other disciplines

F.

Futuring. Today's constraints (e.g. computing speed, communications) may be limiting the generation of creative solutions. Think to the future when these constraints may no longer exist. Remove all possible constraints from the problem statement and solution criteria. G.

Incubate. Take a break. Let your subconscious work on the problem while you do something else. Sometimes all you need is a breather to achieve that final breakthrough! 4. Choose best alternative from the ideas generated (chapter 5) A.

Decision Making 1. Musts 2. Wants 3. Adverse Consequences



B.

Planning 1. 2. 3. 4.

Potential Problem Consequences Preventative Action Contingent Action

5.

Follow Through (chapter 6) A. Gantt Chart B. Deployment Chart C. Evaluation - Is the problem you are solving still relevant? 6.

Evaluate (chapter 7) A. Does the solution satisfy all the stated and implied criteria? B. Is the solution safe to people and property? C. Is the solution ethical? See an example of the OEP Algorithm in action, as applied to the Cobra Problem. Bloom's Taxonomy can help you classify your problem and determine a method of attack.



Previous Open-Ended Problems 1. Design of Reaction Engineering Experiment The experiment is to be used in the undergraduate laboratory and the cost less than $500 to build. The judging criteria are the same as the criteria for the National AIChE Student Chapter Competition. The design is to be displayed on a poster board and explained to a panel of judges. Guidelines for the poster board display are provided by Jack Fishman and are given on the CD ROM. 2. Pharmacokinetics of Cobra Bites In Thailand alone, snakebites are responsible for the deaths of approximately 2,500 people a year. The interaction of snake venom with newly-developed antivenoms in the human body can be modeled as a chemical reaction engineering catalysis problem. Students use this knowledge to create and solve unique snakebite scenarios. Focus: catalysis, multiple reaction kinetics. 3. Effective Lubricant Design Lubricants used in car engines are formulated by blending a base oil with additives to yield a mixture with the desirable physical attributes. In this problem, students examine the degradation of lubricants by oxidation and design an improved lubricant system. The design should include the lubricant system's physical and chemical characteristics, as well as an explanation as to how it is applied to automobiles. Focus: automotive industry, petroleum industry. 4. Peach Bottom Nuclear Reactor The radioactive effluent stream from a newly-constructed nuclear power plant must be made to conform with Nuclear Regulatory Commission standards. Students use chemical reaction engineering and creative problem solving to propose solutions for the treatment of the reactor effluent. Focus: problem analysis, safety, ethics. 5. Underground Wet Oxidation You work for a specialty chemicals company, which produces large amounts of aqueous waste. Your Chief Executive Officer (CEO) read in a journal about an emerging technology for reducing hazardous waste, and you must evaluate the system and its feasibility. Focus: waste processing, environmental issues, ethics. 6. Hydrodesulfurization Reactor Design Your supervisor at Kleen Petrochemical wishes to use a hydrodesulfurization reaction to produce ethylbenzene from a process waste stream. You have been assigned the task of designing a reactor for the hydrodesulfurization reaction. Focus: reactor design. 7. Continuous Bioprocessing Most commercial bioreactions are carried out in batch reactors. The design of a continuous bioreactor is desired

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since it may prove to be more economically rewarding than batch processes. Most desirable is a reactor that can sustain cells that are suspended in the reactor while growth medium is fed in, without allowing the cells to exit the reactor. Focus: mixing modeling, separations, bioprocess kinetics, reactor design. 8. Methanol Synthesis Kinetic models based on experimental data are being used increasingly in the chemical industry for the design of catalytic reactors. However, the modeling process itself can influence the final reactor design and its ultimate prerformance by incorporating different interpretations of experimental design into the basic kinetic models. In this problem, students are asked to develop kinetic modeling methods/approaches and apply them in the development of a model for the production of methanol from experimental data. Focus: kinetic modeling, reactor design. 9. Cajun Seafood Gumbo Most gourmet foods are prepared by batch processes. In this problem, students are challenged to design a continuous process for the production of gourmet-quality Cajun seafood gumbo from an old family recipe. Focus: reactor design.

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Homogeneous Example 1

Type 1 Home Problem Problems with a straight-forward calculation. The following reaction takes place in a CSTR:

Pure A is fed to the reactor under the following conditions:

F Ao = 10 mol/min C Ao = 2 mol/dm 3 X=?

V= 500 dm 3 and k=0.1/min Rate Law: -r A = kCA What is the conversion in the CSTR? Solution to Homogeneous Problem #1 Other Homogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Homogeneous Example 1

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Thoughts on Problem Solving: Heterogeneous Example 1

Type 1 Home Problem Straight forward calculation

F A0 = 10 mol/s X=?

C A0 = 1 mol/dm 3 W = 500kg k = 0.1111 dm 3 s -1 kg -1 What is the conversion of the PFR? Solution to Heterogeneous Problem #1 Other Heterogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Heterogeneous Example 1

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Type 2 Home Problem Problems that require intermediate calculations or manipulations. The following reaction takes place in a CSTR:


F Ao = 10 mol/min C Ao = 2 mol/dm 3 At T=350 K X=0.75

V= ? and k=? Rate Law: -r A = kCA The activation energy: E=20 kcal/mol What is the conversion in a PFR at 325 K with the same volume?



Solution to Homogeneous Problem #2 Other Homogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Homogeneous Example 2



Type 2 Home Problem Intermediate calculations or manipulations.

F A0 = 10 mol/s X1 = 0.8

C A0 = 0.1 mol/dm 3 W1 = ? T = 325 K ; E = 2500 cal

F A0 = 10 mol/s C A0 = 0.1 mol/dm 3 W 2 = W1 T = 350 K What is the conversion of the CSTR? Solution to Heterogeneous Problem #2 Other Heterogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Heterogeneous Example 2


X2 = ?




Type 3 Home Problem Problems that are over-specified. The following irreversible reaction takes place in a CSTR:

It takes place under the following conditions:

F To = 40 mol/min C Ao = 2 mol/dm 3 The feed is 75 mol% A & 25 mol% inerts. X=?

k(400 K) = 0.1/min V = 500 dm 3 and T = 400 K

This reaction follows the first order rate law: -r A = kCA The activation energy (E) is 10 kcal/mol & the Arrhenius constant (A) is 3*104 /min What is the conversion? Solution to Homogeneous Problem #3 Other Homogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10






Type 3 Home Problem Over-specified problems

F A0 = 10 mol/s X=?

C A0 = 1 mol/dm 3 W = 500 kg k = 0.1111 dm 3 s -1 kg-1

If the diameter of the reactor is 40 cm, the calatylst density is 2.9 g/cm3 and the catalyst is replaced once a month, what is the conversion? Solution to Heterogeneous Problem #3 Other Heterogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10

Heterogeneous Example 3



Type 4 Home Problem Problems that are under-specified and require the student to consult other information sources. The following reaction takes place in a membrane reactor:

Given: F Ao =10 mol/min X=?

At T = 298 K

F A=?

At P = 15 psia

F B =?

Diffusivity: DBM =5*10 -5 m 2/sec

F C =? k = 1/min V=0.5 m 3 Diameter (DT ) = 0.1m

A pure feed of formaldehyde (A) enters the reactor. The membrane only allows the hydrogen to pass through it. Rate Law:

Using Polymath, what is the conversion of this system? Also plot the flow rates versus the volume of the membrane reactor. Solution to Homogeneous Problem #4



Other Homogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Homogeneous Example 4



Type 4 Home Page Under-specified problems

Reactant A is fed to a catalyst-packed membrane reactor at a rate of 8 mol/min with an initial concentration of 0.5 mol/dm3 . What is the final conversion of A into C?

kR = 0.749 Boundary layer thickness = 0.1 cm KC = 2.5 Diffusivity = 0.25 cm3/s T = 298 K Bulk Density = 1.8 kg/dm 3 Solution to Heterogeneous Problem #4 Other Heterogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10




Type 5 Home Problem "What if …" problems that promote discussion. Recall the membrane reactor from Homogeneous Example 4. What if there were no membrane and the following reaction were carried out in a simple PFR:

Given: F Ao =10 mol/min

F A=?

At T = 298 K

F B =?

At P = 15 psia

F C =?

X=?

k = 1/min V=0.5 m 3 Diameter (DT ) = 0.1m

Rate Law:

Using Polymath, determine the conversion for this system. Also, plot the flow rates versus the volume of the PFR. What PFR volume is required to achieve the same conversion as the membrane reactor (X=0.86)? Do you expect this PFR to be significantly larger than the membrane reactor from Example 4?



Solution to Homogeneous Problem #5 Other Homogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Homogeneous Example 5



Type 5 Home Problem "What if..." problems

F A0 = 10 mol/s X=?

C A0 = 1 mol/dm 3 W = 500 kg k = 0.1111 dm 3 s -1 kg-1 Bulk density = 1.8 g/cm3 Find the conversion : What if the PFR it was 100dm long and had a total volume of 200dm 3 ? What if it was 4dm wide and 10dm long? What if only 0.5dm wide and 300dm long? What if 4 PFRs 25 dm long were used in parallel instead of one 150dm 3 PFR? Solution to Heterogeneous Problem #5 Other Heterogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10




Type 6 Home Problem Problems, or parts of problems, that are open-ended. The following gas phase reaction takes place in an industrial PFR:

C is the desired product. D is formed in an undesirable side reaction. Given:

F Ao = 50 mol/min

FA = ?

F Bo = 25 mol/min

FB = ? V = 2 m3

F C = 81,000 mol/hr

T = ? and P = ? Additional Information: k1(300 K) = 0.075/sec

Rate Laws:

E1=15000 cal/mol k2(300 K) = 0.0015/sec E2=17500 cal/mol Find the operating temperature (T) and pressure (P) for the PFR that will generate our desired production of C. What are the factors that affect your decision? Solution to Homogeneous Problem #6






Type 6 Home Problem Open-ended problems You are working for a company that is building a new plant. The company is still deciding whether to use a catalystfilled membrane reactor or a PFR for a certain process. The reaction is an equilibrium reaction, which makes some favor a membrane reactor; however, the transport rate constant is small and the membrane reactor is more expensive than the PFR. What would your recommendation be? Other Heterogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Heterogeneous Example 6



Type 7 Home Problem Problems where the student must explore the situation by varying operating conditions or parameters. Let us revisit the membrane reactor (previously addressed in Homogenous Example 4). The following gas phase reaction will take place in our reactor:

Once again, our membrane will allow B to exit, but it will retain A and C. Given:

F Ao = 10 mol/min

X=?

yAo = 1

FA = ?

At T = 400 K

FB = ?

At P = 10 atm

FC = ? V = 100 dm 3 kB = 0.5/min KC = 105 mol/dm 3 k = 0.7 min -1

Rate Law:



Using Polymath, determine the conversion of this system. Then, vary K C, k, and kB as follows:

Is there an optimal set of conditions? Can you explain why those conditions are most effective? Is the membrane reactor a proper system for this reaction? Solution to Homogeneous Problem #7 Other Homogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Homogeneous Example 7



Type 7 Home Problem Vary conditions or parameters

C To = 0.5 mol/dm 3 F Ao = 8 mol/s kR = 0.7 KC = 2.5 Bulk Density = 1.8 kg/dm

T = 298 K kc = 2.5 cm3/s E = 5000 cal delta H = 2500 cal 3

Explore and describe the effects of varying kc, temperature, and entering flowrate. Solution to Heterogeneous Problem #7 Other Heterogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Heterogeneous Example 7



Type 8 Home Problem Problems that challenge assumptions. In Homogeneous Example 7, we calculated the conversion of a membrane reactor, assuming that there was no pressure drop. In this problem, we will investigate pressure drop in our system, and determine it's effect (if any) on conversion. The following gas phase reaction takes place in a membrane reactor:

Once again, our membrane will allow product B to exit, but it will retain reactant A and product C. Given:

F Ao = 10 mol/min

X=?

yAo = 1

FA = ?

At T = 400 K

FB = ?

At P = 10 atm

FC = ? V = 100 dm 3 D = 3 in kB = 0.5/min KC = 105 mol/dm3 k = 0.7/min

Additional Information: m A = m B = m C = 1.8 Pa*s


Rate Law:


rA = 0.0015 kg/dm 3

rB = 0.001 kg/dm 3

rC = 0.00125 kg/dm 3 The membrane dimensions are similar to those of 16 BWG tubing. Setup the Polymath solution to this problem. What is the exiting pressure of our reactor? What is the conversion of this reactor? Is it safe to assume that the effects of pressure drop are negligible? Solution to Homogeneous Problem #8 Other Homogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Homogeneous Example 8



Type 8 Home Problem Problems which challenge assumptions

delta P = 0

F A0 , C A0

X=?

F A0 , C A0 , P 0

X=? P=?

F Ao = 0.1 mol/s C Ao = 10 mol/dm 3 k = 0.1111 dm 3 s -1 kg-1 P o = 10 atm What is the conversion in the first PFR? If the particle size is decreased, the pressure drop becomes an important factor. If the pressure drop is modeled by the equation:

and alpha = 0.0011, what is the new conversion? What is the final pressure?



Solution to Heterogeneous Problem #8 Other Heterogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10




Type 9 Home Problem Problems that promote discussion. Worthless Chemical has been making tirene (B) from butalene (A) (both dark liquids) using a 12 ft3 CSTR followed by a 4.5 ft3 PFR. The entering flow rate is 100 mol/min of butalene (A). The feed is heated to 60 ° C before entering the CSTR. A conversion of 80% (Xi) is typically achieved using this arrangement. One morning, the plant manager, Dr. Pakbed, arrived and found that the conversion had dropped to approximately 36% (Xii ). After inspecting the reactors, the PFR was found to be working perfectly, but a dent was found in the CSTR that may have been caused by something like a fork-lift. He also noted that the CSTR, which normally makes a "woosh" sound was not as noisy as it had been the previous day. The manager calls in four junior engineers (Jane, Steve, Bill, and you) to investigate the problem. Given:

F Ao = 100 mol/min

Xi = 0.8

T = 60 °C

Xii = 0.36

A plot of the inverse of the reaction rate versus conversion is presented below:



I. Make a list of all the things that could cause the drop in conversion. Propose tests to confirm the explanations. Quantify the explanations with numerical calculations, where possible. II. Dr. Pakbed tells you that in order to meet the production schedules down stream, he needs a conversion of a least 70%. Can this conversion be obtained without taking time to fix to the CSTR? Solution to Homogeneous Problem #9 Other Homogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Homogeneous Example 9



Type 9 Home Problem Group discussion

F A0 = 10 mol/s C A0 = 0.1

Xmeasured = 0.5 Xcalculated = 0.7

mol/dm 3 W = 500 kg

Why is the measured conversion different from the calculated conversion? Solution to Heterogeneous Problem #9 Other Heterogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Heterogeneous Example 9



Type 10 Home Problem Problems that develop life-long learning skills. After reading the journal review by Y. T. Shah et al. [AIChE J., 28, 353 (1982)], design the following bubble column reactor. Sixty percent carbon dioxide (CO2 ) in a flue gas is to be removed by bubbling through a solution of sodium hydroxide (NaOH). The reaction is mass-transfer limited. Calculate the reactor size (length and diameter) necessary to remove 99.9% of the carbon dioxide. The flow rate of the flue gas is 0.05 m 3 /sec and the flow rate of the sodium hydroxide is 0.001 m 3 /sec. The reactor must operate in the bubbly flow regime, so please recommend a type of sparger to use. Given:

Bubble Column Reactor vL = 0.001 m 3/sec

DC = ?

vGo = 0.05 m 3/sec

HC = ?

vG = 0.02 m 3/sec

Solution to Homogeneous Problem #10 file:///H:/html/probsolv/tentypes/homo/q10/homoq10.htm[05/12/2011 16:59:30]





Type 10 Home Problem Life-long learning Now that you have developed a little knowledge of heterogeneous reactions, find an article on the web or in a journal that deals with an issue discussed in class, or that covers a topic that interests you. Give a brief synopsis of the article and explain anything new you learned from the article and/or how it applies to what you have learned in class. (Maybe you could keep a journal, too, because you never know when an interesting concept will come in handy later in life.) Other Heterogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Heterogeneous Example 10


authors.htm


H. Scott Fogler

Steven E. LeBlanc

College of Engineering University of Michigan Ann Arbor, Michigan 48109-2136

College of Engineering University of Toledo Toledo, Ohio 43606-3390


file:///H:/html/probsolv/strategy/authors.htm[05/12/2011 16:59:31]

review.htm


Review of The Book Reveiwed by George E. Dieter In ASEE PRISM, December 1995 We tell prospective engineering students that studying engineering will make them good problem solvers; however, we hardly devote any time in the curriculum to teaching them about problem solving in a generic sense, namely one outside the context of a particular discipline. Strategies for Creative Problem Solving goes a long way toward resolving that oversight. In the words of the authors, "Here is a book to help problem solvers improve their street smarts". In the book, Scott Fogler and Steven LeBlanc suggest a five-step problem-solving heuristic: 1) define the problem; 2) generate solutions; 3) decide the course of action; 4) implement the solution; and 5) evaluate the solution. Separate chapters cover each of these steps and include descriptions of pertinent methodologies and tools to use, as well as short examples. For every methodology, the process to follow is described in detail with attractive sidebar cartoons and illustrations. Chief methodologies and tools include: Problem Solving: Present-state and desired-state techniques Generating Solutions: Blockbusting, Osborne's checklist for brainstorming, and the Fishbone diagram for oganizing brainstorming Deciding the Course of Action: Kepner-Tregoe situation analysis, the Pareto diagram, and the decision matrix method Implementing the Solution: The Gantt Chart and using the elements of experimental design Evaluation: An evaluation checklist This is definitely a how-to book. The authors end each chapter with suggestions for further reading and a set of exercises for homework or class participation. Interactive computer software is available that reinforces the concepts in the text. With funding from the National Science Foundation, the publisher was able to send the software to every U.S. engineering dean. Contact Scott Fogler at the University of Michigan for an extra copy of the software, if you need one. (I did not evaluate the software.) Overall, the book has a practical, customer orientation. While it does not follow a full, total quality management (TQM) problem-solving approach, it has a definite TQM flavor. The importance of working in teams is mentioned in an early chapter, but by not following through on the subject in subsequent chapters, the authors miss an important opportunity to strengthen the text. The authors claim Strategies for Creative Problem Solving is intended for students, new graduate practitioners, or anyone who wants to increase his or her problem-solving skills. However, because of the level of the examples and the rather succinct presentation of most topics, it will most likely find its audience among freshman and sophomore engineering students.

file:///H:/html/probsolv/strategy/review.htm[05/12/2011 16:59:32]

review.htm

There are few textbooks of this type available, and because of the relatively brief nature of the book, the subject material can be introduced into a variety of the freshman-orientation courses that abound in our engineering colleges. Of course, for students to benefit from this approach, faculty membes will need to reinforce problem-solving skills in subsequent engineering science and design courses. This book represents a coherent, well-written, nicley illustrated presentation of technical problem solving that should find ready acceptance in many places within engineering education. George E. Dieter is Professor Emeritus of Mechanical Engineering at the University of Maryland and past president of ASEE.


file:///H:/html/probsolv/strategy/review.htm[05/12/2011 16:59:32]

publish.htm


Strategies for Creative Problem Solving was published by Prentice Hall PTR, Englewood Cliffs, New Jersey 07632. Copyright 1995. The publisher offers discounts on this book when ordered in bulk quantities. For more information, contact: Corporate Sales Department PTR Prentice Hall 113 Sylvan Avenue Englewood Cliffs, NJ 07632 Phone: 201-592-2863 FAX: 201-592-2249 To obtain Interactive Computer Modules based on Strategies for Creative Problem Solving, contact the CACHE Corporation at: CACHE Corporation P.O. Box 7939 Austin, TX 78713-7379 FAX: 512-471-7060 E-mail: [email protected]


file:///H:/html/probsolv/strategy/publish.htm[05/12/2011 16:59:32]

CRE - Integrating Factor

The Integrating Factor - What? Why? How? Problem - You want to integrate (1) However, that f(z)y term really messes things up! If you only had an expression of the form (2) things would be much easier, then you could integrate with respect to z and find y(z). How can you combine f(z) and y to get this simplification? First note that is of the form of the derivative of a product, so examine first the product y u, where u is some function of f(z) you still have to define. Recall (3) That's looking close to the left hand side of equation (1), but there is a "u" in front of the dy/dz term, and a du/dz expression where f(z) is. If you had a form of u such that du/dz = u f(z), then you could manipulate equation (3): (4) where the term in brackets is the left hand side of equation (1). You need du/dz = u f(z). Recall (5) If you define

and f(z) = dq/dz (i.e.

, then (6)

This satisfies the condition that du/dz=u f(z). That's the ezpression you needed! Therefore, substituting into equation (4),

, and

(7) where the term in brackets is the left hand side of equation (1). CONCLUSION: If your problem is of the form (1) you can multiply both sides of the equation by the

(which you should be able to evaluate, since you know f(z)), to yield (9) or, substituting from equation (7)

file:///H:/html/course/intfact.htm[05/12/2011 16:59:40]

CRE - Integrating Factor

(10) so that (11)

(12) EXAMPLE

f(t)=k 2, so

From equation (12), the solution is then

The constant can be obtained from the intitial condition that at t=0, CB=0;


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UM Course Syllabus, Exam I Crib Sheet

ChE 344: Exam I Crib Sheet


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Learning Resource CDP2-A Solution a. Over what range of conversions are the plug-flow reactor and CSTR volumes identical? So that it is easier to visualize the solution, we first plot the inverse of the reaction rate versus conversion. This type of plot is often called a "Levenspiel Plot.":

Recalling the mole balance equations for a CSTR and a PFR:

Until the conversion (X) reaches 0.5, the reaction rate is independent of conversion and the reactor volumes will be identical. i.e.



b. What conversion will be achieved in a CSTR that has a volume of 90 L? For now, we will assume that conversion (X) will be less that 0.5. We start with the CSTR mole balance:

Our calculated conversion is extremely small. c. What plug-flow reactor volume is necessary to achieve 70% conversion? This problem will be divided into two parts, as seen below:

1. The PFR volume required in reaching X=0.5 (reaction rate is independent of conversion). file:///H:/html/02chap/html/ahs02-a.htm[05/12/2011 16:59:42]


2. The PFR volume required to go from X=0.5 to X=0.7 (reaction rate depends on conversion).

Finally, we add V2 to V1 and get: Vtot = V1 + V2 = 2.3*1011 m3 d. What CSTR reactor volume is required if effluent from the plug-flow reactor in part (c) is fed to a CSTR to raise the conversion to 90%?



We notice that the new inverse of the reaction rate (1/-rA) is 7*108. We insert this new value into our CSTR mole balance equation:

e. If the reaction is carried out in a constant-pressure batch reactor in which pure A is fed to the reactor, what length of time is necessary to achieve 40% conversion? We will begin with the mole balance on a batch system. Since there is no flow into or out of the system, it can be written as:

From the stoichiometry of the reaction we know that V = Vo(1+eX) and e is 1. We insert file:///H:/html/02chap/html/ahs02-a.htm[05/12/2011 16:59:42]


this into our mole balance equation and solve for time (t):

After integration, we have:

Inserting the values for our variables:

That is 640 years. f. Plot the rate of reaction and conversion as a function of PFR volume. The following graph plots the reaction rate (-rA) versus the PFR volume:



Below is a plot of conversion versus the PFR volume. Notice how the relation is linear until the conversion exceeds 50%.

The volume required for 99% conversion exceeds 4*1011 m 3.



g. Critique the answers to this problem. The rate of reaction for this problem is extremely small, and the flow rate is quite large. To obtain the desired conversion, it would require a reactor of geological proportions (a CSTR or PFR approximately the size of the Los Angeles Basin), or as we saw in the case of the batch reactor, a very long time.




Lectures 3 and 4 Self Test What is the reaction rate law for the reaction

if the reaction is elementary? What is r B? Solution

Solution -r A = kA CA CB1/2 r B/(1/2) = r A /1 file:///H:/html/course/lectures/three/test1.htm[05/12/2011 16:59:43]


r B = r A /2 r B = -(k A /2)C A CB1/2 Back to Lectures 3 and 4

file:///H:/html/course/lectures/three/test1.htm[05/12/2011 16:59:43]

Lectures 3 and 4 -- Click Back #1

Deriving -rA: The forward rate is:

And the reverse rate law is:

The net rate for species A is the sum of the forward and reverse rate laws:

Substituting for rfor and rrev :

Return to Lectures 3 and 4

file:///H:/html/course/lectures/three/click03a.htm[05/12/2011 16:59:44]


Solving for KP : Remember, the expression we're trying to derive is that:

At equilibrium, rnet

0, so:

Solving for KP :

The conditions are satisfied.


file:///H:/html/course/lectures/three/click03b.htm[05/12/2011 16:59:44]


For Gas Phase Flow Systems: From the compressibility factor equation of state:

The total molar flowrate is:

Substituting for F T gives:


file:///H:/html/course/lectures/three/click03c.htm[05/12/2011 16:59:45]


Deriving CA and CB: Remember that the reaction is:

For a gas phase system:

If the conditions are isothermal (T = T0) and isobaric (P = P 0):

And if the feed is equal molar, then:

This leaves us with C A as a function of conversion alone:

Similarly for C B :


file:///H:/html/course/lectures/three/click03d.htm[05/12/2011 16:59:45]


The Equilibrium Constant (KC) and Equilibrium Conversion (Xe) for a Constant Volume System: You are given the reversible reaction:

which takes place in a constant volume batch reactor. The equilibrium constant, KC , for this reaction is:

where C Ae and C Be are:

Substituting for C Ae and C Be gives us:

Substituting known values (CA0 = 0.2 mol/dm 3 and KC = 100 dm 3/mol):

Solving for the equilibrium conversion, Xe, yields: Xe = 0.83


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file:///H:/html/course/lectures/five/click05a.htm[05/12/2011 16:59:46]

Lectures 5 and 6 -- click Back #2

The Equilibrium Constant (KC) and Equilibrium Conversion (Xe) for a Non-Constant Volume System: You are given the reversible reaction:

which takes place in gas phase PFR. Since gas phase reactions almost always involve volume changes, we will have to account for volume changes in our calculations. The equilibrium constant, KC , for this reaction is:

where C Ae and C Be are:

Substituting for C Ae and C Be gives us:

Substituting known values (CA0 = 0.2 mol/dm 3 and KC = 100 dm 3/mol), and realizing that:

we end up with:

file:///H:/html/course/lectures/five/click05b.htm[05/12/2011 16:59:47]


Solving for the equilibrium conversion, Xe, yields: Xe = 0.89


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Lectures 5 and 6 -- Tip #1

Tip for Using Polymath to Determine Xe in a PFR Volume is the independent variable in the PFR design equation, so you will want to vary volume until you achieve the desired conversion. The easiest way to do this is to choose a large volume (in this case, a final volume of 500 dm 3 might be a good place to start), calculate your conversion as a function of volume, and then plot conversion versus volume. Remember, the point you're looking for in the example is the volume where conversion equals 80 percent of the equilibrium conversion, or a conversion of 71.1 percent (i.e., X =0.711). Return to Lectures 5 and 6

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Lectures 5 and 6 -- Polymath Screen Shots -- Equations

Polymath Screen Shots -- Equations


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Lectures 5 and 6 -- Polymath Screen Shots -- X vs. V

Polymath Screen Shots -- Conversion versus Volume


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Lectures 5 and 6 -- Polymath Screen Shots -- Table of V and X

Polymath Screen Shots -- Table of Volume and Conversion


file:///H:/html/course/lectures/five/poly01c.htm[05/12/2011 16:59:48]


Learning Resource Additional Homework Problems CDP3-AB For this problem, the behavior of the beetle can be modeled by the Arrhenius equation. Essentially, the beetle's speed (k) increases exponentially with increases in temperature. We begin with the Arrhenius equation:

In order to calculate how fast the beetle can push the ball, we will need to determine the activation energy (E) and the Arrhenius coefficient (A) of the beetle. We will arrange our data as such: Rate Constant (cm/s)

Temperature (K)

k1 = 6.5

T1 = 300 K

k2 = 13

T2 = 310 K

k3 = 18

T3 = 313 K

k4 = ?

T4 = 314.5 K

To solve the problem graphically and get an approximate answer, we may plot ln(k) vs. 1/T. This plot should form a straight line and will predict the ln(k) for T=314.5 K.



As we can see here, ln(k) for T=314.5 is equal to ~2.95. This corresponds to a k4 value of 19.1 cm/s. Now we will solve the problem numerically to get an exact answer. By dividing k1 by k2 we can eliminate A and solve for E:

Then inserting the values for our variables (k1, k2, T1, T2), we get



Inserting this activation energy into our Arrhenius equation for k3, we can solve for our coefficient (A):

Now that we have solved for our Arrhenius constants, we can calculate the rate (k4) at which the beetle pushes the ball at 314.5 K:

Back to CD Problem 3-AB



Problem 3-A Solution Legal Statement


Problem 3-B Solution

Learning Resource Additional Homework Problems CDP3-B B The first step in setting up our stoichiometric table is to determine the stoichiometric coefficients (a, b, c, d, & e).

After making sure the elements are balanced on both sides of the above equation, the complete equation looks like:

We will assume that our reactor acts like a PFR in setting up our stoichiometric table. The stoichiometric table consists of the species name (I), species symbol (II), the flow rate in (III), the change in flow rate (IV), and the flow rate out (V). We will demonstrate how the table is set up with an example on the trichlorosilane (SiHCl 3). I. Species Our first species is trichlorosilane (SiHCl 3). II. Symbol We will give SiHCl 3 the symbol A. III. Flow Rate In The flow of A into the PFR is F Ao . IV. Change in Flow Rate The change in flow of A will be -F Ao X. Where X is conversion. V. Flow Rate Out The flow of A out of the system is F Ao (1-X).



The complete stoichiometric table is given below: Species

Symbol Flow Rate In

Change in Flow Rate

Flow Rate Out

SiHCl 3

A

F Ao

-F Ao X

F Ao (1-X)

H2

B

F Ao

-F Ao X

F Ao (1-X)

Si

C

0

0.5FAo X

0.5FAo X

HCl

D

0

2F Ao X

2F Ao X

Si 1H2Cl 2

E

0

0.5FAo X

0.5FAo X

The second part of the problem is to sketch the concentration of each species as a function of conversion. If the PFR operates at constant temperature (T) and pressure (P) then we know the total concentration (CT ) by the following relation:

Under constant temperature and pressure, the concentration of each species (Ci ) is:

Knowing that C To = C T . The total molar flow rate (F T ) is:

Where d is:

We do not include component C in this calculation because it is a solid.



Combining the above equations, the concentration of each species can be written as:

The resulting sketch looks like

Back to CD Problem 3-B B





Lectures 7 and 8 -- Click Back

Gas Phase Reaction:


file:///H:/html/course/lectures/seven/click07d.htm[05/12/2011 16:59:50]


Combine:


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Variable Density: from conservation of mass


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Converting from Length (z) to Catalyst Weight (W): Starting with this equation: (1) we begin by multiplying the top and bottom of the left-hand side of equation (1) by the same combination of constants (which is equivalent to multiplying equation (1) by one):

Since the product of is constant, we can take it inside the derivative in the denominator to combine it with our length (z):

We'll convert from reactor length (z) as our dimension to catalyst weight (W) by making use of the equation for catalyst weight:

where:

Then:

With a little rearranging:

which we substitute into equation (1) to get:

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file:///H:/html/course/lectures/seven/click07b.htm[05/12/2011 16:59:52]


Total Molar Flowrate:


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Solve: Combine

Separate Integrate with Limits X = 0 when W = 0


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Turbulent Flow


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Turbulent Flow

Taking the ratio

For constant


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ChE 344 Reactor Photos

Click on any of the items below to see scanned photos and schematics of reactors used in the chemical processing industy. Reactor system in use Amoco. Spherical Ultraforming Unit in use Amoco. 3 Spherical Ultraforming Units in use Amoco. Spherical Ultraforming Unit in series in use Amoco. Hydrotreating Unit in use at Amoco. Cutaway view of a CSTR (Courtesy of Pfaudler Inc.) Batch reactor (Courtesy of Pfaudler Inc.) Stirring apparatus for a batch reactor (Courtesy of Pfaudler Inc.) To see pictures of SASOL Reactors To see pictures of BP Reactors CRE Reactor Photos

Reactor System Used at Amoco

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Spherical Reactor

Three Spherical Reactors



Spherical Reactors Connected in Series

Hydrotreating Unit



Cutaway View of CSTR



Batch Reactor



Batch Reactor Stirring Apparatus





Constant Density:


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Des Plaines River Wetlands Project

DPRWP The Des Plaines River Wetlands Project (DPRWP) The Des Plaines River is located in southern Wisconsin and northeast Illinois (see the image, below). It runs alongside the city of Chicago and finally empties into the Illinois and Kankakee rivers.

The Des Plaines River Wetlands Project consists of several interconnected ponds, as shown in the picture (below). Water is pumped from the river so that it enters the ponds to the left side of pond EW3 and at the top of ponds EW4, EW5, and EW6. The discharge of the wetlands is controlled by weirs. Ponds EW3, EW4, and EW5 all discharge to a common outlet before reentering the river, while the discharge from pond EW6 enters a small lake before reentering the river. (The lake is just out of view in the photo, to the left of pond EW6.)

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The level of the Des Plaines River can rise and fall rather significantly. These photos (below) were taken at roughly the same location on the river. The one on the left was taken in May of 1989 and shows the river under normal conditions. The picture on the right was taken in August of 1988 and shows the river under drought conditions.

The Des Plaines River has been classified as "semipolluted." The mean concentration of its total suspended solids (TSS) was 59 mg/L, which indicates a high level of turbidity. The Des Plaines River Wetlands were able to remove as much as 88% of the sediment from river water passing through them, acting as a highly efficient natural filter for the polluted river water. Then during the summer of 1990, the TSS level rose unexpectedly. The sudden jump was attributed to a large population of carp that had recently moved into the wetlands. As the fish scavenged for food, they stirred up the pond bottoms, causing a resuspension of trapped solids in the water. It may seem cruel, but the carp were intentionally frozen out, in order to return the wetlands to normal.



The wetlands also remove other pollutants from the river water including 65 to 80 percent of the phosphorus and a goodly amount of the atrazine* . The wetlands are more effective in the summer than in the winter, in part because the concentration of phosphorus is very low in the winter. Much of the water that enters the river in the winter is run-off from frozen land, ice, or snow.

How Do Wetlands Relate to Chemical Engineering? Now that you know what a wetland can be used for, how exactly could you, as a chemical engineer, model one of these systems? Trust me, many hours have been spent trying to answer that very question. Since fluid mixing in a wetland is not very good, we won't be able to model one as a CSTR. We can, however, model wetlands as PFRs. Wetlands | DPRWP | Modeling Polymath | References * Atrazine is a herbicide that is used in the area. At times levels in the river peak above the federal drinking water standard. Atrazine was found to degrade on sediments in the wetlands according to a first-order rate law. Therefore, levels of atrazine in the outflow of the wetlands are lower than what the river inputs.


Modeling Wetlands as Plug Flow Reactors

Modeling Wetlands as Plug Flow Reactors Here's a cross-sectional view of one of the ponds from the Des Plaines River Wetlands Project:

Image Courtesy of Lisa Ingall

Let us assume that the degradation of toxic chemicals follows irreversible, first-order, homogeneous kinetics. As the waste water flows through the wetlands, the contents react and some of the water evaporates from the surface at a constant rate, Q (where Q has units of kmoles water / hr-m 2 ). We'll also asume that none of the toxic species are lost to the air by evaporation.

Symbol

Description

Value

W

Width

100 m

vo

Entering volumetric flow rate

2 m 3/hr

L

Length

1000 m

C A0

Entering concentration of toxics

10-5 mol/dm 3

D

Average Depth

0.25 m

Molar Density of Water

55.5 kmol H2O /m 3

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Q

Evaporation Rate

1.00x10-3 kmol/hr m 2

k1

Specific reaction rate

16x10-5 hr-1

First, let's derive an equation for the molar flow rate of toxics, FA , as a function of distance through the wetlands.

F A = F A0 (1-X)

(1)

(2)

(3)

-rA = k1C A

(4)

(5)

Combining equations 1, 3, and 5:

(6)

Then combining equations 2, 4, 5, and 6:

(7)

Rearranging:



(8)

Calculating the definite integrals with the boundary conditions (X=0 at z=0 and X=X at z=z):

(9)

Rearranging and solving the equation for X(z) gives:

(10)

Finally, placing this result back into the original equation for FA0 (Equation 1) leads to the equation that we are looking for:

(11)

Also, if we consider the case with no evaporation or rain (Q=0 and therefore v=v o ), the student can derive the following solution:

(12)

Questions: 1. Now that we have these equations, what would F be at z = 100 m and also z = 1000 m? (Answer) file:///H:/html/web_mod/wetlands/model.htm[05/12/2011 17:00:01]


A

2. What would a plot of conversion and reaction rate as a function of distance look like? (Answer) 3. How could you solve this problem using polymath? Wetlands | DPRWP | Modeling Polymath | References


Modeling Wetlands Using Polymath

Modeling Wetlands Using Polymath The polymath file required to solve this problem could be written as shown below:

Using this as our model, we can generate a lot of data, including plots of: molar flowrate (F A ) versus length (z) conversion (X) versus length (z) reaction rate ( r A ) versus length (z) Wetlands | DPRWP | Modeling Polymath | References

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Wetlands, References

References References Kadlec, Robert H. & Knight, Robert L., Treatment Wetlands. Boca Raton: CRC Press, Inc., 1996. [ All of the pictures in the Wetlands web module were taken by Professor Robert Kadlec or one of his graduate students. They were taken at the Des Plaines River Wetlands, and Professor Kadlec has given us permission to use them. ] Fogler, H. Scott, Elements of Chemical Reaction Engineering, 3rd Ed. Upper Saddle River, NJ: Prentice-Hall, 1998.

Problem History This page was created by Scott J. Conaway during Winter Semeter 1997 as a special project for Professor H. Scott Fogler. These pages were further edited by Brad Lintner for use on the Chemical Reaction Engineering Web Site during Fall Semester 1997, and again during Summer 1998 by Dieter Andrew Schweiss for the CD-ROM to accompany the 3rd Edition of Elements of Chemical Reaction Engineering. Wetlands | DPRWP | Modeling Polymath | References

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Introduction

Introduction What is a membrane reactor? A membrane reactor is really just a plug-flow reactor that contains an additional cylinder of some porous material within it, kind of like the tube within the shell of a shell-and-tube heat exchanger. This porous inner cylinder is the membrane that gives the membrane reactor its name.

The membrane is a barrier that only allows certain components to pass through it. The selectivity of the membrane is controlled by its pore diameter, which can be on the order of Angstroms, for microporous layers, or on the order of microns for macroporous layers.

Why use a membrane reactor? Membrane reactors combine reaction with separation to increase conversion. One of the products of a given reaction is removed from the reactor through the membrane, forcing the equilibrium of the reaction "to the right" (according to Le Chatelier's Principle), so that more of that product is produced. Membrane reactors are commonly used in dehydrogenation reactions (e.g., dehydrogenation of ethane), where only one of the products (molecular hydrogen) is small enough to pass through the membrane. This raises the conversion for the reaction, making the process more economical.

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Introduction

What kinds of membrane reactors are available? Membrane reactors are most commonly used when a reaction involves some form of catalyst, and there are two main types of these membrane reactors: the inert membrance reactor and the catalytic membrane reactor. The inert membrane reactor allows catalyst pellets to flow with the reactants on the feed side (usually the inside of the membrane). It is known as an IMRCF, which stands for Inert Membrane Reactor with Catalyst on the Feed side. In this kind of membrane reactor, the membrane does not participate in the reaction directly; it simply acts as a barrier to the reactants and some products. A catalytic membrane reactor (CMR) has a membrane that has either been coated with or is made of a material that contains catalyst, which means that the membrane itself participates in the reaction. Some of the reaction products (those that are small enough) pass through the membrane and exit the reactor on the permeate side. Main | Introduction | Algorithm Example | Comparison | Credits Reference: Fogler, H. Scott. Elements of Chemical Reaction Engineering, 3rd Ed. Prentice-Hall: Upper Saddle River, NJ, 1998.

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Algorithm

Algorithm The basic algorithm for solving reaction engineering problems is described below. This algorithm is a useful tool, and it can be applied to a wide variety of reactor problems, not just membrane reactor problems. For demonstration purposes, we'll examine a membrane reactor in which the following gas phase reaction occurs:

Product B diffuses through the membrane, but reactant A and product C do not.

1. Mole Balance: For a differential mole balance on A in the catalytic bed at steady state: IN (by flow) - OUT (by flow) + Generation = Accumulation

Dividing by

and taking the limit as

gives:

Similarly, a differential mole balance on C in the catalytic bed at steady state will give: IN (by flow) - OUT (by flow) + Generation = Accumulation

Dividing by


gives:

The steady state, differential mole balance on B looks slightly different, since B is the only species that passes through the membrane: IN (by flow) - OUT (by flow) + Generation - OUT (by diffusion) = Accumulation

where RB is the molar flowrate of B through the membrane per unit volume of the reactor. Dividing by


file:///H:/html/web_mod/membrane/algo.htm[05/12/2011 17:00:04]

gives:

Algorithm

2. Rate Law: The rate of disappearance of reactant A follows the rate law:

where k is the specific reaction rate constant, and KC is the equilibrium constant. Products B and C obey the following rate laws:

3. Transport Law: The transport or flux of species B through the membrane follows the transport law:

where km is a mass transport coefficient for the flow of product B through the membrane.

4. Stoichiometry: For gas-phase reactions:

The subscript o indicates initial conditions and v is the volumetric flow rate. The concentrations, in terms of molar flow rates, are:

Substituting for the volumetric flow rate, we get:


Algorithm

If we make use of the fact that:

we can get our concentrations in terms of the total initial concentration:

Quite often we can make the assumption that the reactor operates isothermally and isobarically:

5. Combine: Substituting the concentration terms into the rate law yields:

where the total molar flow rate is:

and:


Algorithm

Main | Introduction | Algorithm Example | Comparison | Credits


Examples

Sample Membrane Reactor Problem A membrane reactor is used for the reaction but not to reactant A or product C.

, where the membrane is permeable to product B,

Additional Information: P0

6.0 atm

T0

373 K

k KC

0.7 min-1

km FA0

0.2 min-1 15 mol/min

FB0 = F C0

0 mol/min

0.05 mol/dm3

Consider the following questions: 1. What volume is required for the base case membrane reactor? Solution

2. What if the membrane transfer coefficient, km , were 0.002 min -1? Compare plots of molar flowrates versus volume and conversion versus volume for this case with your base case. Solution

3. What if the membrane transfer coefficient, km , were 20.0 min -1? Compare plots of molar flowrates versus volume and conversion versus volume for this case with your base case. Solution

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Examples

4. What if the base case flowrate were changed from 15 mol/min to 5 mol/min? How would this affect the behavior of the membrane reactor? Solution

5. What if the base case flowrate were changed from 15 mol/min to 25 mol/min? How would this affect the behavior of the membrane reactor? Solution Main | Introduction | Algorithm Example | Comparison | Credits

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Comparison, Part 1

Comparison Comparing Membrane Reactors with PFRs, Part 1 The obvious difference between membrane reactors and plug flow reactors is the absence of a membrane in PFRs. The only difference between our treatment of membrane reactors and PFRs is in the term R B. Specifically, km , which reflects the ease or difficulty of transport through a membrane, will be zero for a PFR, since no membrane is present in a PFR. Therefore, R B will also be zero for a PFR. The following comparison is based on the conditions given in the membrane reactor example problems.

Break It Down for Me: List of Equations Summary Tables Membrane Reactor PFR Flowrates versus Volume Membrane Reactor PFR Conversion versus Volume Membrane Reactor PFR Comparison, Part 2

List of Equations The same equations are used by both membrane reactors and PFRs (note that km will be set to zero for a PFR).

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Comparison, Part 1

Summary Table -- Membrane Reactor


Comparison, Part 1

Summary Table -- PFR


Comparison, Part 1

FA, FB, and FC versus Volume -- Membrane Reactor


Comparison, Part 1

FA, FB, and FC versus Volume -- PFR


Comparison, Part 1

Conversion versus Volume -- Membrane Reactor


Comparison, Part 1

Conversion versus Volume -- PFR


Comparison, Part 1

Forward to Comparison, Part 2 Main | Introduction | Algorithm Example | Comparison | Credits


Membrane Reactors -- Credits

Credits This site was originally presented as an Open-Ended Problem in the Winter 1997 Chemical Reaction Engineering Class at the University of Michigan. The students who developed this module were Kim Dillon, Namrita Kumar, Amy Miles, and Lynn Zwica. The module was further expanded and improved by Ellyne Buckingham, Dieter Andrew Schweiss, Anurag Mairal, and H. Scott Fogler for use with the Chemical Reaction Engineering Web Site and CD-ROM. Main | Introduction | Algorithm Example | Comparison | Credits

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review.htm

For this module there are a couple of topics that need to be reviewed or dicussed: 1) Semibatch reactor equations 2) Modifications due to reactive distillation

Mole Balance Shown above is a representation of a typical semibatch reactor. B is fed to A, which is already in the reactor, and there are no outputs (evporation included). To model a semibatch reactor, start with a mole balance on each species :

Starting with species A :

C and D are similar :

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review.htm

Species B is fed to the reactor making its mole balance equation slightly different :

We now have all the equations needed to describe the changes in concentration of the different species in the reactor over time. However, since there is a feed stream, the volume in the reactor will change. So, we need to develop an equation describing how the volume in the reactor changes with time.

Mass Balance Starting with a mass balance and remebering that mass can neither be created nor destroyed :

In terms of the system :

The mass terms in the equation can be replaced by :

giving :


review.htm

Assuming that the system is at constant density leads to :

That is the last equation needed to model the semibatch reactor.

Summary

Mole Balance

Mass Balance

Rate Law

Now we can move on and talk about reactive distillation.


intro.htm

This module is a learning resource for Chapter Four of Elements of Chemical Reaction Engineering. Reactive distillation is used with reversible, liquid phase reactions. Suppose a reversible reaction had the following chemical equation :

For many revesible reactions the equilibrium point lies far to the left and little product is formed :

However, if one or more of the products are removed more of the product will be formed because of Le Chatlier's Principle :

Removing one or more of the products is one of the principles behind reactive distillation. The reaction mixture is heated and the product(s) are boiled off. However, caution must be taken that the reactants won't boil off before the products. For example, Reactive Distillation can be used in removing acetic acid from water. Acetic acid is the byproduct of several reactions and is very usefull in its own right. Derivatives of acetic acid are used in foods, pharmaceuticals, explosives, medicinals and solvents. It is also found in many homes in the form of vinegar. However, it is considered a polutant in waste water from a reaction and must be removed.

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You have been given a summer internship by J. Flabetes Inc., a somewhat well known chemical company located in sunny California. Your last project for the summer is to study a proposed method of removing acetic acid from the waste water. The acetic acid is to be removed by reacting it with methanol in a catalyst filled semibatch reactor :

It has been suggested that reactive distillation be included in the plans, but your boss is kind of skeptical. Your job is to simulate the reaction and see if reactive distillation is a better option than without. We will begin by looking at the semibatch reactor case with no reactive distillation.

No Reactive Distillation

Mole Balance

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Mass Balance

Rate Law

We can use and Ordinary Differential Equation solver such as Polymath to model the reaction.

Polymath Click here for equations

Graph From the results we see that after 120 minutes, 116 moles of acetic acid is left in the reactor from the original 300. This is a conversion of 61.3 %

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ref.htm

CRC Handbook of Chemistry and Physics, 73rd Edition. Ed. David R. Lide. CRC Press. Boca Raton, Florida. 1992 Fogler, H. Scott. Elements of Chemical Reaction Engineering, 3rd Edition. Prentice Hall. Upper Saddle River, NJ. 1998 Howard, Phillip H., and William M. Meylon.. Handbook of Physical Properties of Organic Chemicals. CRC Press. Boca Raton, Florida. 1997. pg 595. Perry's Chemical Engineering Handbook, 6th Edition. Ed. Robert H. Perry and Don Green. McGraw-Hill Inc. New York. 1984 Ullmann's Encyclopedia of Industrial Chemistry. Ed. Wolfgang Gerhartz et. al. VCH Publishers. Deerfield Beach, Florida. 1985. Xu, Z. P., and K. T. Chuang. Kinetics of Acetic Acid Esterification over Ion Exchange Catalysts. The Canadian Journal of Chemical Engineering. Vol. 74 Aug:pg493-500. 1996.

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CRE -- Problem 4-A Solution

CD P4-AB First, we will list the known values: AgClO 4 + CH 3I A+B

CH 3ClO 4 + AgI C+D

Batch Process V = 30 dm 3 C Ao = 0.5 mol/dm 3 C Bo = 0.7 mol/dm 3 rB = -kC B C A3/2 at T = 298 K k = 0.00042 (dm3/mol)3/2 /s Final X = 0.98 Mole Balance

Rate Law rA = rB = -kC B C A3/2 Stoichiometry

Combine Combining the above equations, we arrive at:



Evaluate The final equation, with all values entered, is:

This equation can only be solved numerically, unless you really enjoy integration. The Polymath simultaneous differential equation solver may be used to solve the problem quickly and easily. The Polymath equations used should look like this:



After Polymath solves the equations, it gives a chart of initial, maximum, minimum, and final values:



From these results, we can see that the final time (and therefore the time to reach 98% conversion) is t = 162531 s = 45.15 hr Back to problem 4-a




4.Identify and Name: A. Relative Theories and Equations

Differential Form of PFR equation: see derivation of PFR equation B. Systems Volume of PFR C. Dependent and Independent Variables Independent: VPFR, F A0 , F B0, T Dependent: conversion (X) D. Knowns and Unknowns Knowns: VPFR, vA0 , vB0, F A0 , F B0, T, X see values of knowns Unknowns: Specific reaction rate constant, k E. Inputs and Outputs In: F A0 , F B0 Out: F A, F B 5.Assumptions/Approximation Isothermal: no temperature change No volume change on mixing 6. Specifications This problem is overspecified: heat capacity, viscosity and boiling point of the components.

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7.Algorithm:

Mole Balance on the PFR: evaluate liquid phase PFR equation Rate Law: Stoichiometry: Equal Molar thetaB = 1 step through stoichiometry to arrive at ...

Combine:

step through combination of equations Evaluate: rearrange to solve for k

evaluate k Forward to Part 2 : CSTR & PFR Back to Original Problem



4. Identify and Name:

A. Relative Theories and Equations Algebraic Form of CSTR equation:

derive CSTR volume For reactors in series, such as a PFR after a CSTR, the integral form of a PFR equation becomes:

derive PFR volume B. Systems Volume of PFR Volume of CSTR C. Dependent and Independent Variables Independent: VCSTR , VPFR , FA0, FB0 , T Dependent: k, X1 , X2 D. Knowns and Unknowns Knowns: same as Part I values of knowns Unknowns: conversion from CSTR, X1 conversion from PFR, X2



E. Inputs and Outputs In: same as Part I Out: same as Part I 5.Assumptions/Approximations same as Part I 6.Algorithm: Same as Part I

Combine: steps in combining Evaluate:

evaluating for X1

evaluating for X2 Check units, X is dimensionless. With a CSTR upstream of the PFR, the conversion increased from 0.5 to 0.684. Back to Part 1 : PFR Back to the Original Problem


05FTN.htm

Footnotes for Chapter 5, Professional Reference Shelf 1 A.

C. Norris, Computational Chemistry: An Introduction to Numerical Solution (New York: Wiley, 1981). 2 D. M. Himmelblau, Process Analysis by Statistical Methods (New York: Wiley, 1970), p. 195. 3 Norris, Computational Chemistry, p. 293. 4 G. E. P. Box, W. G. Hunter, and J. S. Hunter, Statistics for Experimenters: An Introduction to Design, Data Analysis, and Model Building (New York: Wiley, 1978). 5 S. N. Deming, Chemtech, p. 118, February 1990. 6 Charles D. Hendrix, Chemtech, p. 167, March 1979. 7 W. Strunk and E. B. White, The Elements of Style (New York: MacMillian, 1979). 8 V. W. Weekman, AIChE J., 20, 833 (1974). 9 Ibid.


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Optimum Yield with Multiple Reactions: To maximize the amount of B produced, we need to differentiate our function for CB. Setting the differential equation equal to zero, we can solve for the time to reach this optimum, t opt.

Solving for t opt gives


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For Liquid Phase Reactions in a PFR: PFR Design Equation

for liquid phase systems then substituting for F A then


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Background Information

Background Information The country of Thailand is half a world away from the United States. It is a land of vast jungles, busy cities, friendly people, and ancient traditions. Unfortunately, it is also the home of some of the most dangerous snakes in the world, and many people die each year in Thailand as a direct result of being bitten by such snakes. (But don't let that stop you from visiting Thailand! It's a beautiful country.) Until recently, the medical world was uncertain about the mechanism by which snake venom attacked the human body. Fortunately, research in this area has progressed rapidly in the past ten years, and medical science has been able to determine what happens when someone is bitten by a poisonous snake. This knowledge has resulted in the development of antivenoms that can save snake bite victims, but only if they are injected in time with an appropriate dosage of the correct antivenom.

A close-up view of a King Cobra (Ophiophagus hannah)! Two of the deadliest snakes in Thailand are the King Cobra (Ophiophagus hannah) and the Siamese Cobra (Naja siamensis). The King Cobra's venom is one of the most lethal and fast-acting neurotoxins (i.e., it attacks the nervous system) found in nature, and a single bite can incapacitate some one's respiratory system within 30 minutes. Death would then swiftly follow, unless an injection of antivenom were immediately available. The bite of the Siamese Cobra, which is primarily neurotoxic, is even deadlier. It can cause paralysis, nausea, and difficulty in breathing. Without treatment, heart and breathing failure results in death 1. 1

This information was graciously provided by Mr. Jaruwan Liwsrisakul.

Human Respiration

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Background Information

Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss

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Effects of Cobra Venom

Effects of Cobra Venom Cobras have several meathods for delivering their deadly venom to their prey. Some cobras can spit their venom into a victim's eyes, causing extreme pain and blindness. However, the most common and well known method of venom delivery is injection into a victim's body through their bite.

A cobra striking! (Quicktime Movie - 2.2MB) NOTE: You will need the QuickTime Plug-in to view this movie.

Cobras belong to the sub-group of snakes known as elapids; there are over 270 species of cobras and their relatives. An elapid's venom contains postsynaptic neurotoxins that spread rapidly in its victim's bloodstream, causing respiratory failure and, eventually, death. Cobra venom is an example of a molecule that prohibits the interaction of acetylcholine molecules (transmitted from nerve endings surrounding the diaphragm muscle) with the receptor sites on the diaphragm muscle. (See the section on Human Respiration for more details). It binds to the receptor sites, blocking them from interacting with acetylcholine molecules. Even worse, the venom molecule will not immediately break down and vacate the receptor site, effectively removing the site from active duty. It has been determined that even if only a third of the receptor sites on your diaphragm become blocked by venom, you will cease breathing. With cobra venoms, this process can take as little as 30 minutes. The only way to counteract the effects of cobra venom (or most other poisonous snake venoms) is to inject the appropriate antivenom shortly after the bite occurs. If antivenom is unavailable, your life can still be saved by putting you on an artificial respirator until the paralysis of the diaphragm muscle wears off.

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Effects of Cobra Venom

(If this watered-down explanation of effects of cobra venom wasn't enough for you, then check out a more-detailed explanation of the effects of cobra venom.) Antivenom Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss

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Engineering Aspects

Engineering Aspects of Cobra Venom Reactions The interaction of the venom and the antivenom with the receptor sites can be modeled as a reaction engineering catalysis problem. Examples of the reactions are given below.

Adsorption of venom onto site:

Adsorption of antivenom onto site:

Reaction of venom with antivenom on site: file:///H:/html/web_mod/cobra/reaction.htm[05/12/2011 17:00:16]

Engineering Aspects

Reaction of antivenom with venom on site:

Reaction of venom and antivenom in blood:

Removal of product and reactants from system:

file:///H:/html/web_mod/cobra/reaction.htm[05/12/2011 17:00:16]

Engineering Aspects

where: V = venom A = antivenom S = unoccupied receptor site VS = site occupied by venom AS = site occupied by antivenom AV = neutralized product from venom/antivenom reaction Developing the Equations Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss

file:///H:/html/web_mod/cobra/reaction.htm[05/12/2011 17:00:16]

Problem Statement

Problem Statement This module on cobra venom and its behavior in the human body was originally given as an open-ended problem, or OEP, during Winter Semester 1994 at the University of Michigan. In the spirit of truely open-ended problems, students were simply asked to investigate the effects of being bitten by a poisonous snake. A base case for the problem and some possible starting suggestions were given to the students, but the emphasis was on creativity in exploring the problem.

Base Case Initially, you should investigate the case where a human is bitten by a poisonous snake, but no antivenom is injected. Plot the fraction of free sites in the body as a function of time. You should be able to verify the time it would take for 1/3 of the receptor sites to be blocked by venom, which would then result in the death of the victim by respiratory failure. Look at the changes that occur when antivenom is injected: How long can you wait to inject the antivenom? How much antivenom should be injected into the victim? Explain the behavior of your graphs.

Other Suggestions After you have investigated the base case, look at several variations of the problem. Some problems that you could investigate are given below; however, this list of problems is not exhaustive. Remember creativity and effort are very important! 1. Can a large dose of antivenom be lethal? What if a large dose of antivenom is injected into someone who has not been bitten by a cobra? Is there a way to reverse the effects of an antivenom overdose? 2. What happens if you are bitten by a cobra twice? 3. What if the antivenom were injected slowly, over a period of 1 to 4 hours? 4. What if you were bitten by a cobra and received proper antivenom treatment, and then, a short time later, you were bitten again? Can antivenom be given? If so, how much should be given? 5. What if the venom from the snake was injected into the muscle instead of directly into the bloodstream? How would the rate of diffusion of venom into the blood affect the time for antivenom treatment? What concentrations of antivenom could be safely administered? 6. Can you find any data or information that provides more accurate rate constants for the reactions? file:///H:/html/web_mod/cobra/problems.htm[05/12/2011 17:00:17]

Problem Statement

Is there a better way to define the system? Investigate the technical aspects of this problem and discuss any changes or improvements. How do the changes you make compare with the information you found when solving the base case? Intimidated yet? Well, don't be. The beauty of an open-ended problem is that the sky's the limit. You can explore any possibility, as long as it's realistically possible to model its behavior. It also helps you develop your creative skills, which are important for real world applications. Take a look at our coverage of the base case solution, and then read our brief look at applying the open-ended problem solving aglorithm to this problem. We encourage you to let your imagination be your guide! Let's apply the Open-Ended Problem Algorithm to this problem! Exploring the Base Case References Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss

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References

Problem History Pharmacokinetics of Snake Bites was created by Susan Stagg to be used as an Open-Ended Problem (OEP) by Professor H. Scott Fogler at the University of Michigan. It was given as an OEP to the undergraduate kinetics class during Winter Semester 1994. This problem was also featured in Open-ended Problems in Chemical Reaction Engineering, a set of problems that accompanied the 2nd edition of Professor Fogler's text, Elements of Chemical Reaction Engineering. The cobra problem was converted into a web document for the Chemical Reaction Engineering Web Site by Gavin Sy and Dieter Andrew Schweiss, who wish to thank Susan Stagg for all of her hard work in creating the cobra problem in the first place. Gavin created the frames set for the CRE Web Site version of the Cobra Problem, entered all of the original text from Susan Stagg's problem, and converted amateur video into the cobra movies. Dieter rewrote the text of the Cobra Problem to make it more "user friendly" (e.g., the background information now starts from a "layperson's" point of view, but allows advanced readers to go into more detail by reading the original text for the problem); interpreted the original equations and created the page that explains where they come from (including all of the images); applied the Open Ended Problem Solving Algorithm to the Cobra Problem; solved a base case for the open-ended problem presented in this document; expanded on the base case with a few examples; put together the whole solution to the base case (including the Polymath screenshots); got bit by a cobra to prove that it only takes 30 minutes for respiratory paralysis to occur (just kidding!); and generally wrapped up all the loose ends. Dieter also rewrote the HTML coding for the CD-ROM version of the Cobra Problem. (Let's just say that I was busy.)

References Cheung, Johnson, and Taylor. "Kinetics of Interaction of v epsilon-flourescein isothiocyanate-lysine-23-cobra alpha toxin with the Acetylcholine Receptor." Biophysics Journal, February 1984, pp. 447-454. Ferreia and Joaa. "Influence of Chemistry in Immobilization of Cobra Venom Phospholipase A 2 : Implications as to Mechanism." Biochemistry, 32, 8099, [1993]. Fogler, H. Scott. Elements of Chemical Reaction Engineering. 3rd Ed., Prentice-Hall, Englewood Cliffs, NJ, 1998. Gill, Paul G. Pocket Guide to Wilderness Medicine & First-Aid. Ragged Mountain Press [McGraw-Hill], Camden, Maine, 1997. Greene, Harry W. Snakes: The Evolution of Mystery in Nature. University of California Press, Berkeley, 1997. Malasit, P. "Prevention and Mechanism of Early (anaphylactic) Antivenom Reactions in Victim of Snake Bites." British Medical Journal. January 4, 1986, p. 292. Ortiz, Angela. "Implications of a Consensus Recognition Site for Phosphatidylcholine Separate from the Active Site in

file:///H:/html/web_mod/cobra/ref.htm[05/12/2011 17:00:17]

References

Cobra Venom Phospholipases A2 ." Biochemistry, 31, 2887-2896, [1992]. Thwin, M. M. "Kinetics of Envenomization with Russel's Viper Venom and of Antivenom in Mice." Toxicon, 26(4), 373-378, [1988].

Links Here are a few links to cobra-related sites that you might try. (NOTE: CD-ROM users will need to be connected to the internet for these links to work correctly.) A National Geographic site on King Cobras. A Federal Food and Drug Administration Page on Treating Venomous Snake Bites. The Cobra Information Site. Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss

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Pseudo Steady State Hypothesis (PSSH): Given that for reaction (3): Assume that reactions (1) and (2) are elementary reactions, such that:

The net reaction rate for

is the sum of the individual reaction rates for

The PSSH assumes that the net rate of

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is zero:

:



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Michaelis-Menten Kinetics:


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derive1

Calculation of number average molecular weight Basis: 100 grams total


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p = fraction of carboxyl groups reacted

Multiplying by M o and rearranging


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click36g


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click36h


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click36i

Overall balance on all polymers (j=1 to ¥ )

(1)

(2)

(3)

(4)

Let's look at the first term (1)

Let's look at the second term (2 )

Let's look at the third term (3)

Now consider all the terms in brackets

then


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click36j

Mole balance on P1

Integrating


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click36k

Mole balance on P2


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Example: Reasons for "f" less than 1.0 1. Chain transfer to initiator

However, even though a radical is generated, this reaction does result in a wastage of initiator, there is no increase in the number of radicals or the amount of monomer being converted to polymer. 2. Side reactions Cage effects

[ ]

Most significant reaction in the cage is

[ ] = cage where radicals are held for some time before they diffuse out. Once outside the cage the radicals can react with monomer [2 f COO•] 2 f COO• Other reactions

of secondary importance

Average lifetime of neighboring radicals is 10–10 to 10–9 seconds with kf = 107 dm 3 /mol/s In the cage: The concentration of radicals in the solvent cage ~ 10 mol/dm3 Out of cage: CMonomer = 0.1 to 10 mol/dm3 CR1 = 10–9 to 10–7 mol/dm3 3. Recombination of primary radicals has no effect on initiator efficiency

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Derive: ka are defined wrt reactant.

The net rate of termination of all radicals is


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07ftn.htm

FOOTNOTES FOR CHAPTER 7 1

H. Scott Fogler, Elements of Chemical Kinetics and Reactor Calculations, Prentice-Hall, New Jersey, 1974. 2

R. G. Carbonell and M. D. Kostin, AIChE J., 18, 1 (1972).

3

H. Theorell et al., Acta Chem. Scand., 9, 1148 (1955).

4 K.

Dalziel, Acta Chem. Scand., 11, 1706 (1957).

5

N. K. Gupta and W. G. Robinson, Biochim. Biophys. Acta, 118, 431 (1966).

6

P. H. Calderbank and M. B. Moo-Young, Trans. Inst. Chem. Eng., 37, 26 (1959).

7

N. Mitsuishi and N. Hirai, J. Chem. Eng. Japan, 2, 217 (1969).

8

Y. Oyama and K. Endoh, Kagaku Koyaku, 19, 2 (1955).

9

P. H. Calderbank and M. B. Moo-Young, Trans. Inst. Chem. Eng., 37, 26 (1959).

10

N. Mitsuishi and N. Hirai, J. Chem. Eng. Japan, 2, 217 (1969).

11D.

W. Hubbard, L. R. Harris, and M. K. Wierenga, Chem. Eng. Prog., 84 (8), p. 55 (1988).

12

S. P. Rogovin, V. E. Sohns, and E. L. Griffin, Ind. Eng. Chem., 53, 37 (1961).

13

D. C. Wang et al., Fermentation and Enzyme Technology, Wiley, New York, 1979.

14

T. J. Bailey and D. Ollis, Biochemical Engineering, 2nd ed., McGraw-Hill, New York, 1987.



Manipulating the Energy Exchange Term

Combining:


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Using the Taylor Series Approximation


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Derive the Steady State Energy Balance (w/o work)

Differentiating with respect to W:

Mole Balance on species i:

Enthalpy for species i:

Differentiating with respect to W:


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Derive

Factor F A0 C P0 and then divide by F A0

For a CSTR: F A0X = -rA V


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Balance on a system volume that is well-mixed:


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For Liquid Phase Flow Through A CSTR: Applying the General Mole Balance Equation to a CSTR yields:

Since

we can substitute for F A0 , F A, and NA to get:

The reactor volume will be constant, so we can divide through by V to get:

And for liquid phase systems

, so:


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sol8-2a.htm

Learning Resources

Example CD8-2 Solution, Part A Second Order Reaction Carried Out Adiabatically in a CSTR

(a) We will solve part (a) by using the nonisothermal reactor design algorithm discussed in Chapter 8. 1. CSTR Design Equation:

2. Rate Law:

3. Stoichiometry: liquid,

4. Combine:

Given conversion (X), you must first determine the reaction temperature (T), and then you can calculate the reactor volume (V). file:///H:/html/08chap/html/sol8-2a.htm[05/12/2011 17:00:29]

sol8-2a.htm

5. Determine T: From the adiabatic energy balance (as applied to CSTRs):

which reduces to:

Substituting for known values and solving for T:

6. Solve for the Rate Constant (k) at T = 380 K:

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7. Calculate the CSTR Reactor Volume (V): Recall that:

Substituting for known values and solving for V:

Continue with solution...

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08ftn.htm

FOOTNOTES FOR CHAPTER 8 1 V.

Balakotaiah and D. Luss, in Chemical Reaction Engineering, Boston ACS Symposium Series 196 (Washington, D.C.: American Chemical Society, 1982), p. 65; M. Golubitsky and B. L. Keyfitz, SIAM J. Math. Anal., 11, 316 (1980); A. Uppal, W. H. Ray, and A. B. Poore, Chem. Eng. Sci., 29, 967 (1974).

2 From

a problem of R. A. Schmitz, Notre Dame University, South Bend, Indiana.



09ftn.htm

FOOTNOTES FOR CHAPTER 9 1 B.

Carnahan, H. A. Luther, and J. O. Wilkes, Applied Numerical Methods (New York: Wiley, 1969).




Deriving an Expression for rs:

A site balance yields:

Grouping the equilibrium constants together into one constant, KP :

And grouping the remaining constants together as a single constant, k:

We get:

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Deriving an Expression for the Concentration of Species A on the Surface Beginning with our expression for the rate of adsorption:

If the surface reaction is limiting, then:

Recall that:

and

Then:

Multiply both sides by C t and we're left with:


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Deriving an expression for r'A: Assume that the surface reaction is limiting, then:

where: = fraction of vacant sites = fraction of sites occupied by species i

Similarly:

where:

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Surface reaction limiting:


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For the homogeneous reaction:

Let:

then


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10ftn.htm

Footnotes for Chapter 10 1 R.

F. H. Ward, Proc. R. Soc. London, A133, 506 (1931)

2 Named

after Irving Langmuir (1881-1957), who first proposed it. He received the Nobel prize in 1932 for his discoveries in surface chemistry.

3 B.

Chapman, Glow Discharge, Wiley, New York, 1980.

4 A.

Demos and H.S. Fogler, AIChE J., 41(3), 658 (1995)

5 W.

E. Kline and H. S. Fogler, Chem. Eng. Sci., 36, 871 (1981).

6 Ibid.




Deriving the Dimensionless Equation: We begin with this equation:

Let's define our dimensionless variables as:

Before we can make any substitutions, we have to do a little setup:

Now we can make our substitutions:

and we can collect our constants into one term:

Our dimensionless equation is then:

where:

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11ftn.htm

Footnotes for Chapter 11, Professional Reference Shelf

1 N.

Arashi, Y. Hishinuma, K. Narato, F. Nakajima, and H. Kuroda, Int. Chem. Eng., 22(3), 489C (1982).

2

For a discussion of the catalytic muffler, see J. Wei in Chemical Reaction Engineering Reviews, H. M. Hulburt, ed., American Chemical Society, Washington, D.C., 1974, p. 1.

3

C. N. Satterfield and D. H. Cortez, Ind. Eng. Chem. Fund. , 9, 613 (1970); D. Roberts and G. R. Gillespie, in Chemical Reaction Engineering II, H. M. Hulburt, ed., Adv. Chem., 133, 600 (1974). 4

T. Shimizu, O. Imai, Y. Sakakibara, and T. Ohrui, Int. Chem. Eng., 22, 329. (1982)



12ftn.htm

Foonotes for Chapter 12, Professional Reference Shelf 1A

number of worked example problems for three-phase reactors can be found in an article by P. A. Ramachandran and R. V. Chaudhari, Chem. Eng., 87(24), 74 (1980).

2 C.

N. Satterfield, AIChE J., 21, 209 (1975); M. Herskowitz and J. M. Smith, AIChE J., 29, 1 (1983); G. A. Hughmark, Ind. Eng. Chem. Fund., 19, 198 (1980); Y. T. Shah, B. G. Kelkar, S. P. Godbole, and W. D. Deckwer, AIChE J., 28, 353 (1982). 3 See 4 P.

Foonote 2

A. Ramachandran and R. V. Chaudhari, Chem. Eng., 87(24), 74 (1980).

5 M.

Herskowitz and J. M. Smith, op. cit.; F. Turek and R. Lange, Chem. Eng. Sci., 36, 569 (1981).

6 Ramachandran 7 M.

and Chaudhari, op. cit.

O. Tarhan, Catalytic Reactor Design, McGraw-Hill, New York, 1983, p. 189.

8 This

material is based on the article by H. S. Fogler and L. F. Brown, in Reactors, ACS Symposium Series 168, H. S. Fogler, ed. (Washington, D.C.: American Chemical Society, 1981), p. 31, which in turn was based on a set of notes by Fogler and Brown.

9 D.

Kunii and O. Levenspiel, Fluidization Engineering (New York: Wiley, 1968).

10 T.

E. Broadhurst and H. A. Becker, AIChE J. 21, 238 (1975).

11C.Y.

Wen and Y.H. Yu, AIChE J. 21, 610 (1966)

12Kunii

and Levenspiel, Fluidization Engineering.

13Ibid. 14J.F.

Davidson and D. Harrison, Fluidized Particles (New York: Cambridge University Press, 1963).

15Kunii


16Davidson 17S.

and Harrison, Fluidized Particles.

Mori and C. Y. Wen, AIChE J. 21, 109 (1975).

18

J. Werther, in ACS Symposium Series 72, D. Luss and V. W. Weekman, eds. (Washington, D.C.: American Chemical Society, 1978).<

19Mori

and Wen, AIChEJ 21.

20Ibid.


12ftn.htm

21Kunii


22Ibid 23

Ibid.

24Ibid. 25L.

Massimilla and R. F. Johnstone, Chem. Eng. Sci. 16 (105) (1961).

26K.

F. Jensen, Chem. Eng. Sci., 42, 923 (1987).

27K.

F. Jensen, J. Electrochem. Soc., 130, 1450 (1983).

28H.

S. Mickley, T. K. Sherwood, and C. E. Reed, Applied Mathematics in Chemical Engineering, McGraw-Hill, New York, 1957, p. 174.



excd12-1.htm

Professional Reference Shelf Example CD12-1 Trickle Bed Reactor The hydrogenation of an unsaturated organic is to be carried out in a trickle bed reactor packed with 0.20-cm-diameter spherical catalyst particles.

The reaction in the pellet is first-order in both hydrogen and the organic. Hydrogen and nitrogen are fed in equimolar portions at a total pressure of 20 atm and a total molar rate of 10 mol/s. The reactor diameter is to be 1.0 m. The superficial liquid mass velocity is 5.0 kg/m 2 s. The corresponding pressure gradient through the bed is 25 kPa/m. As a first approximation, assume that the concentration of organic is constant and the pseudo-first-order specific reaction rate is 3 x 10-5 m 3/ kg cat. s at 400 K. (a) For each transport step, determine its fraction of the total resistance to mass transport and reaction. (b) Calculate the catalyst weight necessary to achieve 55% conversion of the hydrogen. Additional information:

(a) Let A = H2, B = unsaturated organic, and C = saturated organic:


excd12-1.htm

1.Mole balance on H 2 (A):

(CDE121.1)

2. Rate law.Assuming a constant liquid reactant concentration for low conversion of B, (CDE121.2)

with

(CDE121.3)

3 .Stoichiometry. The isothermal gas-phase concentration is


(CDE121.4)

excd12-1.htm

4. Pressure drop:

(CDE121.5)

5. Combining yields

(CDE121.6)

Integrating gives us

(CDE121.7)

6. Evaluating the parameters: A.Solubility

(CDE121.8)

B. Internal diffusion and reaction

(CDE121.9)

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excd12-1.htm

For large values of the Thiele modulus,

(12-1)

(CDE121.11)

C. Gas absorption

(CDE121.12) (TCD121A)


excd12-1.htm

D. Transport from gas-liquid interface to bulk liquid

(CDE121.13)

From the correlation for organic liquids,

(CDE121.14)

It has been noted5 that this correlation gives a mass transfer coefficient that is too low

E. Resistance from bulk liquid to catalyst

(CDE121.15) (TCD12IF)


excd12-1.htm

F. Total and percentage resistances

RT = Rc + Rl + Rg + RR =

(CDE121.16)

Individual resistances:

(b) Calculate catalyst weight

(CDE121.7)

(CDE121.17)


excd12-1.htm

Substitution yields

Solving, for W, we obtain

W = 34,500 kg

The reactor volume corresponding to this catalyst weight is

The total height of the reactor

Four 1-m-diameter towers each 12.2 m in height connected in series will be sufficient.

Checking assumption of constant C B


excd12-1.htm

Consequently, our assumption that the concentration of organic is essentially constant was valid.


12prof2.htm

Professional Reference Shelf CD12.2 Fluidized-Bed Reactors CD12.2.3 Mass Transfer in Fluidized Beds * CD12.2.6 Limiting Situations

CD12.2.3 Mass Transfer in Fluidized Beds

Two types of mass transport are important in fluidized-bed operations. The first is the transport between gas and solid. In some situations this can affect the analysis of fluidized-bed behavior significantly, and in others it might not enter the calculations at all. It will be seen that the treatment of this type of transport is quite similar to gas-solid mass transfer in other types of operations.

The second type of mass transfer is unique to fluidized-bed operations. It concerns the transfer of material between the bubbles and the clouds, and between the clouds and the emulsion (Figures CD12-5, CD12-7, and CD12-8). In almost every type of fluidized-bed operation, there are significant gas-phase concentration differences between the various elements of the fluidized bed. Consequently, calculations involving this type of mass transfer occur in almost every fluidized-bed analysis.

CD12.2.A Gas-Solid Mass Transfer

In the bubble phase of a fluidized bed, the solid particles are sufficiently separated so that in effect there is mass transfer between a gas and single particles. The most widely used correlation for this purpose is the equation of Fröessling 20for mass transfer to single spheres:

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(CD1248)

12prof2.htm

The relative velocity between the solid particle and the gas used in calculating the Reynolds number will be taken as u0 . In the emulsion phase, the equation would be one that applied to fixed-bed operation with a porosity . The in the bed equal to and a velocity of equation recommended by Kunii and Levenspiel 21 is

Transport from gas to single

(CD12-49)

Figure CD12-7 Transfer between bubble, cloud, and emulsion.

Mass transfer coefficients obtained from these relationships may then be combined with mass transfer among the various phases in the fluidized bed to yield the overall behavior with regard to the transport of mass. Owing to the small particle sizes and high surface area per volume of solids used in fluidized beds, the mass transfer from the gas to the solid surface is usually quite rapid, and consequently, it seldom limits the reaction.

CD12.2.3B Mass Transfer between the Fluidized-Bed Phases

For gas interchange between the bubble and the cloud, Kunii and Levenspiel22 defined the mass transfer coefficient K bc (s -1) in the following manner:

(CD12-50)


12prof2.htm

Figure CD12-8 Sketch of flow pattern in a fluidized bed for downflow of emulsion gas, ue/u 0 , <0 or u0 /u mf > 6 to 11. Adapted from D. Kunii and O. Levenspiel, Fluidization Engineering (Melbourne, Fla.: Robert E. Krieger Publishing Co., 1977).

Where C Ab and C Ac are the concentration of A in the bubble and cloud, respectively (mol/dm3), and WAbc represents the number of moles of A transferred from the bubble to the cloud per unit time per unit volume of bubble (mol/ dm 3 s). The concept of basing all mass transfer (and later, all reaction) on the bubble volume proves to simplify the calculations markedly. For the products (e.g., B in A B) the rate of transfer into the bubble from the cloud is given by a similar equation:

WBcb =K cb (C Bc -C Bb)


(CD12-51)

12prof2.htm

The mass transfer coefficient K bc can also be thought of as an exchange volume q between the bubble and the cloud:

where qb is the volume of gas flowing from the bubble to the cloud per unit time per unit volume of bubble, qc the volume of gas flowing from the cloud to the bubble per unit time per unit volume of bubble, and q0 the exchange volume between the bubble and cloud per unit time per unit volume of bubble (i.e., K bc ; q0 =qc =q).

WBcb =qbC Ab -qcC Ac =q0(C Ab -C Ac)

(CD12-52)

Using Davidson's expression for gas transfer between the bubble and the cloud, and then basing it on the volume of the bubble, Kunii and Levenspiel23 obtained this equation for evaluating K bc

Mass transfer between bubble and cloud

(CD12-53)

where is in cm /s, db is in cm, DAB is the diffusivity (cm 2/ s), and g is the gravitational constant (980 cm/s2). We note that

K bc = K cb

and a typical value of K bc is 2 s -1

Similarly, these authors defined a mass transfer coefficient for gas interchange between the cloud and the emulsion:

(CD12-54)

where WAce is the moles of A transferred from the cloud to the emulsion per unit time per unit volume of bubble. Note that even though this mass transfer does not involve the bubble directly, it is still based on the bubble


12prof2.htm

volume.

Using Higbie's penetration theory and his analogy for mass transfer from a bubble to a liquid, Kunii and Levenspiel24 developed an equation for evaluating K ce:

Mass transfer between cloud and emulsion

(CD12-55)

where ub is the velocity of the bubble rise in cm /s and the other symbols are as defined below Equation (CD1253). A typical value of K ce is 1 s -1. K ce can also be thought of as the exchange volume between the cloud and the emulsion.

With knowledge of the mass transfer coefficients, the amount of gas interchange between the phases of a fluidized bed can be calculated and combined to predict the overall mass transfer behavior or reaction behavior of a fluidized-bed process.

CD12.2.4 Reaction Behavior in a Fluidized Bed

In order to use the Kunii-Levenspiel model to predict reaction rates in a fluidized-bed reactor, the reaction-rate law for the heterogeneous reaction per gram (or other fixed unit) of solid must be known. Then the reaction rate in the bubble phase, the cloud, and the emulsion phase, all per unit of bubble volume, can be calculated. Assuming that these reaction rates are known, the overall reaction rate can be evaluated using the mass transfer relationships presented in the preceding section. All this is accomplished in the following fashion. We consider an nth-order constant-volume catalytic reaction. In the bubble phase

in which the reaction rate is defined per unit volume of bubble. In the cloud,

Rate laws


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where ke, kc, and kb are the specific reaction rates in the emulsion, cloud, and bubble, respectively. In the latter two equations, the reaction rate is also defined per unit volume of bubble.

and similarly in the emulsion,

CD12.2.5 Mole Balance on the Bubble, Cloud, and Emulsion Phases

Material balances will be written over an incremental for substance A in each of the three phases height (bubble, cloud, and emulsion) (Figure CD12-9).

Figure CD12-9 Section of a bubbling fluidized bed.

CD12.2.5A Balance on the Bubble Phase

The amount of A entering at is the bubble phase by flow,

A similar expression can be written for the amount of A leaving in the bubble phase in flow at + :


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Dividing by

A balance on A in the bubble phase for steady-state is operation in section

Balance on the bubble


yields

CD12.2.5B Balance on the Cloud Phase

In the material balance on the clouds and wakes in , it is easiest to base all terms on the bubble section volume. The material balance for the clouds and wakes is

Balance on the clouds

(CD12-57)

CD12.2.5C Balance on the Emulsion Phase

The fraction of the bed in the emulsion phase is 1 - - . The material balance for A in the emulsion results in the following expression for the emulsion-phase material balance on A:

Balance on the emulsion

The three material balances thus result in three coupled ordinary differential equations, with one independent variable ( ) and three dependent file:///H:/html/12chap/html/12prof2b.htm[05/12/2011 17:00:39]

12prof2.htm

variables (C Ab, C Ac, C Ae). These equations can be solved numerically. The Kunii-Levenspiel model simplifies these still further by assuming that the derivative terms on the left-hand side of the material balances on the cloud and emulsion are negligible compared with the terms on the righthand side. Using this assumption and letting t = /ub (i.e., the time the bubble has spent in the bed), the three equations take the form

(CD12-59)

Balance equations

(CD12-60)

(CD12-61)

Note

one differential equation and two algebraic equations. In all equations, represents the gram-moles per second reacted in the particular phase per volume of bubbles. CD12.2.5D Partitioning of the Catalyst

To solve these equations, it is necessary to have values of kb, kc, and ke. Three new parameters are defined:

First, the specific reaction rate of solid catalyst, kcat must be known. It is normally determined from laboratory experiments. The term kcat represents the gram-moles reacted per volume of solid catalyst. Then


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Relating the specific reaction rates

(CD12-62)

The term k' is the specific reaction rate per weight of catalyst.

The value of branges between 0.001 and 0.01, with 0.005 being the more typical number. The volume fraction of catalyst in the clouds and wakes . The volume of cloud and wakes per is 1 volume of bubble is

The volume of catalysts in the clouds is c

so the expression for

c

is

(CD12-63)

It turns out that the value of is normally far from insignificant in this expression for c and represents a weakness in the model because there does not yet exist a reliable method for determining . The typical values of c range from 0.3 to 0.4. The value of c can be quite incorrect on occasion, in particular, a value of = 1. The volume fraction of the solids in the emulsion phase is again 1 . The volume of emulsion per volume of bubble is

so the expression for

e

is


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The volume of catalysts in the clouds is

(CD12-64)

e

Typical values of b, c , and e are 0.005, 0.2, and 1.5, respectively. Using the expressions given above, the three balance equations become

For reactors other than first or zero order these equations must be solved numerically.

(CD12-65)

(CD12-66) (CD12-50)

CD12.2.5E Solution to the Balance Equations for a First-Order Reaction

If the reaction is first order, C Ac and C Ae can be eliminated using the two algebraic equations, and the differential equation can be solved analytically for C Ab as a function of t. An analogous situation would exist if the reaction were zero. Except for these two situations, solutions to these two equations must be obtained numerically.

For first-order reactions we can combine the three balance equations into one differential equation which we can then solve to determine the conversion achieved in a fluidized-bed reactor. In addition, the closed-form solution allows us to examine certain limiting situations to determine which operating parameters are most influential in dictating bed performance. Here we can pose and ask a number of what-if questions. To arrive at our fluidized-bed design equation for a first-order reaction, we simply express the concentration of A in both the emulsion C Ae and in the cloud C Ac in terms of the bubble concentration C Ab . First we use the emulsion balance

to solve for C Ae in terms of C Ac. Rearranging Equation (CD12-68)


(CD1268)

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for a first-order reaction (n = 1), we obtain

We now use this equation to substitute for C Ae in the cloud balance:

Solving for C Ac in terms of C Ab gives

(CD1269)

(CD1270)

We now substitute for C Ac in the bubble balance:

Rearranging yields

After some further rearrangement we obtain

(CD12-71)

Overall transport coefficient KR for a firstorder reaction.


(CD12-72)

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(CD12-73)

Design Equation

Expressing C Ab as a function of X, that is,

(CD12-73)

we can substitute to obtain

and integrating yields

The height of the bed necessary to achieve this conversion is

h = tu b

(CD12-75)

The corresponding catalyst weight is (CD12-76)


(CD12-77)

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CD12.2.5F Procedure

Unfortunately, one must use an iterative procedure to calculate the catalyst weight. This predicament is a consequence of the fact that both K R and u b depend on the bubble diameter, which depends on the bed height, Equation (CD12-75). Consequently, one should check the estimated average bubble diameter using the value of h calculated from Equation (CD12-75). A flowchart outlining this procedure is shown in Figure CD12-10. However, either POLYMATH or MatLab can be used to eliminate some of the assumptions used in arriving at Figure CD 12-10. Specifically, one could use the equation for bubble size as a function of height (i.e., Equation CD 12-37) directly in the equations for the transport coefficients (Equations CD 12-53 and CD 12-55) and them use an ODE solver rather than evaluating the bubble size at the midpoint in the column. One could also couple the unsteady state balances on the cloud and emulsion phases (Equations CD 12-57 and CD 12-58) to the mole balance on the bubble rather than neglecting the transient terms in these balances.


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Figure CD12-10 Computational algorithm for fluidized-bed reactor design. Reprinted with permission from H. S. Fogler and L. F. Brown, "Reaction Control and Transport," in Chemical Reactors, ACS Symposium Series 168, H. S. Fogler, ed. (Washington, D.C.: American Chemical Society, 1981).


Back Next

* This material was developed from notes

by Dr. Lee F. Brown and H. Scott Fogler.


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Return to the top of the page.



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Professional Reference Shelf CD12.2 Fluidized-Bed Reactors CD12.2.3 Mass Transfer in Fluidized Beds CD12.2.6 Limiting Situations CD12.2.6 Limiting Situations

As engineers, it is important to deduce how a bed will operate if one were to change operating conditions such as gas flow rate and catalyst particle size. To give some general guidelines as to how changes will affect bed behavior, we consider the two limiting circumstances of reaction control and transport control.

In the Kunii-Levenspiel bubbling-bed model, reaction occurs within the three phases of the bed, and material is continuously transferred between the phases. Two limiting situations thus arise. In one, the interphase transport is relatively fast and transport equilibrium is maintained, causing the system performance to be controlled by the rate of reaction. In the other, the reaction rate is relatively fast and the performance is controlled by interphase transport between the bubbles, clouds, and emulsions. It will be shown that the ammonia oxidation example used above is essentially a reaction-limited system. The overall reaction rate in the bed is proportional to K R, so the reciprocal of K R can be viewed as an overall resistance to the reaction. The different terms and groups on the right-hand side of Equation (CD12-72) can be viewed as individual resistances which can be arranged in series or parallel to give the overall resistance.

(CD12-78)

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(CD12-79)


CD12.2.6A Slow Reaction

In addition to the obvious way of increasing the temperature to increase the conversion, and perhaps some unwanted side reactions, there are other ways that the conversion may be increased when the reaction is slow. From Equation (CD1254) we know that the conversion depends on h, kcat, ub, and K R. We first determine K R under this situation. For a slow reaction, kcat is small compared to K bc and K ce , so that resistance to transport is essentially zero, that is,

(CD12-81)

and

(CD12-82)

then


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Consequently, we see that K R can be increased by decreasing , the volume fraction of bubbles. For the ammonia oxidation example, this would give

or about 11% higher than the value obtained by the more elaborate calculations, which included the transport. This would predict a conversion of 21.4%, very close to the 20% given by the method that includes the transport limitations. Thus the ammonia oxidation system of Massimilla and Johnstone is essentially a reaction-limited system. The conversion and catalyst weight are related by

Substituting for K R yields

Recalling Equation (CD12-46), we have

(CD12-83)

Using Equation (CD12-64) to substitute for

e

, we have

(CD12-84)

Neglecting

b

with respect of the second term yields

(CD12-85)

(CD12-86)

(CD12-87)


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(CD12-46)

In most instances u b is significantly greater than

(1 +

), so that Equation (CD12-68) is approximately

(CD12-88)

Combining Equations (CD12-87) and (CD12-88) gives

Approximate catalyst weight for slow reactions

(CD12-89)

Therefore, one observes that to reduce the catalyst weight should be as close as for a specified conversion, u0 and What-if questions

possible. One can now ask in what ways the catalyst weight may be reduced for a specified conversion. The answer to this question is the same as that to the question: How may one increase the conversion for a fixed catalyst weight?

Example CD12-4 Calculation of Resistances

Example CD12-5 Effect of Particle Size on Catalyst Weight--Slow Reaction

CD12.2.6B Rapid Reaction

To analyze this limiting situation, we shall assume that the particles are sufficiently small so that the effectiveness factor is essentially 1 and that the rate of transfer from the bulk fluid to the individual catalyst particles is rapid compared with the rate of transfer between the fluidization phases. For the case of rapid reaction


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Using these approximations in the equation for K R, which is

one observes that the first term to be neglected is b and we also note that because the reaction is rapid, kcat/K ce is a large number.

Then neglecting the reciprocal of g e with respect to kcat/K cb , K R becomes

Situation 1 will be analyzed in the text; the analysis of situation 2 is left as an exercise. Assuming that very few particles are present in the bubble phase yields

The catalyst weight is given by combining Equations (CD12-77) and (CD12-91):

(CD12-90)

There are two situations one can analyze here:

(CD12-91)


(CD12-92)

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Neglecting with respect to 1 in the numerator gives

Approximate catalyst rate for a rapid reaction

(CD12-93)

On observing that the equation for K bc , Equation (CD1253), is the sum of two terms, A0 and B 0,

one finds that the problem can be further divided. Case A: Case B:

(CD12-53)

Case A is considered here; case B is left as an exercise.

For case A, (CD12-94)

Then

(CD12-95)

Recalling the equation for ub and neglecting other terms in the equation with respect to the velocity of rise of a single bubble,

and


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(CD12-96)

The average bubble diameter is a function of the tower diameter (thus the tower cross-sectional area Ac), height, uo, . As a first approximation, we assume that the and average bubble diameter is some fraction (say, 0.75) of the maximum bubble diameter:

(CD12-97)

Then from Equation (CD12-38) we have

(CD12-98)

which is substituted into Equation (CD12-96) to give

One approximation for fast reactions.

(CD12-99)

Example CD12-6 Effect of Catalyst Weight on Particle Size--Rapid Reaction

SUMMARY

1. Minimum fluidization velocity:

(S12-1)

2. Porosity at minimum fluidization:


(S12-2)

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or

(S12-3)

3. Bubble size:

(S12-4)

where

(S12-5)

For porous plates: (S12-6)

4. Velocity of bubble rise:

5. Bed height÷conversion in first-order reaction:

(S12-7) (S12-8)

6. Mass transfer parameters: a. Between the bubble and the cloud:

(S12-9)

(S12-10)

b. Between the cloud and the emulsion:

(S12-11)

7. Reaction-rate parameters: (S12-12)


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(S12-13)

(S12-14)

(S12-15)

(S12-16)

where

is given by Figure 12-6.

8. Procedure (see Figure CD12-10).

Supplementary Reading 1. ANONYMOUS , Chem. Eng. Prog. 53(10), 50 2. 3. 4.

5. 6.

7. 8.

(Oct. 1957). B LANDING , F. H., Ind. Eng. Chem. 45, 1186 (1953). C ARLSMITH , L. E., and F. B. J OHNSON , Ind. Eng. Chem. 37, 451 (1945). DAVIDSON, J. F., and D. L. KEAIRNS, Fluidization, Proceedings of the 2nd Engineering Foundation Conference. (Cambridge: Cambridge University Press, 1978). GRACE, J. R., and J. M. M ATSEN, Fluidization (New York: Plenum Press, 1980). KEAIRNS, D. L., ed., Fluidization Technology, Vols. I and II (Washington, D.C.: Hemisphere Publishing Co. in association with McGrawHill, 1975, 1976). KUNII , D., and O. LEVENSPIEL , Ind. Eng. Chem. Process Des. Dev. 7, 481 (1968). KUNII , D., and O. LEVENSPIEL , Fluidization


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9. 10. 11. 12. 13. 14. 15.

Engineering, 2nd ed. (Stoneham, Mass.: Butterworth-Heinemann, 1991). M ATHUR, R. B., and N. EPSTEIN , Spouted Beds (San Diego, Calif.: Academic Press, 1974). WEEKMAN, V. W., J R., Ind. Eng. Chem. Process Des. Dev. 7, 91 (1968). WEN , C. Y., and Y. H. YU , AIChE J. 12 , 610 (1966). ZENZ , F. A., Ind. Eng. Chem. Fund. 3, 65 (1964). ZENZ , F. A., Ind. Chem. Eng. Symp. Ser. 30, 136 (1968). ZENZ , F. A., and N. A. WEIL , AIChE J. 4, 472 (1958). Yates, J. G., Fundamentals of Fluidized-bed Chemical Processes (Butterworths, London 1983).

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excd12-2.htm


Example CD12-2 Maximum Solids Holdup

A pilot fluidized bed is to be used to test a chemical reaction. The bed diameter is 91.4 cm. You wish to process 28.3 x 103 cm3/s of gaseous material. The average particle diameter is 100 m. The reactor height is 10 ft. Allowing for a disengaging height of 7 ft, this means that we have a maximum bed height of 91.4 cm. The distributor plate is a porous disk. What is the maximum weight of solids (i.e., holdup) in the bed? Other data:

Color of pellet: brown

Solution: The amount of solids in the reactor is given by

The two parameters that need to be found are

A. Calculation of

and .

(CD1219) (CD1229)

1. Gravity term:

2. Cross-sectional area:


excd12-2.htm

Superficial velocity:

Porosity at minimum fluidization [Equation (CD12-9)]:

B. Calculation of volume fraction of bubbles

(CD1246)

Here we see that we must calculate u mf and ub. Step 1. First the minimum fluidization velocity is obtained from Equation (CD12-25):

(CD1235)

Step 2. To calculate ub we must know the size of the bubble db, that is, (CD12-36):

Step 3. The average size of the bubble, db, is determined by evaluating Equation (CD12-37) at h/2:

where dbm and db0 are given in Equations (CD12-38) and (CD12-39), respectively

Maximum bubble diameter:


(CD1236) (CD1237)

excd12-2.htm

(CD1238)

At the top of the bed (h = 91.4 cm), db = 8.86 cm. For purposes of the Kunii-Levenspiel model, we shall take the bubble diameter to be 5 cm

Step 4. We can now return to calculate the velocity of bubble rise and the fraction of bed occupied by bubbles from Equation (CD12-36). We have

From Figure CD12-6 we see that a 100- m particle corresponds to a value of of 0.5. Substituting this value into Equation (CD12-46), the fraction of the bed occupied by the bubble is

Minimum bubble diameter:

(CD1239)

Solving for db yields

Thus 94% of the bed is in the emulsion phase plus the wakes.

C. Amount of solids holdup,


excd12-2.htm

or

Ws = 678 lb of solid particles



excd12-7.htm

Professional Reference Shelf Example CD12-7

Diffusion Between Wafers

Derive an equation for the reactant gas concentration as a function of wafer radius and then determine the effectiveness factor.

In terms of the diffusing gas phase components, we can write this reaction as

Solution The shell balance on the reactant diffusing between two wafers separated by a distance l shown in Figure CDE127.1 gives

where is the rate of generation of species A per unit wafer surface area. The factor of 2 appears in the generation term because there are two wafer surfaces exposed in each differential volume element. Dividing by , taking the limit as approaches zero, and then rearranging gives

(CDE127.1)

Recalling the constitutive equation for the molar flux WAr in radial coordinates yields


(CDE127.2)

excd12-7.htm

Diffusion between the wafers

Figure 12-7-1

For every one molecule of SiH 2 (i.e., species A) that diffuses in, one molecule of H2 (i.e., species B) diffuses out.

Then

For a first-order reaction,

Substituting Equations (CD12-20) and (CDE12-2.3) into Equation (CDE12-2.1), we get

(CDE127.3) (CD12101)

Diffusion with reaction between wafers

(CDE127.4)

The corresponding boundary conditions are

(CDE127.5) (CDE127.6)


excd12-7.htm

(CDE127.7)

where

where o is a modified Bessel function of the first kind of order zero and K o is a modified Bessel function of the second kind of order zero. The second boundary condition to be finite at = 0. Therefore, B must be zero requires because K o(0) = . Using the first boundary condition, we ; then . The concentration profile in get the space between the wafers is

The boundary conditions are

Equation (CDE12-7.2) is a form of Bessel's equation. The general form of the solution to Bessel's equation is 28 (CDE127.8)

(CDE127.9) (CDE127.10)

(CDE127.11)


excd12-7.htm

(CDE127.12)

The concentration profile along the radius of the wafer disk and the wafer shape are shown in Figure CDE12-7.2 for different values of the Thiele modulus.

Figure CD12-7-2 Radial concentration profile



excd12-7.htm

Legal Statement


excd12-8.htm


Example CD12-8 CVD Boat Reactor

Silicon is to be deposited on wafers in a LPCVD reactor. We want to obtain an analytical solution for the silicon deposition rate and reactant concentration profile for the simplified version of the LPCVD reactor just discussed. Analytical solutions of this type are important in that an engineer can rapidly gain an understanding of the important parameters and their sensitivities, without making a number of runs on the computer. The reaction that is taking place is

Sections of the reactor are shown in Figures CDE12 8.1 and CDE12-8.2.

Figure CDE12-8.1

Figure CDE12-8.2

1. Balances. In forming our shell balance on the annular region we shall assume that there are no radial gradients in the annulus and include the outer tube walls and the boat, which consume some of the reactant by deposition on the walls in the balance. In


excd12-8.htm

addition, we shall neglect any dispersion or diffusion in the axial direction. Balance on reactant A:

(CDE128.1)

(CDE128.2)

2. Rate laws. The rate of silicon deposition, (mol/ 2 dm s), is equal to the rate of depletion of SiH 2.

Mole balance on the reacting gas

Dividing through by 0 gives


(CD12101)

where the units of C A and k are mol/dm 3 and dm/s, respectively. Deposition takes place on the reactor walls, the support, and on the wafer surfaces. The corresponding depletion of reactant gas on each of these surfaces is

(CDE128.3)

Concentration profile and effectiveness factor. From Example CD12-2 we derived the radial profile concentration profile between the wafers as

Radial concentration


(CDE127.9)

excd12-8.htm

The corresponding effectiveness factor was

4. Concentration profile in the annular region. Combining Equations (CDE12-8.2) and (CDE12-8.3) yields

Writing FAz and C AA in terms of conversion, we have

Axial concentration

(CDE127.12)

(CDE128.4)

(CDE128.5)

profile

where C A0 and FA0 refer to the reactant concentration

(CDE128.6)

and molar flow rates at the entrance to the reactor.

Combining Equations (CDE12-8.4) and (CDE12-8.5) gives

Collecting terms, we can obtain an expression involving the Damköhler number, Da:

(CDE128.7)

where

Solving for conversion as a function of distance along the length of the reactor yields


(CDE128.8)

excd12-8.htm

or, in terms of concentration,

The deposition rate as a function of r and z can now be obtained as follows. The deposition rate at a location r and is

(CDE128.11)

(CDE128.9)

(CDE128.10)

First, using Equation (CDE12-2.9) to relate C A(r, z) and C AA(z), we obtain

Next, we use Equation (CDE12-3.9) to determine the rate as a function of distance down the reactor.

The thickness,T, of the deposit is obtained by integrating the deposition rate with respect to time,

where r is the molar density of the material deposited, g mol/cm3.

The 2 accounts for deposition on both sides of the wafer. Integrating, we obtain

(CDE128.12)

The reactant concentration profile and deposition thickness along the length of the reactor are shown schematically in Figure CDE12-8.3 for the case of small values of the Thiele modulus


excd12-8.htm

Figure 12-8.3



cdp12-e.htm

Chapter 12, Figure CDP12-E


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14ftn.htm

FOOTNOTES FOR CHAPTER 14 1 H.

Brenner and D. A. Edwards, Macrotransport Processes, Butterworth-Heinemann, Boston, 1993.

2 K.

B. Bischoff and R. L. Dedrick, J. Theoret. Biol., 29, 63 (1970).



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Thoughts on Problem Solving: PFR/CSTR Example

Example Algorithm for Steps in Solving Closed-Ended Problems 1. Statement The elementary, liquid-phase, irreversible reaction

is to be carried out in a flow reactor. Two reactors are available, an 800 dm 3 PFR that can only be operated at 300 K and a 200 dm 3 CSTR that can be operated at 350 K. The two feed streams to the reactor mix to form a single fee d stream that is equal molar in A and B, with a total volumetric flowrate of 10 dm 3 /min. Which of the two reactors will give us the highest conversion?

Additional Information:

at 300 K, k = 0.07 dm 3/mol-min E = 85000 J/mol-K C A0B = C B0B = 2 mol/dm 3

vA0 = vB0 = 0.5*v0 = 5 dm 3/min 2. Real Problem We have two choices, a PFR operated at 300 K and a CSTR operated at 350 K. Which one do we choose?

3. Sketch

file:///H:/html/probsolv/closed/alg/cre-exam/pfr-cstr/index.htm[05/12/2011 17:00:51]


4. Identify and Name A. Relevant Theories and Equations

Arrhenius Equation: The higher the temperature, the faster the reaction rate. Rate Law: Mole Balances: B. Systems Volume of CSTR Volume of PFR C. Dependent and Independent Variables Independent: V, FA0, T Dependent: X D. Knowns and Unknowns Knowns: k 0 , E, V, n0 , CA0B, CB0B Unknowns: X E. Inputs and Outputs In: FA0 = FB0 , so QB = 1 Out: FA = FA0(1-X), FB = FA0(1-X), FC = FA0X F. What color should we paint the reactor? Not an issue.

5. Assumptions Isothermal, no pressure drop. The CSTR is well mixed. There are no radial variations in the PFR.



6. Specifications There is neither too much redundant information, nor is there too little information given. Therefore, the problem is neither over-specified, nor under-specified.

7. Similar or Related Example Problems This problem has a solution procedure in common with Examples 4-2 and 4-4 in the text.

8. Algorithm CSTR A. Mole Balance

B. Rate Law

C. Stoichiometry - liquid, v = vo ; equal molar \ QB = 1

CA = CA0(1-X) CB = CA0(1-X) D. Combine

(eqn 1)

(eqn 2) E. Evaluate v A0 = 5 dm 3 /min Before mixing CA0B = 2 mol/dm3 FA0 = CA0B* vA0 F

= (5 dm 3 /min)(2 mol/dm3 ) = 10 mol/min


PFR


A0

After mixing v0 = vA0 + vB0 = 5 dm 3 /min + 5 dm 3 /min = 10 dm 3 /min CA0 = 1 mol/dm3 at 350 K, k = 8.447 dm 3 /mol-min 9. Manipulate A. CSTR @ 350 K the combined CSTR equation (eqn 1) can be arranged as

B. PFR @ 300 K

Choose the CSTR, because it gives the highest conversion.

10. Units Check file:///H:/html/probsolv/closed/alg/cre-exam/pfr-cstr/index.htm[05/12/2011 17:00:51]


X is dimensionless

11. Is it reasonable? This is a reasonable conversion. This the end of the PFR/CSTR example. A sample registration exam problem is also available.

Back to Chemical Reaction Engineering Examples


ChE 344 Reactor Photos -- SASOL Reactors

SASOL Reactor Photos

Click on any of the items below to see scanned photos and schematics of reactors used one of SASOL plants . Packed Bed Reactor in use for a Fisher-Tropsch synthesis reaction at Sasol Limited Chemical. Straight Through Transport Reactor (STTR) in use for a Fisher-Tropsch synthesis reaction at Sasol Limited Chemical. Automotive Catalytic Converter designed by SASOL. New generation Sasol Advanced Synthol Reactor in use to enhance a high temperature Fischer-Tropsch process. Slurry Phase Distillate Reactor in use at Sasolburg Side View of Slurry Phase Reactor

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Learn about the Fischer-Tropsch Process

Packed Bed Reactor

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Slurry Phase Distillate Reactor



Side View of Slurry Phase Distillate Reactor



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344 Reactor Photos - British Petroleum Reactors

British Petroleum Reactor Photos Click on any of the items below to see scanned photos and schematics of reactors at one of British Petroleum's (BP's) Oil Refineries. The British Petroleum Home Page is located at www.bp.com. Reformer Reactors in Series used to boost the octane level of pentanes Side View Reformer Reactors in Series Fluidized Catalytic Cracker Fixed Bed Reactor using a colbalt-molybednum catalyst to convert SO 2 to H2 S 3 Stage Converter used to convert H2 S to SO 2 Fluidized Bed Reactor using H2 SO 4 as a catalyst to bond butanes and iso-butanes to make high octane gas Hydrogen Plant that uses a catalyst to convert CH 4 to 98% pure hydrogen Fixed Bed Reactor that converts sulfur in diesel fuel to H2 S

Reformer Reactors in Series

Side View of Reformer Reactors

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Fluidized Catalytic Cracker

Fixed Bed Reactor



3 Stage Converter



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Hydrogen Plant

Fixed Bed Reactor



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CRE -- Reactors Module, Page Two

Reactors Module-part 2 Let's say you're interested in the design details for CSTRS. Clicking on the Equipment Design button brings up a short movie that illustrates the continuous flow of reactants and products through a CSTR:

You can see that CSTRs are basically large tanks with agitators in them, but their design can get quite sophisticated. For example, the stirring blades on the agitators can vary widely:

Return to page one.

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CRE -- Reactors Module, Page Two

Advance to page three.



Aspen Plus

Here is the inlet arrow highlighted by four black dots:

Return to the previous page.

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Open-Ended Problem Solving Algorithm

Open-Ended Problem Solving Algorithm This is an application of the Open-Ended Problem Solving Algorithm, as it applies to the Cobra Problem. NOTE: This page is intended merely as an example. The algorithm is not set in stone, so not all steps will apply to a give problem, as you will see. 1. Write an initial problem statement. 1. What happens when you get bit by a cobra? 2. Make sure you are proceeding to solve the real problem as opposed to the perceived problem. Carry out one or more of the following: A. Find out where the problem came from. B. Explore the problem. C. Apply the Duncker Diagram. D. Use the statement-restatement technique. E. Apply Problem Analysis. 2.B. Base Case -- will you die if bitten by a cobra? how long can you wait for antivenom? what if you use antivenom first and then get bit? 3. Generate solutions. A. Understand what conceptual blocks can occur so that you will be aware of them when they surface. 1. Perceptual 2. Emotional 3. Cultural 4. Environmental 5. Intellectual 6. Expressive 3.A.5. Intellectual blocks -- knowledge of human physiology (everyone's different), understanding of neurotoxins and what they do B. Brainstorm 1. Free association 2. Osbornís Check List 3. Lateral Thinking a. Random Stimulation file:///H:/html/web_mod/cobra/oepsa.htm[05/12/2011 17:00:57]


b. Other People's Views 3.B. Brainstorm -- Re-explore the problem: when you get bit, how much venom is injected? How much antivenom should you use? Fiddle with the rate constants. Acutally call up experts on snake venom for more info. Check the web for sites/links about cobras and India, where cobras are more common (than in the U.S). C. Analogy 1. State the problem 2. Generate analogies 3. Solve the analogy 4. Transfer the analogy to the solution 3.C. Analogy -- In this case we're treating the human body like it's a batch reactor. Could you model it some other way? D. Organize the ideas/solutions that have been generated 1. Fishbone Diagram 3.D. Organize -- group similar ideas together E. Cross Fertilize 1. Draw analogies from other disciplines 3.E. Draw analogies from other disciplines -- how would a biochemistry student solve the problem? F. Futuring. Todayís constraints (e.g. computing speed, communications) may be limiting the generation of a creative solutions. Think to the future when these constraints may no longer exist. Remove all possible constraints from the problem statement and solution criteria. 3.F. Futuring -- future antivenoms may not block receptor sites, could you take antivenom like a flu shot? G. Incubate 3.G. Incubate -- sleep on it, take a shower, etc. 4. Choose best alternative from the ideas generated (chapter 5) A. Decision Making 1. Musts 2. Wants 3. Adverse Consequences 4.A.1. bite victim must survive 2. want to administer minimum amount of antivenom at the optimum time file:///H:/html/web_mod/cobra/oepsa.htm[05/12/2011 17:00:57]


3. too much antivenom is almost as bad as no antivenom, or antivenom given too late B. Planning 1. Potential Problem 2. Consequences 3. Preventative Action 4. Contingent Action 4.B.1. antivenom may not be immediately available 2. potential death of the victim 3. have antivenom in backpack when hiking 4. have an emergency vehicle ready to transport victims to a hospital for better treatment 5. Follow Through A. Gantt Chart B. Deployment Chart C. Evaluation - Is the problem you are solving still relevant? 5. follow through with a proposal for emergency services to be available in high risk snake bite areas; consider funding, location, etc. 6. Evaluate A. Does the solution satisfy all the stated and implied criteria? B. Is the solution safe to people and property? C. Is the solution ethical? 6.A. don't want people to die because of snake bites B. having a snake bite team is better than not having one C. ethical? aren't we trying to save lives? Bloom's Taxonomy can also help you classify your problem and determine a method of attack. Return to the Problem Statement Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss

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National AIChE Student Chapter Competition Rules and Judging Criteria Outline of a University of Michigan entry for this competition 1. "Design, build and operate an apparatus for an undergraduate laboratory experiment that would demonstrate the principles of reaction engineering, and that is novel or perhaps strives to do the improbable (e.g. won't a concrete canoe sink?)." The experiment should be bench scale and of the type currently found in most undergraduate laboratory courses. It is also acceptable that the experiment would be of the type that would be used for a lecture demonstration. 2. The experiments should encompass "Simplicity/ease of communication to non-technical people." The process should be one easily understood by people outside the profession. Either the object of the process, such as manufacture of yogurt, the import ance of the project, such as feeding many people from increasingly scarce resources, or the process itself, should be easily communicated to people without a background in chemical engineering. Media coverage (newspapers/television/radio) is one way to s how success. 3. The competition will be conducted on the honor system. The faculty and graduate students can only act as sounding boards to the students' queries. The faculty cannot be idea generators for the project. The student chapter advisor or depar tment chair must write a cover letter stating to the best of his/her knowledge the students have abided by the rules. The students that work on the project must also sign a statement stating that they have abided by the rules. 4. The competition is to be a team competition with at least 20% of the team being composed of members from each of the junior and senior chemical engineering classes. The minimum number of participants is five and the maximum is fifteen per universi ty. 5. Associated measuring equipment (e.g. pH meter) must be of the type that is readily available at most universities through department ownership or borrowing from other departments in the university. 6. Purchased parts must cost less than $500. This price does not include a PC for data acquisition, or associated measuring or other (e.g. pumps, fittings, vessels) equipment that exists in most chemical engineering undergraduate laboratories. 7. The experiments will be displayed at the regional meeting. A poster board should accompany the apparatus as well as a five to ten page report describing how the idea for the experiment was generated, the underlying principles, the experimental pro cedure, and sample results. In the event the apparatus may not be physically brought to the meeting, a video tape or other means may be used to help understand the experiment. The top one or two winners of the Regional Student Chapters will be eligible to compete in the finals to be held at the annual meeting. 8. Safety of assembling and operating the experiment must be addressed. 9. Any questions about rules interpretation should come from the faculty student chapter advisor or department chair and be directed to either: Professor Martin Abraham

Professor Skip Rochefort

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University of Toledo Chemical Engineering Department 2801 W. Bancroft St. Toledo, OH 43606-3390 Phone: 419-530-8080 or 8092 Fax: 419-530-8086 e-mail: [email protected]

Oregon State University Chemical Engineering Department 103 Gleeson Hall Corvallis, OR 97331-2702 Phone: 541-737-2408 Fax: 541-737-4600 e-mail: [email protected]

10. The student chapter advisor or department chair at the host chapter at the regional conference will select a panel of three judges. The judges can be from industry, faculty, or the student body. The judges cannot be affiliated with any organizati on that has an entry. The number of winners selected to go to the finals will depend upon the number of regional entries. If there are six or fewer entries, then one winner will be selected to advance to the national competition. If there are seven or more entries, then two winners will be selected. The decision of the regional and national judges shall be final. Judging Criteria Creativity/novelty/originality

40 points

Statement of the principle to be demonstrated and clarity in demonstrating that principle.

30 points

Proper description of the safety issues associated with building and operating the experiment.

15 points

Simplicity/ease of communication/media coverage

15 points

Quality of communication Introduction: How was the idea generated? What is the principle that the experiment demonstrates and why is it important? Discussion: Explanation of fundamentals. Procedure: Safety precautions. Results: Describe what you found.

10 points

Opportunity for subsequent laboratory groups to study different variables or outcomes using the same apparatus.

10 points

Ease, desirability, and feasibility of being replicated by another student chapter.

10 points

Physical appearance.

10 points

Participation (more than 16 hours) by someone who is not a chemical engineering major (3 pts. for each nonchemical engineering major) and/or participation by ChE sophomores (2 pts. for each sophomore) - 10 pts. maximum

10 points

TOTAL

150 points

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Homogeneous Example 1, Solution Page

Type 1 Home Problem -- Solution Problems with a straight-forward calculation. The design equation for a CSTR is:

Derive Equation Because the reaction is elementary, the combined mole balance and rate law becomes:

From stoichiometry, this equation becomes:

Solving for X, we get the following equation:

then we substitute numerical values for our variables and calculate a conversion of:

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Back to Homogeneous Example 1




The differential equation for a PFR with catalyst is :

Derive Equation

Assuming that the reaction is elementary since we are not told otherwise, the reaction rate is defined as :

through stoichiometry, this becomes:

Substituting the reaction rate back into the differential equation gives :

which can be integrated from X=0 when W=0 to X=X when W=W :

to give :

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Solving gives a conversion of :

X = 0.96 Back to Heterogeneous Example 1 Heterogeneous Example 1

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Type 2 Home Problem -- Solution Problems that require intermediate calculations or manipulations. In this problem we have to find a way to relate these two reactors. At first, this question looks unsolvable, but then we recall the following relationship between k1 and k2 :

Derive Equation In the Homogeneous Example 1 Solution, we derived the following equation for a CSTR:

Our next step is to group our unknowns on one side of the equation and solve for their product:

Then, we use our relationship between k1 and k2 :



k2V=3.3 Next, we look at our PFR design equation:

Derive Equation We integrate this equation over our limits:

and solve for X:

Recalling our result for k2V that we determined above, we get:



X=0.72 Back to Homogeneous Example 2 Homogeneous Example 2



To solve this problem we need to find out how to relate the two reactors. In both cases we are not given a catalyst weight or a reaction rate constant. We can try to solve for each individually. If that is not possible then we can lump them together and solve for both. From the Type 1 Home Problem solution:

which can be rearranged to form:

Solving for Wk1 :

Now we can solve for catalyst weight and the rate constant for the PFR, but we still need some way to relate the two reactors. Looking at the information given, we see that the two reactors are at different temperatures. The way to realate two equations at different temperatures is through the Arrhenius equation :

Derive Equation



In the problem we are given that W1 = W2. We can replace both with W and multiply both sides of the previous equation by it:

Solving for Wk2 :

Now we have the realation we need between the two reactors. Next, we need to develop an equation to find the conversion in the CSTR. We start with the CSTR design equation:

Derive Equation Unfortunately, the design equation deals with volume where we have been given catalyst weight. To solve this problem we can multiply the design equation by the catalyst density to get the catalyst weight:

When the reaction rate is divided by the catalyst density it becomes the reaction rate with respect to catalyst weight and is designated with a prime:

Simplifying, we get a CSTR design equation with respect to catalyst weight instead of volume:

The reaction rate is :

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Substituting in for the reaction rate of the CSTR equation :

Rearranging and substituting in the numbers from the problem gives :

Solve for epsilon

Solving for conversion : X = -4.85 or X = 0.66 Conversion can never be greater than 1 and it can never be negative so the first answer can not be correct. Therefore:

X = 0.66 Back to Heterogeneous Example 2 Heterogeneous Example 2



Type 3 Home Problem -- Solution Problems that are over-specified. This problem might confuse the student with excess information about the rate constant, but there is a straight-forward solution. We begin with the design equation for a CSTR is:

Derive Equation Following the calculations in the Homogeneous Example 1 Solution, we get the equation for the conversion:

The problem gave us every variable except for FAo. FAo is just the mole fraction of A (x A ) multiplied by the total initial flow rate (F To ).

Then, we substitute for our variables and calculate a conversion of:

Back to Homogenous Example 3






The solution to this problem is the same as the Type 1 Home Problem solution. The diameter and catalyst information is superfluous unless asked to determine profit. Back to problem statement Heterogeneous Example 3


Homogeneous Example 4, Solution Page 1

Type 4 Home Problem Problems that are under-specified and require the student to consult other information sources. PART 1 - Type 4 Solution Our first task is to compile our simultaneous equations that will be used in Polymath. The membrane reactor functions like a PFR, so we start with the PFR design equation.

Derive Equation We then make differential equations for each of our products:

RB is the rate of hydrogen (B) removal through the membrane. It is directly proportional to the concentration of B (CB):

Next we insert our rate law:



The concentration is the molar flow rate divided by the volumetric flow rate. They are written as follows:

We also need to solve for the volumetric flow rate (v):

We can calculate the initial volumetric flow rate (v o ):

At this point, we still have two unknowns, the equilibrium constant (KC) and the diffusion rate constant (k B). We will need to determine these constants before the problem can be solved. Solving for these constants will require us to use resources beyond our text book. Go to PART 2 of the Solution Back to Homogeneous Example 4






Starting with the catalyst differential PFR equation:

Similar balances on components B and C give :

R B is the rate at which B passes through the membrane. It is based on the concentration of B and a mass transfer coefficient :

Plugging this back into the differential equation for B gives :

Back to problem statement

Part 2 of solution




Type 5 Home Problem -- Solution "What if …" problems that promote discussion. There are a few variations in this solution, as compared to the Homogeneous Example 4 Solution. We remove the diffusion term (RB) from the differential equation for the hydrogen (B) flow rate.

We will need to solve the differential equations for each of the species in our PFR simultaneously. The list of equations can be seen in the Polymath screen-shot below:

The conversion can be found by using information from this Polymath output:



Conversion, X = (F A0 - FA )/FA0. So with our data, X = (10 - 2.3)/10 = 0.77, or 77 percent. Next we plot the flow rates versus the reactor volume:



The flow of hydrogen (B) at the end of the reactor is much higher in this plot than in the membrane reactor plot. The extra presence of hydrogen (B) has lowered the reaction rate and the conversion. As with any reversible (equilibrium) reaction, we expect that as the concentration of products increases, the driving force for creating more products will decrease, according to Le Chatelier's Principle. This will prevent the reaction from going to completion, resulting in a drop in conversion. In order to achieve an 86% conversion, we will need to have an outlet flow rate of product (B) that is roughly 8.6 moles/min. [Recall that X = (F A0 - FA )/FA0, so FA = FA0(1 - X).] The following Polymath table tells us how much reactor volume we need to achieve 86% conversion:



Compared to the membrane reactor (V = 0.5 dm 3 ), this reactor (V = 0.72 dm 3 ) requires 44% more volume to achieve the same conversion. Back to Homogeneous Example 5




We start with the catalyst design equation derived in the Type 1 Home Problem solution :

Dividing both sides by the catalyst bulk density gives the conversion with respect to volume:

Volume is the cross sectional area times the length:

Taking the derivative of the volume gives:

In our case the cross sectional area of the reactor stays constant. This reduces the previous equation to

Substituting into the design equation:

The rest of the equations needed are the same as in the Type 1 Home Problem solution. Back to problem statement

Part 2 of solution




Type 6 Home Problem -- Solution Problems, or parts of problems, that are open-ended. We will use Polymath's simultaneous differential equation solver to handle this problem. Our first step to solve this multiple reaction problem is write our mole balances:

Next, we prepare our rate constant equations:

and

From stoichiometry:



and FT=FA +FB+FC+FD Knowing that CTo = P/ RT ,

, and


, we can then set up our Polymath solution:


Before we begin solving our simultaneous differential equations, we need to convert the desired product flow rate into units, to which we can relate:



which in terms of conversion is:

Now, we must perform trial and error calculations with different temperatures (T) and pressures (P). Our dilemma is that we need to achieve the 90% conversion to the product (C) within the volume limitations of our PFR. We need a temperature and a pressure that will produce 22.5 mol/sec of C in 2 m 3 , but that won't convert product (C) to product (D). We found that at 320 K and 25 atm, our PFR exceeds the requirements. Our flow rate of product is 23.1 mol/sec. That translates into a conversion of 92.3%. An example of the results can be seen in the Polymath output below:

It is easier to establish our optimal conditions if we can see the effect of temperature on conversion (see the Conversion versus Temperature graph, below). At a constant pressure of 25 atm, we can see that our conversion reaches a maximum at a temperature of about 320 K. If we increase the temperature beyond 320 K, the rate of our unwanted reaction increases, and we see a drop in conversion.



Although the conversion approaches its limit at 25 atm, an increase in pressure will have a positive effect on the conversion of this system:



However, exceeding 30 atm in pressure would not be advantageous, because the gains in production are not worth the higher expenditures (in capital investment and in operating costs) that would be associated with those operating pressures. Equipment for high pressure applications is much more expensive, the feed would need to be pressurized, etc., and as you'll see in the real world, a project's budget must be economical, or else the project will never get off the ground. Back to Homogeneous Example 6




Type 7 Home Problem -- Solution Problems where the student must explore the situation by varying operating conditions or parameters. We will use the Homogeneous Example 4 Solution as the basis for our solution here, so consult that page for a more extensive derivation. Some of the equations and values will need to be revised for the present situation. We have a new differential equation for B:

We need to insert our new rate law:

Finally, we will need new values for the volumetric flow rate (v o ):

Below are the simultaneous differential equations, as entered in Polymath:



The conversion under these conditions is: X = 0.73 When k = 2, kB = 10, and KC = 107 (their largest values), the conversion is greatest:



X = 0.997 These results make sense. As KC increases, the second term in our rate law (the one that represents the reverse reaction) decreases, so more product will be formed and the conversion will increase. As k increases, the reaction rate will increase, again causing more product to form and the conversion to increase. As kB increases (remember that kB represents how quickly product B can pass through the membrane), the concentration of B in the reactor decreases, which drives the reaction towards the products. Once again, the conversion will increase. A plot of conversion versus the product of our constants (KC*kB*k) is shown below. In each case, two of the constants are maintained at their original value, while the third is varied. Under the given conditions, changes in the equilibrium constant (KC) have almost no effect on the conversion, but the conversion varies strongly with changes in our rate constant (k).



The fact that the equilibrium constant (KC) has almost no effect on the conversion means that this reaction is essentially irreversible. This reaction could be carried out in a PFR with the same effectiveness. This is only true for our higher range of KC. The following graph displays the difference in conversion between a PFR and our membrane reactor over a lower range of equilibrium constants:

Although the difference in conversion looks nearly constant, the percent difference in conversion (see the graph below) increases significantly at lower KC values.



From this information, we conclude that the membrane reactor is most effective (in comparison to other reactors) when the reaction does not favor the products. Back to Homogeneous Example 7 Homogeneous Example 7



The equations used are the same as those found in the Type 4 Home Problem solution. The only difference is that we now make kR and Kc temperature dependent. The equations and knowns can be plugged into POLYM ATH and the numbers varied.

Varying kc - graphs Changing the value of kc changes the rate at which B passes through the membrane. When it is small, little B passes through the membrane so the reaction is not driven to the right. The flowrate of B stays roughly the same as C. When kc is large, B can pass easily through the membrane, driving the reaction to the right. The flow rate of B takes a large dive as the concentration of B becomes high enough for B to start passing rapidly through the membrane. Varying Temperature - graphs Changing the temperature affects the reaction rate constant and the equilibrium constant. At low temperatures the reaction progresses very slowly and little product is formed. When the temperature is high the reaction progresses very quickly and the rea ction is essentially over at a short distance into the reactor. Varying Flowrate - graphs Changing the flowrate changes the rate at which the materials travel through the reactor. When the flowrate is small, the reactants travel slowly through the reactor and have plenty of time to react. As the flow rate increases the materials pass quickly through the reactor and do not have as much time for the reaction to progress. Other than the variables we have already changed, the problem could also ask to vary groups such as kc and KC or file:///H:/html/probsolv/tentypes/hetero/clicks/click7.htm[05/12/2011 17:01:11]


ratios such as kR/k c. Back to problem statement




Type 8 Home Problem -- Solution Problems that challenge assumptions. Our first step is to select an equation to characterize the pressure drop in our system. We have chosen the following extension of the Bernoulli Equation (energy balance):

Define Equation Variable definitions:

Next, we list our simultaneous differential equations:



and our rate law:

Our concentrations can be written as:

The volumetric flow rate (v) will vary with the pressure and molar flow rate:

Now, we need to solve for our constants:



For laminar flow (Re < 2300), the friction factor (f) is inversely proportional to the Reynolds Number (Re):

For the sake of this approximation, we will assume that the density ( r) is the average of the component densities ( r = 0.00125 kg/dm3 ):

Finally, we enter these equations into Polymath:



The new conversion is almost identical to the conversion from our Homogeneous Example 7 Solution. The results are given below:





As you can see, the pressure (denoted by the variable P) did not vary substantially. This supports our previous assumption that the pressure drop for gases flowing through pipes (without packing) can be neglected. Back to Homogeneous Example 8




The solution to the first PFR is the same as the Type 1 Home Problem solution. In the second problem the only assumption that can be made is that the problem is isothermal. The volumeteric flowrate then becomes

This causes the rate law to be

To solve this problem, we can plug in our know values and equations into POLYMATH to solve:

y = P/Po



Graph of Conversion and y vs. Catalyst Weight

X = 0.907 y = 0.072 0.072 = P/P o P = 0.072 P o

P = 0.72 atm Back to problem statement




Type 9 Home Problem -- PART I of the Solution Problems that promote discussion. First, the group analyzes the problem by sizing the reactors graphically on the plot of the inverse reaction rate (1/-r A ) versus the conversion (X). They see how it operated before the damage to the CSTR:

and after the damage to the CSTR:

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From this, the group gathers that the reaction rate (-r A ) in the CSTR dropped significantly. The group then starts to brain-storm on the possible malfunctions that could cause this change in reaction rate. Jane: "Maybe the dent has caused dead-zones (areas where there is incomplete mixing) within the reactor, so that the CSTR is no longer well-mixed. We could take samples from various locations in the CSTR and check the concentration." Steve: "Maybe there was just an error in measuring the conversion. We should check the lab equipment." You: "Maybe the flow to or from the CSTR is no longer optimized. If the flow to the CSTR is too low, then we will not get enough product. If the flow increases, then the conversion drops. We know this from the mole balance on a CSTR. With a constant volume, an increase in FAo would cause a slide to lower conversion."

"We should check the flow meters and the pump between the heat exchanger and the CSTR to make sure they are operating properly." Bill: "The feed temperature could be too low. This would decrease our rate constant (k) and our reaction rate (-r A ), if the reaction is endothermic. If the temperature is too high and the reaction is exothermic, that would also decrease our reaction rate. We need to check and make sure that the heat exchanger and temperature sensors are working properly." file:///H:/html/probsolv/tentypes/homo/q9/homos9i.htm[05/12/2011 17:01:14]


Steve: "The dent may have decreased the volume of the CSTR enough to cause a slide to the left along the graph. We should empty the tank and fix the dent." Jane: "The feed concentration may be too low. That could cause a decrease in the reaction rate and a slide to the left in conversion. We should check the concentration of our feed." You: "Well maybe the agitator in the CSTR is broken. That could be why the reactor makes a different noise. If the fluid in the CSTR were not properly mixed, the reaction rate would decrease." Following this discussion, you go to Dr. Pakbed and tell him your theories. He also believes it is the agitator that is broken, so he orders an inspection of the agitator. After a careful inspection, it turns out that the agitator is indeed the culprit. Unfortunately, Dr. Pakbed cannot shut down the line to have the agitator fixed. He needs the CSTR and PFR to convert 70% of the butalene (A) to tirene (B). He wants you to propose a solution that can be implemented without shutting the line down for a day to fix the agitator. On to PART II of the Solution Back to Homogeneous Example 9 Homogeneous Example 9

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An example of a possible group discussion on this problem: Linda: So what could be some possible reasons for the discrepancy between the measured conversion and the calculated conversion? George: Well, there could be channeling in the reactor, which would make the catalyst less effective. Susan: Also, maybe there isn't really 500 kg of catalyst in the reactor. Maybe there's less than what is specified. Diane: Yes, or there might not be as much precious metal coating on the catalyst as we think. Maybe the catalyst isn't as good as expected. Jeff: Do we know if the measured concentration is correct? That could cause the discrepancy also. Linda: If the flow rate is too high, the conversion will go down as well. Maybe the flow needs to be controlled more effectively. George: What about the pressure drop? It could be too large. Susan: Let's calculate out one of these theories and see what kind of changes could affect the conversion. Back to problem statement



Homogeneous Question 10

Type 10 Home Problem -- Solution Problems that develop life-long learning skills. After reading the bubble column review by Y. T. Shah et al. [AIChE J., 28, 353 (1982)], we notice that there are two ways to attack this problem: #1: Graphical Approach. Use Figure 2.3 to approximate a superficial velocity (uG ) that will result in a conversion of 99.9%. Then size the reactor accordingly. #2: Mole Transfer Approach. Use Table 2.1 (find e G ) and Table 4-a.2 (to find kLa). Then calculate the moles that must be transferred and the volume required for that transfer. Sparger Types

#1 Graphical Approach On page 357 of the article by Shah, there is a plot of carbon dioxide conversion versus initial superficial gas velocity (uGo). The results do vary with the height of the column. Using this graph, we will size a tall reactor (HC = 2.36 m) and a short reactor (HC = 0.41 m). For the 2.36 m tall reactor, we see that the conversion approaches 100% at a superficial velocity (uGo) of about 0.04 m/s. This velocity is within the bubbly flow range. (The limit is usually ~ 0.05 m/s.) Knowing the volumetric gas flow rate (v Go) and the required superficial velocity of our system, we calculate the diameter of our column:

A conversion of 99.9% in a 0.41 m tall column will require a superficial gas velocity of about 0.02 m/s. Once again, we calculate our diameter:

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A comparison of the tall and short columns:

Column Height

Diameter

Column Area

Volume

2.35 m

1.26m

1.25 m 2

2.95 m 3

0.41 m

1.78m

2.5 m 2

1.03 m 3

The conclusion that can be drawn from this data is that shorter columns are more efficient (on a per volume basis) in stripping carbon dioxide from the flue gas.

#2 Mole Transfer Approach Our first step is to select our correlations for gas hold-up (e G ) and mole transfer (k La). We have chosen the following correlations from Akita and Yoshida (1973), because they are valid for carbon dioxide and water systems at relatively large column heights and diameters:

Variable Definitions The driving force for this liquid phase transfer is the difference between the equilibrium concentration of carbon



dioxide (CCO2*) and the actual liquid phase concentration of carbon dioxide (CCO2), which we will assume is zero (the sodium hydroxide instantly neutralizes the aqueous carbon dioxide). The rate of mole transfer per reactor volume can be written as:

Our next step is to determine the rate of mole transfer that is needed to remove 99.9% of the carbon dioxide in the feed gas. We will assume that carbon dioxide is an ideal gas under the conditions of our column.

In order to calculate the mole transfer coefficient (k La), we will need to find the following properties of our fluids:

We will set our initial superficial velocity at the bubbly flow limit (uGo = 0.05 m/sec) and calculate the area and diameter of our column:



For the gas phase hold-up equation, we will set C = 0.2 and assume that the superficial velocity (uG ) is approximately the average of the entering and exiting velocities (uG = 0.035 m/s). Using an equation solver, we get: e G = 0.055 Next, we solve for our mole transfer rate constant:

Going back to the driving force equation, when we solve for the reactor volume (V), we get:



The dimensions of our bubble column are: DC = 1.13 m and HC = 0.15 m The mole transfer correlation predicts that we will need a much smaller column than we would have predicted from the graphical approach. Compare with Graphical Approach Results Our results with the mole transfer approach are not the same as the graphical approach results, and this is probably due to some different assumptions we made in using the mole transfer approach: We assumed that our sodium hydroxide solution had the properties of water. The actual concentration of carbon dioxide in the solution may not have been zero. The presence of carbon dioxide in the solution would have decreased the driving force for the mole transfer and required us to increase the height of our reactor.

Sparger Types Two popular sparger types are perforated plates and porous spargers. We would choose the porous spargers, because they release bubbles with smaller diameters. As the bubble size decreases, the superficial velocity for bubbly flow increases and subsequently, the mole transfer rate increases. Back to Homogeneous Example 10



CRE Reactor Photos

CRE Reactor Photos

This is a 2,500 gallon reactor with a 3 inch half-pipe jacket. The jacket can function as a heating or cooling mechanism for the reaction.

photo provided by Central Fabricators, Inc.

This 1,000 gallon reactor vessel is made of 304SS and has a carbon steel jacket that can handle up to 400 degrees Fahrenheit.

photo provided by Central Fabricators, Inc.

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CRE Reactor Photos


CRE Reactor Photos



Answer to Question One

Answer to Question One FA at z = 100 m and z = 1000 m The answer to our question is as follows: First, since the solution for FA includes the values of FAo, we must find FAo:

At z = 100 m: FA (z=100 m) = .0162 mol/hr At z = 1000 m: FA (z=1000 m) = 9.578*10-5 mol/hr This is a graph of molar flowrate (F A ) vs length (z):

file:///H:/html/web_mod/wetlands/answer1.htm[05/12/2011 17:01:17]


Changing Weather What if the weather changed? What if the ambient air is saturated so there is no evaporation? And what if there is rain, making the evaporation rate negative? Using Q = 0 for no evaporation, and Q = -1*10-3 kmol/hr m 2 , the following plot of the flowrate of contaminants through the wetlands was produced. Notice that the flowrate increases when it rains, because the rain contributes to the amount of water present in the system.



Weather problem courtesy of Lisa Ingall.

Wetlands | DPRWP | Modeling Polymath | References


Answer to Question Two

Answer to Question Two Graph of Conversion (X) vs Length (z)

Graph of Reaction Rate (rA) vs Length (z)


Answer to Question Two



Molar Flowrate (Fa) versus Length (z)

Modeling Wetlands using Polymath A plot of the molar flowrate of toxins (F A ) along the length (z) of the PFR:


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Conversion (X) versus Length (z)

Modeling Wetlands using Polymath Conversion (X) versus Length (z) A plot of the conversion (X) along the length (z) of the PFR:


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Reaction Rate (ra) versus Length (z)

Modeling Wetlands using Polymath Reaction Rate (rA) versus Length (z) A plot of the reaction rate (rA ) along the length (z) of the PFR:


file:///H:/html/web_mod/wetlands/rate.htm[05/12/2011 17:01:19]

Algorithm, Flux Explained

Flux Explained: So what is the origin of this expression for RB ? To answer that question, we must examine the flux, W, of species B through the membrane. Consider the following diagram:

We see that the flux of species B is the same for each region in the diagram; i.e., for bulk side tranport, for transport through the membrane itself, and for shell side transport. The general expression for flux is:

The driving force usually takes the form of a concentration difference, while the transport coefficient is (usually) a constant that represents the ease or difficulty of transport through a given material. The flux of species B from the bulk fluid to the interior membrane interface is given by:

Similary, the flux of species B from the exterior membrane interface to the the shell side fluid is given by:

As we shall soon see, the most important flux expression is the one for the flux of species B through the membrane itself, from the interior membrane interface to the exterior membrane interface:

We want to combine our three flux expressions, but first let's rearrange them:

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Now let's combine them and see what drops out:

Rearranging some more we get:

If we look at the transport coefficients as being analogous to electrical resistances, then we have three "resistances" in the transport of species B through the membrane. We can safely assume that the greatest resistance to transport will occur in the membrane itself. In other words:

Or:

Applying this to our flux expression, we get:

If we make the additional assumption that

, and we note that

Of course, RB is another way of expressing the flux, W, of species B.

Return to Algorithm Main | Introduction | Algorithm Example | Comparison | Credits file:///H:/html/web_mod/membrane/flux.htm[05/12/2011 17:01:20]

, then we get:


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Solution One

Solutions 1. What volume is required for the base case membrane reactor?

Base Case List of Equations:

Summary Table

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Solution One

F A, F B , and F C versus Volume


Solution One

Conversion versus Volume


Solution One

Back to the Examples Forward to Solution Two


Solution Two

Solutions 2. What if the membrane transfer coefficient, km , were 0.002 min -1? Compare plots of molar flowrates versus volume and conversion versus volume for this case with your base case.



Solution Two

Question 2 Summary Table



Solution Two



Solution Two

Back to the Examples Back to Solution One Forward to Solution Three


Solution Three

Solutions 3. What if the membrane transfer coefficient, km , were 20.0 min -1? Compare plots of molar flowrates versus volume and conversion versus volume for this case with your base case.



Solution Three

Question 3 Summary Table



Solution Three

Isolating F B versus Volume


Solution Three



Solution Three

Back to the Examples Back to Solution Two Forward to Solution Four


Solution Four

Solutions 4. What if the base case flowrate were changed from 15 mol/min to 5 mol/min? How would this affect the behavior of the membrane reactor?


Question 4 file:///H:/html/web_mod/membrane/sol-4.htm[05/12/2011 17:01:23]

Solution Four

Summary Table



Solution Four



Solution Four

Back to the Examples Back to Solution Three Forward to Solution Five


Solution Five

Solutions 5. What if the base case flowrate were changed from 15 mol/min to 25 mol/min? How would this affect the behavior of the membrane reactor?


Question 5 file:///H:/html/web_mod/membrane/sol-5.htm[05/12/2011 17:01:24]

Solution Five

Summary Table



Solution Five



Solution Five

Back to the Examples Back to Solution Four


Comparison, Part 2

Comparison Comparing Membrane Reactors with PFRs, Part 2

Changing FA0 Let's take a look at the effects of changing the inlet flowrate, F A0, and how it affects our individual flowrates and the conversion:

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Comparison, Part 2


Comparison, Part 2

As expected, our flowrates increase as F A0 increases. Interestingly though, conversion actually decreases as F A0 increases. Why? Even though more of reactant A enters the reactor as the flowrate increases, it spends less time in the reactor, which causes the decrease in conversion.

Changing km Now let's see what happens when we vary km :


Comparison, Part 2


Comparison, Part 2

The transport coefficient, km , is a measure of how easily a species will pass through a given substance (i.e., the membrane). The lower the value of km , the more difficult it will be for species B to pass through the membrane, so it makes sense that our highest conversion occurs when km is highest. (See the page on flux for more information.) The resulting effect on our flowrates is also no surprise, since the removal of species B through the membrane will drive our reaction equilibrium to the right. In our extreme case of large km , species B is removed as quickly as it is produced. file:///H:/html/web_mod/membrane/compare2.htm[05/12/2011 17:01:25]

Comparison, Part 2

Back to Comparison, Part 1 Main | Introduction | Algorithm Example | Comparison | Credits


Rate Law To illustrate the effects of reactive distillation lets take the equation :

Taking the reaction to be elementary, the rate law is :

In reactive distillation, one of the products is continually removed causing the concentration of the removed product to become very small. As a result the reverse reaction decreases and more product is formed.

Mole Balance Removing a component also has an effect on the mole balance. If D is the only component being removed, its mole balance becomes:

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Mass Balance Again, the overall mass balance must be done. However, this time the flow of D out of the system must be taken into account.

The mass terms in the equation can be replaced by :

giving :

Assuming that the system is at constant density leads to :

Summary


Mole Balance

Mass Balance

Rate Law


Polymath equations file

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file:///H:/html/web_mod/distill/solution/poly/sol1g.htm[05/12/2011 17:01:27]

We can also use reactive distillation with the reaction and see if the conversion improves. Methyl Acetate will be the product that is removed.

Reactive Distillation - Only Methyl Acetate Evaporates

Mole Balance

Mass Balance

Rate Law

Unfortunatly we are not given an evaporation rate for methy acetate, so we must look else where to find one. One such equation is :

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F MeAc XMeAc P v MeAc

Rate at which Methyl Acetate is removed from the reaction mixture by evaporation Mole fraction of reaction mixture that is methyl acetate Vapor pressure of methyl acetate at reaction temperature

P Total

Total pressure of the system

F Total

Rate at which air is bubbled through the reactor. As the species evaporate, they will be collected in the air bubbles.

Now we need to figure out the vapor pressure of methyl acetate. From Perry's Chemical Engineering Handbook, we can develop the equation :

and that for methy acetate :


Figure 1. Schematic of a CSTR and a PFR in series When reactors are connected in series, the effluent stream of the first reactor becomes the feed stream for the second reactor. Let's start with the definition for conversion:

We see that our two conversions, which correspond to the points marked in Figure 1 (above), are:

We can rearrange the equations to get our flow rates:

The general mole balance for a CSTR is:

We assume there are no spatial variations in the rate of reaction. At steady state:

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After some rearranging:

Working in terms of conversion: FA1 = FA0(1-X1 ) Substituting into the equation:

See the PFR derivation, too? Close Window


Figure 1. Schematic of a CSTR and a PFR in series When reactors are connected in series, the effluent stream of the first reactor becomes the feed stream for the second reactor. Let's start with the definition for conversion:

We see that our two conversions, which correspond to the points marked in Figure 1 (above), are:

We can rearrange the equations to get our flow rates:

As we saw for the CSTR derivation:

The derivation for the PFR is a similar and begins with the PFR design equation:

Substituting for FA2, we get:


Since FA0 is a constant, its derivative will be zero and we can remove it from the derivative for FA0X2 :

Then:

After some rearranging we get:

Taking the integral over the PFR boundary conditions gives:

Close Window


The knowns for this problem are as follows: VPFR = 5.33 ft3 F A0 = 1.51 lbmol/hr VCSTR = 100 gal F B0 = 1.51 lbmol/hr

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Evaluating X1: Rearrange equation:

Close Window


Solving for X2:

Close Window


The Layperson's Guide to Human Respiration

The Layperson's Guide to Human Respiration People tend to take their breathing for granted--it's just something that you do. If I were to ask you to take a breath, you could consciously control your breathing, but as soon as your mind began to wander, your body would take over for you again. (Which is good news for you, since if you stop breathing, you'll die!) But what really happens when you breathe? Well, if you think back to your high school biology class, you'll recall that your lungs are like a bellows in the way they draw air into your body and exhale it out again. But your lungs are useless without your diaphragm muscle, which does the pushing and pulling on your lungs to make them work. So when you take a breath, your brain sends an electrical impulse through your nervous system to your diaphragm muscle, telling it to do its thing, but what is "its thing?"

When the order to breathe arrives at your diaphragm muscle, the nerve endings that surround the muscle are triggered and they release chemical signals for your diaphragm. These chemical signals consist of acetylcholine molecules, which are released from transmitter sites in the nerve endings. The acetylcholine molecules bind to receptor sites on the individual fibers of your diaphragm muscle. (A transmitter/receptor pairing is known as a synapse.)

When enough of these chemical signals are received by your muscle fibers, they will stimulate your entire diaphragm file:///H:/html/web_mod/cobra/humresp.htm[05/12/2011 17:01:36]

The Layperson's Guide to Human Respiration

to contract and then relax. During relaxation, the acetylcholine molecules bound to receptor sites will break down and vacate the receptor sites they occupied, so that the contraction/relaxation cycle can start again. This happens every time you take a breath. (If this watered-down explanation of breathing wasn't enough for you, then check out a more-detailed explanation of the mechanism of human respiration.) Venom Effects Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss

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"Cobra Strike" Movie

A cobra strikes! (2.2M) Return to learn more about cobra venom! Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss

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Effects of Cobra Venom in Detail

Effects of Cobra Venom in Detail Cobras have several meathods for delivering their deadly venom to their prey. Some cobras can spit their venom into a victim's eyes, causing extreme pain and blindness. However, the most common and well known method of venom delivery is injection into a victim's body through their bite.

A cobra striking! (Quicktime Movie - 2.2MB) NOTE: You will need the QuickTime Plug-in to view this movie.

Cobras belong to the sub-group of snakes known as elapids; there are over 270 species of cobras and their relatives. An elapid's venom contains postsynaptic neurotoxins that spread rapidly in its victim's bloodstream, causing respiratory failure and, eventually, death. Cobra venom is an example of a molecule that prohibits the interaction of acetylcholine molecules (transmitted from nerve endings surrounding the diaphragm muscle) with the receptor sites on the diaphragm muscle. (See the section on Human Respiration for more details). The venom disrupts the neuromuscular junctions involved in human respiration by reacting with the receptor sites in place of the acetylcholine molecules, thus blocking the receptor sites.

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Effects of Cobra Venom in Detail

Unlike the acetylcholine molecule; however, the venom molecule will not immediately react with the alcohol group of the receptor site, and therefore, it will not brake down and vacate the receptor site. The permanent interaction causes the channels of the receptor site to remain open, and a draining of the electrical impulse occurs. When the impulse is drained, the muscle fiber does not receive sufficient stimulation. Since the muscle response is an additive effect created by multiple muscle fibers, only 1/3 of the receptor sites in the diaphragm need to be blocked for cessation of the muscle function to occur. In this case, the victim usually dies within 30 minutes. The only way to save the life of a victim of a cobra bite is to inject the appropriate antivenom shortly after the patient has been bitten or put him on an artificial respirator. Antivenom Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss

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Mechanism of Antivenom

Mechanism of Antivenom * Antivenom acts to neutralize the poisonous venom of the cobra and causes the venom to be released from the receptor site. Thus, the receptor sites that were previously blocked by venom are now free to interact with the acetylcholine molecule, and normal respiration resumes. The spent antivenom and the neutralized venom are then excreted from the body. Venom composition (and its corresponding toxicity) can vary among cobras from the same species and even from the same litter--it can also vary for an individual cobra during its lifetime--and all of this makes each cobra bite truly unique. In order to insure correct treatment, antibodies specific to each form of cobra venom must be developed. The correct antibodies may be synthesized by injecting horses with a small amount of cobra venom, and then collecting the antibodies produced by the horses' immune systems. Of course, large samples of cobra venom must be collected for this process, and many snake farms around the world make significant amounts of money by harvesting the deadly snake toxin.

Milking a cobra for venom. (QuickTime Movie - 9MB) NOTE: You will need the QuickTime Plug-in to view this movie.

Careful execution of the injection of the antivenom is necessary to avoid any complications that may result from improper treatment. If the amount of antivenom is not sufficient to neutralize all of the venom, a portion of the receptor sites will remain blocked and the person would require the use of artificial respiration machines and electrical impulses to have complete respiration. Due to the size of the antivenom molecule, if given in great excess it may act to shield the receptor site from interaction with the acetylcholine molecule. Thus, the victim would develop symptoms similar to that of being bitten by a snake. Unlike cobra venom; however, the antivenom will eventually be released from the body. The rate of release is very slow, and although there are no proven cases of excess antivenom causing

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Mechanism of Antivenom

death, severe problems such as paralysis have occurred. Engineering Aspects of Cobra Venom Reactions

* [The reader should note, that the correct technical term to use here is antivenin, not antivenom. The authors made the decision to use antivenom rather than antivenin, because this is a non-technical presentation, and the average person is more familiar with the word antivenom. It is also more intuitive to use the word antivenom, when discussing a substance that works against venom.] Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss

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Developing the Equations

Developing the Equations Our development of the equations you will need, in order to model the behavior of venom and/or antivenom in the human body, will be quick and dirty, but also relatively straight-forward. You may want to read through this information more than once to get a feel for the procedure.

Batch Reactor Design Equation:

The design equation for a batch reactor is:

If we assume the reactor is well-mixed, then the reaction rate will be constant over every differential portion of volume, and we'll get:

Finally, combining the number of moles of venom in the blood, NV , with the volume will give us the concentration of venom in the blood:

The development is similar for the other relevant concentrations for antivenom and venomantivenom product in the blood, C A and C P , respectively.

Analogous to Catalytic Reactions? Believe it or not, but the way in which venom and antivenom interact with receptor sites on the diaphragm muscle may be modeled as a catalytic reaction. You may want to read Chapter Ten of Elements of Chemical Reaction Engineering, 3rd edition (or Chapter Six of the 2nd edtion), for coverage of this topic in greater detail. The most convenient way to discuss site coverage by either venom or antivenom is to refer to the fraction of sites that are unoccupied (i.e., free sites), occupied

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by venom, or occupied by antivenom, which are represented by f S , f SV, and f SA, respectively. (Oddly enough, the fraction of sites occupied by the venom-antivenom product, f SP , is never used, since the rate of product leaving a site is considered to be instantaneous.)

Combined Rate Laws and Material Balances: Here we combine the rate laws and the material balances for each of the important variables in our problem.

Fraction of sites occupied by venom:

Fraction of sites occupied by antivenom:

Fraction of free sites:



Concentration of venom in the blood:

Concentration of antivenom in the blood:



Concentration of venomantivenom product in the blood:

where: V = venom A = antivenom P = venom-antivenom product S = unoccupied site S0 = initial number of sites SV = site occupied by venom SA = site occupied by antivenom

Go to the Problems Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss


Exploring the Base Case, Part 1

Exploring the Base Case, Part 1 First we'll do what the problem statement suggested, and we'll try to prove the claim that 1/3 of the fraction of free sites will be covered in 1/2 hour. (Remember that no antivenom will be injected to save this person's life. Cruel, aren't we?) To do this, we'll want to make use of Polymath.

Additional Information: A cobra typically injects 2.0 x 10-7 moles of venom per bite. Based on the fact that the average human has 40 dm 3 of body fluid, the initial concentration of venom in the blood stream is 5.0 x 10-9 M.

Rate Constants:

Initial Conditions: at time = 0, the fraction of receptor sites = 1

Base Case Equations Here are the equations we'll enter, based on the additional information above and the reaction information we were given earlier:

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NOTE: We are making two major assumptions about this reaction system (i.e., the venom and antivenom interacting in a human body): 1. it can be modeled as a batch reactor, and 2. the reactor is well-mixed. These may or may not be appropriate assumptions to make. That is for you to determine in your exploration of this problem. As it turns out, we are left with one term, CS0, which we must solve for first. CS0 is the initial concentration of free sites, and to find it, we must solve backwards from our solution, which demands that 1/3 of the free sites be occupied by venom after 1/2 hour. f S is the fraction of free sites available, so it should be equal to 2/3, or 0.667, after 1/2 hour. With a little trial and error (guessing the value of CS0 and solving for f S until it equals 0.667), we can determine that CS0 must equal 5 x 10-9 M, which just so happens to be equal to the concentration of venom in the blood stream, CV, at time zero.



Next Back to the Problems Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss


Easter Egg, 2nd Printing

Easter Egg for the 2nd Printing This is the second printing of the text and CD, so I decided to put a second easter egg on it (sorry, you'll have to find the other easter egg on your own). My name is Dieter Andrew Schweiss, and after devoting a year of my life to this project for the first printing, and several more months for the second printing, I figure I can get away with adding something that's hidden like this to the CD.

Anime I've been a fan of anime (i.e., Japanese Animation) for years, so I've decided that this printing's easter egg should be anime-related. Various anime shows have been around in the states for years (Kimba the White Lion, StarBlazers, Battle of the Planets, etc.), while other shows have made more recent forays into the U.S. television scene (DragonBall, Pokemon, etc.), so you may have enjoyed an anime show and never realized it. The quality of anime titles ranges from the very good -- Miyazaki films like Kiki's Delivery Service (now available from Disney), My Neighbor Totoro (available from FOX Home Video?), Porco Rosso, Laputa: Castle in the Sky, etc. -- to the very bad (I won't go into that). I highly recommend that you watch sub-titled anime whenever possible, so that you can experience the shows and movies in their original form. But please make sure that you watch good anime (see my comments in the Anime Links section, below).

City Hunter This is an image from a long-running television and manga (Japanese comics) series called City Hunter. The gentleman in the image is the City Hunter himself, Ryo Saeba, while the tom-boy behind him is his current partner, Kaori. City Hunter is a private investigator with some of the smoothest moves in the business. He's a crack shot, well-connected, and very street-smart. (Sort of a mix between James Bond and Dirty Harry.) After his partner dies, Ryo promises to look after Kaori, his former partner's sister, but she insists on taking her brother's place and becomes Ryo's new partner. This places Ryo in quite the predicament, because he falls in lover with her (of course), but he can never admit his feelings (even though Kaori returns them in spades). If he did, then he would feel that he's breaking his promise to his dead partner to protect her, but "leave her alone".

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You see, in spite of the fact that he can be the coolest man alive (when he wants to be), Ryo has a terrible weakness: he's constantly obsessed with sex and he has an outrageous lingerie fetish. This obsession can be very distracting during his investigations, but fortunately Kaori is always there to keep him on track. Since this is an animated series, Kaori's disgust with Ryo often takes the form of a giant wooden hammer (as you can see in the image). That's what's great about this series: it can flip-flop between serious action/drama and hilarious hijinks at the drop of a panty, er, hat. (Don't worry, it's all in good fun. There's never any nudity or sex, although there are certainly some adult situations, but it's a TV show, so nothing ever goes too far.) It's a great action/comedy and I highly recommend it.

Anime Links Here are some links to anime sites on the web, just in case you're intersted in learning more about it. If you're concerned that anime is just disgusting, violent, and pornographic, then you haven't seen good anime. Just as the porno industry is a small part of Hollywood, the "dirty" anime titles are just a small part of the Japanese Animation market. So just like you don't recommend Debby Does Dallas to someone who's never seen a movie before, I don't recommend that pornographic, garbage anime to someone who's never seen Japanese Animation before. Anyway, here are some links to get you started: ANIMANIA -- The University of Michigan Japanese Animation Film Society Anime Web Turnpike -- An excellent starting point! Anime Web Turnpike: Music & Sounds -- I highly recommend this page. (The music is my favorite part of anime.) Anime InfoStation Anime Link Archive Anime and Manga Resources List -- This is the original anime list of links. It's been around forever. AnimeNET Anime Pitstop Yahoo! Anime

TTFN Well that's all for now. Until next time,

Dieter sua cuique voluptas





cb8-2s1.htm

Learning Resources

Example CD8-2 Solution, Clickback #1 Second Order Reaction Carried Out Adiabatically in a CSTR

5. Determine T: From the adiabatic energy balance (as applied to CSTRs):

For this problem:

which leaves us with:

After some rearranging we are left with:

Return

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sol8-2b.htm

Learning Resources

Example CD8-2 Solution, Part B Second Order Reaction Carried Out Adiabatically in a CSTR

(b) For part (b) we will again use the nonisothermal reactor design algorithm discussed in Chapter 8. The first four steps of the algorithm we used in part (a) apply to our solution to part (b). It is at step number 5, where the algorithm changes. 1. CSTR Design Equation:

2. Rate Law:


4. Combine:

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sol8-2b.htm

NOTE: We will find it more convenient to work with this equation in terms of space time, rather than volume:

Given reactor volume (V), you must solve the energy balance and the mole balance simultaneously for conversion (X), since it is a function of temperature (T). 5. Solve the Energy Balance for XEB as a function of T:

From the adiabatic energy balance (as applied to CSTRs):

6. Solve the Mole Balance for XMB as a function of T: We'll rearrange our combined equation from step 4 to give us:

Solving for X gives us:


sol8-2b.htm

Let's simplify a little more, by introducing the Damköhler Number, Da:

We then have:

7. Plot XEB and XMB : You want to plot XEB and XMB on the same graph (as functions of T) to see where they intersect. This will tell you where your steady-state point is. To accomplish this, we will use Polymath (but you could use a spreadsheet). Plot of XEB and XMB versus T We see that our conversion would be about 0.87, at a temperature of 387 K.

Continue with solution...


344 Reactor Photos -- SASOL Reactors

Fischer-Tropsch Reaction The Fischer-Tropsch reaction is used to convert the synthesis of gas into hydrocarbons. The reaction can be written in a simplified form as follows: CO + 2H2 ----> -CH 2 - + H2 O This reaction was discovered in the 1920's and has been used by Sasol for the production of liquid fuels and chemicals from synthesis gas for over forty years. The hydrocarbons are synthesized by a chain growth process, with the length of the chain dependent on the catalyst selectivity and reaction conditions. Two types of Fischer-Tropsch conversion steps have been developed and operated by Sasol. One makes use of the Slurry Phase Reactor to produce waxes and distillate fuels, while the other uses the Advanced Synthol Reactor mainly to produce light olefins and gasoline fractions. Preheated synthesis gas is fed to the bottom of the reactor, where it is distributed into the slurry consisting of liquid wax and catalyst particles. As the gas bubbles upwards through the slurry, it diffuses into the slurry and is converted into more wax by the Fischer-Tropsch reaction. The heat given off by the reaction is removed using cooling coils inside the Slurry Phase Reactor that generate steam. The product wax is separated from the slurry containing the catalyst particles in a proprietary process. The lighter, more volatile fractions leave in a gas stream from the top of the reactor. The gas stream is cooled to recover the lighter cuts and water. The hydrocarbon streams are sent to the product upgrading unit, while the water stream is treated in the water recovery unit. Return to the Sasol Reactor Page

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CRE -- Reactors Module, Page Three

Reactors Module-part 3 The module keeps track of which areas you've already viewed:

But you can always visit an area again. I forgot -- what were the pros and cons of using a CSTR? Let's take a look at the Advantages / Disadvantages section:

Hmmm... It looks like the cost of the reactor will be low, but the conversion on a per volume of reactor basis isn't


CRE -- Reactors Module, Page Three

very good. Perhaps we should look at another reactor type... Well, that's the nickel tour of the Reactors Module. If you own the third edition of Elements the Chemical Reaction Engineering, you will find the module in the Reactors directory of the CD-ROM that accompanies the text. The Reactors Module covers a wide variety of reactors, so we encourage you to take advantage of this learning resource. Return to page two. Return to the Reactor Photos Page.



Homogeneous Example 1, CSTR Design Equation Derivation

Type 1 Home Problem- CSTR Design Equation Problems with straight forward calculations. To derive the CSTR design equation, we begin with the general mole balance:

Assuming that the tank is well-mixed and the reaction rate is constant throughout the reactor, the mole balance can be written:

This equation can then be rearranged to find the volume of the CSTR based on the flow rates and the reaction rate:

From the definition of conversion, FA = FAo(1-X) or FAo - FA = FAoX, so the equation can be rewritten:

Back to the Solution


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Thoughts on Problem Solving: PFR Differential Equation

A tubular reactor consists of a cylindrical pipe and is normally operated at steady state. We consider the flow here to be highly turbulent and the flow field may be modeled by that of plug flow. That is, there is no radial variation in concentration and the reactor is referred to as a plug-flow reactor. General mole balance for PFR:

rearranging gives:

taking the limit as

, we obtain:

multiply both sides by -1, we get:

where by stoichometry F A = F A0 - F A0 X

by substitution

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Thoughts on Problem Solving: PFR Differential Equation

This can be rewritten :

This equation deals with volume where we have been given catalyst weight. To solve this problem we can divide the equation by the catalyst bulk density and arrive at:

When the reaction rate is divided by the catalyst density it becomes the reaction rate with respect to catalyst and is designated with a prime :

So :

Back to previous page Heterogeneous Example 1



By definition the concentration of A is the flow rate of A over the volumetric flow rate :

Since we are dealing with a gas the volumeteric flow rate will change down the PFR according to the equation:

The total flow rate, F T , is equal to the intial flow rate plus the change in flow rate as the reaction occurs :

Substituting in and simplifying :

We will assume that there is no pressure drop and that it is isothermal since no information is given to the contrary:

The flow rate of A at any length along the PFR is the entering flow rate minus the amount converted:

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Substituting both equations into the original gives:




Epsilon is defined as :

yAo is the mole fraction of A that enters. Since A is the only component fed to the reactor :

Sigma comes from the stoichiometeric coefficients:

Putting them together :


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Homogeneous Example 2, Arrhenius Equation

Type 2 Home Problem Problems that require intermediate calculations or manipulations. The reaction rate constant, k, is dependent on temperature according to the Arrhenius equation:

Since the PFR is at a different temperature than the CSTR, we need to relate the two rate constants. The easiest way to do this is to divide the second rate constant by the first rate constant:

Rearranging for k2 gives:

Back to the Solution Homogeneous Example 2

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Type 1 Home Problem -- Solution Problems with a straight-forward calculation. The design equation for a CSTR is:

Derive Equation Because the reaction is elementary, the combined mole balance and rate law becomes:

From stoichiometry, this equation becomes:

Solving for X, we get the following equation:

then we substitute numerical values for our variables and calculate a conversion of:

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Back to Homogeneous Example 1



Homogeneous Example 2, PFR Design Equation Derivation

Type 2 Home Problem Problems that require intermediate calculations or manipulations.

Our tubular reactor consists of a cylindrical pipe and operates at steady-state. The turbulent flow we expect may be modeled by that of a plug flow reactor (PFR). A PFR has no radial variation in concentration. We start with the general steady-state mole balance for the PFR:

Rearranging gives:

Taking the limit as

, we obtain:

From the definition of conversion, F A is:

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Then substituting F A into the differential, we get:

Inserting our rate law (-rA=kC Ao (1-X)) and dividing both sides by F Ao , we get our PFR design equation:

Back to the Solution Homogeneous Example 2



The differential equation for a PFR with catalyst is :

Derive Equation

Assuming that the reaction is elementary since we are not told otherwise, the reaction rate is defined as :

through stoichiometry, this becomes:

Substituting the reaction rate back into the differential equation gives :

which can be integrated from X=0 when W=0 to X=X when W=W :

to give :

Solve for epsilon

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The reaction rate constant, k, is dependent on temperature according to the Arrhenius equation:

Since the PFR is at a different temperature than the CSTR, we need to relate the two rate constants. The easiest way is by dividing the second rate constant by the first:

Rearranging gives:




To find the CSTR design equation we begin with the general mole balance:

Assuming that the tank is well mixed and the reaction rate is constant throughout the reactor the mole balance can be written:


Since FA is the conversion, X, times the entering flow rate, FAo , the equation can be rewritten:




Type 3 Home Problem Problems that are over-specified. To derive the CSTR design equation, we begin with the general mole balance:






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Type 4 Home Problem- PFR Design Equation Derivation Problems that are under-specified. Our tubular reactor consists of a cylindrical pipe and operates at steady-state. The turbulent flow we expect may be modeled by that of a plug flow reactor (PFR). A PFR has no radial variation in concentration. We start with the general steady-state mole balance for the PFR:

rearranging gives:

taking the limit as

, we obtain:

Back to PART 1 of the Solution






Type 4 Home Problem Problems that are under-specified and require the student to consult other information sources. PART 2 - Type 4 Solution In Part 2, we must derive the equilibrium constant (KC) and the diffusion rate constant (k B). Often the exact value that is required is either too cumbersome to calculate or unavailable in the literature. Because we are just looking for estimates of KC and kB, several assumptions will be made. We will start with our derivation of K C. We begin with a thermodynamic equation for K C:

From Perry's Chemical Engineering Handbook 1 , we find the following free energies of formation at 298 K for our components:

We then calculate the change in free energy for our reaction:

and insert the value into our KC equation:

This value for KC favors our products. Our next task is to calculate our diffusion rate constant (k B). To calculate it, we will need to introduce three new dimensionless constants--the Reynolds Number (Re), the Schmidt Number (Sc), and the Sherwood Number(Sh):

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A correlation from Fundamentals of Momentum, Heat and Mass Transfer 2 by Welty, Wicks, and Wilson (or simply WWW), we can relate kB to Re and Sc through the Sherwood number (Sh):

This equation is valid for 2000 < Re < 70000 and 1 < Sc < 2260. First, we will approximate the bulk properties of our fluid by assuming that throughout the reactor, our fluid is half A and half C. From WWW, we found m C=1.785*10 -5 kg /(m*s) (1 cP=0.1 Pa*s or kg/(m*s)), and from Perry's (Table 3311), we found the viscosity of acetone. For the sake of these calculations we will assume that the viscosities of acetone and formaldehyde (A) are similar, m A =0.75*10 -5 kg/(m*s).

The density of the bulk fluid will also be calculated using the 50-50% assumption:

We need to calculate the superficial velocity of our fluid (u bulk ). To do so we will assume the volumetric flow rate (v o ) stays constant throughout the reactor:

Now, we calculate Re:

In order to determine Sc, we need to find the kinematic viscosity of the hydrogen (B). In WWW, we find that uB = 1.096*10-4 sec/m 2 . We then calculate the Schmidt number (Sh):



Using the Sherwood correlation, we solve for kB':

Unfortunately, kB' is the rate constant per membrane surface area. We need the diffusion rate constant (k B) per volume of the reactor. This conversion can be performed quite easily. We multiply by the perimeter and divide by the crosssectional area:

The diffusion rate constant per minute: kB = 4.2/min Go to Part 3 of the Solution - Polymath Back to PART 1 of the Solution Homogeneous Example 4 REFERENCES: 1. Perry, R.H. and Green, D.W. (Editors). Perry's Chemical Engineering Handbook, New York: MacGraw-Hill Inc., 1984. 2. Welty, J.R., Wicks, C.E., & Wilson, R.E. Fundamentals of Momentum, Heat and Mass Transfer, New York: Wiley & Sons, 1984.


Solution for Type 4 Home Problem

Now that we have found the flow rates of the components in the reactor we can turn our attention to the rate law. The reaction rate for A going to B and C is :

Since the reaction is reversible we have to take into account the conversion of B and C into A :

The overall reaction rate becomes the amount of A converted into B and C minus the amount of B and C that become A :

This equation can be simplified to :

where

and

kR = kA1 To find the concentrations of the components, we start with the volumeteric flow rate. Since this reaction involves gases it will not remain constant but can be found using the equation :

Since we aren't told otherwise, we will assume no pressure drop and isothermal conditions :

This can be plugged into the definition of concentration for a species, i :



which can be simplified to :

Back to Part 1 of solution

On to final part of solution




To find the needed cross sectional areas and lengths :

Scenario 1 : Length = 100 dm, Volume = 200 dm3 Ac : 2 dm 2 Length : 100dm Conversion : 0.91

Scenario 2 : Diameter = 4 dm, Length = 10 dm Ac : 12.6 dm 2 Length : 10 dm Conversion : 0.81

Scenario 3 : Diameter = 0.5 dm, Length = 300 dm Ac : 0.20 dm 2 Length : 300 dm Conversion : 0.59

Scenario 4 : Length = 25 dm (X4), Total Volume = 150 dm3 Ac : 1.5 dm 2 Length : 25 Conversion : 0.85 Back to part 1 of solution







Starting with the catalyst differential PFR equation:

Knowing that:

We can rewrite the differential equation:

The differential equation for component B is similar to A except that it is positive because it is a product the rate law:

It must also be taken into account that B is diffusing through the membrane walls. This will change the equation:

The rate will be governed by the equation :

kc prime is the transport coefficient with respect to catalyst weight. We are not given a value for kc prime, but we are given the diffusivity and the boundary layer thickness. Searching through material it can be found that



To find kc prime just divide kc by the catalyst bulk density. Plugging this back into the differential equation for the flow rate of B:

C doesn't diffuse through the membrane so:

Back to previous page

Part 2 of solution




Large kc X = 0.998

Graph of Flow Rates vs. Catalyst Weight

Small kc X = 0.91








Large T X = 0.988


Small T X = 0.036








Large FAo X = 0.91


Small FAo X = 0.92







Homogeneous Example 8, Derivation of Pressure Drop in Pipes

Type 8 Home Problem -- Derivation of Pressure Drop in Pipes Problems that challenge assumptions. Variable definitions:

First, we start with our differential equation for pressure drop along a pipe:

Inserting

, and assuming G is constant throughout the length of the pipe, we rearrange things to get:

Integrating over the limits(

), and assuming f is constant along the pipe, we have:

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Homogeneous Example 8, Derivation of Pressure Drop in Pipes

We expect the pressure drop to be very small, so we will neglect the second term on the right-hand side. We are left with:

Rearranging again, we get:

Back to the Homogeneous Example 8 Solution Homogeneous Example 8

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Type 7 Home Problem -- Solution Problems where the student must explore the situation by varying operating conditions or parameters. We will use the Homogeneous Example 4 Solution as the basis for our solution here, so consult that page for a more extensive derivation. Some of the equations and values will need to be revised for the present situation. We have a new differential equation for B:

We need to insert our new rate law:

Finally, we will need new values for the volumetric flow rate (v o ):

Below are the simultaneous differential equations, as entered in Polymath:



The conversion under these conditions is: X = 0.73 When k = 2, kB = 10, and KC = 107 (their largest values), the conversion is greatest:



X = 0.997 These results make sense. As KC increases, the second term in our rate law (the one that represents the reverse reaction) decreases, so more product will be formed and the conversion will increase. As k increases, the reaction rate will increase, again causing more product to form and the conversion to increase. As kB increases (remember that kB represents how quickly product B can pass through the membrane), the concentration of B in the reactor decreases, which drives the reaction towards the products. Once again, the conversion will increase. A plot of conversion versus the product of our constants (KC*kB*k) is shown below. In each case, two of the constants are maintained at their original value, while the third is varied. Under the given conditions, changes in the equilibrium constant (KC) have almost no effect on the conversion, but the conversion varies strongly with changes in our rate constant (k).



The fact that the equilibrium constant (KC) has almost no effect on the conversion means that this reaction is essentially irreversible. This reaction could be carried out in a PFR with the same effectiveness. This is only true for our higher range of KC. The following graph displays the difference in conversion between a PFR and our membrane reactor over a lower range of equilibrium constants:

Although the difference in conversion looks nearly constant, the percent difference in conversion (see the graph below) increases significantly at lower KC values.



From this information, we conclude that the membrane reactor is most effective (in comparison to other reactors) when the reaction does not favor the products. Back to Homogeneous Example 7 Homogeneous Example 7



Type 9 Home Problem -- PART II of the Solution Problems that promote discussion. You decide to reverse the order of the reactors. You believe that if the PFR precedes the CSTR, the CSTR can achieve a final conversion of greater than 70%.

Our first step is to find the conversion of the PFR (Xi) through graphical integration. We know that the volume of our PFR is 4.5 ft3 . Using Simpson's Rule [see Appendix A.4, Elements of Chemical Reaction Engineering, (3rd Ed.)], we find that our PFR can obtain a conversion of 21% (Xi=0.21).

We then use trial and error to find the point on the graph (inverse of reaction rate (1/-rA) versus conversion (X)) that provides us with the appropriate CSTR volume. We find that our CSTR can achieve a final conversion of 73%.

The final plot can be seen below:

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Back to PART I of the Solution Homogeneous Example 9

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Homogeneous Example 10, Variable Definitions

Type 10 Home Problem -- Variable Definitions Problems that develop life-long learning skills.

Symbol (units)

Definition Gas hold-up: fraction of the column volume that is gas Gas hold-up equation coefficient Gravitational acceleration Diameter of bubble column Fluid density

Gas-liquid surface tension

Diffusivity of CO 2 in water Dynamic viscosity Kinematic viscosity (m/r)

Rate constant of mole transfer per interfacial area Bubble-liquid interfacial area Rate of mole transfer necessary to achieve conversion

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Homogeneous Example 10, Variable Definitions

Back to the Homogeneous Example 10 Solution


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To download the Polymath Equations to a PC: 1. Click on the equation below that you desire 2. When the browser can't figure out what to do with the file you should be given an option of saving the file. You will want to pick that option. 3. Pick where you want to put the file on your local hard drive

No Reactive Distillation Reactive Distillation - Only Methyl Acetate Evaporates Reactive Distillation - All Species Evaporate

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The critical pressure and temperature needed to find the reduced pressure and temperature can be found in the CRC Handbook of Chemistry and Physics:

Tc = 506.5 K Pc = 4750 kPa These equations can now be plugged into Polymath.

Polymath Click here for equations

Graph In this case only 18 moles of acetic acid is left giving a conversion of 94 %. This is much better than the case with no reactive distillation. In real life the other products will also evaporate. We can develop similar equations for the other species to see how their evporation effects the reaction.

Reactive Distillation - All Species Evaporate

Mole Balance

Mass Balance

Rate Law

Physical properties and vapor pressure equations for acetic acid, methanol, and water



The Mechanism of Human Respiration in Detail

The Mechanism of Human Respiration in Detail Human respiration is dependent upon the interaction of acetylcholine molecules with acetylcholine esterase receptors on the diaphragm muscle. Each time a person takes a breath, nerve endings that contain "sacks" of acetylcholine are stimulated. Each sack has roughly 1 x 1014 acetylcholine molecules inside. The sacks move toward the end of the nerve and eventually strike the wall of the nerve. The force of the collision causes the sacks to release the acetylcholine molecules into the neuromuscular junction or synapse.

The acetylcholine molecule has a positive nitrogen group which is attracted to the negative charge of an acetylcholine esterase receptor site on the diaphragm.

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The attraction between the molecule and a receptor site causes a bridging to occur, and a channel for impulses from the nerve to the muscle is opened. Each interaction with the receptor site causes the channel to open for approximately 400 microseconds . The opening of the channel allows for the transmission of an electrical impulse that stimulates the contraction of the muscle fiber. Many of these neuromuscular interactions combine to create a uniform muscle response; i.e., a contraction of the diaphragm, which is the driving force behind human respiration. Each breath a human takes is a result of the interaction described above. The acetylcholine molecule contains an ester group which reacts with the alcohol group of the receptor site. This reaction is responsible for the degradation of the acetylcholine molecule.

Once the molecule is broken down, a reaction with water occurs and the receptor site releases the molecules. Once the molecules are released, the impulse channel closes and the receptor site is free to interact with another acetylcholine molecule.



The molecules that are released from the receptor site are then used by the body to form new acetylcholine molecules that are again stored in the sacks in the nerve ending.

Since the body produces the acetylcholine molecule, the process is cyclic in nature and self-sustaining. The process will continue to occur until something prohibits the interaction with the receptor site and stops the formation of the acetylcholine molecule. Venom Effects Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss


"Milking A Cobra" Movie

Milking a cobra for venom. (9M!) Return to learn more about antivenom! Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss

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Exploring the Base Case, Part 2 Now let's take a look at a plot of the fraction of free sites versus time:

It might be helpful to also take a look at the summary table for the base case from Polymath:



As we can see, 1/3 of the receptor sites, corresponding to an f S of 0.667, are indeed covered after 1/2 hour (t = 0.5 hours). Next Previous Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss


cb8-2s2.htm

Learning Resources


4. Combine:

Space time is defined as:

After some rearranging:

Substituting:

Return


cb8-2s3.htm

Learning Resources


6. Solve the Mole Balance for XMB as a function of T:

Rearranging gives:

Our equation for X has taken the form of a quadratic equation, so we solve for X accordingly:

After some final rearranging we get:

Return


poly-b1.htm

Learning Resources

Example CD8-2 Solution, Polymath for Part B Second Order Reaction Carried Out Adiabatically in a CSTR

Our plot of XEB and XMB shows that our steady state operating point will be at:

X = 0.87 and T = 387 K

Our corresponding Polymath program looks like this:

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poly-b1.htm

NOTE: Our use of d(T)/d(t)=2 in the above program is merely a way for us to generate a range of temperatures as we plot conversion as a function of temperature.

Return

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sol8-2c.htm

Learning Resources

Example CD8-2 Solution, Part C Second Order Reaction Carried Out Adiabatically in a CSTR

(c) For part (c) we will simply modify the Polymath program we used in part (b), setting our initial temperature to 280 K. All other equations remain unchanged. 1. CSTR Design Equation:

2. Rate Law:


4. Combine:

Given reactor volume (V), you must solve the energy balance and the mole balance simultaneously for conversion (X), since it is a

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function of temperature (T). 5. Solve the Energy Balance for XEB as a function of T:

6. Solve the Mole Balance for XMB as a function of T:

7. Plot XEB and XMB : You want to plot XEB and XMB on the same graph (as functions of T) to see where they intersect. This will tell you where your steady-state point is. To accomplish this, we will use Polymath (but you could use a spreadsheet). Plot of XEB and XMB versus T We see that our conversion would be about 0.75, at a temperature of 355 K.

Return to problem statement

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Type 1 Home Problem- CSTR Design Equation Problems with straight forward calculations. To derive the CSTR design equation, we begin with the general mole balance:






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Type 1 Home Problem Problems with a straight-forward calculation. The following reaction takes place in a CSTR:


F Ao = 10 mol/min C Ao = 2 mol/dm 3 X=?

V= 500 dm 3 and k=0.1/min Rate Law: -r A = kCA What is the conversion in the CSTR? Solution to Homogeneous Problem #1 Other Homogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Homogeneous Example 1

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Type 4 Home Problem Problems that are under-specified and require the student to consult other information sources. PART 3 - Type 4 Solution - Polymath Here is the list of equations:

This is the Polymath output file:

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We will use this file to calculate the conversion. The conversion (X) equals the difference between the initial and final flow rates of A divided by the initial flow rate:

Finally, we plot the flow rates of the components along the reactor:



Back to PART 2 of the Solution Homogeneous Example 4



There is only one more thing needed before the problem can be solved : the mass transfer coefficient. Since it is not given in the problem we must look elsewere for an equation that will fit the information given in the problem. On page 535 of "Fundamentals of Momentum, Heat and Mass Transfer"* we find that :

We need a mass tranfer coefficient with respect to catalyst weight. That can be obtainded by dividing k by the catalyst bulk density. All of the equations and knowns can now be plugged into an ODE solver such as POLYMATH and solved :



Graph of Flow rates vs. Catalyst Weight

Graph of conversion vs. Catalyst Weight

Conversion = 0.91

* "Fundamentals of Momentum, Heat, and Mass Transfer" by James Welty, Charles Wicks, and Robert



Wilson. Copyright 1984 by John Wiley & Sons, Inc. Back to Part 2 of solution





Thoughts on Problem Solving > Ten Types of Home Problems

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Type 7 Home Problem Problems where the student must explore the situation by varying operating conditions or parameters. Let us revisit the membrane reactor (previously addressed in Homogenous Example 4). The following gas phase reaction will take place in our reactor:

Once again, our membrane will allow B to exit, but it will retain A and C. Given:

F Ao = 10 mol/min

X=?

yAo = 1

FA = ?

At T = 400 K

FB = ?

At P = 10 atm

FC = ? V = 100 dm 3 kB = 0.5/min KC = 105 mol/dm 3 k = 0.7 min -1

Rate Law:



Using Polymath, determine the conversion of this system. Then, vary K C, k, and kB as follows:

Is there an optimal set of conditions? Can you explain why those conditions are most effective? Is the membrane reactor a proper system for this reaction? Solution to Homogeneous Problem #7 Other Homogeneous Problems 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 Homogeneous Example 7


d(Nmeoh)/d(t)=rhac*V+Fmeoho d(V)/d(t)=vo d(Nhac)/d(t)=rhac*V d(Nmeac)/d(t)=-rhac*V d(Nh2o)/d(t)=-rhac*V Fmeoho=3 Cmeoh=Nmeoh/V Chac=Nhac/V Cmeac=Nmeac/V Ch2o=Nh2o/V T=350 Cmeoho=5 k=(8.88*10**8)*exp(-7032.1/T) K=5.2*exp((-8000/1.978)*((1/298)-(1/T))) vo=Fmeoho/Cmeoho rhac=-k*((Cmeoh*Chac)-((Cmeac*Ch2o)/K)) t(0)=0 Nmeoh(0)=0 V(0)=150 Nhac(0)=300 Nmeac(0)=0 Nh2o(0)=0 t(f)=120 r local hard drive
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d(Nmeoh)/d(t)=rhac*V+Fmeoho d(V)/d(t)=vo-Fmeac*74/943.4 d(Nhac)/d(t)=rhac*V d(Nmeac)/d(t)=-rhac*V-Fmeac d(Nh2o)/d(t)=-rhac*V Fmeoho=3 Cmeoh=Nmeoh/V Chac=Nhac/V Cmeac=Nmeac/V Ch2o=Nh2o/V T=350 Cmeoho=5 Xmeac=Nmeac/(Nmeoh+Nhac+Nmeac+Nh2o) Ptotal=101.3 Ftotal=100 k=(8.88*10**8)*exp(-7032.1/T) Tr=T/506.5 K=5.2*exp((-8000/1.978)*((1/298)-(1/T))) vo=Fmeoho/Cmeoho Pvmeac=4750*exp(10.703-(11.0088/Tr)-5.4361*(ln(Tr))+0.3058*Tr**6) Fmeac=Xmeac*(Pvmeac/Ptotal)*Ftotal rhac=-k*((Cmeoh*Chac)-((Cmeac*Ch2o)/K)) t(0)=1e-08 Nmeoh(0)=0 V(0)=150 Nhac(0)=300 Nmeac(0)=0 Nh2o(0)=0 t(f)=120 Distillation - All Species Ev


d(Nmeoh)/d(t)=rhac*V+3-Fmeoh d(V)/d(t)=vo-((Fmeac*74.08)+(Fmeoh*32.04)+(Fhac*60.05)+(Fh2o*18))/943.4 d(Nhac)/d(t)=rhac*V-Fhac d(Nmeac)/d(t)=-rhac*V-Fmeac d(Nh2o)/d(t)=-rhac*V-Fh2o d(I)/d(t)=Fhac vo=3/5 X=(300-Nhac-I)/300 Cmeoh=Nmeoh/V Chac=Nhac/V Cmeac=Nmeac/V Ch2o=Nh2o/V T=325 Ptotal=101.3 Ftotal=100 Nt=Nmeoh+Nhac+Nmeac+Nh2o Xmeac=Nmeac/Nt k=(8.88*10**8)*exp(-7032.1/T) K=5.2*exp((-8000/1.978)*((1/298)-(1/T))) Xmeoh=Nmeoh/(Nt) Xhac=Nhac/Nt Xh2o=Nh2o/Nt Pvmeoh=8092*exp(14.413-(14.8248/(T/512.6))-8.8032*(ln(T/512.6))+0.4118*(T/512.6)**6) Pvhac=5786*exp(12.446-(12.8016/(T/592.7))-7.1135*(ln(T/592.7))+0.3556*(T/592.7)**6) Pvh2o=22060*exp(11.06-(11.376/(T/647.1))-5.9233*(ln(T/647.1))+0.316*(T/647.1)**6) Pvmeac=4750*exp(10.703-(11.0088/(T/506.5))-5.4361*(ln(T/506.5))+0.3058*(T/506.5)**6) Fmeoh=Xmeoh*(Pvmeoh/Ptotal)*Ftotal Fmeac=Xmeac*(Pvmeac/Ptotal)*Ftotal Fhac=Xhac*(Pvhac/Ptotal)*Ftotal Fh2o=Xh2o*(Pvh2o/Ptotal)*Ftotal rhac=-k*((Cmeoh*Chac)-((Cmeac*Ch2o)/K)) t(0)=1e-08 Nmeoh(0)=0 V(0)=150 Nhac(0)=300 Nmeac(0)=0 Nh2o(0)=0 I(0)=0 t(f)=38 kinetics/Web_Mod/Distill/Solution/pics m




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Physical Acetic Acid Properties Tc = 592.7 K Pc = 5786 kPa MW = 60.05 Methanol Tc = 512.6 K Pc = 8092 kPa MW = 32.04

Vapor Pressure Data

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Water Tc = 647.1 K Pc = 22060 kPa MW = 18.01

Polymath CLick here for equations

Graph In this scenario all of the products and reactants are gone after 38 minutes. You might think this is the best conversion, but remember that both the reactants and the products were evaporating so not all the acetic acid reacted. In fact, the calculate d conversion is 15%. The conversion is not the only thing that can be studied in the different cases. Try exploring as many of the questions below to get a better understanding of how reactive distillation and other parameters effect the reaction. We have set the equations so that the reaction constant and the equilibrium constant are dependant on temperature. What happens with the reaction at high temperatures? Low? The boiling point of methyl acetate is 330 K. What if the temperature is below that point? What if it is above the boiling point of water? The rate at which the species evaporate are affected by the system pressure. What happens as the pressure increases or decreases? If the pressure in the reactor increased, what other parameters could you change to try to bring the reaction back to i ts original state? What is the effect on conversion if you increase the rate at which air is bubbled through the reactor? Sketch your answer then use Polymath to check your hypothesis.



Exploring the Base Case, Part 3 So what happens if we do inject antivenom (and hopefully save the life of the snake-bite victim in the process)? The list of equations we'll use will be the same one we used initially, with one exception: our concentration of antivenom, CV, at time zero will be 5 x 10-9 M. This is an arbitrary value that we have chosen for CV, based on an extra tidbit of information that we have; namely, that the normal antivenom dose is typically 3.6 x 10-9 M (although it can be as high as 9.1 x 10-9 M). With this in mind, we decided to make the antivenom dose equal to the venom dose. Please realize that this is an idealized situation, where the amount of venom injected by a hypothetical cobra is known. Such knowledge would be difficult to determine for an actual snake-bite.



Let's take a look at the resulting plot of the fraction of free sites versus time:



Summary Table It looks like our snake-bite victim will recover just fine. Next Previous Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss


poly-c1.htm

Learning Resources

Example CD8-2 Solution, Polymath for Part C Second Order Reaction Carried Out Adiabatically in a CSTR

Our plot of XEB and XMB shows that we will have three possible operating points at:

X = 0.062 at T = 286 K (stable) X = 0.38 at T = 318 K (unstable) X = 0.75 at T = 355 K (stable)

It looks like the higher conversion, stable operating point (X = 0.75 at T = 355 K) will be our desired steady state. Our corresponding Polymath program looks like this:

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poly-c1.htm

NOTE: Our use of d(T)/d(t)=2 in the above program is merely a way for us to generate a range of temperatures as we plot conversion as a function of temperature.

Return

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Exploring the Base Case, Part 4 But what if you don't have antivenom immediately available? Let's set the scene: You and your friends are camping somewhere in the jungles of Indonesia, where cobras are quite common (well, more common than in the United States, at any rate). Suddenly, your friend, Jim, screams, "I'm bit! Oh, the humanity!", and you find yourself rushing to his aid. You see a cobra slithering off into the underbrush, so you know he's not pulling your leg. Creepy, eh? Anyways, if you happen to have some antivenom with you, then you can inject it immediately, in which case you'll get the plot from the previous page. Here, we're talking about the case where you don't have antivenom with you. Again, we use Polymath, but we have to take a few extra steps to get the results we want. The following plot combines several of our models, with antivenom being initially administered to our friend, Jim, at 5 minute intervals.

As you can see, Jim would probably be fine, as long as we could get him some antivenom within about 27.5 minutes. After that, he could be in some serious trouble.

Your Turn file:///H:/html/web_mod/cobra/base4.htm[05/12/2011 17:03:04]


Well, that's all we have for our exploration of the base case. Remember, this is an open-ended problem, so there are many aspects of snake bites that we've purposely left unexplored. (What effect does the victim's body size have on our model of a snake bite, etc.) Let your imagination be your guide, and explore the cobra problem on your own! Previous Return to Problem Statement Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg Web Page by Gavin Sy & Dieter Andrew Schweiss


Exploring the Base Case, Part 4 Addendum

Special Steps in Polymath Why bother with the process we're about to describe? Well, Polymath can be used to solve for the data that you want, but ploting it using Polymath is another thing. We found that we could plot the data more easily using a spreadsheet, than by using Polymath. So, here goes: You'll want to enter your equations, like we did for the base case, where CA was zero at time zero. Then set the final time to some arbitrary value. (For this problem, we'll be taking increments of 5 minutes -- 0 min, 5 min, 10 min, etc.)

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Have Polymath solve your equations for this time range. When presented with a summary table of results, choose the option to save them to a DOS file. You will be able to solve up to three values versus time. We chose to only save f S data versus time.



Hit SHIFT-ENTER to make changes to the equations. You'll see the equations screen again. One of the options at the bottom of the screen will be to start the problem again from the current conditions. Select that option. You will be presented with a chance to set a new time final. Don't do that yet! Instead, hit ESC. You can then move your cursor using the UP/DOWN arrows to the line for d(Ca)/d(t). Type "i" to set the initial value of CA to 5 X 10-9 M (or whatever value you wish to use). Then type "f" to set the final time value (a value of 2 for two hours works well). Have Polymath solve your equations for this new time range. You will again be presented with a summary table. Choose the option to save the f S data versus time again. Don't save it using the same file name as before! Choose a different filename. Open your favorite spreadsheet application and import/parse your data from Polymath. You'll need to combine your data from before you added antivenom with your data from after you added antivenom. Then you can plot your full f S versus time curve. Return Introduction | Background | Venom Effects Reactions | Problems | References Original Work by Susan Stagg file:///H:/html/web_mod/cobra/base4-a.htm[05/12/2011 17:03:04]


Web Page by Gavin Sy & Dieter Andrew Schweiss


CD_ROM_Elements of Chemical Reaction Engineering_3ed

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