Mass Transfer Operations for the Practicing Engineer (Essential Engineering Calculations Series)

Mass Transfer Operations for the Practicing Engineer

Mass Transfer Operations for the Practicing Engineer Louis Theodore Francesco Ricci

Copyright # 2010 by John Wiley & Sons, Inc. All rights reserved Published by John Wiley & Sons, Inc., Hoboken, New Jersey Published simultaneously in Canada No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470, or on the web at www.copyright.com. Requests to the publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/permission. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representations or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. The advice and strategies contained herein may not be suitable for your situation. You should consult with a professional where appropriate. Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. For general information on our other products and services or for technical support, please contact our Customer Care Department within the United States at (800) 762-2974, outside the United States at (317) 572-3993 or fax (317) 572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print may not be available in electronic formats. For more information about Wiley products, visit our web site at www.wiley.com. Library of Congress Cataloging-in-Publication Data: Theodore, Louis. Mass transfer operations for the practicing engineer / Louis Theodore, Francesco Ricci. p. cm. Includes Index. ISBN 978-0-470-57758-5 (hardback) 1. Engineering mathematics. 2. Mass transfer. I. Ricci, Francesco. II. Title. TA331.T476 2010 530.40 7501512—dc22 2010013924 Printed in the United States of America 10 9

8 7

6 5

4 3 2

1

To Ann Cadigan and Meg Norris: for putting up with me (LT) and To my mother Laura, my father Joseph, and my brother Joseph Jr: for reasons which need not be spoken (FR)

Contents

Preface

xv

Part One Introduction 1. History of Chemical Engineering and Mass Transfer Operations References

5

2. Transport Phenomena vs Unit Operations Approach References

3

7

10

3. Basic Calculations

11

Introduction 11 Units and Dimensions 11 Conversion of Units 15 The Gravitational Constant gc 17 Significant Figures and Scientific Notation References 18 4. Process Variables Introduction 19 Temperature 20 Pressure 22 Moles and Molecular Weight Mass, Volume, and Density Viscosity 25 Reynolds Number 28 pH 29 Vapor Pressure 31 Ideal Gas Law 31 References 35

17

19

23 25

vii

viii

Contents

5. Equilibrium vs Rate Considerations Introduction 37 Equilibrium 37 Rate 38 Chemical Reactions References 40

37

39

6. Phase Equilibrium Principles

41

Introduction 41 Gibb’s Phase Rule 44 Raoult’s Law 45 Henry’s Law 53 Raoult’s Law vs Henry’s Law 59 Vapor – Liquid Equilibrium in Nonideal Solutions Vapor – Solid Equilibrium 64 Liquid – Solid Equilibrium 68 References 69

61

7. Rate Principles

71

Introduction 71 The Operating Line Fick’s Law 73 Diffusion in Gases Diffusion in Liquids

72 75 79

Mass Transfer Coefficients

80

Individual Mass Transfer Coefficients 81 Equimolar Counterdiffusion 83 Diffusion of Component A Through Non-diffusing Component B

Overall Mass Transfer Coefficients

87

Equimolar Counterdiffusion and/or Diffusion in Dilute Solutions Gas Phase Resistance Controlling 89 Liquid Phase Resistance Controlling 89 Experimental Mass Transfer Coefficients 90

References

84

88

93

Part Two Applications: Component and Phase Separation Processes 8. Introduction to Mass Transfer Operations Introduction

97

97

Contents

Classification of Mass Transfer Operations Contact of Immiscible Phases 98 Miscible Phases Separated by a Membrane Direct Contact of Miscible Phases 102

Mass Transfer Equipment

ix

97 101

102

Distillation 103 Absorption 104 Adsorption 104 Extraction 104 Humidification and Drying 105 Other Mass Transfer Unit Operations The Selection Decision 106

105

Characteristics of Mass Transfer Operations

107

Unsteady-State vs Steady-State Operation 108 Flow Pattern 109 Stagewise vs Continuous Operation 116

References

117

9. Distillation

119

Introduction 119 Flash Distillation 120 Batch Distillation 127 Continuous Distillation with Reflux

133

Equipment and Operation 133 Equilibrium Considerations 140 Binary Distillation Design: McCabe–Thiele Graphical Method 142 Multicomponent Distillation: Fenske –Underwood–Gilliland (FUG) Method 161 Packed Column Distillation 184

References

185

10. Absorption and Stripping Introduction 187 Description of Equipment Packed Columns Plate Columns

187 189

189 196

Design and Performance Equations—Packed Columns Liquid Rate 200 Column Diameter 207 Column Height 210 Pressure Drop 224

200

x

Contents

Design and Performance Equations—Plate Columns Stripping 235 Packed vs Plate Tower Comparison 241 Summary of Key Equations 242 References 243 11. Adsorption

227

245

Introduction 245 Adsorption Classification Activated Carbon Activated Alumina Silica Gel 249 Molecular Sieves

247

248 248 249

Adsorption Equilibria Freundlich Equation Langmuir Isotherms

250 253 253

Description of Equipment 257 Design and Performance Equations Regeneration 283 References 291

264

12. Liquid – Liquid and Solid – Liquid Extraction Introduction 293 Liquid – Liquid Extraction

293

294

The Extraction Process 294 Equipment 295 Solvent Selection 298 Equilibrium 300 Graphical Procedures 301 Analytical Procedures 304

Solid – Liquid Extraction (Leaching) Process Variables 313 Equipment and Operation 315 Design and Predictive Equations

References

312

317

325

13. Humidification and Drying Introduction 327 Psychrometry and the Psychrometric Chart Humidification 339

327 327

Contents Equipment 341 Describing Equations

Drying

xi

343

347

Rotary Dryers Spray Dryers

References

352 361

369

14. Crystallization

371

Introduction 371 Phase Diagrams 373 The Crystallization Process 379 Crystal Physical Characteristics 382 Equipment 391 Describing Equations 393 Design Considerations 397 References 404 15. Membrane Separation Processes

407

Introduction 407 Reverse Osmosis 408 Describing Equations

Ultrafiltration

420

Describing Equations

Microfiltration

421

427

Describing Equations

Gas Permeation

428

432

Describing Equations

References

414

433

437

16. Phase Separation Equipment

439

Introduction 439 Fluid – Particle Dynamics 442 Gas– Solid (G – S) Equipment 446 Gravity Settlers 447 Cyclones 449 Electrostatic Precipitators Venturi Scrubbers 457 Baghouses 461

454

xii

Contents

Gas– Liquid (G– L) Equipment Liquid – Solid (L – S) Equipment

465 467

Sedimentation 467 Centrifugation 471 Flotation 472

Liquid – Liquid (L – L) Equipment 475 Solid – Solid (S – S) Equipment 477 High-Gradient Magnetic Separation Solidification 477

References

Part Three

477

479

Other Topics

17. Other and Novel Separation Processes

483

Freeze Crystallization 484 Ion Exchange 484 Liquid Ion Exchange 484 Resin Adsorption 485 Evaporation 485 Foam Fractionation 486 Dissociation Extraction 486 Electrophoresis 486 Vibrating Screens 487 References 488 18. Economics and Finance Introduction 489 The Need for Economic Analyses Definitions 491

489 489

Simple Interest 491 Compound Interest 491 Present Worth 492 Evaluation of Sums of Money 492 Depreciation 493 Fabricated Equipment Cost Index 493 Capital Recovery Factor 493 Present Net Worth 494 Perpetual Life 494 Break-Even Point 495 Approximate Rate of Return 495

Contents

xiii

Exact Rate of Return 495 Bonds 496 Incremental Cost 496

Principles of Accounting Applications 499 References 511

496

19. Numerical Methods Introduction Applications References

513

513 514 531

20. Open-Ended Problems

533

Introduction 533 Developing Students’ Power of Critical Thinking Creativity 534 Brainstorming 536 Inquiring Minds 536 Failure, Uncertainty, Success: Are They 537 Related? Angels on a Pin 538 Applications 539 References 547 21. Ethics

549

Introduction 549 Teaching Ethics 550 Case Study Approach 551 Integrity 553 Moral Issues 554 Guardianship 556 Engineering and Environmental Ethics Future Trends 559 Applications 561 References 563

557

22. Environmental Management and Safety Issues Introduction 565 Environmental Issues of Concern Health Risk Assessment 568 Risk Evaluation Process for Health

534

566 570

565

xiv

Contents

Hazard Risk Assessment

571

Risk Evaluation Process for Accidents

Applications References

572

574 591

Appendix Appendix A. Units A.1 A.2 A.3 A.4 A.5 A.6 A.7

The Metric System 595 The SI System 597 Seven Base Units 597 Two Supplementary Units 598 SI Multiples and Prefixes 599 Conversion Constants (SI) 599 Selected Common Abbreviations

595

603

Appendix B. Miscellaneous Tables

605

Appendix C. Steam Tables

615

Index

623

Preface

Mass transfer is one of the basic tenets of chemical engineering, and contains many practical concepts that are utilized in countless industrial applications. Therefore, the authors considered writing a practical text. The text would hopefully serve as a training tool for those individuals in academia and industry involved with mass transfer operations. Although the literature is inundated with texts emphasizing theory and theoretical derivations, the goal of this text is to present the subject from a strictly pragmatic point-of-view. The book is divided into three parts: Introduction, Applications, and Other Topics. The first part provides a series of chapters concerned with principles that are required when solving most engineering problems, including those in mass transfer operations. The second part deals exclusively with specific mass transfer operations e.g., distillation, absorption and stripping, adsorption, and so on. The last part provides an overview of ABET (Accreditation Board for Engineering and Technology) related topics as they apply to mass transfer operations plus novel mass transfer processes. An Appendix is also included. An outline of the topics covered can be found in the Table of Contents. The authors cannot claim sole authorship to all of the essay material and illustrative examples in this text. The present book has evolved from a host of sources, including: notes, homework problems and exam problems prepared by several faculty for a required one-semester, three-credit, “Principles III: Mass Transfer” undergraduate course offered at Manhattan College; L. Theodore and J. Barden, “Mass Transfer”, A Theodore Tutorial, East Williston, NY, 1994; J. Reynolds, J. Jeris, and L. Theodore, “Handbook of Chemical and Environmental Engineering Calculations,” John Wiley & Sons, Hoboken, NJ, 2004, and J. Santoleri, J. Reynolds, and L. Theodore, “Introduction to Hazardous Waste Management,” 2nd edition, John Wiley & Sons, Hoboken, NJ, 2000. Although the bulk of the problems are original and/or taken from sources that the authors have been directly involved with, every effort has been made to acknowledge material drawn from other sources. It is hoped that we have placed in the hands of academic, industrial, and government personnel, a book that covers the principles and applications of mass transfer in a thorough and clear manner. Upon completion of the text, the reader should have acquired not only a working knowledge of the principles of mass transfer operations, but also experience in their application; and, the reader should find himself/herself approaching advanced texts, engineering literature and industrial applications (even unique ones) with more confidence. We strongly believe that, while understanding the basic concepts is of paramount importance, this knowledge may xv

xvi

Preface

be rendered virtually useless to an engineer if he/she cannot apply these concepts to real-world situations. This is the essence of engineering. Last, but not least, we believe that this modest work will help the majority of individuals working and/or studying in the field of engineering to obtain a more complete understanding of mass transfer operations. If you have come this far and read through most of the Preface, you have more than just a passing interest in this subject. We strongly suggest that you try this text; we think you will like it. Our sincere thanks are extended to Dr. Paul Marnell at Manhattan College for his invaluable help in contributing to Chapter 9 on Distillation and Chapter 14 on Crystallization. Thanks are also due to Anne Mohan for her assistance in preparing the first draft of Chapter 13 (Humidification and Drying) and to Brian Bermingham and Min Feng Zheng for their assistance during the preparation of Chapter 12 (Liquid – Liquid and Solid – Liquid Extraction). Finally, Shannon O’Brien, Kathryn Scherpf and Kimberly Valentine did an exceptional job in reviewing the manuscript and page proofs.

April 2010

FRANCESCO RICCI LOUIS THEODORE

NOTE: An additional resource is available for this text. An accompanying website contains over 200 additional problems and 15 hours of exams; solutions for the problems and exams are available at www.wiley.com for those who adopt the book for training and/or academic purposes.

Part One

Introduction The purpose of this Part can be found in its title. The book itself offers the reader the fundamentals of mass transfer operations with appropriate practical applications, and serves as an introduction to the specialized and more sophisticated texts in this area. The reader should realize that the contents are geared towards practitioners in this field, as well as students of science and engineering, not chemical engineers per se. Simply put, topics of interest to all practicing engineers have been included. Finally, it should also be noted that the microscopic approach of mass transfer operations is not treated in any required undergraduate Manhattan College offering. The Manhattan approach is to place more emphasis on real-world and design applications. However, microscopic approach material is available in the literature, as noted in the ensuing chapters. The decision on whether to include the material presented ultimately depends on the reader and/or the approach and mentality of both the instructor and the institution. A general discussion of the philosophy and the contents of this introductory section follows. Since the chapters in this Part provide an introduction and overview of mass transfer operations, there is some duplication due to the nature of the overlapping nature of overview/introductory material, particularly those dealing with principles. Part One chapter contents include: 1 History of Chemical Engineering and Mass Transfer Operations 2 Transport Phenomena vs Unit Operations Approach 3 Basic Calculations 4 Process Variables 5 Equilibrium vs Rate Considerations 6 Phase Equilibrium Principles 7 Rate Principles Topics covered in the first two introductory chapters include a history of chemical engineering and mass transfer operations, and a discussion of transport phenomena vs unit operations. The remaining chapters are concerned with introductory engineering principles. The next Part is concerned with describing and designing the various mass transfer unit operations and equipment. Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

1

Chapter

1

History of Chemical Engineering and Mass Transfer Operations A discussion on the field of chemical engineering is warranted before proceeding to some specific details regarding mass transfer operations (MTO) and the contents of this first chapter. A reasonable question to ask is: What is Chemical Engineering? An outdated, but once official definition provided by the American Institute of Chemical Engineers is: Chemical Engineering is that branch of engineering concerned with the development and application of manufacturing processes in which chemical or certain physical changes are involved. These processes may usually be resolved into a coordinated series of unit physical “operations” (hence part of the name of the chapter and book) and chemical processes. The work of the chemical engineer is concerned primarily with the design, construction, and operation of equipment and plants in which these unit operations and processes are applied. Chemistry, physics, and mathematics are the underlying sciences of chemical engineering, and economics is its guide in practice.

The above definition was appropriate up until a few decades ago when the profession branched out from the chemical industry. Today, that definition has changed. Although it is still based on chemical fundamentals and physical principles, these principles have been de-emphasized in order to allow for the expansion of the profession to other areas (biotechnology, semiconductors, fuel cells, environment, etc.). These areas include environmental management, health and safety, computer applications, and economics and finance. This has led to many new definitions of chemical engineering, several of which are either too specific or too vague. A definition proposed here is simply that “Chemical Engineers solve problems”. Mass transfer is the one subject area that somewhat uniquely falls in the domain of the chemical engineer. It is often presented after fluid flow(1) and heat transfer,(2) since fluids are involved as well as heat transfer and heat effects can become important in any of the mass transfer unit operations. Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

3

4

Chapter 1 History of Chemical Engineering and Mass Transfer Operations

A classical approach to chemical engineering education, which is still used today, has been to develop problem solving skills through the study of several topics. One of the topics that has withstood the test of time is mass transfer operations; the area that this book is concerned with. In many mass transfer operations, one component of a fluid phase is transferred to another phase because the component is more soluble in the latter phase. The resulting distribution of components between phases depends upon the equilibrium of the system. Mass transfer operations may also be used to separate products (and reactants) and may be used to remove byproducts or impurities to obtain highly pure products. Finally, it can be used to purify raw materials. Although the chemical engineering profession is usually thought to have originated shortly before 1900, many of the processes associated with this discipline were developed in antiquity. For example, filtration operations were carried out 5000 years ago by the Egyptians. MTOs such as crystallization, precipitation, and distillation soon followed. During this period, other MTOs evolved from a mixture of craft, mysticism, incorrect theories, and empirical guesses. In a very real sense, the chemical industry dates back to prehistoric times when people first attempted to control and modify their environment. The chemical industry developed as did any other trade or craft. With little knowledge of chemical science and no means of chemical analysis, the earliest chemical “engineers” had to rely on previous art and superstition. As one would imagine, progress was slow. This changed with time. The chemical industry in the world today is a sprawling complex of raw-material sources, manufacturing plants, and distribution facilities which supply society with thousands of chemical products, most of which were unknown over a century ago. In the latter half of the nineteenth century, an increased demand arose for engineers trained in the fundamentals of chemical processes. This demand was ultimately met by chemical engineers. The first attempt to organize the principles of chemical processing and to clarify the professional area of chemical engineering was made in England by George E. Davis. In 1880, he organized a Society of Chemical Engineers and gave a series of lectures in 1887 which were later expanded and published in 1901 as A Handbook of Chemical Engineering. In 1888, the first course in chemical engineering in the United States was organized at the Massachusetts Institute of Technology by Lewis M. Norton, a professor of industrial chemistry. The course applied aspects of chemistry and mechanical engineering to chemical processes.(3) Chemical engineering began to gain professional acceptance in the early years of the twentieth century. The American Chemical Society had been founded in 1876 and, in 1908, it organized a Division of Industrial Chemists and Chemical Engineers while authorizing the publication of the Journal of Industrial and Engineering Chemistry. Also in 1908, a group of prominent chemical engineers met in Philadelphia and founded the American Institute of Chemical Engineers.(3) The mold for what is now called chemical engineering was fashioned at the 1922 meeting of the American Institute of Chemical Engineers when A. D. Little’s committee presented its report on chemical engineering education. The 1922 meeting marked the official endorsement of the unit operations concept and saw the approval of a

History of Chemical Engineering and Mass Transfer Operations

5

“declaration of independence” for the profession.(3) A key component of this report included the following: Any chemical process, on whatever scale conducted, may be resolved into a coordinated series of what may be termed “unit operations,” as pulverizing, mixing, heating, roasting, absorbing, precipitation, crystallizing, filtering, dissolving, and so on. The number of these basic unit operations is not very large and relatively few of them are involved in any particular process. . . An ability to cope broadly and adequately with the demands of this (the chemical engineer’s) profession can be attained only through the analysis of processes into the unit actions as they are carried out on the commercial scale under the conditions imposed by practice.

It also went on to state that: Chemical Engineering, as distinguished from the aggregate number of subjects comprised in courses of that name, is not a composite of chemistry and mechanical and civil engineering, but is itself a branch of engineering. . .

A time line diagram of the history of chemical engineering between the profession’s founding to the present day is shown in Figure 1.1.(3) As can be seen from the time line, the profession has reached a crossroads regarding the future education/curriculum for chemical engineers. This is highlighted by the differences of Transport Phenomena and Unit Operations, a topic that is treated in the next chapter.

REFERENCES 1. P. ABULENCIA and L. THEODORE, “Fluid Flow for the Practicing Engineer,” John Wiley & Sons, Hoboken, NJ, 2009. 2. L. THEODORE, “Heat Transfer for the Practicing Engineer,” John Wiley & Sons, Hoboken, NJ, 2011 (in preparation). 3. N. SERINO, “2005 Chemical Engineering 125th Year Anniversary Calendar,” term project, submitted to L. Theodore, 2004. 4. R. BIRD, W. STEWART, and E. LIGHTFOOT, “Transport Phenomena,” 2nd edition, John Wiley & Sons, Hoboken, NJ, 2002.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

6

George Davis provides the blueprint for a new profession with 12 lectures on Chemical Engineering in Manchester, England

The Massachusetts Institute of Technology begins “Course X”, the first fouryear Chemical Engineering program in the United States

1888

Pennsylvania University begins its Chemical Engineering curriculum

1892

Figure 1.1 Chemical engineering time line.(3)

George Davis proposes a “Society of Chemical Engineers” in England

1880

Tulane begins its Chemical Engineering curriculum

1894

The American Institute of Chemical Engineers is formed

1908

William H. Walker and Warren K. Lewis, two prominent professors, establish a School of Chemical Engineering Practice

1916

The Massachusetts Institute of Technology starts an Independent Department of Chemical Engineering

1920

Manhattan College begins its Chemical Engineering curriculum. Adoption of the R. Bird et al. “Transport Phenomena” approach(4)

1960

ABET; stresses once again the emphasis on the practical/design approach

1990

Unit Operations vs Transport Phenomena; the profession at a crossroad

Today

Chapter

2

Transport Phenomena vs Unit Operations Approach The history of Unit Operations is interesting. As indicated in the previous chapter, chemical engineering courses were originally based on the study of unit processes and/or industrial technologies. However, it soon became apparent that the changes produced in equipment from different industries were similar in nature, i.e., there was a commonality in the mass transfer operations in the petroleum industry as with the utility industry. These similar operations became known as Unit Operations. This approach to chemical engineering was promulgated in the Little report discussed earlier, and has, with varying degrees and emphasis, dominated the profession to this day. The Unit Operations approach was adopted by the profession soon after its inception. During the 130 years (since 1880) that the profession has been in existence as a branch of engineering, society’s needs have changed tremendously and so has chemical engineering. The teaching of Unit Operations at the undergraduate level has remained relatively unchanged since the publication of several early- to mid-1900 texts. However, by the middle of the 20th century, there was a slow movement from the unit operation concept to a more theoretical treatment called transport phenomena or, more simply, engineering science. The focal point of this science is the rigorous mathematical description of all physical rate processes in terms of mass, heat, or momentum crossing phase boundaries. This approach took hold of the education/curriculum of the profession with the publication of the first edition of the Bird et al. book.(1) Some, including both authors of this text, feel that this concept set the profession back several decades since graduating chemical engineers, in terms of training, were more applied physicists than traditional chemical engineers. There has fortunately been a return to the traditional approach to chemical engineering, primarily as a result of the efforts of ABET (Accreditation Board for Engineering and Technology). Detractors to this pragmatic approach argue that this type of theoretical education experience provides answers to what and how, but not necessarily why, i.e., it provides a greater understanding of both fundamental physical and chemical processes. However, in terms Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

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8

Chapter 2 Transport Phenomena vs Unit Operations Approach

of reality, nearly all chemical engineers are now presently involved with the why questions. Therefore, material normally covered here has been replaced, in part, with a new emphasis on solving design and open-ended problems; this approach is emphasized in this text. The following paragraphs attempt to qualitatively describe the differences between the above two approaches. Both deal with the transfer of certain quantities (momentum, energy, and mass) from one point in a system to another. There are three basic transport mechanisms which can potentially be involved in a process. They are: 1 Radiation 2 Convection 3 Molecular Diffusion The first mechanism, radiative transfer, arises as a result of wave motion and is not considered, since it may be justifiably neglected in most engineering applications. The second mechanism, convective transfer, occurs simply because of bulk motion. The final mechanism, molecular diffusion, can be defined as the transport mechanism arising as a result of gradients. For example, momentum is transferred in the presence of a velocity gradient; energy in the form of heat is transferred because of a temperature gradient; and, mass is transferred in the presence of a concentration gradient. These molecular diffusion effects are described by phenomenological laws.(1) Momentum, energy, and mass are all conserved. As such, each quantity obeys the conservation law within a system: 9 9 8 9 8 9 8 8 < quantity = < quantity = < quantity = < quantity = þ generated in ¼ accumulated  out of into ; ; : ; : ; : : in system system system system

(2:1)

This equation may also be written on a time rate basis: 9 9 8 9 8 9 8 8 rate rate = = < < rate = < rate = <  out of þ generated in ¼ accumulated into ; ; : ; : ; : : in system system system system

(2:2)

The conservation law may be applied at the macroscopic, microscopic, or molecular level. One can best illustrate the differences in these methods with an example. Consider a system in which a fluid is flowing through a cylindrical tube (see Fig. 2.1) and define the system as the fluid contained within the tube between points 1 and 2 at any time. If one is interested in determining changes occurring at the inlet and outlet of a system, the conservation law is applied on a “macroscopic” level to the entire system. The resultant equation (usually algebraic) describes the overall changes occurring to the system (or equipment). This approach is usually applied in the Unit Operation

Transport Phenomena vs Unit Operations Approach 1

9

2 Fluid out

Fluid in

1

2

Figure 2.1 Flow system.

(or its equivalent) courses, an approach which is highlighted in this text and its two companion texts.(2,3) In the microscopic/transport phenomena approach, detailed information concerning the behavior within a system is required; this is occasionally requested of and by the engineer. The conservation law is then applied to a differential element within the system that is large compared to an individual molecule, but small compared to the entire system. The resulting differential equation is then expanded via an integration in order to describe the behavior of the entire system. The molecular approach involves the application of the conservation laws to individual molecules. This leads to a study of statistical and quantum mechanics— both of which are beyond the scope of this text. In any case, the description at the molecular level is of little value to the practicing engineer. However, the statistical averaging of molecular quantities in either a differential or finite element within a system can lead to a more meaningful description of the behavior of a system. Both the microscopic and molecular approaches shed light on the physical reasons for the observed macroscopic phenomena. Ultimately, however, for the practicing engineer, these approaches may be valid but are akin to attempting to kill a fly with a machine gun. Developing and solving these equations (in spite of the advent of computer software packages) is typically not worth the trouble. Traditionally, the applied mathematician has developed differential equations describing the detailed behavior of systems by applying the appropriate conservation law to a differential element or shell within the system. Equations were derived with each new application. The engineer later removed the need for these tedious and error-prone derivations by developing a general set of equations that could be used to describe systems. These have come to be referred to by many as the transport equations. In recent years, the trend toward expressing these equations in vector form has gained momentum (no pun intended). However, the shell-balance approach has been retained in most texts where the equations are presented in componential form, i.e., in three particular coordinate systems—rectangular, cylindrical, and spherical. The componential terms can be “lumped” together to produce a more concise equation in vector form. The vector equation can be, in turn, re-expanded into other coordinate systems. This information is available in the literature.(1,4)

10

Chapter 2 Transport Phenomena vs Unit Operations Approach

ILLUSTRATIVE EXAMPLE 2.1 Explain why the practicing engineer/scientist invariably employs the macroscopic approach in the solution of real world problems. SOLUTION: The macroscopic approach involves examining the relationship between changes occurring at the inlet and the outlet of a system. This approach attempts to identify and solve problems found in the real world, and is more straightforward than and preferable to the more involved microscopic approach. The microscopic approach, which requires an understanding of all internal variations taking place within the system that can lead up to an overall system result, simply may not be necessary. B

REFERENCES 1. R. BIRD, W. STEWART, and E. LIGHTFOOT, “Transport Phenomena,” John Wiley & Sons, Hoboken, NJ, 1960. 2. L. THEODORE, “Heat Transfer for the Practicing Engineer,” John Wiley & Sons, Hoboken, NJ, 2011 (in preparation). 3. P. ABULENCIA and L. THEODORE, “Fluid Flow for the Practicing Engineer,” John Wiley & Sons, Hoboken, NJ, 2009. 4. L. THEODORE, “Introduction to Transport Phenomena,” International Textbook Co., Scranton, PA, 1970.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

3

Basic Calculations INTRODUCTION This chapter provides a review of basic calculations and the fundamentals of measurement. Four topics receive treatment: 1 Units and Dimensions 2 Conversion of Units 3 The Gravitational Constant, gc 4 Significant Figures and Scientific Notation The reader is directed to the literature in the Reference section of this chapter(1 – 3) for additional information on these four topics.

UNITS AND DIMENSIONS The units used in this text are consistent with those adopted by the engineering profession in the United States. For engineering work, SI (Syste`me International) and English units are most often employed. In the United States, the English engineering units are generally used, although efforts are still underway to obtain universal adoption of SI units for all engineering and science applications. The SI units have the advantage of being based on the decimal system, which allows for more convenient conversion of units within the system. There are other systems of units; some of the more common of these are shown in Table 3.1. Although English engineering units will primarily be used, Tables 3.2 and 3.3 present units for both the English and SI systems, respectively. Some of the more common prefixes for SI units are given in Table 3.4 (see also Appendix A.5) and the decimal equivalents are provided in Table 3.5. Conversion factors between SI and English units and additional details on the SI system are provided in Appendices A and B.

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

11

12

meter centimeter foot

foot

foot

American Engineering

British Engineering

Length

SI egs fps

System

Table 3.1 Common Systems of Units

second

second

second second second

Time

slug

pound

kilogram gram pound

Mass

pound (force)

pound (force)

Newton dyne poundal

Force

British thermal unit, horsepower . hour British thermal unit, foot pound (force)

Joule erg, Joule, or calorie foot poundal

Energy

Kelvin, degree Celsius Kelvin, degree Celsius degree Rankine, degree Fahrenheit degree Rankine, degree Fahrenheit degree Rankine, degree Fahrenheit

Temperature

Table 3.2 English Engineering Units Physical quantity

Name of unit

Symbol for unit

Length Time Mass Temperature Temperature (alternative) Moles Energy Energy (alternative) Force Acceleration Velocity Volume Area Frequency Power Heat capacity

foot second, minute, hour pound (mass) degree Rankine degree Fahrenheit pound mole British thermal unit horsepower . hour pound (force) foot per second square foot per second cubic foot square foot cycles per second, Hertz horsepower, Btu per second British thermal unit per (pound mass . degree Rankine) pound (mass) per cubic foot pound (force) per square inch pound (force) per square foot atmospheres bar

ft s, min, h lb 8R 8F lbmol Btu hp . h lbf ft/s2 ft/s ft3 ft2 cycles/s, Hz hp, Btu/s Btu/lb . 8R

Density Pressure

lb/ft3 psi psf atm bar

Table 3.3 SI Units Physical unit

Name of unit

Length Mass Time Temperature Temperature (alternative) Moles Energy Force Acceleration Pressure Pressure (alternative) Velocity Volume Area Frequency Power Heat capacity Density Angular velocity

meter kilogram, gram second Kelvin degree Celsius gram mole Joule Newton meters per second squared Pascal, Newton per square meter bar meters per second cubic meters, liters square meters Hertz Watt Joule per kilogram . Kelvin kilogram per cubic meter radians per second

Symbol for unit m kg, g s K 8C gmol J, kg . m2/s2 N, kg . m/s2, J/m m/s2 Pa, N/m2 bar m/s m3, L m2 Hz, cycles/s W, kg . m2 . s3, J/s J/kg . K kg/m3 rad/s

14

Chapter 3 Basic Calculations

Table 3.4 Prefixes for SI Units Multiplication factors 18

1,000,000,000,000,000,000 ¼ 10 1,000,000,000,000,000 ¼ 1015 1,000,000,000,000 ¼ 1012 1,000,000,000 ¼ 109 1,000,000 ¼ 106 1,000 ¼ 103 100 ¼ 102 10 ¼ 101 0.1 ¼ 1021 0.01 ¼ 1022 0.001 ¼ 1023 0.000 001 ¼ 1026 0.000 000 001 ¼ 1029 0.000 000 000 001 ¼ 10212 0.000 000 000 000 001 ¼ 10215 0.000 000 000 000 000 001 ¼ 10218

Prefix

Symbol

exa peta tera giga mega kilo hecto deka deci centi milli micro nano pico femto atto

E P T G M k h da d c m m n p f a

Table 3.5 Decimal Equivalents Inch in fractions A. 4ths and 8ths 1/8 1/4 3/8 1/2 5/8 3/4 7/8 B. 16ths 1/16 3/16 5/16 7/16 9/16 11/16 13/16 15/16 C. 32nds 1/32 3/32

Decimal equivalent

Millimeter equivalent

0.125 0.250 0.375 0.500 0.625 0.750 0.875

3.175 6.350 9.525 12.700 15.875 19.050 22.225

0.0625 0.1875 0.3125 0.4375 0.5625 0.6875 0.8125 0.9375

1.588 4.763 7.938 11.113 14.288 17.463 20.638 23.813

0.03125 0.09375

0.794 2.381 (Continued)

Conversion of Units

15

TABLE 3.5 Continued Inch in fractions

Decimal equivalent

Millimeter equivalent

0.15625 0.21875 0.28125 0.34375 0.40625 0.46875 0.53125 0.59375 0.65625 0.71875 0.78125 0.84375 0.90625 0.96875

3.969 5.556 7.144 8.731 10.319 11.906 13.494 15.081 16.669 18.256 19.844 21.431 23.019 24.606

5/32 7/32 9/32 11/32 13/32 15/32 17/32 19/32 21/32 23/32 25/32 27/32 29/32 31/32

Two units that appear in dated literature are the poundal and slug. By definition, one poundal force will give a 1 pound mass an acceleration of 1 ft/s2. Alternatively, 1 slug is defined as the mass that will accelerate 1 ft/s2 when acted upon by a 1 pound force; thus, a slug is equal to 32.2 pounds mass.

CONVERSION OF UNITS Converting a measurement from one unit to another can conveniently be accomplished by using unit conversion factors; these factors are obtained from a simple equation that relates the two units numerically. For example, from 12 inches (in) ¼ 1 foot (ft)

(3:1)

the following conversion factor can be obtained: 12 in=1 ft ¼ 1

(3:2)

Since this factor is equal to unity, multiplying some quantity (e.g., 18 ft) by this factor cannot alter its value. Hence 18 ft (12 in=1 ft) ¼ 216 in

(3:3)

Note that in Equation (3.3), the old units of feet on the left-hand side cancel out leaving only the desired units of inches. Physical equations must be dimensionally consistent. For the equality to hold, each additive term in the equation must have the same dimensions. This condition can be and should be checked when solving engineering problems. Throughout the

16

Chapter 3 Basic Calculations

text, great care is exercised in maintaining the dimensional formulas of all terms and the dimensional homogeneity of each equation. Equations will generally be developed in terms of specific units rather than general dimensions, e.g., feet, rather than length. This approach should help the reader to more easily attach physical significance to the equations presented in these chapters. Consider now the example of calculating the perimeter, P, of a rectangle with length, L, and height, H. Mathematically, this may be expressed as P ¼ 2L þ 2H. This is about as simple a mathematical equation that one can find. However, it only applies when P, L, and H are expressed in the same units. Terms in equations must be consistent from a “magnitude” viewpoint.(3) Differential terms cannot be equated with finite or integral terms. Care should also be exercised in solving differential equations. In order to solve differential equations to obtain a description of the pressure, temperature, composition, etc., of a system, it is necessary to specify boundary and/or initial conditions (B a/o IC) for the system. This information arises from a description of the problem or the physical situation. The number of boundary conditions (BC) that must be specified is the sum of the highest order derivative for each independent differential equation. A value of the solution on the boundary of the system is one type of boundary condition. The number of initial conditions (IC) that must be specified is the highest order time derivative appearing in the differential equation. The value for the solution at time equal to zero constitutes an initial condition. For example, the equation d 2 CA ¼ 0; dz2

CA ¼ concentration

(3:4)

requires 2 BCs (in terms of the position variable z). The equation dCA ¼ 0; t ¼ time dt

(3:5)

requires 1 IC. And finally, the equation @CA @2 C A ¼D 2 @t @y

(3:6)

requires 1 IC and 2 BCs (in terms of the position variable y). ILLUSTRATIVE EXAMPLE 3.1 Convert units of acceleration in cm/s2 to miles/yr2. SOLUTION:

The procedure outlined on the previous page is applied to the units of cm/s2.

        1 cm 36002 s2 242 h2 3652 day2 1 in 1 ft 1 mile s2 2:54 cm 12 in 5280 ft 1 yr2 1 h2 1 day2 ¼ 6:18  109 miles=yr2 Thus, 1.0 cm/s2 is equal to 6.18  109 miles/yr2.

B

Significant Figures and Scientific Notation

17

THE GRAVITATIONAL CONSTANT gc The momentum of a system is defined as the product of the mass and velocity of the system: Momentum ¼ (mass)(velocity) (3:7) A commonly employed set of units for momentum are therefore lb . ft/s. The units of the time rate of change of momentum (hereafter referred to as rate of momentum) are simply the units of momentum divided by time, i.e., Rate of momentum ;

lb  ft s2

(3:8)

The above units can be converted to units of pound force (lbf ) if multiplied by an appropriate constant. As noted earlier, a conversion constant is a term that is used to obtain units in a more convenient form; all conversion constants have magnitude and units in the term, but can also be shown to be equal to 1.0 (unity) with no units (i.e., dimensionless). A defining equation is lb  ft (3:9) 1 lbf ¼ 32:2 2 s If this equation is divided by lbf, one obtains 1:0 ¼ 32:2

lb  ft lbf  s2

(3:10)

This serves to define the conversion constant gc. If the rate of momentum is divided by gc as 32.2 lb . ft/lbf . s2—this operation being equivalent to dividing by 1.0—the following units result:    lb  ft lbf  s2 Rate of momentum ; lb  ft s2 (3:11) ; lbf One can conclude from the above dimensional analysis that a force is equivalent to a rate of momentum.

SIGNIFICANT FIGURES AND SCIENTIFIC NOTATION(3) Significant figures provide an indication of the precision with which a quantity is measured or known. The last digit represents, in a qualitative sense, some degree of doubt. For example, a measurement of 8.32 inches implies that the actual quantity is somewhere between 8.315 and 8.325 inches. This applies to calculated and measured quantities; quantities that are known exactly (e.g., pure integers) have an infinite number of significant figures. The significant digits of a number are the digits from the first nonzero digit on the left to either (a) the last digit (whether it is nonzero or zero) on the right if there is a

18

Chapter 3 Basic Calculations

decimal point, or (b) the last nonzero digit of the number if there is no decimal point. For example: 370 370. 370.0 28,070 0.037 0.0370 0.02807

has 2 has 3 has 4 has 4 has 2 has 3 has 4

significant significant significant significant significant significant significant

figures figures figures figures figures figures figures

Whenever quantities are combined by multiplication and/or division, the number of significant figures in the result should equal the lowest number of significant figures of any of the quantities. In long calculations, the final result should be rounded off to the correct number of significant figures. When quantities are combined by addition and/or subtraction, the final result cannot be more precise than any of the quantities added or subtracted. Therefore, the position (relative to the decimal point) of the last significant digit in the number that has the lowest degree of precision is the position of the last permissible significant digit in the result. For example, the sum of 3702., 370, 0.037, 4, and 37. should be reported as 4110 (without a decimal). The least precise of the five numbers is 370, which has its last significant digit in the tens position. The answer should also have its last significant digit in the tens position. Unfortunately, engineers and scientists rarely concern themselves with significant figures in their calculations. However, it is recommended—at least for this chapter— that the reader attempt to follow the calculational procedure set forth in this section. In the process of performing engineering calculations, very large and very small numbers are often encountered. A convenient way to represent these numbers is to use scientific notation. Generally, a number represented in scientific notation is the product of a number (,10 but . or ¼ 1) and 10 raised to an integer power. For example, 28,070,000,000 ¼ 2:807  1010 0:000 002 807 ¼ 2:807  106 A positive feature of using scientific notation is that only the significant figures need appear in the number.

REFERENCES 1. D. GREEN and R. PERRY (eds), “Perry’s Chemical Engineers’ Handbook,” 8th edition, McGraw-Hill, New York City, NY, 2008. 2. J. REYNOLDS, J. JERIS, and L. THEODORE, “Handbook of Chemical and Environmental Engineering Calculations,” John Wiley & Sons, Hoboken, NJ, 2004. 3. J. SANTOLERI, J. REYNOLDS, and L. THEODORE, “Introduction to Hazardous Waste Incineration,” 2nd edition, John Wiley & Sons, Hoboken, NJ, 2000.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

4

Process Variables INTRODUCTION The authors originally considered the title “State, Physical, and Chemical Properties” for this chapter. However, since these three properties have been used interchangeably and have come to mean different things to different people, it was decided to employ the title “Process Variables.” The three aforementioned properties were therefore integrated into this all-purpose title and eliminated the need for differentiating between the three. This chapter provides a review of some basic concepts from physics, chemistry, and engineering in preparation for material that is covered in later chapters. All of these topics are vital in some manner to mass transfer operations. Because many of these topics are unrelated to each other, this chapter admittedly lacks the cohesiveness of chapters covering a single topic. This is usually the case when basic material from widely differing areas of knowledge such as physics, chemistry, and engineering are surveyed. Though these topics are widely divergent and covered with varying degrees of thoroughness, all of them will find use later in this text. If additional information on these review topics is needed, the reader is directed to the literature in the reference section of this chapter. ILLUSTRATIVE EXAMPLE 4.1 Discuss the traditional difference between chemical and physical properties. SOLUTION: Every compound has a unique set of properties that allows one to recognize and distinguish it from other compounds. These properties can be grouped into two main categories: physical and chemical. Physical properties are defined as those that can be measured without changing the identity and composition of the substance. Key properties include viscosity, density, surface tension, melting point, boiling point, and so on. Chemical properties are defined as those that may be altered via reaction to form other compounds or substances. Key chemical properties include upper and lower flammability limits, enthalpy of reaction, autoignition temperature, and so on.

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

19

20

Chapter 4 Process Variables

These properties may be further divided into two categories—intensive and extensive. Intensive properties are not a function of the quantity of the substance, while extensive properties depend on the quantity of the substance. B

The remainder of the chapter addresses a variety of process variables.

TEMPERATURE Whether in the gaseous, liquid, or solid state, all molecules possess some degree of kinetic energy, i.e., they are in constant motion—vibrating, rotating, or translating. The kinetic energies of individual molecules cannot be measured, but the combined effect of these energies in a very large number of molecules can. This measurable quantity is known as temperature, which is a macroscopic concept only and as such does not exist on the molecular level. Temperature can be measured in many ways; the most common method makes use of the expansion of mercury (usually encased inside a glass capillary tube) with increasing temperature. (However, in most industrial systems, thermocouples or thermistors are more commonly employed.) The two most commonly used temperature scales are the Celsius (or Centigrade) and Fahrenheit scales. The Celsius scale is based on the boiling and freezing points of water at 1 atm pressure; to the former, a value of 1008C is assigned, and to the latter, a value of 08C. On the older Fahrenheit scale, these temperatures correspond to 2128F and 328F, respectively. Equations (4.1) and (4.2) show the conversion from one temperature scale to the other:

where

8F ¼ 1:8(8C) þ 32

(4:1)

8C ¼ (8F  32)=1:8

(4:2)

8F ¼ a temperature on the Fahrenheit scale 8C ¼ a temperature on the Celsius scale

Experiments with gases at low-to-moderate pressures (up to a few atmospheres) have shown that, if the pressure is kept constant, the volume of a gas and its temperature are linearly related (Charles’ law, see later section) and that a decrease of 0.3663% or (1/273) of the initial volume is experienced for every temperature drop of 18C. These experiments were not extended to very low temperatures, but if the linear relationship were extrapolated, the volume of the gas would theoretically be zero at a temperature of approximately 22738C or 24608F. This temperature has become known as absolute zero and is the basis for the definition of two absolute temperature scales. (An absolute scale is one that does not allow negative quantities.) These absolute temperature scales are known as the Kelvin (K) and Rankine (8R) scales; the Kelvin scale is defined by shifting the Celsius scale by 2738C so that 0 K is equal to 22738C. Equation (4.3) shows the relationship described above: K ¼ 8C þ 273

(4:3)

21

Temperature

The Rankine scale is defined by shifting the Fahrenheit scale 4608, so that 8R ¼ 8F þ 460

(4:4)

The relationships among the various temperature scales are shown in Figure 4.1.

212

672

Boiling point, H2O (1 atm)

100

373

32 0

492 460

Freezing point, H2O (1 atm)

0 –18

273 255

–460

0

–273

0

°F

Absolute zero °R

°C

K

Figure 4.1 Temperature scales.

ILLUSTRATIVE EXAMPLE 4.2 Perform the following temperature conversions: 1 Convert 558F to (a) Rankine, (b) Celsius, and (c) Kelvin 2 Convert 558C to (a) Fahrenheit, (b) Rankine, and (c) Kelvin

SOLUTION: 1 (a) (b) (c) 2 (a) (b) (c)

8R ¼ 8F þ 460 ¼ 55 þ 460 ¼ 515 8C ¼ 59 (8F  32) ¼ 59 (55  32) ¼ 12:8 K ¼ 59 (8F þ 460) ¼ 59 (55 þ 460) ¼ 286 8F ¼ 1.8(8C) þ 32 ¼ 1.8(55) þ 32 ¼ 131 8R ¼ 1.8(8C) þ 492 ¼ 1.8(55) þ 492 ¼ 591 K ¼ 8C þ 273 ¼ 55 þ 273 ¼ 328

B

22

Chapter 4 Process Variables

PRESSURE Molecules possess a high degree of translational kinetic energy in the gaseous state, which means they are able to move quite freely throughout the body of the gas. If the gas is in a container of some type, the molecules are constantly bombarding the walls of the container. The macroscopic effect of this bombardment by a tremendous number of molecules—enough to make the effect measurable—is called pressure. The natural units of pressure are force per unit area. In the example of the gas in a container, the unit area is a portion of the inside solid surface of the container wall and the force, measured perpendicularly to the unit area, is the result of the molecules hitting the unit area and losing momentum during the sudden change of direction. There are a number of different methods used to express a pressure measurement. Some of them are natural units (i.e., based on a force per unit area) and include pound (force) per square inch (abbreviated lbf/in2 or psi) and dyne per square centimeter (dyn/cm2). Others are based on a fluid height such as inches of water (in H2O) or millimeters of mercury (mm Hg). These latter units are convenient when the pressure is indicated by a difference between two levels of a liquid as in a manometer or barometer. Barometric pressure and atmospheric pressure are synonymous and measure the ambient air pressure. Standard barometric pressure is the average atmospheric pressure at sea level, 458 north latitude at 328F. It is used to define another unit of pressure called the atmosphere (atm). Standard barometric pressure is 1 atm and is equivalent to 14.696 psi and 29.921 in Hg. As one might expect, barometric pressure varies with weather and altitude. Measurements of pressure by most gauges indicate the difference in pressure either above or below that of the atmosphere surrounding the gauge. Gauge pressure is the pressure indicated by such a device. If the pressure in the system measured by the gauge is greater than the pressure prevailing in the atmosphere, the gauge pressure is expressed positively; if lower than atmospheric pressure, the gauge pressure is a negative quantity; the term vacuum designates a negative gauge pressure. Gauge pressures are often identified by the letter “g” after the pressure unit, e.g., psig (pounds per square inch gauge) is a gauge pressure in psi units. Since gauge pressure is the pressure relative to the prevailing atmospheric pressure, the sum of the two yields the absolute pressure, indicated by the letter “a” after the unit, e.g., psia (pounds per square inch absolute): P ¼ Pa þ Pg where

(4:5)

P ¼ absolute pressure (psia) Pa ¼ atmospheric pressure (psia) Pg ¼ gauge pressure (psig)

The absolute pressure scale is absolute in the same sense that the absolute temperature scale is absolute, i.e., a pressure of zero psia is the lowest possible pressure theoretically achievable—a perfect vacuum.

Moles and Molecular Weight

23

ILLUSTRATIVE EXAMPLE 4.3 Consider the following pressure calculations. 1 A liquid weighing 100 lb held in a cylindrical column with a base area of 3 in2 exerts what pressure at the base in lbf/ft2? 2 If the pressure is 35 psig (pounds per square inch gauge), what is the absolute pressure? SOLUTION: 1 Refer to Chapter 3. F ¼ mg=gc ¼ 100 lb (1 lbf =lb) ¼ 100 lbf As discussed in Chapter 3, gc is a conversion factor equal to 32.2 lb . ft/lbf . s2; g is the gravitational acceleration, which is equal, or close to, 32.2 ft/s2 on Earth’s surface. Therefore, P ¼ F=S ¼ 100 lbf =3 in2 ¼ 33:33 lbf =in2 ¼ 4800 lbf =ft2 where S ¼ surface area onto which force is applied. 2 P ¼ Pa þ Pg ¼ 14:7 þ 35 ¼ 49:7 psia

B

MOLES AND MOLECULAR WEIGHT An atom consists of protons and neutrons in a nucleus surrounded by electrons. An electron has such a small mass relative to that of the proton and neutron that the weight of the atom (called the atomic weight) is approximately equal to the sum of the weights of the particles in its nucleus (the protons and neutrons). Atomic weight may be expressed in atomic mass units (amu) per atom or in grams per gram atom. One gram atom contains 6.02  1023 atoms (Avogadro’s number). The atomic weights of the elements are available in the literature.(1,2) The molecular weight (MW) of a compound is the sum of the atomic weights of the atoms that make up the molecule. Units of atomic mass units per molecule (amu/ molecule) or grams per gram mole (g/gmol) are used for molecular weight. One gram mole (gmol) contains an Avogadro’s number of molecules. For the English system, a pound mole (lbmol) contains 454  6.023  1023 molecules. Molal units are used extensively in mass transfer calculations as they greatly simplify material balances where chemical reactions are occurring. For mixtures of substances (gases, liquids, or solids), it is also convenient to express compositions in mole fractions or mole percentages instead of mass fractions. The mole fraction is

24

Chapter 4 Process Variables

the ratio of the number of moles of one component to the total number of moles in the mixture. Equations (4.6) – (4.9) express these relationships: mass A molecular weight of A mA nA ¼ (MW)A

moles of A ¼

moles A total moles nA yA ¼ n mass A mass fraction A ¼ total mass mA wA ¼ m volume A volume fraction A ¼ total volume VA vA ¼ V

(4:6)

mole fraction A ¼

(4:7)

(4:8)

(4:9)

The reader should note that, in general, mass fraction (or percent) is NOT equal to mole fraction (or percent). ILLUSTRATIVE EXAMPLE 4.4 A 55-gal tank contains 20.0 lb of water. 1 How many pound moles of water does it contain? 2 How many gram moles does it contain? 3 How many molecules does it contain? SOLUTION:

The molecular weight of the water is MW ¼ (2)(1:008) þ (15:999) ¼ 18:015 g=gmol ¼ 18:015 lb=lbmol

  lbmol ¼ 1:11 lbmol water 1 (20:0 lb) 18:015 lb    453:593 g gmol ¼ 503:6 gmol water 2 (20:0 lb) 1 lb 18:015 g   6:023  1023 molecules 3 (503:6 gmol) ¼ 3:033  1026 molecules 1 gmol

B

Viscosity

25

MASS, VOLUME, AND DENSITY The density (r) of a substance is the ratio of its mass to its volume and may be expressed in units of pounds per cubic foot (lb/ft3), kilograms per cubic meter (kg/m3), etc. For solids, density can be easily determined by placing a known mass of the substance in a liquid and determining the displaced volume. The density of a liquid can be measured by weighing a known volume of the liquid in a volumetric flask. For gases, the ideal gas law, to be discussed later in this chapter, can be used to calculate the density from the pressure, temperature, and molecular weight of the gas. Unlike gases, the densities of pure solids and liquids are relatively independent of temperature and pressure and can be found in standard reference books.(1,2) The specific volume (v) of a substance is its volume per unit mass (ft3/lb, m3/kg, etc.) and is therefore the inverse of its density. Two key densities that the practicing engineer should be familiar with are that for air and water. Although the effect of temperature and pressure can be obtained directly from the ideal gas law, the following equation can be used to estimate the density of air at atmospheric conditions(1):

rair (kg=m3 ) ¼ 1:30  4:68  103 (T)  1:40  105 (T)2 ; T ; 8C

(4:10)

As one might expect, the effect of temperature on liquid water is negligible. However, the following equation may be employed to account for any temperature variation(1):

rH2 O (kg=m3 ) ¼ 999:85 þ 6:1474  102 (T)  8:3633  103 (T)2 þ 6:6805  105 (T)3  4:3869  107 (T)4 þ 1:3095  109 (T)5 ; T ; 8C

(4:11)

The reader should note that the density of a gas is denoted by r, rV, or rG in this text. The term r will generally be employed, but the other notations appear in certain situations for clarity. In addition, the density of a liquid is represented as rL.

VISCOSITY Viscosity is a property associated with a fluid’s resistance to flow. More precisely, this property accounts for energy losses that result from shear stresses occurring between different portions of the fluid which are moving at different velocities.(3–5) The absolute viscosity (m) has units of mass per length . time; the fundamental unit is the poise (P), which is defined as 1 g/cm . s. This unit is inconveniently large for many practical purposes and viscosities are frequently given in centipoises (0.01 poise), which is abbreviated to cP. The viscosity of pure water at 68.68F is 1.00 cP. In English units, absolute viscosity is expressed either as pounds (mass) per foot . second (lb/ft . s) or pounds per foot . hour (lb/ft . h). The absolute viscosity depends primarily on temperature and to a lesser degree on pressure. The kinematic viscosity (n) is the absolute viscosity divided by the density of the fluid and is useful in certain fluid

26

Chapter 4 Process Variables

flow problems. The units for this quantity are length squared per time, e.g., square foot per second (ft2/s) or square meters per hour (m2/h). A kinematic viscosity of 1 cm2/s is called a stoke (S). For pure water at 708F, n ¼ 0.983 cS (centistokes). Because fluid viscosity changes rapidly with temperature, a numerical value of viscosity has no significance unless the temperature is specified. Liquid viscosity is usually measured by the amount of time it takes for a given volume of liquid to flow through an orifice. The Saybolt universal viscometer is the most widely used device in the United States for the determination of the viscosity of fuel oils and liquid wastes. It should be stressed that Saybolt viscosities, which are expressed in Saybolt seconds (SSU), are not even approximately proportional to absolute viscosities except in the range above 200 SSU; hence, converting units from Saybolt seconds to other units requires the use of special conversion tables. As the time of flow through a viscometer decreases, the deviation becomes more marked. In any event, viscosity is an important property because of possible flow problems with viscous liquids. The viscosities of air at atmospheric pressure and water are presented as functions of temperature in Tables 4.1 and 4.2, respectively. Viscosities of other substances are available in the literature(6) and the Appendix. Several simple equations for directly calculating viscosities are presented below. Equation (4.12) is used to calculate the viscosity of air:

mair (N  s=m2 ) ¼ 1:71  105 þ 5:0  108 T; T ; 8C

(4:12)

Regarding the viscosity of water, Equation (4.13) applies when the temperature is greater than 08C and less than or equal to 208C: log mH2 O (kg=m  s) ¼

1301 998:333 þ 8:1855(T  20) þ 5:85  103 (T  20)2  1:30233;

T ; 8C

(4:13)

This equation may be rearranged to give 

mH2 O ¼ 10



1301  1:30233 998:333þ8:1855(T20)þ5:85103 (T20)2

Table 4.1 Viscosity of Air at 1 Atmosphere a T (8C) 0 18 40 54 74 229 a

Viscosity, micropoise (mP) 170.8 182.7 190.4 195.8 210.2 263.8

1 P ¼ 100 cP ¼ 106 mP; 1 cP ¼ 6.72  1024 lb/ft . s.

(4:14)

Viscosity

27

Table 4.2 Viscosity of Water T (8C)

Viscosity, centipoise (cP)

0 5 10 15 20 25 30 35 40 50 60 70 80 90 100

1.792 1.519 1.308 1.140 1.000 0.894 0.801 0.723 0.656 0.594 0.469 0.406 0.357 0.317 0.284

The viscosity of the water in the 208C –1008C range may be calculated from Equation (4.15) ! mH2 O@T 1:3272(20  T)  1:053  103 (T  20)2 ¼ ; T ; 8C (4:15) log mH2 O@208C T þ 105 The reader should note that, as with density, the viscosity of a gas is denoted by m, mV, and mG in this text. In addition, the viscosity of a liquid is represented as mL. ILLUSTRATIVE EXAMPLE 4.5 What is the kinematic viscosity of a gas whose specific gravity (SG ¼ r=rH2 O ) and absolute viscosity are 0.8 and 0.02 cP, respectively? SOLUTION:    0:02 cP 6:720  104 lb=ft  s ¼ 1:344  105 lb=ft  s m¼ 1 1 cP

r ¼ (SG)(rref ) ¼ (0:8)(62:43 lb=ft3 ) ¼ 49:94 lb=ft3 n ¼ m=r ¼ (1:344  105 lb=ft  s)=(49:94 lb=ft3 ) ¼ 2:691  107 ft2 =s

ILLUSTRATIVE EXAMPLE 4.6 Calculate the viscosity of water at 108C.

B

28

Chapter 4 Process Variables

SOLUTION: Equation (4.13) can be used over the temperature range of 08C to 208C. Since the desired temperature is 108C, log mH2 O ¼

1301  1:30233 998:333 þ 8:1855(T  20) þ 5:85  103 (T  20)2

(4:13)

Substituting T ¼ 108C, log mH2 O ¼

mH2 O

1301  1:30233 998:333 þ 8:1855(108C  20) þ 5:85  103 (108C  20)2

¼ 0:11633 ¼ 1:307 centipoise ¼ 1:307  103 kg=m  s

B

REYNOLDS NUMBER The Reynolds number, Re, is a dimensionless number that indicates whether a fluid flowing is in the laminar or turbulent flow regime. Laminar flow is characteristic of fluids flowing slowly enough so that there are no eddies (whirlpools) or macroscopic mixing of different portions of the fluid. (Note: In any fluid, there is always molecular mixing due to the thermal activity of the molecules; this is distinct from macroscopic mixing due to the swirling motion of different portions of the fluid.) In laminar flow, a fluid can be imagined to flow like a deck of cards, with adjacent layers sliding past one another. Turbulent flow is characterized by eddies and macroscopic currents. In practice, moving gases are generally in the turbulent region.(5) For flow in a pipe, a Reynolds number above 2100 is an indication of turbulent flow. The Reynolds number is dependent on the fluid velocity, density, viscosity, and some length characteristic of the system or conduit; for pipes, this characteristic length is the inside diameter: Re ¼ Dvr=m ¼ Dv=n where Re ¼ Reynolds number D ¼ inside diameter of the pipe (ft) v ¼ fluid velocity (ft/s) r ¼ fluid density (lb/ft3) m ¼ fluid viscosity (lb/ft . s) n ¼ fluid kinematic viscosity (ft2/s) Any consistent set of units may be used with Equation (4.16).

(4:16)

pH

29

ILLUSTRATIVE EXAMPLE 4.7 Calculate the Reynolds number for a fluid flowing through a 5-inch diameter pipe at 10 fps (feet per second) with a density of 50 lb/ft3 and a viscosity of 0.65 cP. Is the flow turbulent or laminar? SOLUTION:

By definition Re ¼ Dvr=m

(4:16)

Substitution yields  Re ¼

      50 lb 10 ft 5 in 1 ft 1 1 cP s 1 12 in 0:65 cP 6:720  104 lb=ft  s ft3

¼ (50 lb=ft3 )(10 ft=s)(5=12 ft)=(0:65  6:72  104 lb=ft  s) ¼ 477,000 The Reynolds number is .2100; therefore, the flow is turbulent.

B

pH An important chemical property of an aqueous solution is pH. The pH measures the acidity or basicity of a solution. In a neutral solution, such as pure water, the hydrogen (Hþ) and hydroxyl (OH 2 ) ion concentrations are equal. At ordinary temperatures, this concentration is CHþ ¼ COH ¼ 107 gmol=L where

(4:17)

CHþ ¼ hydrogen ion concentration COH ¼ hydroxyl ion concentration

In all aqueous solutions, whether neutral, basic, or acidic, a chemical equilibrium or balance is established between these two concentrations, so that Keq ¼ CHþ COH ¼ 1014

(4:18)

where Keq ¼ equilibrium constant. The numerical value for Keq given in Equation (4.18) holds for room temperature and only when the concentrations are expressed in gram mole per liter (gmol/L). In acid solutions, CHþ . COH ; in basic solutions, COH predominates. The pH is a direct measure of the hydrogen ion concentration and is defined by pH ¼ log CHþ

(4:19)

30

Chapter 4 Process Variables

Thus, an acidic solution is characterized by a pH below 7 (the lower the pH, the higher the acidity), a basic solution by a pH above 7, and a neutral solution by a pH of 7. It should be pointed out that Equation (4.19) is not the exact definition of pH but is a close approximation. Strictly speaking, the activity of the hydrogen ion, aHþ , and not the ion concentration belongs in Equation (4.19). For a discussion of chemical activities, the reader is directed to the literature.(2,6) ILLUSTRATIVE EXAMPLE 4.8 Calculate the hydrogen ion and the hydroxyl ion concentrations of an aqueous solution if the pH of the solution is 1.0. SOLUTION:

For a pH of 1.0, apply Equation (4.19): pH ¼ log(CHþ ) CHþ ¼ 10pH ¼ 101 ¼ 0:1 gmol=L CHþ  COH ¼ 1014 COH ¼

1014 1014 ¼ 1 CHþ 10

¼ 1013 gmol=L

B

ILLUSTRATIVE EXAMPLE 4.9 Process considerations require pH control in a 50,000-gal storage tank used for incoming waste mixtures (including liquid plus solids) at a hazardous waste incinerator. Normally, the tank is kept at neutral pH. However, the operation can tolerate pH variations from 6 to 8. Waste arrives in 5000-gal shipments. Assume that the tank is completely mixed, contains 45,000 gal when the shipment arrives, the incoming acidic waste is fully dissociated, and that there is negligible buffering capacity in the tank. What is the pH of the most acidic waste shipment that can be handled without the need for neutralization? SOLUTION: The pH of the most acidic waste shipment that can be handled without neutralization is calculated as follows: 5000 gal of waste with a [Hþ] ¼ X is diluted by 45,000 gal at pH ¼ 7 or [Hþ] ¼ 1027. The minimum pH of 6 that can be tolerated is equivalent to a [Hþ] ¼ 1026. From an ion balance: [Hþ ] ¼ 106 ¼ (5000=50,000)X þ (45,000=50,000)(107 )    50,000 45,000(107 ) 106  X¼ 5000 50,000 ¼ 9:1  106 pH ¼ 5:04

B

Ideal Gas Law

31

VAPOR PRESSURE Vapor pressure is an important property of liquids, and, to a much lesser extent, of solids. If a liquid is allowed to evaporate in a confined space, the pressure in the vapor space increases as the amount of vapor increases. If there is sufficient liquid present, a point is eventually reached at which the pressure in the vapor space is exactly equal to the pressure exerted by the liquid at its own surface. At this point, a dynamic equilibrium exists in which vaporization and condensation take place at equal rates and the pressure in the vapor space remains constant. The pressure exerted at equilibrium is equal to the vapor pressure of the liquid. The magnitude of this pressure for a given liquid depends on the temperature, but not on the amount of liquid present. Solids, like liquids, also exert a vapor pressure. Evaporation of solids (sublimation) is noticeable only for those with appreciable vapor pressures. This is reviewed in more detail in Chapter 6.

IDEAL GAS LAW(7) Observations based on physical experimentation often can be synthesized into simple mathematical equations called laws. These laws are never perfect and hence are only an approximate representation of reality. The ideal gas law (IGL) was derived from experiments in which the effects of pressure and temperature on gaseous volumes were measured over moderate temperature and pressure ranges. This law works well in the pressure and temperature ranges that were used in taking the data; extrapolations outside of the ranges have been found to work well in some cases and poorly in others. As a general rule, this law works best when the molecules of the gas are far apart, i.e., when the pressure is low and the temperature is high. Under these conditions, the gas is said to behave ideally, i.e., its behavior is a close approximation to the so-called perfect or ideal gas: a hypothetical entity that obeys the ideal gas law perfectly. For engineering calculations, the ideal gas law is often assumed to be valid since it generally works well (usually within a few percent of the correct result) up to the highest pressures and down to the lowest temperatures used in many industrial applications. The two precursors of the ideal gas law were Boyle’s and Charles’ laws. Boyle found that the volume of a given mass of gas is inversely proportional to the absolute pressure if the temperature is kept constant: P1 V1 ¼ P2 V2

(4:20)

where V1 ¼ volume of gas at absolute pressure P1 and temperature T and V2 ¼ volume of gas at absolute pressure P2 and temperature T. Charles found that the volume of a given mass of gas varies directly with the absolute temperature at constant pressure: V1 V2 ¼ T1 T2

(4:21)

32

Chapter 4 Process Variables

where V1 ¼ volume of gas at pressure P and absolute temperature T1 and V2 ¼ volume of gas at pressure P and absolute temperature T2. Boyle’s and Charles’ laws may be combined into a single equation in which neither temperature nor pressure need be held constant: P1 V1 P2 V2 ¼ T1 T2

(4:22)

For Equation (4.22) to hold, the mass of gas must be constant as the conditions change from (P1, T1) to (P2, T2). This equation indicates that for a given mass of a specific gas, PV/T has a constant value. Since, at the same temperature and pressure, volume and mass must be directly proportional, this statement may be extended to PV ¼C mT

(4:23)

where m ¼ mass of a specific gas and C ¼ constant that depends on the gas. Moreover, experiments with different gases showed that Equation (4.23) could be expressed in a far more generalized form. If the number of moles (n) is used in place of the mass (m), the constant is the same for all gases: PV ¼R nT

(4:24)

where R ¼ universal gas constant. Equation (4.24) is called the ideal gas law. Numerically, the value of R depends on the units used for P, V, T, and n (see Table 4.3). Other useful forms of the ideal gas law are shown in Equations (4.25) and (4.26). Equation (4.25) applies to gas flow rather than to gas confined in a container(7): Pq ¼ n_ RT

(4:25)

where q ¼ gas volumetric flow rate (ft3/h), P ¼ absolute pressure (psia), n_ ¼ molar flow rate (lbmol/h), T ¼ absolute temperature (8R), and R ¼ 10.73 psia . ft3/ lbmol . 8R. Equation (4.26) combines n and V from Equation (4.25) to express the law in terms of density: P(MW) ¼ rRT (4:26) where MW ¼ molecular weight of the gas (lb/lbmol) and r ¼ density of the gas (lb/ft3). Volumetric flow rates are often not given at the actual conditions of pressure and temperature but at arbitrarily chosen standard conditions (STP, standard temperature and pressure). To distinguish between flow rates based on the two conditions, the letters “a” and “s” are often used as part of the unit. The units acfm and scfm represent actual cubic feet per minute and standard cubic feet per minute, respectively. The ideal gas law can be used to convert from standard to actual conditions, but,

Ideal Gas Law

33

Table 4.3 Values of R in Various Units R 10.73 0.7302 21.85 555.0 297.0 0.7398 1545.0 24.75 1.9872 0.0007805 0.0005819 500.7 1.314 998.9 19.32 62.361 0.08205 0.08314 8314 8.314 82.057 1.9872 8.314

Temperature scale

Units of V

Units of n

Units of P

Units of PV (energy)

8R 8R 8R 8R 8R 8R 8R 8R 8R 8R 8R 8R K K K K K K K K K K K

ft3 ft3 ft3 ft3 ft3 ft3 ft3 ft3 – – – – ft3 ft3 ft3 L L L L m3 cm3 – –

lbmol lbmol lbmol lbmol lbmol lbmol lbmol lbmol lbmol lbmol lbmol lbmol lbmol lbmol lbmol gmol gmol gmol gmol gmol gmol gmol gmol

psia atm in Hg mm Hg in H2O bar psfa ft H2O – – – – atm mm Hg psia mm Hg atm bar Pa Pa atm – –

– – – – – – – – Btu hp . h kW . h cal – – – – – – – – – cal J

since there are many standard conditions in use, the STP being used must be known. Standard conditions most often used are shown in Table 4.4. The reader is cautioned on the incorrect use of acfm and/or scfm. Employing standard conditions is a convenience; when predicting the performance of or designing equipment, the actual conditions must be employed. Designs based on standard conditions can lead to disastrous results, with the unit usually underdesigned. For example, for a gas stream at Table 4.4 Common Standard Conditions System SI Universal scientific Natural gas industry American engineering Environmental industry

Temperature

Pressure

Molar volume

273 K 08C 608F 328F 608F 708F

101.3 kPa 760 mm Hg 14.7 psia 1 atm 1 atm 1 atm

22.4 m3/kmol 22.4 L/gmol 379 ft3/lbmol 359 ft3/lbmol 379 ft3/lbmol 387 ft3/lbmol

34

Chapter 4 Process Variables

21408F, the ratio of acfm to scfm (standard temperature ¼ 608F) is 5.0. Equation (4.27), which is a form of Charles’ law, can be used to correct flow rates from standard to actual conditions:   Ta (4:27) qa ¼ qs Ts where qa ¼ volumetric flow rate at actual conditions (ft3/h), qs ¼ volumetric flow rate at standard conditions (ft3/h), Ta ¼ actual absolute temperature (8R), and Ts ¼ standard absolute temperature (8R). The reader is again reminded that absolute temperatures and absolute pressures must be employed in all ideal gas law calculations. In engineering practice, mixtures of gases are more often encountered than single or pure gases. The ideal gas law is based on the number of molecules present in the gas volume; the kind of molecules is not a significant factor, only the number. This ideal law applies equally well to mixtures and pure gases alike. Dalton and Amagat both applied the ideal gas law to mixtures of gases. Since pressure is caused by gas molecules colliding with the walls of the container, it seems reasonable that the total pressure of a gas mixture is made up of pressure contributions due to each of the component gases. These pressure contributions are called partial pressures. Dalton defined the partial pressure of a component as the pressure that would be exerted if the same mass of the component gas occupied the same total volume alone at the same temperature as the mixture. The sum of these partial pressures equals the total pressure: P ¼ pa þ pb þ pc þ    þ pn ¼

n X

pi

(4:28)

i¼1

where P ¼ total pressure, n ¼ number of components, and pi ¼ partial pressure of component i. Equation (4.28) is known as Dalton’s law. Applying the ideal gas law to one component (A) only, pA V ¼ nA RT

(4:29)

where nA ¼ number of moles of component A. Eliminating R, T, and V between Equations (4.24) and (4.29) yields pA nA ¼ ¼ yA P n or pA ¼ yA P where yA ¼ mole fraction of component A. ILLUSTRATIVE EXAMPLE 4.10 What is the density of air at 758F and 14.7 psia? The molecular weight of air is 29.

(4:30)

References SOLUTION:

35

This example is solved using the ideal gas law: m PV ¼ nRT ¼ RT (MW) P(MW) m ¼ ¼r RT V

or

r¼

P(MW) RT

(4:26)

Substituting,

r¼

(14:7 psia)(29 lb=lbmol) (10:73 ft3  psi=lbmol  8R)(75 þ 460)

¼ 0:0743 lb=ft3

B

ILLUSTRATIVE EXAMPLE 4.11 The exhaust gas flow rate from an absorber is 1000 scfm. All of the gas is vented through a small stack that has an inlet area of 1.0 ft2. The exhaust gas temperature is 3008F. What is the velocity of the gas through the stack inlet in feet per second? Assume standard conditions to be 708F and 1.0 atm. Neglect the pressure drop across the stack. SOLUTION: Note that since the gas is vented through the stack to the atmosphere, the pressure is 1.0 atm. Calculate the actual flow rate, in acfm, using Charles’ law: qa ¼ qs

  Ta Ts

(4:27)

Substituting ¼ 1000

  460 þ 300 460 þ 70

¼ 1434 acfm Calculate the velocity of the gas: v¼

qa 1434 ¼ 1:0 S

¼ 1434 ft=min

B

REFERENCES 1. R. C. WEAST (ed), “CRC Handbook of Chemistry and Physics,” 80th edition, CRC Press, Boca Raton, FL, 1999.

36

Chapter 4 Process Variables

2. S. MARON and C. PRUTTON, “Principles of Physical Chemistry,” 4th edition, MacMillan, New York City, NY, 1970. 3. R. BIRD, W. STEWART, and E. LIGHTFOOT, “Transport Phenomena,” 2nd edition, John Wiley & Sons, Hoboken, NJ, 2002. 4. L. THEODORE, “Transport Phenomena for Engineers,” International Textbook Co., Scranton, PA, 1970. 5. P. ABULENCIA and L. THEODORE, “Fluid Flow for the Practicing Engineer,” John Wiley & Sons, Hoboken, NJ, 2009. 6. D. Green and R. Perry (eds), “Perry’s Chemical Engineers’ Handbook,” 8th edition, McGraw-Hill, New York City, NY, 2008. 7. L. THEODORE, F. RICCI, and T. VAN VLIET, “Thermodynamics for the Practicing Engineer,” John Wiley & Sons, Hoboken, NJ, 2009.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

5

Equilibrium vs Rate Considerations INTRODUCTION There are two important factors that need to be taken into consideration when analyzing mass transfer operations: equilibrium and rate. Although these two subjects have been segmented and treated separately below, both need to be considered together when analyzing a mass transfer device. Because of the importance of equilibrium and rate, both receive treatment in the next two chapters. However, the question often asked is: Which is the more important of the two when discussing mass transfer operations? This is best answered by noting that most—but not all—mass transfer calculations assume equilibrium conditions to apply. A correction factor or an efficiency term is then included to adjust/upgrade the result/prediction to actual (as opposed to equilibrium) conditions in order to accurately describe the phenomena in question. As one would expect, the correction factor or efficiency is usually based on experimental data, past experience, similar designs, or just simply good engineering judgement. Although this method of analysis has been maligned by many theoreticians and academicians, the approach has merit and has been routinely used by practicing engineers.(1) This pragmatic approach is primarily employed in this text/reference book.

EQUILIBRIUM A state of equilibrium exists when the forward and reverse rates of a process are equal. From a macroscopic point of view, there is no change with time. Equilibrium represents a limiting value for the practicing engineer as demonstrated in the following example. Whenever a substance distributes itself between two other materials, the system will attempt to approach equilibrium. Thus, if NH3 gas in air is brought into contact with water at 608F, the NH3 will begin and continue to dissolve in the water until

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

37

38

Chapter 5 Equilibrium vs Rate Considerations

its concentration in the water has reached a maximum value at that temperature. This condition represents equilibrium, and no further solution of the NH3 will occur unless this equilibrium is disturbed. Alternatively, if water containing NH3 is brought into contact with air containing no NH3, the NH3 will escape from the water and pass into the gas phase until the value of the concentrations in the two phases has reached the equilibrium value. This particular scenario will be revisited in Chapter 10, a chapter which is concerned with absorption and stripping. If one wishes to remove NH3 from an air mixture by using water, it is obvious that equilibrium can set a limit on the maximum amount of NH3 that can be removed from the air to the water. It can also set the lower limit on the amount of water necessary for a particular degree of NH3 removal. Therefore, equilibrium is a vital factor in the design and operation of these type of mass transfer systems. In the situation described above, equilibrium is fortunately simple to represent, and will find application in absorption as well as other systems. Occasionally, the relationships are complicated. Specific cases of importance will be discussed in Chapters 6 and 10.

RATE The companion to equilibrium in mass transfer operations is rate. The transfer of a substance from one phase to another obviously requires time. The rate of transfer is proportional to the surface of contact between the phases, the resistance to the transfer, and the driving force present for mass transfer. This phenomena is described in Equations (5.1) and (5.2). Employing the macroscopic approach, a transfer process, whether it be mass, energy, or momentum, can be simply described as shown below: Rate of Transfer ¼

(Driving Force)(Area Available for Transfer) (Resistance to Transfer)

(5:1)

For mass transfer, the equation becomes Rate of Mass Transfer ¼

(Concentration Driving Force)(Area Available for Mass Transfer) (Resistance to Mass Transfer)

(5:2)

Other things being equal, Equations (5.1) and (5.2) indicate that the rate of transfer can be increased by: 1 increasing the area, 2 increasing the driving force, and 3 decreasing the resistance. These three factors are almost always considered in the design of mass transfer equipment. As one might intuitively expect, any increase in the transfer rate leads to a more compact mass transfer device that is generally more economical. A more detailed presentation on rate considerations is provided in Chapter 7.

Chemical Reactions

39

Consider again the comments provided earlier in the section on equilibrium concerning the NH3 – air – H2O system. No mention was made of the rate of transfer of NH3 at that time. But in line with Equation (5.2), during the initial stages of the transfer process, the NH3 concentration in the air is high while its concentration in water is low (or even zero). This produces a high concentration difference driving force that leads to a high NH3 transfer rate. However, later in the transfer process, the NH3 concentrations in both phases tend to “equilibrate”, leading to a lower concentration difference driving force, and a lower transfer rate. When the system ultimately reaches/achieves equilibrium, the driving force becomes zero and the rate of transfer also becomes zero. With all mass transfer processes, the true driving force for mass transfer is the chemical potential of the substance to be transferred. Just as temperature acts as a thermal potential for heat transfer, every substance has a chemical (or “mass”) potential which “drives” it from one phase into another. A state of equilibrium is only achieved in a non-reacting system when the temperature, pressure, and chemical potential of every species is equal in all phases. While the study of chemical potential is best left to a thermodynamicist, the subject of mass transfer tends to use concentration as a substitute for representing the chemical potential. Obviously, the rate as well as equilibrium play a role in the transfer process. Both effects need to be considered in designing, predicting performance, and operating mass transfer equipment. Which is more important? It depends. Normally, equilibrium information is required. Moreover, if the rate of transfer is extremely high, rate considerations can in some instances be neglected.

CHEMICAL REACTIONS With regard to chemical reactions, there are two important questions that are of concern to the engineer: 1 how far will the reaction go? 2 how fast will the reaction go? Chemical thermodynamics provides the answer to the first question; however, it tells us nothing about the second. Reaction rates fall within the domain of chemical kinetics and will not be treated in this text.(1) To illustrate the difference and importance of both of the above questions on an engineering analysis of a chemical reaction, consider the following process.(2,3) Substance A, which costs 1 cent/ton, can be converted to B, which is worth $1 million/lb, by the reaction A $ B. Chemical thermodynamics will provide information on the maximum amount of B that can be formed. If 99.99% of A can be converted to B, the reaction would then appear to be economically feasible, from a thermodynamic point of view. However, a rate or kinetic analysis might indicate that the reaction is so slow that, for all practical purposes, its rate is vanishingly small. For example, it might take 106 years to obtain a 1026% conversion of A. The reaction is then economically unfeasible. Thus, it can be seen that both equilibrium and rate/kinetic effects

40

Chapter 5 Equilibrium vs Rate Considerations

must be considered in an overall engineering analysis of a chemical reaction. The same principle applies to gaseous mass transfer separation, e.g., absorption.(4) Equilibrium and rate are both important factors to be considered in the design and prediction of performance of equipment employed for chemical reactions. The rate at which a reaction proceeds will depend on the departure from equilibrium, with the rate at which equilibrium is established essentially dependent on a host of factors. As can be expected, this rate process would cease upon the attainment of equilibrium. If one is conducting a chemical reaction in which reactants go to products, the products will be formed at a rate governed in part by the concentration of the reactants and reaction conditions such as the temperature and pressure. Eventually, as the reactants form products and the products react to form reactants, the net rate of reaction must equal zero. At this point, equilibrium will have been achieved. ILLUSTRATIVE EXAMPLE 5.1 As part of his investigation, Detective Theodore paid a visit to the office of his missing bookmaker—the location where the bookmaker was last seen. In examining the premises, he notices a half-empty 5 inch tall coffee mug on the missing bookmaker’s desk. Based on his experience and the stain left in the mug, Detective Theodore estimated that the coffee mug was initially full to 0.5 inches from the brim at the time of the bookmaker’s disappearance and that the bookmaker had been missing for approximately two weeks. Qualitatively explain how this brilliant sleuth reached this conclusion. SOLUTION: Detective Theodore instinctively realized that the time it took for the coffee in the mug to evaporate would reflect how long the bookmaker had been missing. In mass transfer terms, this case reduced to the evaporation of a single component, i.e., essentially pure water into a stagnant gas (air) at room temperature and 1 atm of pressure. After combining principles to be discussed later in this book, this extraordinary detective was able to calculate the time when his bookmaker disappeared. B

REFERENCES 1. 2. 3. 4.

L. THEODORE: personal notes, 2008. L. THEODORE: personal notes, 1978. L. THEODORE, “Chemical Reaction Kinetics,” A Theodore Tutorial, East Williston, NY, 1994. L. THEODORE, “Air Pollution Control Equipment Calculations,” John Wiley & Sons, Hoboken, NJ, 2009.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

6

Phase Equilibrium Principles INTRODUCTION Equilibrium principles play an important role in designing and predicting the performance of many mass transfer processes. In fact, several mass transfer calculations are based primarily on the application of equilibrium principles and equilibrium data. Distillation is a prime example; virtually every calculational procedure is based on vapor – liquid equilibrium data, e.g., acetone – water. Similarly, adsorption engineering applications almost always utilize vapor (gas) – solid equilibrium data, e.g., acetone – activated carbon. It is for this reason that phase equilibrium principles in general and specific equilibrium systems are reviewed in this chapter. The term phase, for a pure substance, indicates a state of matter, i.e., solid, liquid, or gas. For mixtures, however, a more stringent connotation must be used, since a totally liquid or solid system may contain more than one phase (e.g., a mixture of oil and water). A phase is characterized by uniformity or homogeneity, which means that the same composition and properties must exist throughout the phase region. At most temperatures and pressures, a pure substance normally exists as a single phase. At certain temperatures and pressures, two or perhaps even three phases can coexist in equilibrium. This is shown on the phase diagram for water (see Fig. 6.1). Regarding the interpretation of this diagram, the following points should be noted: 1 The line between the gas and liquid phase regions is the boiling point and dew point line and represents equilibrium between the gas and liquid. 2 The boiling point of a liquid is the temperature at which its vapor pressure is equal to the external pressure. The temperature at which the vapor pressure is equal to 1atm is the normal boiling point. 3 The line between the solid and gas phase regions is the sublimation point and deposition point line and represents equilibrium between the solid and gas. 4 The line between the solid and liquid phase regions is the melting point and freezing point line and represents equilibrium between the liquid and solid.

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

41

Chapter 6 Phase Equilibrium Principles

Pressure, atm

42

220

Critical point

Liquid

Solid 1.0 0.006

Triple point

Vapor

100 0 0.0098 Temperature, °C

374

Figure 6.1 Phase diagram for water.

5 The point at which all three equilibrium lines meet (i.e., the one pressure and temperature where solid, liquid, and gas phases can all coexist) is the triple point. 6 The liquid– gas equilibrium line is bounded on one end by the triple point and the other end by the critical point. The critical temperature (the temperature coordinate of the critical point) is defined as the temperature above which a gas or vapor cannot be liquefied by the application of pressure alone. The term vapor, strictly speaking, is used only for a condensable gas, i.e., a gas below its critical temperature, and should not be applied to a noncondensable gas. It should also be pointed out that the phase diagram for water (Fig. 6.1) differs from that of other substances in one respect—the freezing point line is negatively sloped; for other substances, the slope of this line is positive. This is a consequence of the fact that liquid water is denser than ice. Mass transfer calculations rarely involve single (pure) components. Phase equilibria for multicomponent systems are considerably more complex, mainly because of the addition of composition variables. For example, in a ternary (three-component) system, the mole fractions of two of the components are pertinent variables along with temperature and pressure. In a single-component system, dynamic equilibrium between two phases is achieved when the rate of molecular transfer from one phase to the second equals that in the opposite direction. In multicomponent systems, the equilibrium requirement is more stringent—the rate of transfer of each component must be the same in both directions.

Introduction

43

Relationships governing the equilibrium distribution of a substance between two phases, particularly gas and liquid phases, are the principal subject matter of phase-equilibrium thermodynamics. As noted above, these relationships form the basis of calculational procedures that are employed in the design and the prediction of performance of several mass transfer processes.(1) For many, mass transfer has come to be defined as the tendency of a component in a mixture to travel from a region of high concentration to one of low concentration. For example, if an open beaker with some alcohol in it is placed in a room in which the air is relatively dry, alcohol vapor will diffuse out through the column of air in the beaker. Thus, there is mass transfer of alcohol from a location where its concentration is high ( just above the liquid surface) to a place where its concentration is low (at the top of the beaker). If the gas mixture in the beaker is stagnant (void of any motion), the transfer is said to occur by molecular diffusion. If there is a bulk mixing of the layers of gas in the beaker, mass transfer is said to occur by the mechanism of either forced or natural convection, or both. These two mechanisms are analogous to the transfer of heat by conduction and convection.(2) Analogies with fluid flow are not nearly as obvious.(3) In discussing the principles and applications of mass transfer, the presentation will primarily consider binary mixtures, although some multicomponent mixture issues are addressed. Binary mixtures serve as an excellent starting point for training and educational purposes. However, real world applications often involve more than two components. The problem of transferring materials from one phase to another is encountered in many engineering operations. There are generally three classes of phases encountered in practice: gas, liquid, and solid. If mass transfer occurs between two phases, with one phase being treated and the other performing the treatment, a total of (3)2 or nine possible combinations of operations with two phases is possible. These mini-scenarios are listed in Table 6.1.

Table 6.1 Mass Transfer Scenarios Scenario 1 2 3 4 5 6 7 8 9

Phase undergoing treatment

Phase performing treatment

Gas Gas Gas Liquid Liquid Liquid Solid Solid Solid

Gas Liquid Solid Gas Liquid Solid Gas Liquid Solid

44

Chapter 6 Phase Equilibrium Principles

GIBB’S PHASE RULE The general subject of phases and equilibrium is only one of several topics highlighted in this chapter. Equilibrium in a multiphase system is subject to certain restrictions. These restrictions can be expressed in an equation form, defined as the Phase Rule. The state of a P– V –T system is established when its temperature and pressure and the compositions of all phases are fixed. However, these variables are not all independent for equilibrium states, and fixing a limited number of them automatically establishes the others. This number of independent variables is given by Gibb’s Phase Rule and is defined as the number of degrees of freedom of the system. This number of variables must be specified in order to fix the state of a system at equilibrium. The following nomenclature is employed in the discussion to follow: P ¼ number of phases C ¼ number of chemical components r ¼ number of independent chemical reactions; components minus elements F ¼ number of degrees of freedom The phase-rule variables are temperature, pressure, and C21 mole fractions in each phase. The number of these variables is therefore 2 þ (C21)P. The masses of the phases are not phase-rule variables because they have no effect on the state of the system. It should also be noted that the temperature and pressure have been assumed to be uniform throughout the system for an equilibrium state. The total number of independent equations (by componential balance) is (P 21)C þ r. These equations are functions of temperature, pressure, and compositions; therefore, they represent relations connecting the phase-rule variables. Since F is the difference between the number of variables and the number of equations: F ¼ 2 þ (C  1)P  (P  1)C  r F ¼2PþCr

(6:1)

The variable r can be determined by subtracting the number of elements in the system from the number of components. Note that r ¼ 0 in a non-reacting system. ILLUSTRATIVE EXAMPLE 6.1 A gaseous system at equilibrium consists of CO2, H2, H2O, CH4, and C2H6. Determine the degrees of freedom. SOLUTION:

Determine the number of phases, P, and the number of components, P¼1

C¼5

Determine the number of independent chemical reactions, r. Number of elements ¼ 3; C, H, and O Number of components ¼ 5

Raoult’s Law

45

Therefore, r ¼53¼2 Write the equation for the Gibb’s Phase Rule and calculate the degrees of freedom F ¼2PþCr

(6:1)

Substituting ¼21þ52¼4 Since F ¼ 4, one is free to specify, e.g., T, P and two mole fractions in an equilibrium mixture of these five chemical species, provided nothing else is arbitrarily set. B

In mass transfer applications, the most important equilibrium phase relationship (as noted above) is that between liquid and vapor. Raoult’s law and Henry’s law theoretically describe liquid – vapor behavior and under certain conditions are applicable in practice. Raoult’s law and Henry’s law are the two equations most often used in the introductory study of phase equilibrium—specifically, within the boundaries of ideal vapor – liquid equilibrium (VLE). Both Raoult’s and Henry’s law help one understand the equilibrium properties of liquid mixtures. The next two sections review these laws. Phase equilibrium examines the physical properties of various classes of mixtures, and analyzes how different components affect each other within those mixtures. There are four key classes of mixtures (see Table 6.1): 1 vapor– liquid 2 vapor– solid 3 liquid– solid 4 liquid– liquid Topics (2) and (3) receive superficial treatment later in this chapter. It is the authors’ opinion that the two most important traditional mass transfer operations are distillation and absorption. As such, the presentation to follow will primarily address principles that directly apply to these two mass transfer processes. For example, material on Raoult’s law and equilibrium relationships will be employed in Chapter 9, Distillation. Material on Henry’s law and mass transfer coefficients will be directly applied to absorption processes in Chapter 10.

RAOULT’S LAW Raoult’s law states that the partial pressure of each component ( pi) in a solution is proportional to the mole fraction (xi) of that component in the liquid mixture being studied. The “proportionality constant” is its vapor pressure ( p0i ). Therefore, for component i in a mixture, Raoult’s law can be expressed as: pi ¼ p0i xi

(6:2)

46

Chapter 6 Phase Equilibrium Principles

where pi is the partial pressure of component i in the vapor, p0i is the vapor pressure of pure i at the same temperature, and xi is the mole fraction of component i in the liquid. This expression may be applied to all components so that the total pressure P is given by the sum of all the partial pressures. If the gas phase is ideal, Dalton’s law applies pi ¼ yi P and one may write P¼

X

xi p0i

(6:3)

i

Therefore, Equation (6.2) can then be written as follows:  0 p yi ¼ xi i P

(6:4)

where yi is the mole fraction of component i in the vapor and P is the total system pressure. For example, the mole fraction of water vapor in air that is saturated, and in equilibrium contact with pure water (x ¼ 1.0), is simply given by the ratio of the vapor pressure of water at the system temperature divided by the system pressure. While it is true that some of the air dissolves in the water (making xw , 1.0), one usually neglects this to simplify calculations. These equations primarily find application in distillation, and to a lesser extent, absorption and stripping. The basis of Raoult’s law can be best understood in molecular terms by considering the rates at which molecules leave and return to the liquid. Raoult’s law illustrates how the presence of a second component, say B, reduces the rate at which component A molecules leave the surface of the liquid, but it does not inhibit the rate at which they return.(4) In order to finally deduce Raoult’s law, one must draw on additional experimental information about the relation between the vapor pressures and the composition of the liquid. Raoult himself obtained this data from experiments on mixtures of closely related liquids in order to develop his law. Most mixtures obey Raoult’s law to some extent, however small it may be. The mixtures that closely obey Raoult’s law are those whose components are structurally similar—and therefore Raoult’s law is most useful when dealing with these types of solutions. Mixtures that obey Raoult’s law for the entire composition range are called ideal solutions. A graphical representation of this behavior can be seen in Figure 6.2. Many solutions do deviate significantly from Raoult’s law. However, the law is obeyed even in these cases for the component in excess (which, in this case, would be the solvent) as it approaches purity. Therefore, if the solution of solute is dilute, the properties of the solvent can be approximated using Raoult’s law. Raoult’s law, however, does not have universal applications. There are several aspects of this law which give it limitations. First, it assumes that the vapor is an ideal gas, which is not necessarily the case. Second, Raoult’s law only applies to ideal solutions, and, in reality, there are no ideal solutions. Also, a third problem to be aware of is that Raoult’s law only really works for solutes which do not change their nature when they dissolve (i.e., they do not ionize or associate). A correction factor which accounts for liquid-phase deviations from Raoult’s law is developed later in this chapter.

Raoult’s Law

47

p'A

Total pressure, P

Partial pressure

p'B

Partial pressure of A, pA

Partial pressure of B, pB 0

Mole fraction of A

1

Figure 6.2 Graphical representation of the vapor pressure of an ideal binary solution, i.e., one that obeys Raoult’s law for the entire composition range.

Raoult law applications require vapor pressure information. Vapor pressure data is available in the literature.(1) However, there are two equations that can be used in lieu of actual vapor pressure information—the Clapeyron equation and the Antoine equation. The Clapeyron equation is given by ln p0 ¼ A  (B=T)

(6:5)

where p0 and T are the vapor pressure and temperature, respectively. The Antoine equation is given by ln p0 ¼ A  B=(T þ C)

(6:6)

Table 6.2 Approximate Clapeyron Equation Coefficients

Acetaldehyde Acetic anhydride Ammonium chloride Ammonium cyanide Benzyl alcohol Hydrogen peroxide Nitrobenzene Nitromethane Phenol Tetrachloroethane 

T in K, p0 in mm Hg.

A

B

18.0 20.0 23.0 22.9 21.9 20.4 18.8 18.5 19.8 17.5

3.32  103 5.47  103 10.0  103 11.5  103 7.14  103 5.82  103 5.87  103 4.43  103 5.96  103 4.38  103

48

Chapter 6 Phase Equilibrium Principles Table 6.3 Antoine Equation Coefficients

Acetone Benzene Ethanol n-Heptane Methanol Toluene Water 

A

B

C

14.3916 13.8594 16.6758 13.8587 16.5938 14.0098 16.2620

2795.82 2773.78 3674.49 2911.32 3644.30 3103.01 3799.89

230.00 220.07 226.45 216.64 239.76 219.79 226.35

T in 8C, p0 in kPa.

Note that for both equations, the units of p0 and T must be specified for given values of A and B and/or C. Values for the Clapeyron equation coefficients—A and B—are provided in Table 6.2 for some compounds. Some Antoine equation coefficients— A, B, and C—are listed in Table 6.3. Additional values for these coefficients for both equations are available in the literature.(1,4) ILLUSTRATIVE EXAMPLE 6.2 The Clapeyron equation coefficients for acetone have been experimentally determined to be A ¼ 15:03 B ¼ 2817 with p0 and T in mm Hg and K, respectively. Estimate its vapor pressure at 08C. SOLUTION:

The Clapeyron equation is given by ln p0 ¼ A  (B=T)

(6:5) 0

Employ the correct units. Substituting, the vapor pressure, p , of acetone at 08C is ln p0 ¼ 15:03  2817=(0 þ 273) ¼ 4:7113 0 p ¼ 111:2 mm Hg

B

ILLUSTRATIVE EXAMPLE 6.3 The Antoine coefficients for acetone are: A ¼ 14:3916 B ¼ 2795:82 C ¼ 230:00 with p0 and T in the same units noted in the previous example. Use the Antoine equation to estimate the vapor pressure of acetone at 08C.

Raoult’s Law SOLUTION:

49

The Antoine equation is given by ln p0 ¼ A  B=(T þ C)

(6:6)

Substituting, the vapor pressure of acetone at 08C predicted by the Antoine equation is ln p0 ¼ 14:3916  2795:82=(0 þ 230:00) ¼ 2:236 p0 ¼ 9:35 kPa

B

The reader should note that the Clapeyron equation generally overpredicts the vapor pressure at or near ambient conditions. The Antoine equation is widely used in industry and usually provides excellent results. Also note that, contrary to statements appearing in the Federal Register and some Environmental Protection Agency (EPA) publications, vapor pressure is not a function of pressure.

ILLUSTRATIVE EXAMPLE 6.4 If the vapor pressure of acetone at a given temperature is 71 mm Hg, calculate the maximum vapor phase concentration in mole fraction units at 1 atm total pressure. SOLUTION: The maximum concentration of a component in a noncondensable gas obeying Raoult’s law is given by its vapor pressure p0 divided by the total pressure, P. Any increase in concentration will result in the condensation of the component in question. The maximum mole fraction of acetone in air at 08C and 1 atm is therefore: ymax ¼

p0 P

(6:2)

Substituting, ymax ¼

71 ¼ 0:0934 760

B

It is important to note that vapor mixtures do not condense at one temperature as a pure vapor does. The temperature at which a vapor begins to condense as the temperature is lowered is defined as the dew point. It is determined by calculating the temperature at which a given vapor mixture is saturated. The bubble point of a liquid mixture is defined as the temperature at which it begins to vaporize. The bubble point may also be viewed as the temperature at which the last vapor condenses, while the dew point is the temperature at which the last liquid vaporizes.(1) These examples are based on a given and constant pressure, and are thus referred to as the dew point and bubble point temperatures. Calculations based on holding the temperature constant lead to the dew point and bubble point pressures. Dew point and bubble point calculations enable one to obtain vapor – liquid equilibrium (VLE) relationships for a binary mixture; these are often provided as a P-x, y diagram (with the temperature constant) or a T-x, y diagram (with the pressure constant), or both. VLE data can be generated assuming Raoult’s law applies, and there are two types of diagrams of interest: the aforementioned P-x, y and T-x, y.

50

Chapter 6 Phase Equilibrium Principles

Additional details and procedures for obtaining these graphs are provided in the literature(1); algorithms are also available. The next two Illustrative Examples serve as an introduction to VLE calculations which assume that Raoult’s law applies.

ILLUSTRATIVE EXAMPLE 6.5 A liquid stream contains 5 gmol % ethane (A) and 95 gmol % n-hexane (B) at 258C. The vapor pressure of ethane at 258C is 4150 kPa and the vapor pressure of n-hexane at 258C is 16.1 kPa. If the pressure is such that this is a saturated liquid, what is the pressure and what is the composition of the first vapor to form? SOLUTION:

For a two component (A –B) system, Equation (6.3) reduces to P ¼ xA p0A þ xB p0B

(6:3)

Substituting, P ¼ (0:05)(4150) þ (0:95)(16:1) ¼ 222:8 kPa ¼ 32:3 psia From Equation (6.4), yA ¼ xA p0A =P Substituting, yA ¼ (0:05)(4150)=222:8 ¼ 0:931 yB ¼ 1  0:931 ¼ 0:069 This is an example of a bubble point pressure calculation (since the temperature is fixed) for a liquid mixture. B

ILLUSTRATIVE EXAMPLE 6.6 The vapor pressures of carbon tetrachloride and ethyl acetate at 278C are: carbon tetrachloride (A) ¼ 111 mm Hg ethyl acetate (B) ¼ 92 mm Hg Calculate the composition of the liquid and vapor phases in equilibrium at this temperature and a total pressure of 100 mm Hg. SOLUTION:

Substitute into Equation (6.3) for the total pressure P ¼ p0A xA þ p0B (1  xA ) 100 ¼ 111xA þ 92(1  xA ) xA ¼ 0:421

Raoult’s Law

51

The vapor phase composition is given by yA ¼ p0A xA =P yA ¼ (111)(0:421)=100

(6:4)

yA ¼ 0:467

B

ILLUSTRATIVE EXAMPLE 6.7 Using the vapor– liquid equilibrium (VLE) data for the ethanol–water system provided in Table 6.4, generate a T-x, y diagram. Table 6.4 Vapor–Liquid Equilibrium Data for Ethanol –Water at 1.0 atm T (8C)

xETOH

yETOH

212 192 186 181 180 177 176 174 173 172 171

0.000 0.072 0.124 0.238 0.261 0.397 0.520 0.676 0.750 0.862 1.000

0.000 0.390 0.470 0.545 0.557 0.612 0.661 0.738 0.812 0.925 1.000

SOLUTION: The T-x, y diagram is plotted in Figure 6.3. The top curve ( y vs T ) represents saturated vapor and the bottom curve (x vs T ) is saturated liquid. B

Temperature, °C

215

200 Vapor

185 Liquid 170

0

0.5 x, y (ethanol)

Figure 6.3 T-x, y diagram for the ethanol–water system.

1.0

B

52

Chapter 6 Phase Equilibrium Principles

Finally, the reader should note that all the phase equilibrium calculations in this chapter are based on the assumption that the vapor and liquid are in equilibrium. The above presentation is also limited to two-component systems. An approach to multiphase (two or more coexisting liquid phases) VLE is available in the literature.(1)

ILLUSTRATIVE EXAMPLE 6.8 The following P-x, y data is provided in Table 6.5 for the methanol (m) –water (w) system at 408C. Plot a P-x, y diagram. Also provide an x–y plot.

Table 6.5 P-x, y Data for Methanol –Water System (T ¼ 408C)

SOLUTION:

P, kPa

xm

ym

7.356 10.16 12.97 15.77 18.58 21.39 24.19 27.00 29.81 32.61 35.42

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.000 0.349 0.546 0.674 0.762 0.828 0.878 0.918 0.951 0.977 1.000

Applicable plots are provided in Figures 6.4 and 6.5.

B

41.5 36.5

P (kPa)

31.5 26.5 21.5 16.5 11.5 6.5 1.5 0.0

0.1

0.2

0.3 0.4 0.5 0.6 0.7 Mole fraction (methanol)

Figure 6.4 P-x, y diagram for methanol–water at 408C (Raoult’s law).

0.8

0.9

1.0

Henry’s Law

53

1.0

Mole fraction of vapor methanol, ym

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Mole fraction of liquid methanol, xm

Figure 6.5 x –y diagram for methanol– water (408C).

B

Another application of interest arises in gas absorption operations. The equilibrium of interest is that between a nonvolatile absorbing liquid (solvent) and a solute gas. The solute is ordinarily removed from its mixture in a relatively large amount of a carrier gas which does not dissolve in the absorbing liquid. Therefore, it is often possible, and frequently the case when considering the removal and/or recovery of a gaseous component by absorption, to assume that only the component in question is transferred between phases. Both the solubility of the nondiffusing (inert) gas in the liquid and the presence of vapor from the liquid in the gas phase are usually neglected. The important variables to be considered then are the pressure, temperature, and the concentrations of the component in the liquid and the gas phase. The temperature and pressure may be fixed and the concentration(s) of the component(s) in the various phases are defined from phase equilibrium relationships. An equation that may be employed to relate the equilibrium concentration of the absorbed species in the liquid phase is the aforementioned Henry’s law.(5)

HENRY’S LAW Unfortunately, relatively few mixtures follow Raoult’s law. Henry’s law is another empirical relationship used for representing data on many systems. Henry’s law states that the partial pressure of the solute is proportional to its mole fraction, but the proportionality constant is not the vapor pressure of the pure substance (as it is

54

Chapter 6 Phase Equilibrium Principles

in Raoult’s law). Instead, the proportionality constant is some empirical constant, denoted HB. Therefore, for some component B of a solution, Henry’s law(5) can be written as: pB ¼ HB xB

(6:7)

The value of the Henry’s law constant, HB, is found to be temperature dependent. The value generally increases with increasing temperature. As a consequence, the solubility of gases generally decreases with increasing temperature. In other words, the dissolved gases in a liquid can be driven off by heating the liquid. Mixtures which obey Henry’s law are known as ideal-dilute solutions. The above equation has also been written, for component A in this instance, as yA ¼ mA xA

(6:8)

where mA is once again an empirical constant. In order to fully understand the concepts behind Henry’s law, one can once again examine the basic physical properties of a dilute solution on a molecular level. In a dilute solution, the solvent molecules are in an environment which is not that much different from the one they experience in a pure liquid. The solute molecules, however, are in a completely different environment than that of the pure solute state. Because of this, the solvent behaves as a slightly modified pure liquid, whereas the solute behaves entirely differently from its pure state. In this case, the rate of escape of solute molecules will be proportional to their concentration in the solution, and solute will accumulate in the gas until the return rate is equal to the rate of escape. This return rate will be proportional to the partial pressure of solute with a very dilute gas. It is important to remember that if the solute and solvent are very similar in structure, the solute obeys Raoult’s law. It may be rigorously proven that all non-ionic binary solutions (at relatively low pressures such that the vapor phase is ideal) obey Equation (6.7) as component “B” approaches infinite dillution. Moreover, when the dilute component (B) follows Henry’s law, the other component (A) must obey Raoult’s law over the same range of composition. The results obtained from Henry’s law for the mole fraction of dissolved gas is valid for the liquid layer just beneath the interface, but not necessarily the entire liquid. The latter will be the case only when thermodynamic equilibrium is established throughout the entire liquid body. There, the use of Henry’s law is limited to dilute gas – liquid solutions, i.e., liquids with a small amount of gas dissolved in them. The linear relationship of Henry’s law does not apply in the case when the gas is highly soluble in the liquid. Henry’s law has been found to hold experimentally for all dilute solutions in which the molecular species is the same in the solution as in the gas. One of the most conspicuous and apparent exceptions to this is the class of electrolytic solutions, or solutions in which the solute has ionized or dissociated. As in the case of Raoult’s law, Henry’s law in this particular case, does not hold. Several constants for Henry’s law are provided in Table 6.6. However, equations are available to estimate this constant.(1)

Henry’s Law

55

Table 6.6 Henry’s Law Constants for Gases in Water at Approximately 258C H, (bar)21

Gas Acetylene Air Carbon dioxide Ethane Ethylene Methane

1350 73,000 1700 31,000 11,500 42,000

ILLUSTRATIVE EXAMPLE 6.9 Explain why there is a temperature increase when a gas is “dissolved” in a liquid. SOLUTION: Since gases usually liberate heat when they dissolve in liquids, thermodynamics reveals (as noted earlier) that an increase of temperature will result in a decrease in solubility. This is why gases may be readily removed from solution by heating. B

Another important factor influencing the solubility of a gas is pressure. As is to be expected from kinetic considerations, compression of the gas will tend to increase its solubility. ILLUSTRATIVE EXAMPLE 6.10 Given Henry’s law constant for and the partial pressure of H2S, determine the maximum mole fraction of H2S that can be dissolved in solution. Data are provided below. Partial pressure of H2S ¼ 0.01 atm Total pressure ¼ 1.0 atm Temperature ¼ 608F Henry’s law constant, H H2 S ¼ 483 atm=mole fraction (1atm, 608F) SOLUTION:

Write the equation describing Henry’s law. pH2 S ¼ H H2 S xH2 S pH2 S ¼ yH2 S P

(6:7)

Calculate the maximum mole fraction of H2S that can be dissolved in solution xH2 S ¼ pH2 S =H H2 S ¼ 0:01=483 ¼ 2:07  105

B

56

Chapter 6 Phase Equilibrium Principles (a)

(b)

Air

¨NH3

Air – NH3

H2O

H2O

t<0

t =0

(c)

(d)

Air – NH3

Air – NH3

H2O

H2O–NH3

t* ≈ 0

t > t* > 0

(e)

yNH3

(f)

Air – NH3

Air – NH3

H2O–NH3

H2O–NH3

t=∞

t=∞

xNH3

Figure 6.6 Ammonia absorption system at T and P.

To illustrate the application of Henry’s law to the aforementioned absorption process,(1) consider the air – water– ammonia system at the T and P pictured in Figure 6.6a– f. In this system, NH3 is added to the air at time t ¼ 0 (b). The NH3 slowly proceeds to distribute itself (c, d) until “equilibrium” is reached in (e). The mole fractions in both phases are measured in (f). This data is represented as point (1) in Figure 6.7. If the process in Figure 6.6 is repeated several times with additional quantities of NH3, additional equilibrium points will be generated. These points appear in Figure 6.8.

Henry’s Law

57

y

1 x

Figure 6.7 x –y equilibrium point.

y

m x

Figure 6.8 x –y equilibrium diagram.

Although the plot in Figure 6.8 curves upwards, the data approaches a straight line of slope m at low values of x. It is this region where it is assumed that Henry’s law applies. One of the coauthors(6) of this text believes that as x ! 0, m ! 0. This conclusion, and its ramifications, has received further attention in the literature.(6) ILLUSTRATIVE EXAMPLE 6.11 Convert the ammonia –water equilibrium data given in Table 6.7 to an x–y plot at 308C and 1 atm. Evaluate Henry’s law constant for this system. Over what range of liquid mole

58

Chapter 6 Phase Equilibrium Principles Table 6.7 Equilibrium Data for Ammonia –Water System; 308C, 1 atm x, mole fraction NH3 in liquid 0 0.0126 0.0167 0.0208 0.0258 0.0309 0.0405 0.0503 0.0737 0.0960 0.1370 0.1750 0.2100 0.2410 0.2970

pNH3 , mm Hg 0 11.5 15.3 19.3 24.4 29.6 40.1 51.0 79.7 110.0 179.0 260.0 352.0 454.0 719.0

Table 6.8 Ammonia-Water Equilibrium (x –y) Data; 308C, 1 atm xNH3 0 0.0126 0.0167 0.0208 0.0258 0.0309 0.0405 0.0503 0.0737 0.0960 0.1370 0.1750 0.2100 0.2410 0.2970

yNH3 ¼

pNH3 P

0 0.015 0.0210 0.0254 0.0321 0.03894 0.05276 0.06710 0.10486 0.145 0.236 0.342 0.463 0.596 0.945

Raoult’s Law vs Henry’s Law

59

Henry’s law graph 1.0 y, mole fraction of NH3 in vapor

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

x, mole fraction of NH3 in liquid

Figure 6.9 Ammonia–water equilibrium at 308C and 1 atm.

fraction will Henry’s law predict the equilibrium ammonia vapor content to within 5% of the experimental data? SOLUTION: The partial pressure of ammonia is converted below to mole fraction in the vapor. See Table 6.8. These results are plotted in Figure 6.9. Henry’s law constant from the graph is approximately 1.485 (¼ 0.141/0.095) at x ¼ 0.095 based on linear regression. Since yactual ¼ 0:148 ycalculated ¼ 1:485(0:095) ¼ 0:141   0:141 100 ¼ 95:27% Percent agreement ¼ 0:148 Thus, from x ¼ 0 to x ¼ 0.095, Henry’s law equation, y ¼ 1.485x, predicts the equilibrium vapor content to within 5% of the experimental data. B

RAOULT’S LAW VS HENRY’S LAW(7) A basic difference between Raoult’s and Henry’s laws is that Raoult’s law applies to the solvent, while Henry’s law applies to the solute. In ideal solutions, both the solute and the solvent obey Raoult’s law. However, in ideal dilute solutions, the solute obeys Henry’s law whereas the solvent obeys Raoult’s law. In Figure 6.10, one can see the differences between Raoult’s law and Henry’s law in graphical form, in addition to the behavior of a real solution. It provides information on how two components

60

Chapter 6 Phase Equilibrium Principles

p'B

Real solution

Id e

al

dil

ut e

so

lut ion

(H en ry

’s l

aw )

Pressure

HB

ult’s

) law

o

on luti

(Ra

o

al s

Ide

0

Mole fraction of B, xB

1

Figure 6.10 Graphical representation of the differences between Raoult’s and Henry’s law; component B represents the solute.

are distributed between the vapor and liquid phase, e.g., acetone –water, while Henry’s law provides how a component, e.g., acetone will be distributed between a gas and liquid phase, e.g., air and water. There are several other significant differences between these two laws which must be recognized. Raoult’s law is much more theory-based, since it only applies to ideal situations. Henry’s law, on the other hand, is more empirically based, making it a more general and practical law. In addition to providing a link between the mole fraction of the solute and its partial pressure, Henry’s law constants may also be used to calculate gas solubilities. This plays an important role in biological functions, such as the transport of gases in the bloodstream. A knowledge of Henry’s law constants for gases in fats and lipids is important in discussing respiration.(4,7) For example, consider scuba diving. In this recreational activity, air is supplied at a higher pressure so as to allow the pressure within the diver’s chest to match the pressure exerted by the surrounding water. This water pressure of the ocean increases at approximately 1 atm per 10 meters of depth. However, air inhaled at higher pressures makes nitrogen more soluble in fats and lipids rather than water—causing nitrogen to enter the central nervous system, bone marrow, and fat reserves of the body. The nitrogen then bubbles out of its lipid solution if the diver rises to the surface too quickly, causing a condition called decompression sickness, also known as the bends. This condition can be fatal. The nitrogen gas bubbles can block arteries and cause unconsciousness as they rise to the brain. Another interesting application of Henry’s law is in the treatment of carbon monoxide poisoning. In a hyperbaric oxygen chamber, oxygen is raised to an elevated

Vapor– Liquid Equilibrium in Nonideal Solutions

61

partial pressure. When an individual with carbon monoxide poisoning steps inside this chamber, there is a steep pressure gradient between the partial pressure of arterial blood’s oxygen, and the partial pressure of the oxygen in freshly inhaled air. In this way, oxygen floods quickly into arterial blood, allowing a rapid re-supply of oxygen to the bloodstream. As noted above, Henry’s law and Raoult’s law have several similarities and differences, and each has their restrictions. The knowledge of both of these laws is invaluable not only academically, but practically as well.

VAPOR– LIQUID EQUILIBRIUM IN NONIDEAL SOLUTIONS(1) In the case where liquid solutions cannot be considered ideal, Raoult’s law will give highly inaccurate results. For these nonideal liquid solutions, various alternatives are available. These are considered in most standard thermodynamics texts and will be treated to some extent below. One important case will be mentioned because it is frequently encountered in distillation, namely, where the liquid phase is not an ideal solution, but the pressure is low enough so that the vapor phase behaves as an ideal gas. In this case, the deviations from ideality are localized in the liquid and treatment is possible by quantitatively considering deviations from Raoult’s law. These deviations are taken into account by incorporating a correction factor, g, into Raoult’s law. The purpose of g, defined as the activity coefficient, is to account for the departure of the liquid phase from ideal solution behavior. It is introduced into Raoult’s law equation (for component A) as follows: yA P ¼ pA ¼ gA xA p0A

(6:9)

The activity coefficient is a function of liquid phase composition and temperature. Phase-equilibria problems of the above type are often effectively reduced to evaluating g. Vapor – liquid equilibrium calculations performed with Equation (6.9) are slightly more complex than those made with Raoult’s law. The key equation then becomes X X X ( yi P) ¼ gi xi p0i (6:10) P¼ pi ¼ When applied to a two-component (A– B) system, Equation (6.10) becomes P ¼ yA P þ yB P ¼ gA xA p0A þ gB xB p0B

(6:11)

yA P ¼ pA ¼ gA xA p0A yB P ¼ pB ¼ gB xB p0B

(6:12) (6:13)

so that

Information on methods to determine the activity coefficient(s) follows.

62

Chapter 6 Phase Equilibrium Principles

Theoretical developments in the molecular thermodynamics of non-ideal liquid solution behavior are often based on the concept of local composition. Within a liquid solution, local compositions, different from the overall mixture composition, are presumed to account for the short-range order and nonrandom molecular orientations that result from differences in molecular size and intermolecular forces. The concept was introduced by G. M. Wilson in 1964 with the publication of a model of solution behavior, since known as the Wilson equation.(8) The success of this equation in the correlations of vapor– liquid equilibrium data prompted the development of several alternative local-composition models. Perhaps the most notable of these is the NRTL (Non-Random-Two Liquid) equation of Renon and Prausnitz.(9) The application of both approaches (briefly discussed below) has received extensive treatment in the literature.(1,10) The Wilson equation contains just two adjustable parameters for a binary system (EAB and EBA), and the activity coefficients are written as:   EAB EBA (6:14)  ln gA ¼ ln(xA þ xB EAB ) þ xB xA þ xB EAB xB þ xA EBA   EAB EBA  (6:15) ln gB ¼ ln(xB þ xA EBA )  xA xA þ xB EAB xB þ xA EBA For infinite dilution, these equations become: ln g1 A ¼ ln(EAB ) þ 1  EBA

(6:16)

ln g1 B ¼ ln(EBA ) þ 1  EAB

(6:17)

Note that g1 i is to be applied for any component i as xi ! 0. Also note that the Wilson parameters, EAB and EBA, must always be positive numbers. The temperature dependence of the Wilson parameter is given by: Eij ¼

Vj aij ; i=j exp Vi RT

(6:18)

where Vj and Vi are the molar volumes at temperature T of pure liquids j and i, respectively, and 2aij is a constant independent of composition and temperature. Thus, the Wilson equation has built into it an approximate temperature dependence for the parameters. Wilson model coefficients are provided in Table 6.9 for a number of binary systems.(8,10) The NRTL equation, containing three parameters for a binary system, is: "  # 2 GBA GAB tAB 2 ln gA ¼ xB tBA þ (6:19) xA þ xB GBA ðxB þ xA GAB Þ2 "  # 2 GAB GBA tBA 2 þ (6:20) ln gB ¼ xA tAB xB þ xA GAB ðxA þ xB GBA Þ2

Vapor– Liquid Equilibrium in Nonideal Solutions

63

Table 6.9 Wilson Equation Parameters(8,10)

System

VA VB (cm3/gmol)

Wilson equation (cal/gmol)

Acetone Water

74.05 18.07

291.27

1448.01

Methanol Water

40.73 18.07

107.38

469.55

1-Propanol Water

75.14 18.07

775.48

1351.90

Water 1,4-Dioxane

18.07 85.71

1696.98

2219.39

Methanol Acetonitrile

40.73 66.30

504.31

196.75

Acetone Methanol

74.05 40.73

2161.88

583.11

Methyl acetate Methanol

79.84 40.73

231.19

813.18

Methanol Benzene

40.73 89.41

1734.42

183.04

Ethanol Toluene

58.68 106.85

1556.45

210.52

aAB

aBA

The parameters G and t can be obtained using the following equations GAB ¼ exp(atAB )

(6:21)

GBA ¼ exp(atBA )

(6:22)

Furthermore, bAB RT bBA ¼ RT

tAB ¼ tBA

(6:23)

where a, bAB, bBA are parameters specific to a particular pair of species, independent of composition and temperature. The infinite-dilution values of the activity coefficients are given by: ln g1 A ¼ tBA þ tAB exp(atAB ) ln g1 B ¼ tAB þ tBA exp(atBA )

(6:24) (6:25)

Tabulated values of the parameters for the NRTL model can be found in Table 6.10.(9,10)

64

Chapter 6 Phase Equilibrium Principles Table 6.10

NRTL Equation Parameters(9,10) NRTL equation (cal/gmol) bBA

a

631.05

1197.41

0.5343

2253.88

845.21

0.2994

1-Propanol Water

500.40

1636.57

0.5081

Water 1,4-Dioxane

715.96

548.90

0.2920

Methanol Acetonitrile

343.70

314.59

0.2981

Acetone Methanol

184.70

222.64

0.3084

Methyl acetate Methanol

381.46

346.54

0.2965

Methanol Benzene

730.09

1175.41

0.4743

Ethanol Toluene

713.57

1147.86

0.5292

System

bAB

Acetone Water Methanol Water

VAPOR –SOLID EQUILIBRIUM This section is concerned with a discussion of vapor – solid equilibria. The relation between the amount of substance adsorbed by an adsorbent (solid) and the equilibrium partial pressure or concentration of the substance at constant temperature is called the adsorption isotherm. The adsorption isotherm is the most important and by far the most often used of the various vapor – solid equilibria data which can be measured. Note that, if the substance being adsorbed (adsorbate) is above its critical temperature, it is technically considered a “gas”. Under such conditions, the term “gas –solid” equilibrium is more appropriate. Most available data on adsorption systems are determined at equilibrium conditions. Adsorption equilibrium is the set of conditions at which the number of molecules arriving on the surface of the adsorbent equals the number of molecules that are leaving. An adsorbent bed is said to be “saturated with vapors” and can remove no more vapors from the exhaust stream. Equilibrium determines the maximum amount of vapor that may be adsorbed on the solid at a given set of operating conditions. Although a number of variables affect adsorption, the two most important ones in determining equilibrium for a given system are temperature and pressure.

Vapor–Solid Equilibrium

65

Capacity weight, % (1 lb CCl4/100 lb C)

100

Figure 6.11

32°F

77°F

F

140°

F

212°

10

3.0 2.0 1.5 1.0 0.0001

0.001

F

300°

0.01 Partial pressure, psia

0.1

1.0

Adsorption isotherms for carbon tetrachloride on activated carbon.

Three types of equilibrium graphs and/or data are used to describe adsorption systems: isotherm at constant temperature, isobar at constant pressure, and isostere at constant amount of vapors adsorbed. As noted above, the most common and useful adsorption equilibrium data is the adsorption isotherm. The isotherm is a plot of the adsorbent capacity vs the partial pressure of the adsorbate at a constant temperature. Adsorbent capacity is usually given in weight fraction (or percent) expressed as grams of adsorbate per 100 g of adsorbent. Figure 6.11 shows a typical example of an adsorption isotherm for carbon tetrachloride on activated carbon. Graphs of this type are used to estimate the size of adsorption systems.(1,6) Attempts have been made to develop generalized equations that can predict adsorption equilibrium from physical data. This is very difficult because adsorption isotherms take many shapes depending on the forces involved. Isotherms may be concave upward, concave downward, or “S”-shaped. To date, most of the theories agree with data only for specific adsorbate-systems and are valid over limited concentration ranges. Two additional adsorption equilibrium relationships are the aforementioned isostere and the isobar. The isostere is a plot of the ln p vs 1/T at a constant amount of vapor adsorbed. Adsorption isostere lines are usually straight for most adsorbate – adsorbent systems. The isostere is important in that the slope of the isostere (approximately) corresponds to the heat (enthalpy) of adsorption. The isobar is a plot of the amount of vapors adsorbed vs temperature at a constant partial pressure. However, in the design of most engineering systems, the adsorption isotherm is by far the most commonly used equilibrium relationship. Several models have been proposed to describe this equilibrium phenomena. Freundlich proposed the following equation to represent the variation of the amount of adsorption per unit area or unit mass with partial pressure, Y ¼ kp1=n

(6:26)

66

Chapter 6 Phase Equilibrium Principles

where Y is the weight or volume of gas (or vapor) adsorbed per unit area or unit mass of adsorbent and p is the equilibrium partial pressure of the adsorbate. The k and n are empirical constants dependent on the nature of the solid and adsorbate, and on the temperature. Equation (6.26) may be rewritten as follows. Taking logarithms of both side,   1 log p (6:27) log Y ¼ log k þ n If log Y is now plotted against log p, a straight line should result with slope equal to 1/n and an ordinate intercept equal to log k. Although the requirements of the equation are met satisfactorily at lower pressures, the experimental points curve away from the straight line at higher pressures, indicating that this equation does not have general applicability in reproducing adsorption of gases (or vapors) by solids. A much better equation for isotherms was deduced by Langmuir from theoretical considerations. The final form is given in Equation (6.28). Y¼

ap 1 þ bp

(6:28)

Equation (6.28) may be rewritten as   p 1 b ¼ þ p Y a a

(6:29)

Since a and b are constants, a plot of p/Y vs p should yield a straight line with slope equal to b/a and an ordinate intercept equal to 1/a. By assuming that adsorption on solid surfaces takes place with the formation of secondary, tertiary, and finally multilayers upon the primary monomolecular layer, researchers Brunauer, Emmett, and Teller(11) derived the relation   P 1 c1 p ¼ þ (6:30) v( p0  p) vm c vm c p0 where v is the volume, reduced to standard conditions of gas (or vapor) adsorbed at system pressure P and T; p is the partial pressure of adsorbate; T is the temperature; p0 is the vapor pressure of the adsorbate at temperature T; vm is the volume of gas (or vapor), reduced to standard conditions, adsorbed when the surface is covered with a unimolecular layer; and, c is the constant at any given temperature. The constant c is approximately given by c ¼ e(E1 EL )=RT

(6:31)

where E1 is the enthalpy of adsorption of the first adsorbed layer and EL is the enthalpy of liquefaction of the gas (or vapor). The most useful theory from an engineering design viewpoint, in trying to predict adsorption isotherms, is the Polanyi potential theory. The Polanyi theory states that the adsorption potential is a function of the reversible isothermal work done by

Vapor–Solid Equilibrium

67

the system. Polyani(12) as well as Dubinin and co-workers(13–15) showed that the adsorption isotherms of various vapors can be represented by the following equations ln(W) ¼ ln(W 0 )  k(E=b)2

(6:32)

E ¼ (RT) ln( p0 = p)

(6:33)

where

and W is the volume of condensed adsorbate per gram of carbon (m3/g); W0 is the active pore volume of carbon (m3/g); k is a constant related to pore structure (cal/ gmol)22; b is an affinity coefficient which permits comparison of the adsorption potential of the (test) adsorbate to a reference adsorbate; R is the gas constant, 1.987 cal/gmol . K; T is the absolute temperature (K); p is the equilibrium partial pressure of adsorbate; and p0 is the vapor pressure of adsorbate at T. The above equations were suggested by Dubinin for the case when the pores of the adsorbent are comparable in size to the adsorbed molecules. These can be used to determine the equilibrium adsorption isotherm of any given vapor from the adsorption isotherm of a reference vapor, provided that the value of the affinity coefficient, b, of this vapor is available. The adsorption of nonpolar and weakly polar vapors is dominated by dispersion forces. For this situation, Dubinin proposed that b be determined from the ratio of the molar volume, V, of a test solvent to that of reference solvent used to obtain the values of W0 and k for the given carbon. ILLUSTRATIVE EXAMPLE 6.12 The representation of adsorption data at a constant temperature that indicates the amount adsorbed vs the partial pressure of the adsorbate is called an 1 2 3 4

Isobar Isostere Isotherm Isokinetic

SOLUTION: Adsorption equilibrium data at a constant temperature are referred to collectively as an isotherm. The correct answer is therefore (3). B

ILLUSTRATIVE EXAMPLE 6.13 Determine the equilibrium capacity of an activated carbon bed during the adsorption phase for carbon tetrachloride (CCl4) at the following operating conditions: Airflow rate ¼ 12,000 cfm at 778F Concentration of CCl4 air (inlet) ¼ 410 ppmv System at atmospheric pressure Steam regeneration at 2128F Refer to Figure 6.11.

68

Chapter 6 Phase Equilibrium Principles

SOLUTION: The only pertinent data provided are the operating temperature, pressure, and CCl4 concentration during the adsorption step. At 410 ppmv, the partial pressure is given by p ¼ (410=106 )(14:7);

P ¼ 1 atm ¼ 14:7 psia

¼ 6:03  103 psia Referring to Figure 6.11, the equilibrium capacity at this partial pressure at a temperature of 778F is approximately 40%, or 40 lb CCl4/100 lb activated carbon. B

LIQUID– SOLID EQUILIBRIUM The simplest system in liquid– solid equilibria is one in which the components are completely miscible in the liquid state and the solid phase consists of a pure component. These systems find application in water purification processes. The equilibrium equations and relationships presented in the vapor – solid equilibrium section generally apply to these systems as well. Details are available in the literature.(16) ILLUSTRATIVE EXAMPLE 6.14 Adsorption processes are often used as a follow-up to chemical wastewater treatment to remove organic reaction products that cause taste, odor, color, and toxicity problems. The equilibrium relationship between adsorbents (solid materials that adsorb organic matter, e.g., activated carbon) and adsorbates (substances that are bound to the adsorbents, e.g., benzene) may be simply expressed as q ¼ Kcn

(6:34)

where q ¼ amount of organic matter adsorbed per amount of adsorbent, c ¼ concentration of organic matter in water, n ¼ experimentally determined constant, and K ¼ equilibrium distribution constant. Based upon the data given in Table 6.11 that were obtained from a laboratory sorption experiment, a college student who worked for a company as an intern was asked to evaluate the K and n values for a certain type of activated carbon to be used to remove undesirable by-products formed during chemical treatment. What are the approximate values of K and n that the intern should have generated? Table 6.11 Sorption Data Collected in Laboratory Experiments; Illustrative Example 6.14 c (mg/L) 50 100 200 300 400 650

q (mg/g) 118 316 894 1640 2530 5240

References

3.6

69

log(q) = 1.48 log(c) – 0.456 r 2 = 0.9999

log(q)

3.2

2.8

2.4

2.0 1.6

2.0

2.4

2.8

log(c)

Figure 6.12

Log– log plot of sorption data; Illustrative Example 6.14.

SOLUTION: The equilibrium relationship given in the problem statement can be linearized by taking the log of both sides of the equation. This yields the following equation: log(q) ¼ log(K) þ n log(c) A plot of log(q) versus log(c) yields a straight line if this relationship can be used to represent the experimental data. The slope of this line is equal to n, while the intercept is log(K ). The experimental data analyzed using the equation above are plotted in Figure 6.12 and shows that the data fit this linearized isotherm quite well. The equation generated from a regression analysis(17) indicates that n ¼ 1:48 log(K) ¼ 0:456

or

K ¼ 0:35

provided the units of c and q are mg/L and mg/g, respectively.

B

REFERENCES 1. L. THEODORE, F. RICCI, and T. VAN VLIET, “Thermodynamics for the Practicing Engineer,” John Wiley & Sons, Hoboken, NJ, 2009. 2. L. THEODORE, “Heat Transfer for the Practicing Engineer,” John Wiley & Sons, Hoboken, NJ, 2011 (in preparation). 3. P. ABULENCIA and L. THEODORE, “Fluid Flow for the Practicing Engineer,” John Wiley & Sons, Hoboken, NJ, 2009. 4. P. ATKINS and J. DE PAULA, “Atkins’ Physical Chemistry,” 8th edition, W. H. Freeman and Co., San Francisco, CA, 2006. 5. W. HENRY, Publication unknown, 1803. 6. L. THEODORE, “Air Pollution Control Equipment Calculations,” John Wiley & Sons, Hoboken, NJ, 2009. 7. A. MOHAN, adapted from homework assignment submitted to L. Theodore, 2007. 8. G. WILSON, J. Am. Chem. Soc., 86, 27–130, 1964. 9. H. RENON and J. PRAUSNITZ, AIChE J., 14, 135– 144, 1968.

70

Chapter 6 Phase Equilibrium Principles

10. J. SMITH, H. VAN NESS, and M. ABBOTT, “Chemical Engineering Thermodynamics,” 6th edition, McGraw-Hill, New York City, NY, 2001. 11. S. BRUNAUER, P. H. EMMETT, and E. TELLER, J. Am. Chem. Soc., 60, 309, 1938. 12. K. POLANYI, Trans. Faraday Soc., 28, 316, 1932. 13. J. DUBININ, Chem. Rev., 60, 235, 1960. 14. A. KADLEC and J. DUBININ, Coll. Int. Sci., 31, 479, 1969. 15. J. DUBININ, “Chemistry and Physics of Carbon,” Vol. 2, P. L. Walker (ed), Marcel Dekker, New York City, NY, 51, 1966. 16. D. GREEN and R. PERRY, “Perry’s Chemical Engineers’ Handbook,” 8th edition, McGraw-Hill, New York City, NY, 2008. 17. S. SHAEFER and L. THEODORE, “Probability and Statistics Applications for Environmental Science,” CRC Press/Taylor & Francis Group, Boca Raton, FL, 2007.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

7

Rate Principles INTRODUCTION The rate transfer process can be described by the product of three terms (see Chapter 5): 1 the area available for transfer 2 the driving force for transfer 3 the (reciprocal of the) resistance to the transfer process In effect, the rate process in equation form is rate ¼

(area)(driving force) (resistance)

(7:1)

For mass transfer (MT) applications, Equation (7.1) becomes ðrate of MTÞ ¼

ðarea available for MTÞðdriving force for MTÞ (resistance to MT)

(7:2)

Rate principles may be applied at molecular, microscopic, or macroscopic levels. These three approaches were previously discussed in a generic sense in Chapter 2. For mass transfer operations, the molecular rate process will employ the diffusivity in Fick’s Law,(1) while the microscopic approach will employ overall and individual mass transfer coefficients. Both the molecular and microscopic approaches are reviewed in this chapter and every attempt has been made to relate the molecular and microscopic approaches to each other. Relating the microscopic treatment of mass transfer operations to the macroscopic treatment is not as simple, but information is available in the literature.(2,3) As noted in Chapter 2, macroscopic approaches often produce algebraic equations that have withstood the test of time. For example, the macroscopic equation for calculating the height of an absorber (see Chapter 10) is simply given by the

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

71

72

Chapter 7 Rate Principles

product of two terms: HOG and NOG. HOG is related to the rate process while NOG is related to equilibrium.

THE OPERATING LINE The NH3 –air– H2O discussion presented in Chapter 6 is now revisited in an attempt to shed some light on rate considerations. Refer first to Figure 6.8 in the previous chapter. The initial state and final (equilibrium) state of the system pictured in Figure 7.1 are designated with a square point and a circular point, respectively. If V and L represent the moles of flowing air and water, respectively, and y and x the mole fraction of NH3 in the air and H2O, respectively, an NH3 mole balance written between the initial (0) and final (1) states gives (7:3) L x0 þ Vy0 ¼ Lx1 þ Vy1 (The reader should note that the terms V and G are used interchangeably in this text as well as in the literature.) This assumes that if the mole fraction of NH3 is small in both phases (L and V are essentially constant), this equation may be rearranged to give L(x1  x0 ) ¼ V( y0  y1 ) L y0  y1 ¼ V x1  x0

(7:4)

or 

L y1  y0 ¼ V x1  x0

(7:5)

There are two important points to be made regarding Figure 7.1. 1 The dashed line may be thought of as an “operating line” since it describes the operating x, y values during the system’s transition from point 0 to 1.

0 y

a –L V

b c

1 x

Figure 7.1 Equilibrium-operating line plot.

Fick’s Law

a

73

0

y 1

b

c

x

Figure 7.2 Equilibrium-operating line plot.

2 The vertical displacement of any point from the equilibrium line provides a direct measure of the driving force (and thus the rate) of the system’s attempt to achieve equilibrium. In effect, the driving force is maximum at point 0 and zero at point 1. Equation (7.4) plots out as a straight line as shown with dashes in Figure 7.1. If ab/ bc ¼ 2.0, the mole ratio of liquid to gas is correspondingly 2.0. If the initial state is as shown in Figure 7.2 and ab/bc ¼ 0.5, then the liquid to gas ratio is also 0.5. The important point made in the above analysis is that the rate of the NH3 transfer process is linearly related to the vertical displacement of the point representing the state of the system from the equilibrium point directly below and, as indicated above, the rate ultimately becomes zero when the operating point reaches the equilibrium point. While this type of macroscopic rate/equilibrium approach receives treatment in Part II—Applications, what follows keys on the aforementioned molecular and microscopic considerations.

FICK’S LAW Molecular diffusion results from the motion of molecules. At any instant, the individual molecules in a fluid are moving in random directions at speeds varying from low to high values. The molecules move at random, frequently colliding with one another. Because of the frequent collisions, the molecular velocities are continually changing in both direction and magnitude. Diffusion is more rapid at higher temperatures due to greater molecular velocities. For gases, it is more rapid at low pressures because the average distance between the molecules is greater and the collisions are less frequent. If a solution is not uniform in concentration, the solution is gradually brought into uniformity by diffusion; the molecules move from an area of high concentration to one

74

Chapter 7 Rate Principles

of low concentration. The rate at which a solute travels depends on the concentration gradient which exists in the solution. This gradient applies across adjacent regions of high and low concentrations. However, a quantitative measure of rate is needed to describe what is occurring. The rate of diffusion can be described in terms of a molar flux term, with units of moles/(area)(time), and with the area being measured as that which the solute diffuses through. In a nonuniform solution containing only two components, both must diffuse if uniformity is to occur. This leads to the use of two fluxes to describe the motion of one of the components N, the flux relative to a fixed location and J, the flux of a component relative to the average molar velocity of all components. The first of these is of importance in the design of equipment, but the second is more characteristic of the nature of the component. For example, the rate at which a fish swims upstream against the flowing current is analogous to N, while the velocity of the fish relative to the stream is more characteristic of the swimming ability of the fish and is analogous to J. The diffusivity, or diffusion coefficient, DAB, of component A in solution B, which is a measure of its diffusive mobility, is defined as the ratio of its flux, JA, to its concentration gradient and is given by JA ¼ DAB

@CA @z

(7:6)

This is Fick’s first law(1) written for the z direction. The concentration gradient term represents the variation of the concentration, CA, in the z direction. The negative sign accounts for diffusion occurring from high to low concentrations. The diffusivity is a characteristic of the component and its environment (temperature, pressure, concentration, etc.). This equation is analogous to the flux equations(2,3) defined for momentum transfer (in terms of the previously defined viscosity) and for heat transfer (in terms of the thermal conductivity). The diffusivity is usually expressed with units of (length)2/time or moles/time . area. This coefficient, as well as Fick’s law, will receive additional treatment later in this chapter. ILLUSTRATIVE EXAMPLE 7.1 Express the diffusivity in English units. SOLUTION:

Based on its definition, the units of the diffusivity may be expressed as either ft2 =hr

or lbmol=hr  ft The latter units are derived by simply multiplying ft2/hr by the molar density, i.e., lbmol/ft3. B

Imagine if two fluids are placed side by side in a container separated by a partition.(2) As pictured in Figure 7.3, fluid A is on the left-hand side and fluid B is on

Fick’s Law

A

B

A

Before

B

A

+B

During

A

75

+B

After

Figure 7.3 Diffusion process.

the right-hand side. When the partition is removed, the two fluids begin to diffuse (A towards B and B towards A). The diffusion process occurs because of a finite concentration driving force, which is the concentration gradient between the two containers. Diffusion stops when the concentration is uniform throughout the total mixture, i.e., there is no concentration gradient or driving force. However, imagine that there has been a net mass movement to the right. If the direction to the right is taken as positive, the flux of A (noted as NA), relative to a fixed position is positive, while the flux of B, NB, is negative. At steady state, the net flux is NA þ NB ¼ N

(7:7)

The movement of A is made up of two parts, namely, that resulting from the bulk movement of A in N (i.e., xAN), and that resulting from the diffusion of A through B. This latter effect is defined as JA. The sum of these two effects is then NA ¼ xA N þ JA

(7:8)

Employing Fick’s first law leads to NA ¼ xA (NA þ NB )  DAB

@CA @z

(7:9)

The next sub-section presents three different cases of steady-state molecular diffusion in gases.(2,3)

Diffusion in Gases 1 Diffusion of A through non-diffusing B. This can be illustrated by the example of ammonia (A) being absorbed from air (B) into water. Since air does not dissolve appreciably in water, only ammonia diffuses. For this case, NB ¼ 0 and NA is constant and the following relationship can be written NA ¼1 NA þ NB

(7:10)

It can be shown that for a system where component A diffuses through nondiffusing B, the diffusion rate can be calculated from   DG,AB P NA ¼ (7:11) ( pA1  pA2 ) RTz p B,M

76

Chapter 7 Rate Principles

pB2  pB1 is the log mean partial pressure difference driving lnð pB2 = pB1 Þ force of component B and pB1 ¼ P 2 pA1, pB2 ¼ P 2 pA1, pB1 is the partial pressure of component B at the liquid– vapor interface, pB2 is the partial pressure of component B at distance z from the interface, pA1 is the partial pressure of component A at the interface, pA2 is the partial pressure of component A at distance z from the interface, P is the total system pressure, DG, AB is the diffusivity of component A through B, T is the system absolute temperature, z is the distance from the interface for state 2, and R is the ideal gas constant. For addition analyses of these systems, the interested reader is referred to the original works of Treybal(2) and Geankopolis.(4) where p B,M ¼

2 Steady state equimolar counter-diffusion. For this case, both A and B are diffusing, as occurs in distillation operations. Again, a detailed analysis of this system ultimately leads to   DG,AB NA ¼ ( pA1  pA2 ) (7:12) RTz 3 Steady-state diffusion in multicomponent mixtures. An example of this is the diffusion of oxygen in a non-diffusing mixture of methane and hydrogen. The estimation of diffusion in multicomponent systems is very complicated, but it can usually be handled by defining an effective diffusivity, DAM. The effective diffusivity of one component can be calculated based on its diffusivity with each of the other constituents. It can be shown that DAM can be represented by the following equation DAM ¼ Xi

1

yi i¼B D Ai

(7:13)

where DAi is the binary diffusivity of component A in each of the components present in the system and yi is the mole fraction of component i on an A free basis, e.g., for a system consisting of components A, B, and C, yB is the mole fraction of B based on B and C only. Thus, the calculation of the effective diffusivity of oxygen in a mixture of methane and hydrogen is dependent on the diffusivity of oxygen in methane and oxygen in hydrogen. ILLUSTRATIVE EXAMPLE 7.2 Calculate the rate of diffusion of oxygen (A) in a non-diffusing mixture of carbon monoxide (B) and carbon dioxide (C ), which has a volume ratio of 3 : 1, respectively. The system temperature and pressure are 258C and 1  105 N/m2 at the interface, respectively. The oxygen partial pressure is 15,000 N/m2 at the interface and 7500 N/m2 at a distance of 3 mm from the interface. It can be assumed that the partial pressure of the non-diffusing mixture is the difference between the total pressure and the oxygen partial pressure at each location. In addition, DAB ¼ 0.185 cm2/s and DAC ¼ 0.139 cm2/s.

Fick’s Law

77

SOLUTION: The describing equation for the diffusivity for this system is given by Equation (7.13) DAM ¼ Xi

1

i¼B

yi DAi

Write the mole fraction of each non-diffusing component on an A-free basis yB ¼ 0:75 yC ¼ 0:25 Calculate the effective diffusivity for the mixture 1 1 ¼ 0:171 cm2 =s yi ¼  yB   yC  0:75 0:25 þ þ i¼B D Ai DAB DAC 0:185 0:139

DAM ¼ Xi

1

Obtain the log mean pressure difference for the mixture p B,M ¼

pB2  pB1 (105  7500)  (105  15,000)   ¼ ¼ 8:87  104 N=m2   pB2 (105  7500) ln ln pB1 (105  15,000)

Calculate the diffusion of oxygen in the mixture using Equation (7.11), being careful to maintain consistent units   DG,AB P (0:171=1000)(105 ) ( pA1  pA2 ) ¼ NA ¼ (15,000  7500) RTz p B,M (8:314)(298)(0:003)(8:87  104 ) ¼ 1:95  102 gmol=m2  s

B

ILLUSTRATIVE EXAMPLE 7.3 Ethylene is diffusing at a constant rate through a 2 mm-thick stagnant layer of nitrogen. Conditions are such that at one boundary of the stagnant layer, the gas contains 60% by volume ethylene. The ethylene concentration at the other boundary can be considered negligible. The total pressure is one atmosphere and the temperature 258C. The diffusivity for the mixture is 0.163 cm2/s. Determine the rate of diffusion of ethylene through the nitrogen layer. SOLUTION: This is another case of steady-state diffusion through a second non-diffusing gas; hence, Equation (7.11) is again applicable.   DG,AB P ( pA1  pA2 ) NA ¼ RTz pB,M Pertinent data are now rewritten (A ¼ ethylene, B ¼ nitrogen) DAB ¼ 0.163 cm2/s P ¼ 1.0 atm R ¼ 82.07 cm3 . atm/gmol . K

78

Chapter 7 Rate Principles T ¼ 298 K pA1 ¼ (0:6)(1:0) ¼ 0.6 atm pA2 ¼ 0.0 atm pB1 ¼ 1.0 2 0.6 ¼ 0.4 atm pB2 ¼ 1.0 atm z ¼ 2 mm ¼ 0.2 cm

Substituting yields p B,M ¼

pB2  pB1 1  0:4   ¼   ¼ 0:655 atm pB2 1 ln ln 0:4 pB1

Substitute into Equation (7.11).  NA ¼  ¼

 DG,AB P ( pA1  pA2 ) RTz p B,M

 (0:163)(1) (0:6) (82:07)(298)(0:2)(0:655)

¼ 3:047  105 gmol=cm2  s

B

The diffusivity, or diffusion coefficient, D, was defined previously as the proportionality constant in the rate equation for mass transfer (Fick’s law) and it is a property of the system that is dependent on temperature, pressure, and the nature of the components. Reliable diffusion data is difficult to obtain, particularly over a wide range of temperatures. Table 7.1 lists diffusion coefficients for a few pairs of gases that have been investigated. The diffusion coefficient, DG (cm2/s), at a temperature TC (8C) and pressure P (atm), may be determined from the data in Table 7.1 at state “0” and the following equation   TC þ 273:2 b 1 (7:14) DG ¼ DG0 TC0 þ 273:2 P Additional values for diffusivities may be found in the literature.(5) When experimentally determined diffusivity data is not readily available, several estimation Table 7.1 Diffusion Coefficients Gas

Air

CO2

H2

N2

O2

TC0, 8C

b

Air CO CO2 H2 N2

– – 0.138 0.611 –

– 0.137 – 0.550 0.144

0.611 0.651 – – 0.674

– 0.192 – 0.674 –

0.178 0.185 0.139 0.697a 0.181

0 0 0 0 0

1.75 1.75 2.00 1.75 1.75

a

Temperature (TC0) for H2 in O2 is 208C.

Fick’s Law

79

techniques based on the kinetic theory of gases are available. However, the gas diffusivity may be calculated directly by using the following expression: pffiffiffiffiffiffiffiffiffi B0 T 1:5 MW (7:15) DG ¼ 2 V PsAB where DG ¼ gas diffusivity, cm2/s, B0 ¼ molecular weight parameter, dimensionless, MW ¼ weighted average molecular weight, dimensionless, sAB ¼ molar volume parameter, dimensionless, and V ¼ parameter, dimensionless. The term B0 can be found using the equation: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 B ¼ 0:00217  0:00050 þ MWA MWB 0

(7:16)

where MWA ¼ molecular weight of A and MWB ¼ molecular weight of B.

Diffusion in Liquids No purely theoretical generalized correlation of liquid phase diffusivities has yet been found, but certain empirical equations are available. This probably reflects the inadequacy of kinetic theory when applied to liquids. It is therefore preferable to use experimental data for liquid phase diffusivities. Table 7.2 provides a number of typical values for liquid phase diffusivities. Additional liquid diffusivities are available in the literature.(2,5) Investigations on diffusion in liquids are not as extensive as those on diffusion in gases, with less experimental data available. The rate of diffusion in liquids may take a long time to reach equilibrium unless agitated. This is, in part, explained by the fact that there is a much closer spacing of the molecules in a liquid, thereby retarding the movement of solute. Thus, molecular attractions become more important. Also note that diffusivity values in liquids are therefore smaller than in gases. In the absence of an adequate theory for diffusion in liquids, it is usually assumed that Fick’s law is obeyed and that the equations developed in the subsection for diffusion in gases can also be applied to diffusion in liquids. Two situations are considered below. Table 7.2 Liquid Diffusivities at Atmospheric Pressure Solute NH3 NaCl Ethanol Acetic Acid

Solvent

Temperature, 8C

Diffusivity, m2/s  109

Water Water Water Water

5 18 10 12.5

1.24 1.26 0.5 0.82

80

Chapter 7 Rate Principles

1 Steady-state diffusion of A through non-diffusing B DL,AB  r  NA ¼ (xA1  xA2 ) z x B,M MW avg 2 Steady-state equimolar counterdiffusion DL,AB  r  (xA1  xA2 ) NA ¼ z MW avg

(7:17)

(7:18)

where r is the mass density of the solution, and MW is the molecular weight of the solution. When no experimental data is available, an estimation of the liquid diffusivity, DL, can be obtained using various empirical approaches, one of which is the Wilke –Chang equation(6) 117:3  1018 (wMWB )0:5 DL,AB ¼ (7:19) mv0:6 A where DL,AB is the diffusivity of A in a very dilute solution in solvent B (m2/s), MWB is the molecular weight of solvent (g/gmol), T is the temperature (K), m is the solution viscosity (kg/m . s), vA is the solute molar volume at the normal boiling point (m3/kmol), and w is the association factor for the solvent (w ¼ 2.26 for water, 1.5 for ethanol, and 1.0 for solvents like benzene and ethyl ether). A modified form of Equation (7.19) that can be used to obtain the diffusivity is DL ¼

13:26  105 m1:14 v0:589 A

(7:20)

where DL ¼ liquid diffusivity, cm2/s vA ¼ molar volume of solute, cm3/mol

MASS TRANSFER COEFFICIENTS In the discussion of diffusion in the previous section, the emphasis was placed on the molecular transport in fluids that were stagnant or in laminar flow. However, in many cases, these diffusion processes are too slow, and more rapid diffusion or transport is required. Quite often, to speed up this diffusion, the fluid velocity is increased so that turbulent transport occurs. When a fluid flows past a surface under such conditions that the fluid is in turbulent flow, the actual velocity of small parcels or lumps of fluid cannot be described as simply as in laminar flow. Since fluid flows in smooth streamlines in laminar flow, its behavior can usually be described mathematically. However, there are no orderly streamlines or equations to describe fluid behavior in turbulent motion. However, there are large eddies or “chunks” of fluid which move rapidly in a seemingly random fashion. This eddy transfer, or turbulent diffusion, is very fast in comparison

Mass Transfer Coefficients

81

to the relatively slow process of molecular diffusion, where each solute molecule must move by random motion through the fluid. When a fluid flows past a surface under conditions such that turbulence generally prevails, a thin laminar-type sublayer film exists adjacent to the surface. The mass transfer in this region occurs by molecular diffusion since little or no eddies are present. Since this is a slow process, a large concentration gradient or decrease in concentration across this laminar film occurs. Adjacent to this is the transition or buffer region. Here, some eddy activity exists and the transfer occurs by the sum of molecular and turbulent diffusion. In this region, there is a gradual and non-abrupt transition from the total transfer occurring by almost pure molecular diffusion at one end to mainly turbulent at the other end. The concentration decrease is much less in this region. Although most of the transfer is by turbulent or eddy diffusion, molecular diffusion still occurs, but it contributes little to the overall transfer. The concentration decrease is very small here since the rapid eddy movement evens out any gradients tending to exist. Many approaches to the turbulent (convective) mass transfer problem exist: film theory, combined film-surface-renewal theory, boundary layer theory, empirical approaches, etc. One former theory has somehow managed to survive the test of time, having been successful in interpreting the results of most two-phase mass transfer operations of industrial importance. Film theory(7) (as applied by Whitman) postulates the existence of an imaginary stagnant film next to the interface whose resistance to mass transfer is equal to the total mass transfer resistance of the system. The difficulty with this theory is in the calculation of the effective film thickness. Other theories are briefly detailed below. 1 Surface-renewal theory assumes that a clump of fluid far from the interface: (a) moves to the interface without transferring mass, (b) sits there stagnant, transferring mass by molecular diffusion for a time short enough such that little change in the concentration profile occurs in the clump, and (c) then moves away from the interface without transferring mass en route and mixes with the bulk fluid instantly. This theory is somewhat more satisfactory in general than film theory. 2 Boundary layer theory rests on the solution of a set of simplified differential equations which are approximations to a more nearly correct set of differential equations.(8) 3 Empirical approaches, which are merely data correlations, serve for specific cases, but give little information about extrapolation. Further details regarding any of these approaches are available in the literature.(8)

Individual Mass Transfer Coefficients In the mass transfer operation of gas absorption, two insoluble phases are brought into contact in order to permit the transfer of a solute from one phase to the other (e.g., and as discussed earlier, ammonia can be absorbed from an air– ammonia mixture into

82

Chapter 7 Rate Principles

a water stream without air dissolving appreciably in the water). Concern for this application is with the simultaneous application of the diffusion mechanism for each phase to the combined system. It has already been shown that the rate of diffusion within each phase is dependent on the concentration gradient existing within it. At the same time, the concentration gradients of a two-phase system are indicative of the departure from equilibrium which exists between the phases. Since this departure from equilibrium provides the driving force for diffusion, the rates of diffusion in terms of the driving forces may now be studied. In view of Whitman’s two-film theory, it is assumed that at the gas-liquid interface the principal diffusion resistances occur in a thin film of gas and a thin film of liquid. The rates of diffusion in these two films will describe the mass transfer operation. The diffusion coefficient, D, in Fick’s law is inversely proportional to the concentration of the inert material, cB, in the liquid film through which material A must diffuse. Replacing D with (k/cB) yields   k NA ¼ (7:21) (cAI  cAL ) zcBM where cBM is the log mean concentration difference of the inert material across the film, k is a proportionality constant, and N is the amount of material transferred per unit area per unit time, or mols/area . time. The practical application of Equation (7.21) is based on the assumption that z, the film thickness, is a constant that represents an effective average value throughout the length of the contact path. Also, cBM is considered to be constant since many mass transfer processes usually involve fairly dilute mixtures and solutions. Equation (7.21) can therefore be written as NA ¼ kL (cAI  cAL )

(7:22)

NA ¼ kG ( pAG  pAI )

(7:23)

or where kL is the liquid mass transfer coefficient based on concentration, kG is the gas mass transfer coefficient based on partial pressure, cAI is the interfacial (surface) concentration of component A, cAL is the bulk liquid concentration of component A, pAI is the interfacial partial pressure of component A, pAG is the partial pressure of component A. Equation (7.22) expresses the transfer of N molecules of solute (A) through the liquid film under a concentration driving force, and Equation (7.23), the transfer of the same number of molecules of solute through the gas film under a partial pressure driving force. For certain simplified cases of molecular diffusion, equations can be derived to determine precisely the rate at which mass is being transferred. For example, the equation below was presented in the previous section   DG,AB P (7:11) ( pA1  pA2 ) NA ¼ RTz p B,M

Mass Transfer Coefficients

83

The bracketed term above is an exact definition for the individual mass transfer coefficient corresponding to the steady-state situation of one component diffusing through a nondiffusing second component. In principle, it is then not necessary to calculate any other mass transfer coefficient for laminar flow since molecular diffusion prevails and exact equations are available. However, in general, obtaining such analyticallyderived expressions are difficult; in most cases encountered in practice, it is impossible since turbulent mass transfer, which becomes quite complex, usually prevails. Only the more frequently encountered situations of diffusion will now be discussed, namely, equimolar counterdiffusion and diffusion of one component through another non-diffusing component.(2,4)

Equimolar Counterdiffusion In absorption operations, the absorbing medium may evaporate into the gas being treated, resulting in the simultaneous diffusion of both gases in opposite directions. The diffusion of each gas is affected by the presence of the molecules of the other gas and hindered if the gases are diffusing in opposite directions. When such a situation exists, the diffusion of equal moles per unit area per unit time occurs in opposite directions. This is referred to as equimolar counterdiffusion. For the case of equimolar counterdiffusion, the concentration profile shown in Figure 7.4 can be plotted graphically as in Figure 7.5. In this latter figure, point P represents the bulk phase compositions yAG and xAL, and point M represents the concentrations yAI and xAI at the interface. The equations for the flux of component A, when A is diffusing from a gas to a liquid and there is equimolar counterdiffusion of component B from the liquid, are given by NA ¼ ky0 ( yAG  yAI ) ¼ kx0 (xAI  xAL )

(7:24)

where k0x is the liquid mass transfer coefficient based on mole fraction, k0y is the gas mass transfer coefficient based on mole fraction, xAI and yAI are the bulk liquid and gas interfacial mole fraction, respectively, and xAL and yAG are the bulk liquid and Liquid phase solution of A in liquid L

Gas phase mixture of A in gas G

yAG yAI xAI xAL

NA

Interface

Distance from interface

Figure 7.4 Concentration (mole fraction) profile of solute A diffusing from one phase to another.

Chapter 7 Rate Principles

Mole fraction of solute in the gas

84

Equilibrium curve

P

yAG

D

Slope = – kx¢ /ky¢ Slope = m≤

yAI

M Slope = m¢ C

yA*

xAL

xAI

xA*

Mole fraction of solute in the liquid

Figure 7.5 Overall concentration differences.

gas mole fractions, respectively. Note that the prime with the individual mass transfer coefficient, i.e., ky0 , is a reminder that the transfer process involves equimolar counterdiffusion. The values ( yAG 2 yAI) and (xAI 2 xAL) are the differences in concentration, or driving forces, in each phase. For example, ( yAG 2 yAI) is the driving force in the vapor phase since yAG represents the average vapor concentration at a distance from the liquid vapor interface which has a composition yAI. Rearranging Equation (7.24) gives 

kx0 ( yAG  yAI ) ¼ ky0 (xAL  xAI )

(7:25)

Hence, the slope of PM in Figure 7.5 is (k0x =k0y ). This means that if the two mass transfer coefficients are known, then the interfacial compositions can be determined by the line PM. While the bulk concentrations yAG and xAL can ordinarily be determined experimentally (from equilibrium relationships), the concentrations at the interface cannot, and Equation (7.25) can instead be used.

Diffusion of Component A Through Non-diffusing Component B In the case of component A diffusing through nondiffusing component B, the concentrations can also be plotted as in Figure 7.5 where P represents the bulk phase compositions and M, the interface. The equation for A diffusing through a stagnant gas and through a stagnant liquid can be shown to be: NA ¼

ky0 kx0 ( yAG  yAI ) ¼ (xAI  xAL ) (1  yA )IM (1  xA )IM

(7:26)

Mass Transfer Coefficients

85

where the subscript “IM” represents the bulk flow correction factor for nondilute liquid and gas phases. The correction terms can be calculated from the following two equations: (1  yA )IM ¼

(1  yAI )  (1  yAG ) (1  yAI ) ln (1  yAG )

(7:27)

or, (1  xA )IM ¼

(1  xAL )  (1  xAI ) (1  xAL ) ln (1  xAI )

(7:28)

Equation (7.26) may be combined with Equations (7.27) and (7.28) to give: 

kx0 =(1  xA )IM ( yAG  yAI ) ¼ (xAL  xAI ) ky0 =(1  yA )IM

(7:29)

The slope of the line PM for the case of A diffusing through stagnant B is given by the left-hand side of Equation (7.29). The slope of Equation (7.29) differs from that of Equation (7.25) for equimolar counterdiffusion by the bulk flow correction terms, (1 2 yA)IM and (1 2 xA)IM. When A is diffusing through nondiffusing B and the solutions are dilute, the bulk flow correction terms are approximately unity, and Equation (7.25) can be used instead of Equation (7.29). It is for this reason that Equation (7.25) is often employed even if the transfer process involves A diffusing through nondiffusing B. Also note that the subscripts L and G, e.g., kL and kG, are employed when the rate is expressed in terms of the concentration and partial pressure, respectively. The use of Equation (7.29) to obtain the slope is, by necessity, trial-and-error because the left-hand side contains the interfacial concentrations yAI and xAI which are being sought. A first trial estimate can be obtained using Equation (7.25). With these estimates for yAI and xAI, a value of the left-hand side of Equation (7.29) is computed and a new slope drawn to obtain new values of yAI and xAI (read off the equilibrium line). The second trial is repeated until the values of yAI and xAI do not change significantly with successive trials. Three trials will usually suffice. ILLUSTRATIVE EXAMPLE 7.4 An air– ammonia mixture is being treated with a water stream in a 10 ft pilot scale absorber in a lab test under atmospheric conditions (708F and 1atm). The absorber was set-up in such a way to measure the partial pressure of each component at various locations in the column. It was found that at a distance of 5 feet up the column the liquid contained 0.4 wt% ammonia while the partial pressure of the ammonia was 10 mm Hg. Estimate the interfacial gas and liquid phase concentrations. Based on previous experiments, the ratio of the liquid mass transfer coefficient (2kL/ kG) was found to be 21.0 atm/lbmol . ft3. Solubility data for this system are presented in Table 7.3.

86

Chapter 7 Rate Principles Table 7.3 Ammonia Data NH3 partial pressure, atm

Liquid concentration, lbmol NH3/ft3

0.0045 0.0097 0.0119 0.0158 0.0201 0.0255 0.0309

SOLUTION:

0.0180 0.0367 0.0440 0.0586 0.0733 0.0916 0.1100

Write the describing equation for this system. See Equation (7.24). NA ¼ ky0 ( yAG  yAI ) ¼ kx0 (xAI  xAL )

Construct an equilibrium diagram from the data in Table 7.3, as shown in Figure 7.6. Based on the data provided yNH3 ¼ 10=760 ¼ 0:0136; pNH3 ¼ 0:0136 atm The corresponding liquid concentration is cNH3 ¼ (0:004=17)62:4 ¼ 0:0147 lbmol=ft3 These two conditions are represented as a square point on Figure 7.6. A line extended from that point with a slope of 21.0 intersects the equilibrium line (see triangle point) at approximately: pAI ¼ 0:0056 atm B

cAI ¼ 0:023 lbmol=ft3

Partial pressure of ammonia (atm)

0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 0

0.02

0.08 0.04 0.06 Concentration of ammonia (lbmol/ft3)

Figure 7.6 Equilibrium plot for ammonia–water.

0.1

0.12

Overall Mass Transfer Coefficient

87

ILLUSTRATIVE EXAMPLE 7.5 Referring to Illustrative Example 7.4, if the mass transfer coefficients (kG) is 12.5 lbmol/h . ft2 . atm, estimate the molar flux of the ammonia. SOLUTION:

Apply Equation (7.24). First note that pAI ¼ yAI P

so that, pAI P Since P ¼ 1 atm and the solution is dilute, one may write yAI ¼

NA ¼ ky ( yAG  yAI ) ¼ kG ( pAG  pAI ) ¼ 12:5(0:01316  0:0056) ¼ 0:0945 lbmol=h  ft2

B

OVERALL MASS TRANSFER COEFFICIENT It is generally more convenient to utilize an overall coefficient for the gas and liquid phases rather than the individual coefficients since it is not possible to measure the partial pressure and concentration at the interface ( pA1 or yA1, and cA1 or xA1, respectively). The preferred procedure is to express the overall coefficient in terms of the individual coefficients. For this approach, it is common to employ overall mass transfer coefficients based on the overall driving force between pAG (or yAG) and cAL (or xAL). The overall coefficients may be defined on the basis of the gas film, KG, or the liquid film KL, by the equations NA ¼ K G ( pAG  pA ) ¼ K y ( yAG  yA ) NA ¼ K L (cA  cAL ) ¼ K x (xA  xAL )

(7:30) (7:31)

where pA is the partial pressure in equilibrium with cAL, and cA is the concentration in equilibrium with pAG. Consider again the situation shown in Figures 7.4 and 7.5 using the concentration (mole fraction) driving force. The equilibrium-distribution curve for the system is unique at a fixed temperature and pressure. Then, yA , since it is in equilibrium with xAL, is as good a measure of xAL as xAL itself, and, moreover, it is on the same basis as yAG. In this situation, the entire two-phase mass transfer effect may then be determined from expressions such as those given in Equations (7.30) and (7.31). In this manner, the ratio of the resistance of either phase to the total resistance is given by the ratio of the driving force through that phase to the “total” driving force across both films. In effect, 1=ky0 yAG  yAI ¼ 1=Ky yAG  yA

(7:32)

with a similar equation on the liquid side. This analysis is further expanded in the next three subsections.

88

Chapter 7 Rate Principles

Equimolar Counterdiffusion and/or Diffusion in Dilute Solutions When equimolar counterdiffusion is occurring, or when the solutions are quite dilute, the following equation applies: NA ¼ k 0y ( yAG  yAI ) ¼ k0x (xAI  xAL )

(7:33)

From the geometry of Figure 7.5, ( yAG  yA ) ¼ ( yAG  yAI ) þ ( yAI  yA ) ( yAG 

yA )

0

¼ ( yAG  yAI ) þ m (xAI  xAL )

(7:34) (7:35)

where m0 is the slope of the chord CM. Substituting for the concentration differences, one obtains NA NA NA m0 ¼ 0 þ 0 K 0y ky kx

(7:36)

1 1 m0 0 ¼ 0 þ 0 K y ky kx

(7:37)

or

Equation (7.37) demonstrates the relationship between the individual mass transfer coefficients and the overall mass transfer coefficient. The left-hand side of Equation (7.37) can be looked upon as the total resistance based on the overall gas driving force, which is equal to the sum of the gas film resistance (1=k0y ) and the liquid film resistance (m0 =k 0x ). In a similar manner, from the geometry of Figure 7.5, (xA  xAL ) ¼ (xA  xAI ) þ (xAI  xAL )

(7:38)

The slope between the points M and D is m00 ¼

yAG  yAI xA  xAI

(7:39)

Then, yAG  yAI ¼ xA  xAI m00

(7:40)

and it can be readily shown that 1 1 1 ¼ þ K 0x m00 k0y k0x

(7:41)

As before, the left-hand side of Equation (7.41) is the total resistance and is equal to the sum of the individual resistances.

Overall Mass Transfer Coefficient

89

Gas Phase Resistance Controlling Assuming that the numerical values of k0x and k0y are roughly equal, the importance of the slope of the equilibrium curve chords can readily be demonstrated. However, if m0 is very small, so that the equilibrium curve in Figure 7.5 is nearly flat, then only a very small concentration of yA in the gas will give a relatively large value of xA in equilibrium with the liquid. This indicates that gas solute A is very soluble in the liquid phase, and hence, the term m0 =k0x in Equation (7.37) becomes very small or negligible. Then 1 1 0  0 K y ky

(7:42)

and the major resistance is said to be in the gas phase, or the “gas phase is controlling.” Also, yAG  yA  yAG  yAI

(7:43)

Under such circumstances, even fairly large percentage changes in k 0x will not significantly affect K 0y , and efforts to increase the rate of mass transfer would best be directed toward decreasing the gas-phase resistance, e.g., by increasing the gas phase turbulence or using equipment that specifically will have a high turbulence in the gas phase.

Liquid Phase Resistance Controlling In a similar manner, when m00 is very large or the solute A is very insoluble in the liquid, with k 0x and k 0y again very roughly equal, then the term 1=(m00 k 0y ) becomes very small and 1 1 0  0 K x kx

(7:44)

The major resistance to mass transfer is then in the liquid, the “liquid phase is controlling,” and xA  xAL  xAI  xAL

(7:45)

In such cases, efforts to affect large changes in the rate of mass transfer are best directed toward conditions influencing the liquid coefficient, k 0x , i.e., increasing the turbulence in the liquid phase. For cases where k0x and k0y are not nearly equal, Figure 7.5 shows that it will be the relative size of the ratio k0x =k0y and of m0 (or m00 ) which will determine the location of the controlling mass transfer resistance. Table 7.4 lists some cases of specific films controlling a particular mass transfer operation.(9,10)

90

Chapter 7 Rate Principles Table 7.4 Controlling Films for Various Systems Gas film 1. Absorption of ammonia in water 2. Absorption of ammonia in aqueous ammonia 3. Stripping of ammonia from aqueous ammonia 4. Absorption of water vapor in strong acids 5. Absorption of sulfur trioxide in strong sulfuric acid 6. Absorption of hydrogen chloride in water 7. Absorption of hydrogen chloride in weak hydrochloric acid 8. Absorption of 5 vol% ammonia in acids 9. Absorption of sulfur dioxide in alkali solutions 10. Absorption of sulfur dioxide in ammonia solutions 11. Absorption of hydrogen sulfide in weak caustic 12. Evaporation of liquids 13. Condensation of liquids Liquid film 1. Absorption 2. Absorption 3. Absorption 4. Absorption 5. Absorption

of carbon dioxide in water of oxygen in water of hydrogen in water of carbon dioxide in weak alkali of chlorine in water

Both gas and liquid film 1. Absorption of sulfur dioxide in water 2. Absorption of acetone in water 3. Absorption of nitrogen oxide in strong sulfuric acid

Experimental Mass Transfer Coefficients In many instances, investigators have found that their data correlate well on the basis of some empirical relationship quite different from the foregoing formalized treatment, and that these relationships are useful in design work for systems to be reviewed in Part Two. Examples of such relationships are provided in the literature along with other important data pertaining to the reported experimental studies.(11) Experimental data are often correlated in terms of dimensionless numbers such as the Schmidt number (Sc ¼ m/rDAB) and the Reynolds’s number (Re ¼ Lvr/m). In the absence of experimental mass transfer data, many correlations are available which may be used to estimate the mass transfer coefficient for the system being studied. In practice, when choosing a correlation, one should make every effort to match as closely as possible the system conditions under which the correlation was formulated. Some of the correlations, primarily applicable in gas absorption mass transfer processes, are available in the literature.(2,4,5) In general, they apply to absorption columns loaded with various types of packing. These various types of absorption columns and the different packings utilized will be discussed in Chapter 10.

Overall Mass Transfer Coefficient

91

ILLUSTRATIVE EXAMPLE 7.6 The absorption of ammonia into water from an air –ammonia mixture was studied at 408F and at a total pressure of 2.0 atm. The average value of k 0y was estimated to be 0.40 lbmol/h . ft2 . mol fraction and that of k 0L to be 1.10 lbmol/h . ft2 . mol/ft3. The equilibrium partial pressure of ammonia was approximated by Henry’s law to be p ¼ 0:246c where p is the partial pressure of ammonia (mm Hg) and c is the molar concentration of ammonia in the liquid (lbmol/ft3). Estimate the overall gas mass transfer coefficient in lbmol/h . ft2 . atm. SOLUTION: First, convert k0y to a pressure basis in atm. From Equations (7.23) and (7.24), and noting that pi ¼ yiP kG0 ¼

ky0 0:40 ¼ 0:20 lbmol=h  ft2  atm ¼ 2:0 P

One may now modify and apply Equation (7.37) to obtain K 0G . Note the equilibrium coefficient 0.246 essentially represents Henry’s constant, m0 . 1 1 m0 1 0:246 þ ¼ 5:224 ¼ 0 þ 0 ¼ 0 KG kG kL 0:20 1:10 Therefore, KG0 ¼

1 ¼ 0:1914 lbmol=h  ft2  atm 5:224

B

ILLUSTRATIVE EXAMPLE 7.7 Refer to Illustrative Example 7.6 and determine the relative magnitude of the resistances of the gas and liquid phases. 0 ¼ 5.0, the relative magnitude of the gas phase resistance is the SOLUTION: Since 1/k G overall resistance, R, is

1=kG0 5:0 ¼ 0:957 ¼ 95:7% ¼R¼ 5:224 1=KG0 The relative resistance of the liquid is therefore 4.3%.

B

ILLUSTRATIVE EXAMPLE 7.8 In Illustrative Example 7.6, the average gas concentration at one point in the equipment is 1% ammonia by volume, and the water is pure. For this condition, calculate the interfacial composition for both phases, draw to scale a graph of p vs c, plot the equilibrium curve, the point representing the gas and liquid compositions, and the point representing the interface compositions. Mark the driving forces in each phase and the overall gas phase driving force. Compute the rate of absorption, in lbmol NH3/h . ft2 using kG, kL, and KG.

92

Chapter 7 Rate Principles To pA = 0.02

pA, NH3 atm

0.003

To pA = 0.02 Gas driving force

0.002 Overall gas driving force

A

0.001 pAI Liquid driving force

0.001

0.002

0.003

cAI

0.004

3

c, NH3 lbmol/ft

Figure 7.7 Figure for Illustrative Example 7.8.

SOLUTION: and kL ¼ k L0

0 Start with Equations (7.22) and (7.23), noting that for a dilute solution kG ¼ k G

NA ¼ kG ( pAG  pAI ) ¼ kL (cAI  cAL ) with pAG ¼ 0.01(2.0) ¼ 0.02 atm and cAL ¼ 0: substituting yields 0:20(0:02  pAI ) ¼ 1:10(cAI  0) In addition, at the interface (see previous Illustrative Example) pAI ¼ 0:246cAI The solution of these two equations, or the intersection (point A) of the two lines in Figure 7.7, leads to pAI ¼ 0:000856 atm From the equilibrium relation, cAI ¼

pAI 0:000856 ¼ 0:00348 lbmol=ft3 ¼ 0:246 0:246

The graph is provided in Figure 7.7. The rate of absorption may be calculated from any one of three equations: NA ¼ kG ( pAG  pAI ) ¼ 0:20(0:0191) ¼ 0:00383 lbmol=h  ft2 NA ¼ kL (cAI  cAL ) ¼ 1:10(0:00348) ¼ 0:00383 lbmol=h  ft2 NA ¼ KG ( pAG  pA ) ¼ 0:19(0:02) ¼ 0:00383 lbmol=h  ft2

B

References

93

ILLUSTRATIVE EXAMPLE 7.9 Compare the results generated for the flux from the three equations. SOLUTION:

As can be seen, and as expected, there is excellent agreement.

B

REFERENCES 1. A. FICK, Pogg. Ann., XCIV, 59, 1855. 2. R. TREYBAL, “Mass Transfer Operations,” 1st edition, McGraw-Hill, New York City, NY, 1955. 3. C. BENNETT and J. MEYERS, “Momentum, Heat, and Mass Transfer,” McGraw-Hill, New York City, NY, 1962. 4. C. GEANKOPLIS, “Transport Processes and Unit Operations,” Allyn and Bacon, Boston, MA, 1978. 5. D. GREEN and R. PERRY (eds), “Perry’s Chemical Engineers’ Handbook,” 8th edition, McGraw Hill, New York City, NY, 2008. 6. C. WILKE and P. CHANG, Am. Inst. Chem. Eng. J., New York City, NY, 1, 264, 1955. 7. W. WHITMAN, G. Chem. Met. Eng., 29, 147, 1923. 8. R. BIRD, W. STEWART, and E. LIGHTFOOT, “Transport Phenomena,” 2nd edition, John Wiley & Sons, Hoboken, NJ, 2004. 9. O. DWYER and B. DODGE, Ind. Eng. Chem., 33, 485, New York City, NY, 1941. 10. M. LARA, “Tower Packings and Packed Tower Design,” 2nd edition, U.S. Stoneware Co., Akron, OH, 1953. 11. F. ZENZ, personal communication, 1971.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Part Two

Applications: Component and Phase Separation Processes Part Two of this book is concerned with applications. It contains nine chapters and each serves a unique purpose in an attempt to treat nearly all the important aspects of mass transfer. A major objective of this part is to prepare the reader to solve realworld engineering and design problems that involve mass transfer unit operations. There are various mass transfer operations. The three that are most encountered in practice are distillation, absorption, and adsorption. As such, they receive the bulk of the treatment in the early chapters. Other operations reviewed include, liquid – liquid and liquid – solid extraction, humidification and drying, crystallization, membrane separation processes, and phase separation processes. A brief introduction to each topic is provided below. In distillation (Chapter 9), a liquid mixture at its boiling point is brought into contact with a saturated vapor mixture of the same components with a different concentration. The components are transferred between the phases until equilibrium is established or the phases are separated. In gas absorption (Chapter 10), a component in the gas phase is dissolved by a liquid phase in contact with it. The opposite of gas absorption is stripping, where a component of the liquid phase is transferred to the gas phase. In adsorption (Chapter 11), a component of a gas or liquid is retained on the surface of a solid adsorbent such as activated carbon. In extraction (Chapter 12), one makes use of solubility differences in different liquid phases. In humidification and drying (Chapter 13), water or another liquid is vaporized, and the required heat of vaporization must be transferred to the liquid. Dehumidification is the opposite of humidification, and is defined as the condensation of water from air, or, in general,

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

95

Chapter

8

Introduction to Mass Transfer Operations INTRODUCTION This first chapter in Part Two provides introductory information on mass transfer operations. Three important topics are reviewed: 1 Classification of mass transfer operations 2 Mass transfer equipment 3 Characteristics of mass transfer operations The chapters that follow treat specific mass transfer operations. As noted in the introduction to Part Two, individual chapters are devoted to several mass transfer topics, including what the authors consider the three main operations: distillation, absorption, and adsorption.

CLASSIFICATION OF MASS TRANSFER OPERATIONS Mass transfer operations are generally applicable to processes that essentially involve either componential or physical change, or both. A substantial number of these operations are concerned with changing the composition of solutions and mixtures through methods that do not involve chemical reactions; these are defined as componential separation processes. It is also often desirable to separate the original substance into its component parts by phase. Such separations may be entirely mechanical, such as the separation of a solid from a liquid during filtration or the classification of a granular solid into fractions of different particle size by screening. On the other hand, if the operation(s) involve the aforementioned changes in the composition of solutions, they are defined as componential separation operations, and it is with these processes that the bulk of this Part of the book is primarily concerned. Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

97

98

Chapter 8 Introduction to Mass Transfer Operations

It is also useful to classify mass transfer operations and to cite examples of each. These separations may be brought about by three principle mechanisms. There are nine different phase combinations (see Table 6.1 provided earlier). The individual mass transfer operations falling under these nine classifications are numerous and exceedingly diversified. However, since the underlying principles in all these operations and the general methods of applying them are often the same, these cases will be first grouped together for a study of the factors common to all. Later chapters in this Part will provide specific details on these processes. These nine classifications are now considered in relation to the three mechanisms below.(1–3) 1 Contact of two immiscible phases, with mass transfer or diffusion through the interface between the phases. 2 Indirect contact of miscible phases separated by a permeable or semipermeable membrane, with diffusion through the membrane. 3 Direct contact of miscible phases. These three topics receive treatment below.

Contact of Immiscible Phases The bulk of the real-world industrial mass transfer operations reside in this category of immiscible phases. It is primarily these mass transfer processes that are addressed in this Part. The various categories are briefly described below. Gas– Gas Since all gases are completely soluble in each other, this category of operation cannot generally be practically realized. Gas– Liquid If all components of the system are present in appreciable amounts in both gas and liquid phases, an operation known as distillation may be employed. In this instance, a gas phase can be formed from the liquid phase by the application of heat. For example, if a liquid solution of acetic acid and water is partially vaporized by heating, it is found (as discussed in Chapter 6) that the newly created vapor phase and the residual liquid both contain acetic acid and water, but in proportions that are different for the two phases and different from those in the original solution. If the vapor and residual liquid are separated physically from one another and the vapor is condensed, two liquid solutions, one “richer” and the other “poorer” in acetic acid, are obtained. In this way, a certain degree of separation of the original components is accomplished. Conversely, should a vapor mixture of the two substances be partially condensed, the newly formed liquid phase and the residual vapor will differ in composition. An interdiffusion of both components between the phases eventually establishes their final composition in both instances. It should be noted; however, that distillation is only feasible when the components to be separated have appreciably differing boiling points.

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However, all the components of the solutions discussed above may not be present in appreciable amounts in both the gas and liquid phases. If the liquid phase is a pure liquid containing one component and the gas contains two or more, the operation can be classified as humidification or dehumidification, depending upon the direction of the transfer. For example, the contact of dry air with liquid water results in evaporation of some of the water into the air (humidification of the air). Conversely, the contact of very moist air with pure liquid water may result in the condensation of part of the moisture in the air (dehumidification). Relatively little air dissolves in the water in both cases, and for most practical purposes it is assumed that only water vapor diffuses from one phase to the other. Both phases may also be solutions, each containing only one common component that is distributed between phases. For example, if a mixture of ammonia and air is brought into contact with liquid water (see Chapter 6), a large portion of the ammonia, but relatively little air, will dissolve in the liquid, and in this way the air – ammonia mixture may be separated. This operation is known as gas absorption. On the other hand, if air is brought into contact with an ammonia – water solution, some of the ammonia leaves the liquid and enters the gas phase. This operation is known as desorption, or stripping, a common wastewater treatment method. To complete this classification, consider the case where the gas phase contains but one component and the liquid several, as in evaporation of a saltwater solution by boiling. Here the gas phase contains only water vapor, since the salt is essentially nonvolatile. Such operations do not depend on concentration gradients, but rather on the rate of heat transfer to the boiling solution, and are consequently not considered diffusional separations. Should the salt solution be separated by diffusion of the water into an air stream, however, the operation then becomes one of desorption, or stripping. Gas– Solid It is convenient to classify mass transfer operations in the gas – solid category according to the number of components that appear in the two phases. If a solid solution were to be partially vaporized without the appearance of a liquid phase, the newly formed vapor phase and the residual solid would each contain all the original components but in different proportions. The operation is then fractional sublimation. As in distillation, the final compositions are established by interdiffusion of the components between the phases. Although such an operation is theoretically possible, it is usually not practical because of the inconvenience of dealing with solid phases in this manner. However, all components may be present in both phases. If a solid that is moistened with a volatile liquid is exposed to a relatively dry gas, the liquid leaves the solid and diffuses into the gas, an operation generally known as drying, or sometimes as desorption. An example of this process is the drying of laundry by exposure to air. There are many industrial counterparts such as the drying of lumber or the removal of moisture from a wet filter cake by exposure to dry gas. In these cases, the transfer is, of course, from the solid to the gas phase. If the diffusion takes place in the opposite direction, the operation is known as gaseous adsorption. For example, if a mixture of water vapor and air is brought into contact with silica gel, the water vapor diffuses to the solid, which retains it, and the air is thus dried.

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In other instances, a gas mixture may contain several components, each of which is adsorbed on a solid but to different extents ( fractional adsorption). For example, if a mixture of propane and propylene gases is brought into contact with a molecular sieve, the two hydrocarbons are both adsorbed, but to different extents, thus leading to a separation of the gas mixture. In a case in which the gas phase is a pure vapor, such as in the sublimation of a volatile solid from a mixture with one that is non-volatile, the operation is dependent more on the rate of application of heat than on the concentration gradient. This process is essentially a nondiffusional separation. The same is true of the condensation of a vapor to a pure solid, where the rate of condensation depends on the rate of heat removal. Liquid – Liquid Separations involving the contact of two insoluble liquid phases are known as liquidextraction operations. A simple example is a familiar laboratory procedure: if an acetone – water solution is shaken in a separatory funnel with carbon tetrachloride and the liquids are allowed to settle, a large portion of the acetone will be found in the carbon-tetrachloride-rich phase and will thus have been separated from the water. A small amount of the water will also have been dissolved by the carbon tetrachloride, and a small amount of the latter will have entered the water layer, but these effects are relatively minor and can usually be neglected. In another example, a solution of acetic acid and acetone may be separated by adding the solution to an insoluble mixture of water and carbon tetrachloride. After shaking and settling, both acetone and acetic acid will be found in both liquid phases, but in different proportions. Such an operation is known as fractional extraction. Liquid – Solid Fractional solidification of a liquid, where the solid and liquid phases are both of variable composition containing all components but in different proportions, is theoretically possible but is not ordinarily carried out because of practical difficulties in handling the solid phase and the very slow transfer rates in the solid. Cases involving the distribution of a substance between the solid and liquid phases are common, however. Dissolution of a component from a solid mixture by a liquid solvent is known as leaching (sometimes called solvent extraction). The leaching of gold from ore by cyanide solutions and the leaching of cottonseed oil from cottonseeds by hexane are two examples of this class of leaching process. Diffusion is, of course, from the solid to the liquid phase. If the concentration gradient driving diffusion is in the opposite direction, the operation is known as liquid adsorption. Thus, colored material that contaminates impure cane sugar solutions may be removed by contacting the liquid solutions with activated carbon, whereupon the colored substances are retained on the surface of the solid carbon and removed from solution. When the solid phase is a pure substance and the liquid solution is being separated, the operation is called crystallization; but as ordinarily carried out, crystallization rates are more dependent on heat-transfer rates than on solution concentrations. The

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reverse operation is dissolution. No known operation is included in the category involving a pure liquid phase. Solid – Solid Because of the extraordinarily slow rates of diffusion within solid phases, there is at present no major practical industrial separation operation in this (solid –solid) category.

Miscible Phases Separated by a Membrane In operations involving miscible phases separated by a membrane, the membrane is necessary to prevent intermingling of the phases. It must be permeable differently to the components of the solutions if diffusional separations are to be possible. Three different phases combinations are briefly discussed below. Additional information can be found in Chapter 15. Gas– Gas The operation in the gas – gas category is known as gaseous diffusion, gas permeation, or effusion. If a gas mixture whose components are of different molecular weights is brought into contact with a porous diaphragm, the various components of the gas will diffuse through the pores at different rates. This leads to different compositions on opposite sides of the diaphragm and, consequently, to separation of the gas mixture. In this manner, large-scale separation of the isotopes of uranium, in the form of uranium hexafluoride, can be carried out. Liquid – Liquid The separation of a crystalline substance from a colloid, by contact of the solution with a liquid solvent with an intervening membrane permeable only to the solvent and the dissolved crystalline substance, is known as dialysis. For example, aqueous beet sugar solutions containing undesired colloidal material are freed of the latter by contact with water with an intervening semipermeable membrane. Sugar and water diffuse through the membrane, but the larger colloidal particles cannot. Fractional dialysis for separating two crystalline substances in solution makes use of the difference in membrane permeability of the substances. If an electromotive force is applied across the membrane to assist in the diffusion of charged particles, the operation is electrodialysis. If a solution is separated from the pure solvent by a membrane that is permeable only to the solvent, the solvent diffuses into the solution—an operation that has come to be defined as osmosis. This is not a separation operation, of course, but if the flow of solvent is reversed by superimposing a pressure to oppose the osmotic pressure, the process is labeled reverse osmosis. Solid – Solid The operation in the solid– solid category has found little, if any, practical application in the chemical process industry.

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Direct Contact of Miscible Phases Operations involving direct contact of miscible phases are not generally considered practical industrially except in unusual circumstances because of the difficulty in maintaining concentration gradients without mixing of the fluid. Thermal diffusion involves the the formation of a concentration difference within a single liquid or gaseous phase by the imposition of a temperature gradient upon the fluid, thus making possible a separation of the components of the solution. This process has been used, for example, in the separation of uranium isotopes in the aforementioned form of uranium hexafluoride. If a condensable vapor, such as steam, is allowed to diffuse through a gas mixture, it will preferentially carry one of the components along with it, thus making a separation by the operation known as sweep diffusion. If the two zones within the gas phase, where the concentrations are different, are separated by a screen containing relatively large-size openings, the operation is called atmolysis. If a gas mixture is subjected to a very rapid centrifugation, the components will be separated because of the slightly different forces acting on the various molecules, owing to their different masses. The heavier molecules thus tend to accumulate at the periphery of the centrifuge. ILLUSTRATIVE EXAMPLE 8.1 Discuss why the “doughboys” in World War I employed gas masks to prevent problems with poisonous gas releases. SOLUTION: The gas masks caused the ambient air being drawn in for breathing purposes to pass through a canister filled with activated carbon, which is a highly porous, granular or pelleted form of carbon. The activated carbon readily adsorbed the organic molecules that constituted the poisonous gases released but essentially did not adsorb oxygen and nitrogen, which passed through the canister freely. Thus, the gas mask can be thought of as a small scale version of an industrial gas adsorption unit. During the war, these charcoal and other solid adsorbents were employed in chemical warfare. The use of these materials for adsorption of gases became widely introduced in industrial plants as a result of improvements made during the war in the manufacture of activated charcoal, silica gel and other highly active adsorbents. B

MASS TRANSFER EQUIPMENT There are numerous types of mass transfer equipment employed in industry. Some of the more common devices found in mass transfer processes are briefly introduced below. A comprehensive review, including design and predictive equations, follows in Chapters 9– 16. The reader is referred to the cited literature for additional details on not only those listed below but also those not covered in this section.

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Distillation Distillation is probably the most widely used separation process in the chemical industry. Its application ranges from the rectification of alcohol, which has been practiced since antiquity, to the fractionation of crude oil. The separation of liquid mixtures by distillation is based on differences in volatilities among components; the greater the relative volatilities, the easier the separations. In continuous distillation, vapor flows up the column and liquid flows countercurrently down the column (see Fig. 8.1). The vapor and liquid are brought into contact on plates or inert packing material. A portion of the condensate from the condenser is returned to the top of the column to provide liquid that flows down the column. Similarly, at the base of the column some liquid is vaporized in the reboiler and returned to provide the vapor flow. In the stripping section, which lies below the feed, the more volatile components are stripped from the liquid. Above the feed, in the enrichment or rectifying section, the concentration of the more volatile components is increased in the vapor phase. Figure 8.1 shows a distillation column producing two product streams, referred to as tops and bottoms, from a single feed, as well as with multiple feeds and with sidestreams withdrawn at points throughout the column (see Fig. 8.1b). This does not alter the basic operation, but it does complicate the analysis of the process to some extent. If the process requirement is to strip a volatile component from a relatively non-volatile solvent, the rectifying section may be omitted, and the column is then called a stripping column. (a)

(b)

Condenser

Condenser

Top product Reflux

Reflux Multiple feeds

Feed

Top product

Side streams

Reboiler

Reboiler Bottom product

Bottom product

Figure 8.1 Common distillation column configurations. (a) Basic column. (b) Multiple feeds and side streams.

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In some operations where the top product is required as a vapor, the liquid condensed is sufficient only to provide the reflux (return) flow to the column, and the condenser is referred to as a partial condenser. When the liquid is totally condensed, the liquid returned to the column will have the same composition as the top product. In a partial condenser, the reflux is in equilibrium with the vapor leaving the condenser. Virtually pure top and bottom products can be achieved by using multiple distillation stages or, sometimes, additional columns. A detailed treatment of distillation can be found in the next chapter.

Absorption The process of absorption conventionally refers to the intimate contacting of a gaseous mixture with a liquid so that part of one or more of the constituents of the gas will dissolve in the liquid. The contact usually takes place in some type of packed column. Packed columns are used for the continuous contact between liquid and gas. The countercurrent packed column is the most common type of unit encountered in gaseous pollutant control for the removal of undesirable gas, vapor, or odor. This type of column has also found widespread application in the chemical industry. The gas stream moves upward through the packed bed against an absorbing or reacting liquid that is introduced at the top of the packing. This results in the highest possible recovery efficiency. Since the concentration in the gas stream decreases as it rises through the column, there is constantly fresher liquid available for contact. This provides a maximum average driving force for the transfer process throughout the bed. The opposite of gas absorption is stripping (or desorption) where a component in the liquid phase is transferred to the gas phase. The reader is directed to Chapter 10 of this text where additional information is provided.

Adsorption In the adsorption process, one or more components in a mixture are preferentially removed from the mixture by a solid (referred to as the adsorbent). Adsorption is influenced by the surface area of the adsorbent, the nature of the compound being adsorbed, the pressure of the operating system (gaseous application), and the temperature of operation. These are important parameters to be aware of when designing or evaluating an adsorption process. The adsorption process is normally performed in a column. The column can be run as either a packed or fluidized bed. The adsorbent, after it has reached the end of its useful life, can either be discarded or regenerated. This operation can be applied to either a gas mixture or a liquid mixture. The reader is directed to Chapter 11 where additional information is provided.

Extraction Extraction (referred to by some as leaching) encompasses liquid– liquid as well as solid– liquid systems. Liquid – liquid extraction involves the transfer of solutes from

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one liquid phase into another liquid solvent. It is normally conducted in mixer – settler stages and agitated-tower contacting equipment, or packed or spray towers. In the pharmaceutical industries, antibiotics in an aqueous fermentation solution are often removed with an organic solvent. Liquid – solid extraction, in which a liquid solvent is passed over a solid phase to remove some solute, is carried out in fixed-bed, moving-bed, or agitated-solid columns. A bench-scale example of leaching is the coffee pot, where boiling and/or hot water dissolves the soluble coffee from the insoluble coffee grounds. See Chapter 12 for a detailed treatment of both liquid– liquid and solid – liquid extraction processes.

Humidification and Drying Some mass transfer operations involve simultaneous heat and mass transfer. In the humidification process, water or another liquid is vaporized, and the required heat of vaporization must be transferred to the liquid. Dehumidification is the condensation of water vapor from air, or, in general, the condensation of any vapor from a gas. Drying involves the removal of relatively small amounts of water or organic liquids from a solid phase. The aforementioned humidification process, on the other hand, may be thought of as the addition of water to a gaseous phase. Drying can be contrasted to evaporation processes that remove large amounts of water from liquid solutions. In many applications, such as in corn processing, drying equipment follows an evaporation step to provide an ultra high solids content product stream. Drying, in either a batch or continuous process, removes liquid as a vapor by passing warm gas (usually air) over, or indirectly heating, the solid phase. The drying process is carried out in one of three basic dryer types. The first is a continuous tunnel dryer. In a continuous dryer, supporting trays with wet solids are moved through an enclosed system while warm air blows over the trays. Similar in concept to the continuous tunnel dryer, rotary dryers consist of an inclined rotating hollow cylinder. The wet solids are fed in one side and hot air is usually passed countercurrently over the wet solids. The dried solids then pass out the opposite side of the dryer unit. The final type of dryer is a spray dryer. In spray dryers, a liquid or slurry is sprayed through a nozzle, and fine droplets are dried by a hot gas passed either cocurrently, countercurrently, or co/countercurrently past the falling droplets. This unit has found wide application in air pollution control. Despite the differing methods of heat transfer, the continuous tunnel, as well as indirect, rotary, and spray dryers can all reduce the moisture content of solids to less than 0.01% when designed and operated properly.(2) See Chapter 13 for additional details on both humidification and drying.

Other Mass Transfer Unit Operations In crystallization (Chapter 14), solid crystals are formed from a super-saturated solution, so as to purify and control the physical characteristics of the crystals. Examples include the crystallization of sugar from solution and the crystallization of metal salts in the treatment of metal ore solutions. Fractionation of components of a solution by freeze crystallization (see Chapter 17) is another example; applications

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include the separation of xylene isomers and seawater desalination. Membrane separation (Chapter 15) technology provides a physical barrier that selectively allows one component through while stopping the others. Finally, Chapter 16 concludes this Part by examining phase—as opposed to—component separation processes. ILLUSTRATIVE EXAMPLE 8.2 Discuss the major differences between gaseous adsorption and liquid adsorption. SOLUTION: Adsorption involves the contact of a fluid (gas or liquid) with a rigid particulate phase which has the property of selectively taking up and storing one or more solute species originally contained in the fluid. Gaseous adsorption is analogous to the condensation of gas molecules at the adsorbent interface, whereas liquid adsorption may be viewed as a crystallization process. B

ILLUSTRATIVE EXAMPLE 8.3 Provide qualitative differences between gaseous absorption and gaseous stripping. SOLUTION: Gas absorption is a mass transfer operation in which soluble components of a gas mixture are dissolved in a liquid. Gaseous stripping is the reverse operation. It is employed when it is desired to transfer volatile components from a liquid mixture into a gas. Both processes make use of special equipment and surface area to bring liquid and gas streams into close contact. B

ILLUSTRATIVE EXAMPLE 8.4 Describe the main differences between gaseous adsorption and gaseous absorption. SOLUTION: As noted in Illustrative Example 8.2, gaseous adsorption refers to a process where one or more constituents from a mixture are sequestered by a solid substance. The process is influenced by several variables including the surface area of the adsorbent, the nature of the compound being adsorbed, and the system temperature and pressure. As noted in Illustrative Example 8.3, gaseous absorption refers to the intimate contacting of a mixture of gases with a liquid so that part of one or more of the constituents of the gas will dissolve in the liquid. The process usually occurs in a packed column operated in a countercurrent mode. B

The Selection Decision(2) The engineer or scientist faced with the problem of separating the components of a solution must ordinarily choose among several of the aforementioned equipment. Although the choice is usually limited because of the particular physical characteristics of the materials to be handled, the necessity for making this decision nevertheless almost always exists. Until the equipment and the fundamentals of the various

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operations are clearly understood, no basis for such a decision is available. Therefore, it would be prudent to establish the nature of the alternatives at this time. The individual may sometimes choose between using a diffusional operation of the sort discussed in the chapters that follow or a purely mechanical separation method discussed later in this Part. For example, in the separation of a desired mineral from its ore, it may be possible to use either the diffusional operation of leaching with a solvent or the purely mechanical method of flotation. Vegetable oils may be separated from the seeds in which they occur by extraction or by leaching with a solvent. A vapor may be removed from a mixture with a noncondensable gas by the mechanical operation of compression or by the diffusional operations of gas absorption or adsorption. Sometimes, both mechanical and diffusional operations are used, especially where the former are incomplete, as in processes for recovering vegetable oils wherein extraction is followed by leaching. A more commonplace example is the wringing of water from wet laundry followed by air drying. It is characteristic that at the end of a solely mechanical operation, the substance removed is pure, whereas if removed by diffusional separation methods, it is associated with/transferred to another substance. Frequently, a choice may be made between a diffusional operation and a chemical reaction to bring about a separation. For example, water may be separated from other gases either by absorption in a liquid solvent or by chemical reaction with ferric oxide. Chemical methods ordinarily destroy the substance removed, whereas diffusional methods permit its eventual recovery in an unaltered form without great difficulty. There are also choices to be made within diffusional operations. For example, a gaseous mixture of oxygen and nitrogen may be separated by preferential adsorption of the oxygen on activated carbon, by absorption, by distillation, or by gaseous effusion. A liquid solution of acetic acid may be separated by distillation, by liquid extraction with a suitable solvent, by absorption with a suitable solvent, or by adsorption with a suitable adsorbent. The principal basis for choice in most cases is economics: the method that costs least (on an annual basis) is usually the one selected. Other factors, including legal and/or environmental constraints, also influence the decision. The simplest operation, although it may not be the least costly, is sometimes desired because it will be troublefree. Sometimes, a method will be discarded because imperfect or incomplete knowledge of design methods or the unavailability of data for design will not permit results to be guaranteed. Favorable previous experience with a particular method is often given strong consideration. Cost, however, remains one of the prime factors, and is considered in detail later in Part III—Chapter 18.

CHARACTERISTICS OF MASS TRANSFER OPERATIONS Treybal(3) has discussed the two major characteristic methods of carrying out mass transfer operations. One consideration is the nature of the flow of the phases (whether in steady or unsteady state). However, the flow pattern, i.e., whether the flow direction of the phases is parallel, countercurrent, or cross, must also be considered in the

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analysis.(3) As important is the method of contacting the phases, i.e., whether it is stagewise or continuous-contact.(3) These considerations are briefly described below.

Unsteady-State vs Steady-State Operation A steady-state process is one in which there is no change in conditions (pressure, temperature, composition, etc.) or rates of flow with time at any given point in the system. All other processes are unsteady-state. In a batch process, a given quantity of matter is placed in a container, and a change occurs by chemical and/or physical means. At the end of the process, the container (or adjustment containers to which material may have been transferred) holds the products. In a continuous process, materials are continuously fed to a piece of equipment or to several pieces in series, and products are continuously removed from one or more points. A continuous process may or may not be steady-state. A coal-fired power plant, for example, operates continuously. However, because of the wide variation in power demand between peak and slack periods, there is an equally wide variation in the rate at which the coal is fired. For this reason, power plant problems may require the use of average data over long periods of time. If one examines batch operations, all the phases are stationary from a point of view external to the system or on a “forward flow” basis, although internally there may be relative motion of the phases. A batch reactor, with no flow into or out of the reactor unit during the course of the reaction is one such example. Perhaps a more appropriate mass transfer example is the familiar laboratory extraction procedure involving contact of a solution with an immiscible solvent in a separatory funnel. This a batch operation since, once the liquids are in place, there is no further flow of liquid into or out of the vessel until the operation is completed. The solute diffuses from the solution into the solvent during the course of the extraction and the concentrations in both phases must therefore change with time. Provided the time of contact is sufficient, the maximum change in concentration which is possible occurs when equilibrium exists between the phases, although in practice the operation may be stopped before this occurs. The entire operation is then said to be equivalent to one stage (or step). Many mass transfer operations may be carried out in this general fashion. In a semibatch operation, one phase is stationary while the other flows continuously into and/or out of the system. A semibatch reactor, with either (but not both) flow into and out of the unit is a simple example. A mass transfer application is the case of a drier where a quantity of wet solid is placed in an air stream which flows continuously into and out of the drier, carrying away the vaporized moisture. The concentration of moisture in the solid and in the exiting air stream must, of course, change with time. Ultimately, if sufficient time is permitted, the stationary phase will come to equilibrium with the influent phase. It is a characteristic of steady-state operation that system variables at any position in the system remain constant with the passage of time. This requires continuous, invariable flow of all phases into and out of the system and a maintenance of the flow regime within the system. An example here would be a continuous steady-state

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tubular flow reactor. It should be noted that most mass transfer operations are (or are assumed to be) steady-state and continuous, and unless otherwise noted, the reader should assume this to be the case.

Flow Pattern Flow options arise irrespective of whether the unit is operated in the steady- or unsteady-state mode. There are the three basic flow patterns that arise in practice: parallel, countercurrent, and cross. Each of these flow mechanisms are discussed qualitatively below with details provided by Kelly.(4) In parallel flow, the phases move through the unit in the same direction, entering and leaving together. The net effect insofar as concentrations are concerned is ultimately the same as that for a batch operation: if the phases are in contact long enough, the maximum concentration change will correspond to a state of equilibrium between the effluent phases. Consider the single-stage cocurrent contacting process represented in Figure 8.2. A steady-state material balance can be used to monitor the separation taking place in the process. Only the transfer of a single component will be considered here. Treating the average compositions of the flows into and out of the system, one can write both an overall (Equation 8.1) and a componential balance (Equation 8.2). V0 þ L0 ¼ V1 þ L1

(8:1)

V0 y0 þ L0 x0 ¼ V1 y1 þ L1 x1

(8:2)

The terms V (gas) and L (liquid) refer to the molar flow rates entering and leaving the stage, and x and y are the mole fractions of the transferring component in V and L. Note that these equations may also be written on a mass basis. Also note that the terms V and G (representing the gas flow rate) are used interchangeably in both this text and the literature.

y0, V0

y1, V1 Single-

Stage

Process x0, L0

Figure 8.2 Single stage cocurrent contacting process.

x1, L 1

Chapter 8 Introduction to Mass Transfer Operations

y, mole or mass fraction

110

Slope = –

y0

L V

Operating line

Equilibrium curve y1 x1 x0 x, mole or mass fraction

Figure 8.3 Operating diagram: single-stage device.

The separation taking place in a single stage process can be represented on an operating diagram which can be drawn in conjunction with phase equilibrium information available for the system. In this example, and for any case where the inlet streams to the process are mixed together, the contacting pattern is said to be cocurrent. An operating diagram for the steady-state mass transfer of a single component between these two phases is provided in Figure 8.3 where it is assumed V0 ¼ V1 ¼ V and L0 ¼ L1 ¼ L. This diagram shows the transfer of a single component from phase V to phase L. For transfer in the other direction, the equilibrium line remains the same, but the operating line would be located and appear below the equilibrium relationship from the case shown above. As the end of the operating line representing the exit stream from the process approaches the equilibrium curve, the single-stage process approaches what has come to be defined as an ideal or theoretical stage. If the operating line reaches the equilibrium curve, the single-stage device is a theoretical stage. It is also important to note that the displacement of an operating point on the operating line from the equilibrium line provides a direct measure of both the driving force and the rate of transfer. Thus, the rate of transfer is highest at x0, y0 and zero at x1, y1. The operating line connecting points x0, y0, and x1, y1 describes the actual operating state or condition during the transfer process from point 0 to 1. The shape of the equilibrium curve on the operating diagram arises from the phase equilibria of the system which must reflect changes in temperature, pressure, ionic strength, etc., that occur in the single-stage process. The shape of the operating line reflects changes in the quantity of material in streams V and L as mass (often referred to as the solute) is transferred from one phase to another. To reduce the degree of curvature of the operating line, the use of mole ratios on a solute free basis may be employed instead of mole fractions. This conversion can be obtained by dividing the number of moles of a transferring species (the solute) by the number of moles of those components that are not transferred. Thus, V0 y0 ¼

Vs y0 ¼ Vs Y 0 1  y0

(8:3)

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where Vs is the gas nontransferring (solute-free) portion of V0 and

Y0 ¼

moles of transferring component moles of nontransferring component

(8:4)

A similar expression can be written for liquid stream L. A component balance can be used to derive an operating line expression in terms of these mole or mass ratios: Vs (Y0  Y1 ) ¼ Ls (X1  X0 )

(8:5)

Vs (Y0  Y1 ) ¼ Ls (X0  X1 )

(8:6)

or

An X2Y operating line based on Equation (8.6) would therefore be a straight line passing through points (X0, Y0) and (X1, Y1) with a slope of 2Ls/Vs. However, for most applications, the solute-free mole fraction is not employed, particularly when mole fractions are low. In countercurrent flow, the contacted phases flow in opposite directions through the equipment. For example, in gas absorption the gas to be “washed” may flow upward through a tower while the “washing” liquid flows downward through the gas. The reverse transfer occurs in gas stripping, where the gas does the “washing.” For the maximum possible transfer, one of the effluent phases will come to equilibrium with the other influent phase, although in practice this condition is never met. If the exiting streams from a single-stage device are in equilibrium, the singlestage is defined as a theoretical stage. It is usually desirable to use several or multiple stages for a given separation. When multiple stages are used, some thought must be given to the pattern of contacting the two phases. Cocurrent contacting (see Fig. 8.2), where the inlet stream from one phase is mixed with the inlet stream of another phase, can provide, at best, the equivalent of only one theoretical stage no matter how many actual stages are used. To maximize driving forces throughout a particular process, countercurrent contacting is often used. In this contacting mode, the inlet stream for one phase is contacted with the outlet stream of the other phase. A two-stage contacting system for both cocurrent and countercurrent is pictured in Figure 8.4.(4) The following componential mole balance on a solute free basis is (a) Y0, Vs X1, Ls

Y1, Vs Stage 1

Stage 2 X2, Ls

Y2, Vs X3, Ls

(b) Y0, Vs X1, Ls

Y1, Vs Stage 1

Stage 2 X2, Ls

Y2, Vs X3, Ls

Figure 8.4 Two-stage contacting process. (a) Cocurrent operation. (b) Countercurrent operation.

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applied to the overall countercurrent process pictured in Figure 8.4b. Vs Y0 þ Vs X3 ¼ Vs Y2 þ Ls X1

(8:7)

Y, mole or mass ratio

The operating line for countercurrent contacting is shown in Figure 8.5. Here, the diagram shows the transfer from phase V to phase L. If transfer were in the opposite direction as in a stripping operation, the operating line would, as indicated earlier, be located below the equilibrium line. In general, for a given number of stages, countercurrent contacting yields the highest mass transfer efficiency. The reason for this is that the average mass transfer driving force across the device is greater than would be the case with cocurrent contacting.

Y0 Slope =

Operating line

Ls Vs

Y2 Equilibrium curve

X3

X1 X, mole or mass ratio

Figure 8.5 Operating diagram: countercurrent contacting operation.

In crossflow, the phases flow at right angles to each other, as in the case of the air and water in some atmospheric water-cooling towers. The maximum possible concentration change occurs if one of the effluent streams comes to equilibrium with the other influent stream. Crosscurrent contacting, which is intermediate in mass transfer efficiency to cocurrent and countercurrent contacting, is shown in Figure 8.6 for a two-stage system.(5) X0, Ls

Y0, V s

Stage 1

X0, Ls

Y1, Vs

X1, Ls

Figure 8.6 Two-stage crossflow contacting process.

Stage 2

X2, Ls

Y2, Vs

Y, mole or mass ratio

Characteristics of mass transfer operations

Operating line

Y0

Slope =

113

Ls Vs

Equilibrium curve

Y1

Y2 X0

X2

X1

X, mole or mass ratio

Figure 8.7 Operating diagram: crosscurrent contacting.

Although the liquid L feed to both stages is the same, this need not be the case; the feed rate to each stage can be different as can the composition of that feed stream. To draw the operating line, the following solute free balances can be written: Vs (Y0  Y1 ) ¼ Vs (X1  X0 ); stage 1 Vs (Y1  Y2 ) ¼ Vs (X2  X0 ); stage 2

(8:8) (8:9)

The componential balance for each stage is essentially similar to that for the cocurrent process. However, the stages are coupled. For this case, the operating diagram is presented in Figure 8.7. Crosscurrent contacting is not used as commonly as cocurrent and countercurrent contacting but is employed in extraction, leaching, drying, and air pollution control applications.

ILLUSTRATIVE EXAMPLE 8.5 To illustrate the relative efficiencies of the various contacting modes of operation discussed above, Kelly(4) considered the following example. Suppose there are two discrete stages that can mix and separate phases, and that the stages can be connected cocurrently, countercurrently, or crosscurrently for gas –liquid contacting. Find the contacting mode that will provide the maximum removal of a single transferring component from a gas phase, V, if a fixed amount of solvent, L, is to be used. Assume the inlet liquid flow, L ¼ 20 units/time, inlet liquid composition ¼ 0.0 of the transferring component, inlet gas flow, V ¼ 20 units/time, inlet gas composition ¼ Y0 of the transferring (solute) component on a fractional basis, only one component is transferred between the gas and liquid, each stage can be considered to be an ideal stage, and equilibrium data in terms of mole ratios on a solute free basis is given by Y ¼ mX, where X and Y are the mole ratios of the transferring component to the nontransferring component.

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Chapter 8 Introduction to Mass Transfer Operations

SOLUTION: Consider first a single cocurrent stage, as pictured in Figure 8.4a. The ratio of flows of nontransferring components (the liquid to gas ratio on a solute-free basis) is: Ls Ls Ls ¼ ¼ Vs V  VY0 V(1  Y0 ) The outlet mole ratios, X1 and Y1, can be found through graphical construction on an operating diagram by constructing a line of slope 2Ls/Vs from (X0, Y0) to the equilibrium line since this is a theoretical stage. Since the inlet mole ratio, X0 ¼ 0.0, an analytical expression can also be developed to find X1 and Y1. By material balance, Vs Y0 þ Ls X0 ¼ Vs Y1 þ Ls X1

(8:5)

But, X1 ¼ Y1/m from the given equilibrium relationship. Noting that X0 ¼ 0.0 and rearranging, one obtains Y1 ¼

Y0 Y0 ¼ 1 þ (Ls =Vs m) 1 þ A

where A is defined as the absorption factor, A ¼ Ls/Vsm. Cocurrent contacting, even for a cascade of stages, yields at best one theoretical stage. Since the streams leaving the first stage are in equilibrium, the addition of a second cocurrent stage will not result in any further mass transfer. Thus, the addition of another cocurrent stage does nothing to enhance the transfer of solute as will be the case no matter how many stages are added. Consider countercurrent contacting with two stages, as pictured in Figure 8.4b. A solute balance around stage 1 yields the following result: Y1 ¼

Y0 þ (Ls =Vs )X2 1þA

A balance around stage 2 gives a similar result: Y2 ¼

Y1 þ (Ls =Vs )X3 1þA

Recognizing that X3 ¼ 0.0 and that (Ls/Vs)X2 is the same as AY2, leads to the following expression: Y2 ¼

Y0 1 þ A þ A2

For crosscurrent contacting with two stages (see Fig. 8.6), a material balance around stage 1, with X0 ¼ 0.0, leads to the following expression for Y1 Y1 ¼

Y0 1 þ A=2

Y2 ¼

Y1 1 þ A=2

Similarly, for stage 2, with X0 ¼ 0.0,

Characteristics of mass transfer operations

115

By combining the two equations, Y2 can be expressed as a function of Y0: Y2 ¼

Y0 (1 þ A=2)2

One may now summarize the results for the outlet concentration from each process. Y0 1þA Y0 Crossflow, two stages: Y ¼ ¼ (1 þ A=2)2

Cocurrent, single stage: Y ¼

Countercurrent, two stages: Y ¼

Y0 A2 1þAþ 4



Y0 1 þ A þ A2

The results clearly indicate that maximum separation (or the minimum Y ) is achieved with countercurrent flow. A similar analysis of these flow modes is available for chemical reactions.(6,7) B

ILLUSTRATIVE EXAMPLE 8.6 Formally derive the expression Y2 ¼

Y0 1 þ A þ A2

for the two-stage countercurrent unit described in Illustrative Example 8.5. SOLUTION:

It was previously shown that for stage 1 Y1 ¼

Y0 þ (Ls =Vs )X2 1þA

Y2 ¼

Y1 þ (Ls =Vs )X3 1þA

and for stage 2

Noting that X3 ¼ 0 and (Ls/Vs )X2 ¼ AY2 leads to the equation Y1 ¼

Y0 þ AY2 1þA

and Y2 ¼

Y1 1þA

Replacing Y1 in the latter equation with that in the former equation leads to Y2 ¼

Y0 þ AY2 (1 þ A)2

116

Chapter 8 Introduction to Mass Transfer Operations

or Y2 þ 2Y2 A þ Y2 A2 ¼ Y0 þ AY2 Rearranging leads to the final result Y2 (1 þ A þ A2 ) ¼ Y0 or Y2 ¼

Y0 1 þ A þ A2

B

Stagewise vs Continuous Operation Stagewise operation is considered first. If two insoluble phases are allowed to come into contact so that the various diffusing components of the mixture distribute themselves between the phases, and if the phases are then mechanically separated, the entire operation—as defined earlier—is said to constitute one stage. Thus, a stage is the unit in which contacting occurs and where the phases are separated; and, a single-stage process is one where this operation is naturally conducted once. As an example, the laboratory batch extraction in a separatory funnel which was described earlier may be cited. However, the operation may be carried on in a continuous as well as in a batchwise fashion. Should a series of stages be arranged so that the phases are frequently contacted and separated once in each stage, the entire multistage assemblage is called a cascade and the phases may move through the cascade in parallel, countercurrent, or crossflow mode. In order to establish a standard for the measurement of performance, the ideal, or theoretical, or equilibrium stage referred to earlier is defined as one where the effluent phases are in equilibrium, so that (any) longer time of contact will bring about no additional change of composition. Thus, at equilibrium, no further net change of composition of the phases is possible for a given set of operating conditions. (In actual industrial equipment, it is usually not practical to allow sufficient time, even with thorough mixing, to attain equilibrium.) Therefore, an actual stage does not accomplish as large a change in composition as an equilibrium stage. For this reason, the fractional stage efficiency is defined as the ratio of a composition change in an actual stage to that in an equilibrium stage. Stage efficiencies for industrial equipment range between a few percent to that approaching 100 percent.(5) The approach to equilibrium realized in any stage is then defined as the fractional stage efficiency. In the case of continuous-contact operation, the phases flow through the equipment in continuous intimate contact throughout the unit, without repeated physical separation and contacting. The nature of the method requires the operation to be either semibatch or steady-state, and the resulting change in compositions may be equivalent to that given by a fraction of an ideal stage or by more than one stage.

References

117

Note that equilibrium between two phases at any position in the equipment is generally never completely established. The essential difference between stagewise and continuous-contact operation may then be summarized. In the case of the stagewise operation, the flow of matter between the phases is allowed to reduce the concentration difference. If allowed to contact for long enough, equilibrium can be established after which no further transfer occurs. The rate of transfer and the time (of contact) then determine the stage efficiency realized in any particular application. On the other hand, in the case of the continuous-contact operation, the departure from equilibrium is deliberately maintained and the transfer between the phases may continue without interruption. Economics plays a significant role in determining the most suitable method. ILLUSTRATIVE EXAMPLE 8.7 Consider the following two coffee operations. Classify whether they are (a) steady- or unsteadystate; (b) batch, semibatch, or continuous; (c) stagewise or continuous-contact; (d) single- or multistage. 1 Coffee is prepared by allowing a portion of hot/boiling water to flow once through a bed of ground coffee beans. 2 Coffee is prepared by stirring a sample of ground coffee in a container with hot/boiling water until the desired concentration in the solution is reached and the solid residue filtered from the liquid. The operation is repeated three times with fresh coffee beans. SOLUTION Operation 1—Unsteady-state, batch, can be classified either as continuous or stagewise (depending on interpretation and definition), and single-staged (if stagewise). Operation 2—Unsteady-state, continuous, stagewise, and multistaged. B

REFERENCES 1. R. DUPONT, T. BAXTER, and L. THEODORE, “Environmental Management,” CRC Press, Boca Raton, FL, 1998. 2. R. DUPONT, L. THEODORE, and K. GANESAN, “Pollution Prevention: The Waste Management Approved for the 21st Century,” CRC Press, Boca Raton, FL, 2000. 3. R. TREYBAL, “Mass Transfer Operations,” 1st Edition, McGraw-Hill, New York City, NY, 1955. 4. R. KELLY (Section Author), “General Processing Calculations,” edited by R. Rousseau, “Handbook of Separation Process Technology,” John Wiley & Sons, Hoboken, NJ, 1987. 5. G. BROWN and ASSOCIATES, “Unit Operations,” John Wiley & Sons, Hoboken, NJ, 1950. 6. L. THEODORE, “Chemical Reaction Kinetics,” A Theodore Tutorial, East Williston, NY, 1995. 7. J. REYNOLDS, J. JERIS, and L. THEODORE, “Handbook of Chemical and Environmental Engineering Calculations,” John Wiley & Sons, Hoboken, NJ, 2004.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

9

Distillation INTRODUCTION This first mass transfer operation chapter receives both preferential and unique treatment relative to the other chapters in Part II. It is no secret that the subject matter of distillation is solely located in the domain of the chemical engineer. In many ways, it is a topic which separates (no pun intended) chemical engineering from other engineering and applied science disciplines. As such, the authors made a conscious decision to provide preferential treatment to distillation and to include developmental material which may be lacking in other chapters. It was also decided to expand upon the concepts of binary distillation, and to include an introductory analysis of multicomponent distillation operations. Distillation may be defined as the separation of the components of a liquid feed mixture by a process involving partial vaporization through the application of heat. In general, the vapor evolved is recovered in liquid form by condensation. The more volatile (lighter) components of the liquid mixture are obtained in the vapor discharge at a higher concentration. The extent of the separation is governed by two important factors: the properties of the components involved and by the physical arrangement of the unit used for distillation. In continuous distillation, a feed mixture is introduced to a column where vapor rising up the column is contacted with liquid flowing downward (which is provided by condensing the vapor at the top of the column). This process removes or absorbs the less volatile (heavier) components from the vapor, thus effectively enriching the vapor with the more volatile (lighter) components. This occurs in the section above the feed stream which is referred to as the enriching or rectification section of the column. The product (liquid or vapor) removed from the top of the column is rich in the more volatile components and is defined as the distillate. The section below the feed stream is referred to as the stripping section of the column. In this section, the liquid is stripped of the lighter components by the vapor produced in a reboiler at the bottom of the column. The liquid that is removed from the bottom of the column is called the bottoms, which is richer in the heavier components.

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

119

120

Chapter 9 Distillation

Distillation columns are used throughout industry when mixtures (primarily in liquid form) must be separated. One such example is the petroleum industry. In such an application, crude oil is fed into a large distillation column and different fractions (oil mixtures of varying composition and volatility) are taken out at different heights in the column. Each fraction, such as jet fuel, home heating oil, gasoline, etc., is used by both industry and the consumer in a variety of ways. The separation achieved in a distillation column depends primarily on the relative volatilities of the components to be separated, the number of contacting trays (plates) or packing height, and the ratio of liquid and vapor flow rates. Distillation columns are rarely designed with packing in large scale production because of the liquid distribution problems that arise with large diameter units and the enormity of the height of many columns. However, where applicable, towers filled with packing are competitive in cost, and are particularly useful in cases where the pressure drop must be low and/or the liquid holdup must be small. Packed towers, an overview of which is available at the end of this chapter, are occasionally used for bench-scale or pilot plant work. In contrast, use of trayed towers extends to many areas of the chemical industry. The sections to be covered in this chapter are as follows: 1 Flash Distillation 2 Batch Distillation 3 Continuous Distillation with Reflux i Equipment and Operation ii Equilibrium Considerations iii Binary Distillation Design: McCabe – Thiele Graphical Method iv Multicomponent Distillation: Fenske – Underwood – Gilliland Method During the preparation of this chapter, the authors were ably assisted in many ways by a number of graduate students in Manhattan College’s chemical engineering program. Several of the illustrative examples provided in this chapter were drawn, in part, from the original work of Marnell.(1)

FLASH DISTILLATION The separation of a volatile component from a liquid process can be achieved by means of flash distillation. This operation is referred to as a “flash” since the more volatile component of a saturated liquid mixture rapidly vaporizes upon entering a tank or drum which is at a lower pressure and/or higher temperature than the incoming feed. If the feed is a sub-cooled liquid, a pump and heater may be required to elevate the pressure and temperature, respectively, to achieve an effective flash (see Fig. 9.1). As the feed enters the flash drum, it impinges against an internal deflector plate which promotes liquid– vapor separation of the feed mixture. The composition of the feed stream (F) is given by the mole fractions belonging to the set fzig. Similarly, the compositions of the vapor and liquid product streams are given by fyig and fxig, respectively. Alternatively, some processes require that a vapor stream be cooled, so

Flash Distillation

121

Figure 9.1 Flash distillation system.

as to partially condense the least volatile components in the stream. This process is referred to as a partial condensation, and the following development may be adapted to apply to such processes. As a result of the flash, the vapor phase will be mostly composed of the more volatile components. Typically, flash distillation is not an efficient means of separation when used only once. However, when several flash units are placed in series, much purer products may be achieved. Moreover, it can be a necessary and economical means of separating two or more components with discernible relative volatilities, as is often necessary in the petroleum industry. In a very real sense, an individual flash distillation unit may be seen as analogous to a single tray in a distillation column (as will be explained later in this chapter). Relative volatility (see Chapter 6) describes the tendency of a particular compound to vaporize, relative to another compound. As is expected, a more volatile component is more likely to vaporize from a mixture (as compared to a compound of lesser volatility) when the mixture’s temperature is raised or the pressure is lowered. Relative volatility serves as a quantitative comparison of the volatility difference between two compounds of interest. When a liquid solution may be considered ideal, Raoult’s law applies, and hence the relative volatility (a) of component A in an A/B binary mixture may be defined in terms of each component’s vapor pressure ( p0i )

aAB ¼ where

KA p0 ¼ A0 KB pB

(9:1)

aAB ¼ relative volatility of A with respect to B [dimensionless] Ki ; yi/xi ¼ phase equilibrium constant of component i [dimensionless]

Note that the phase equilibrium constant, Ki, is a function of several thermodynamic variables, namely the system’s temperature, pressure, and the compositions of each phase. For non-ideal conditions, all three of these considerations must be taken into account, and K values must be determined via fugacity calculations. However, within the scope of this text, one may assume that K is solely a function of temperature

122

Chapter 9 Distillation

and pressure.(2) The concept of relative volatility is of the utmost importance in an operation such as distillation, where two or more components are to be separated based on their differences in boiling point (which is directly correlated to differences in volatility). The subsequent development refers to the graphical solution of flash distilling a binary liquid mixture. As with many process operations, an overall material balance can be written to describe the system illustrated in Figure 9.1. An overall mole balance yields F ¼LþV

(9:2)

Based on the above, a componential mole balance can be written for component i, as shown in Equation (9.3): Fzi ¼ L xi þ Vyi

(9:3)

where zi ¼ mole fraction of component i in the feed stream [dimensionless] xi ¼ mole fraction of component i in the liquid stream [dimensionless] yi ¼ mole fraction of component i in the vapor stream [dimensionless] Since only two components are present in the liquid feed, the subscript i will be omitted, assuming that all mole fractions refer to the lighter component A. As a matter of convenience, Equation (9.3) may be rearranged in terms of the vapor composition to represent a straight line of the form y ¼ mx þ b as follows y¼

L F xþ z V V

(9:4)

where L=V ¼ the slope of the operating line (m) ðF=VÞz ¼ the y-intercept of the operating line (b). (Note that V/F is the fraction of the feed stream which is vaporized) The above equation defines an operating line for this system, similar to that discussed in Chapter 8. Since the liquid and vapor streams are assumed to be in thermodynamic equilibrium (as is customarily practiced in quick-sizing mass transfer units), the flash is defined as an equilibrium stage. The equation therefore relates the liquid and vapor composition leaving the flash drum. One should also note that the vapor molar flow is normally represented by V, not G, in standard distillation notation; however, both are used interchangeably for the study of other mass transfer operations in certain chapters to follow. As shown in Figure 9.2, a plot of the equilibrium data, the y ¼ x line, and the operating line set up a procedure to calculate the unknown variables for this system, typically the outlet liquid and vapor compositions. The use of the y ¼ x line simplifies the graphical solution method and intersects the operating line at the feed composition, z. Thus, at this point, y ¼ x ¼ z. The unknown compositions in the vapor and liquid product streams can then determined by the intersection of the operating line and the equilibrium curve.

Flash Distillation

123

Figure 9.2 Graphical analysis of flash distillation.

ILLUSTRATIVE EXAMPLE 9.1 Dr. Ethyl Ester, an organic chemistry professor performed a flash distillation experiment for her students. A 10 kmol/h liquid feed mixture consisted of 20 mol% ethanol and 80 mol% water at 1 atm. While the professor was able to determine that 30 mol% of the feed vaporized in a small flash drum, she lacked the equipment to measure the liquid and vapor compositions. Determine the liquid and vapor compositions, as well as the percent ethanol recovery in the vapor stream. Equilibrium data for the ethanol/water system were calculated via the Wilson equation(2) at 1 atm, and an x, y diagram is provided in Figure 9.3. SOLUTION: Superimposed on Figure 9.3 is the y ¼ x line. The describing equation for the operating line is y¼

L F xþ z V V

The slope of the operating line may be obtained directly from V ¼ 0:3 F V ¼ 0:3F

(9:4)

124

Chapter 9 Distillation

Figure 9.3 Ethanol/Water equilibrium diagram (via Wilson equation, 1 atm). But since V ¼ F – L L ¼ 0:7F L V 0:7F ¼ 0:3F

Slope ¼ 

¼ 2:33 The y-intercept is calculated in a similar fashion F ¼ 3:33 V F yint ¼ z V ¼ (3:33)(0:20) ¼ 0:666 The operating line also appears on the plot. The ethanol liquid and vapor compositions which result from the flash are found from the intersection of the operating line and equilibrium curve: x ¼ 0:10; (1  x) ¼ 0:90;

y ¼ 0:44 (1  y) ¼ 0:56

Now that the liquid and vapor stream compositions are fixed, the liquid and vapor flow rates may be determined by solving Equations (9.2) and (9.3) simultaneously: F ¼LþV Fzi ¼ L xi þ Vyi

(9:2) (9:3)

Flash Distillation

125

Substituting (V ¼ F2L) into Equation (9.3) and substituting mole fractions for ethanol, (10)(0:20) ¼ (L)(0:10) þ (10  L)(0:44) And thus, L ¼ 7:1 kmol=h V ¼ 2:9 kmol=h The percent recovery of ethanol in the vapor stream is calculated as follows: 

 Vyi (100); i ¼ ethanol Fzi   (2:9)(0:44) ¼ (100) (10)(0:20)

%(RECOVERY)i ¼

¼ 63:8% Therefore, 63.8% (by mole) of the ethanol feed is recovered in the vapor product.

B

An iterative procedure is required in order to solve a multicomponent (more than two components) flash problem, the details of which are available in the literature.(2,3) While at least three distinct equations may be derived for multicomponent flash distillation via material balances, one equation in particular is the most convenient. The reason that this equation is so convenient is that it is a monotonically decreasing function (and therefore only has one root, which precludes any extraneous answers). This equation is referred to as the Rachford – Rice (RR) equation, shown in Equation (9.5) below for a mixture of N components, f (C) ¼

N X i¼1

zi (Ki  1) ¼0 1 þ (Ki  1)C

(9:5)

where C ¼ V/F ¼ molar fraction of feed which leaves as vapor [dimensionless] One should note that the vapor flow rate for a flash distillation is given by V ¼ FC, whereas the liquid flow rate is L ¼ F(12 C). Assuming the feed composition is known, Equation (9.5) may be used for either of two purposes: 1 To calculate the fraction of feed which will vaporize (C) or condense (12 C) at a given flash temperature and pressure. 2 To determine the phase equilibrium constants (Ki) for each species by guessing the flash temperature and/or pressure at a desired value of C. Regardless of which parameter is to be calculated (C or Ki), an iterative calculation is carried out until the sum in Equation (9.5) is sufficiently close to zero. Once the RR equation has been solved, the compositions of the liquid and vapor streams are calculated by utilizing material balances, and the definition of the phase

126

Chapter 9 Distillation

equilibrium constant, Ki ; yi/xi. The resulting equations are given below: xi ¼

zi 1 þ (Ki  1)C

(9:6)

yi ¼

zi K i 1 þ (Ki  1)C

(9:7)

In order to solve for the compositions of the liquid and vapor streams, one stream is chosen (V or L), mole fractions are calculated for that stream via Equation (9.6) or Equation (9.7), and the mole fractions of the subsequent stream are determined via the equilibrium constant, Ki ; yi / xi. Alternatively, the compositions of both streams can be calculated directly from Equations (9.6) and (9.7). For near-ideal mixtures, such as light hydrocarbons of the same homologous series at low pressures, Raoult’s law may be assumed to apply. In this case, the K value of any component i is the ratio of the vapor pressure of i to the total pressure. Alternatively, one may employ the DePriester Charts.(2)

ILLUSTRATIVE EXAMPLE 9.2 Calculate the vapor ratio, C, and the vapor and liquid stream compositions for a multicomponent flash operation. The feed mole fractions and phase equilibrium constants (at the flash temperature and pressure) are provided in Table 9.1. Given that the feed flow is 1250 lbmol/h, what are the liquid and vapor molar flow rates? Table 9.1 Multicomponent Flash Data for Illustrative Example 9.2 Component

Mole fraction (zi)

Equilibrium ratio (Ki)

0.28 0.24 0.24 0.08 0.16

2.93 1.55 0.87 0.49 0.138

1 2 3 4 5

SOLUTION: By utilizing the Rachford –Rice flash equation, it is noted that all values have been specified except for the vapor ratio, C. Since the sum of all terms must be equal to zero, the value of the vapor ratio may be calculated by an iterative, trial and error procedure. This may be performed by hand or in a program with iterative solver algorithms.

f (C) ¼

N X i¼1

zi (Ki  1) ¼0 1 þ (Ki  1)C

Upon iteration, C ¼ 0:5647

(9:5)

Batch Distillation

127

The values of xi are calculated via Equation (9.6), and the specified values of Ki were then used to determine yi. The results are shown in Table 9.2. Also note V ¼ FC and L ¼ F(12 C), Table 9.2 Liquid and Vapor Compositions for Illustrative Example 9.2 Component 1 2 3 4 5 SUM

xi

yi ¼ Kixi

0.134 0.183 0.259 0.112 0.312 1.000

0.393 0.284 0.225 0.055 0.043 1.000

V ¼ (1250)(0:5647) ¼ 706 lbmol=h L ¼ (1250)(1  0:5647) ¼ 544 lbmol=h

B

BATCH DISTILLATION Although batch distillations are generally more costly than their continuous counterparts, there are certain applications in which batch distillation is the method of choice. Batch distillation is typically chosen when it is not possible to run a continuous process due to limiting process constraints, the need to distill other process streams, or because the low frequency use of distillation does not warrant a unit devoted solely to a specific product. A relatively efficient separation of two or more components may be accomplished through batch distillation in a pot or tank. Although the purity of the distilled product varies throughout the course of batch distillation, it still has its use in industry. As shown in Figure 9.4, a feed is initially charged to a tank, and the vapor generated by boiling the liquid is withdrawn and enters a condenser. The condensed product is collected as distillate, D, with composition xD, and the liquid remaining in the pot, W, has composition xW. Total and componential material balances around a batch distillation unit are shown below: F ¼W þD FxF ¼ WxW þ DxD

(9:8) (9:9)

Note that W, xW , D, and xD all vary throughout the distillation. A convenient method for mathematically representing a binary batch distillation process is known as the Rayleigh equation. This equation relates the composition and amount of material remaining in the batch to initial feed charge, F, and composition, xF: xðF   W dx (9:10) ln ¼  F y x xW

where y ¼ mole fraction of vapor in equilibrium with liquid of composition x.

128

Chapter 9 Distillation

Figure 9.4 Batch distillation diagram.

At the desired distillate composition, the distillation is stopped. At this time, the moles of residue remaining in the still is denoted Wfinal, with composition xW, final. As such, Equation (9.10) may be re-written as shown below: 2 3 xðF dx 7 6 (9:11) W final ¼ F exp4 5 y  x xW, final

This equation may be solved numerically by plotting 1/( y 2 x) vs x and integrating between the limits xW, final and xF to determine the area (A) under the curve (see Fig. 9.5). The above can therefore be written as W final ¼ F exp [A]

(9:12)

where A ¼ area under the curve. A convenient method for determining the area under a given curve is provided by Simpson’s 3-point rule, as described below in Equation (9.13), xðF

f (x) dx

A¼ xW, final xðF

¼

x  i dx xF  xW, final h W, final þ xF ¼ f (x þ f (x ) þ 4f ) W, final F 6 2 y  x

(9:13)

xW, final

where f (x) ¼ 1=( y  x). Instead of relying on Simpson’s rule, a more accurate answer may often be calculated by fitting the data points with a polynomial regression as in Figure 9.5. The polynomial may then be integrated directly.

Batch Distillation

129

The above integration requires that the function 1/( y 2 x) be evaluated at the appropriate x values. Typically, this is a trial-and-error procedure which involves guessing the composition of the material remaining in the tank, xW, final, and then determining if Equation (9.11) is satisfied.(4) (See Chapter 19 and Illustrative Example 19.3 for additional details on numerical integration.) ILLUSTRATIVE EXAMPLE 9.3 A solution containing 50 moles of benzene and 50 moles of toluene is batch distilled at a constant pressure until such time that only 50 moles of liquid is left. Determine the composition of the liquid residue at the conclusion of the distillation process. Benzene/toluene equilibrium data is provided in Table 9.3. (Note that, as is common practice in binary distillation, all mole fractions refer to the lighter component, which in this case is benzene.) SOLUTION:

Write the Rayleigh equation for simple batch distillation. xðF   W final dx ln ¼ y  x F

(9:10)

xW, final

The initial concentration of benzene (mole fraction) is: xF ¼ 50=(50 þ 50) ¼ 0:50 In addition,     W final 50 ¼ ln(0:5) ¼ 0:693 ¼ ln ln 50 þ 50 F A trial-and-error solution is now required to determine the benzene mole fraction in the liquid. All values are known in Equation (9.10) except for the integral itself, for which xW ¼ xW, final. The area under the curve is determined by guessing values of xW,final until the

Table 9.3 Benzene/Toluene Equilibrium Data x 0.000 0.116 0.228 0.336 0.440 0.541 0.639 0.734 0.825 0.914 1.000

y

y 2 x

1/( y 2 x)

0.000 0.240 0.418 0.533 0.660 0.747 0.817 0.875 0.924 0.965 1.000

0.000 0.124 0.190 0.217 0.221 0.205 0.178 0.142 0.098 0.051 0.000

– 8.06 5.26 4.61 4.52 4.88 5.62 7.04 10.2 19.6 –

130

Chapter 9 Distillation

Figure 9.5 Benzene/Toluene equilibrium Rayleigh diagram. right-hand side of Equation (9.10) is equal to 20.693. If a numerical package is not available, one may plot 1/( y 2 x) vs x as in Figure 9.5. The equation appearing in Figure 9.5 is a polynomial regression fit to the data points, which can facilitate analytical integration. Upon integration of f (x) at varying lower bounds, it can be shown that when xW,final ¼ 0.35, the area under the curve is approximately equal to 0.693, which satisfies Equation (9.10). (This calculation is left as an exercise for the reader.) Therefore, xW, final  0:35 (benzene) 1  xW, final  0:65 (toluene)

B

ILLUSTRATIVE EXAMPLE 9.4 A chemical manufacturer plans to use methanol as a key solvent in a new batch production process. After conducting an economic analysis, the plant manager has determined that it may be cost effective to use existing equipment to separate methanol from a process water stream which is currently part of another process in the plant. His calculations indicate that the existing equipment can handle a charge of 150 lbmol of aqueous stream which could have a maximum of 75 mol% methanol. The distillate collected must be at least 85% pure in order to be fed directly to the main reactor of the new process. However, he needs to know how much methanol can be recovered from the methanol/water mixture, as well as the amount and concentration of the material (residue) remaining in the tank. Perform the integration found in the Raleigh equation using Simpson’s three-point rule. Equilibrium data for the methanol/water mixture is provided in Table 9.4. SOLUTION: Write the Rayleigh equation for simple batch distillation, as provided in Equation (9.11): 2 6 W final ¼ F exp4

xðF xW, final

3 dx 7 5 y  x

Batch Distillation

131

Table 9.4 Methanol/Water Equilibrium Data at 1 atm x

y

0.80 0.77 0.70 0.65 0.60 0.55 0.50 0.45 0.40

0.915 0.895 0.870 0.848 0.830 0.802 0.780 0.754 0.730

As described in the previous example, plot 1/( y 2 x) vs. x as shown in Figure 9.6. As with the previous example, one may obtain a regression equation for the data and integrate to find the area under the curve. Alternatively, one may use Simpson’s rule by guessing a value for xW, final and calculating the area. Both procedures require trial and error. Setting the initial methanol mole fraction to 75%, xF ¼ 0:75 and guess xW, final ¼ 0:59

Figure 9.6 Batch distillation analysis.

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Employ Simpson’s rule with f (x) ¼

1 y  x

Determine f (x) at the limits of integration by reading the approximate ordinate values from Figure 9.6: f (xW, final ) ¼ f (xF ) ¼

1  4:3 y  xW, final 1  7:4 y  xF

Also, f

x

W, final

þ xF 

1  5:5 y  (xW, final þ xF )=2

¼

2

Therefore, A¼

0:16 [4:3 þ 4(5:5) þ 7:4] 6

¼ 0:899 Calculate the moles of material remaining in the tank and the amount and composition of the distillate collected: W ¼ F exp (A) ¼ 150 exp (0:899) ¼ 61:0 In addition, D¼FW ¼ 150  61 ¼ 89 Thus, FxF  WxW, final D (150)(0:75)  (61)(0:59) ¼ 89

xD ¼

¼ 0:860 Check if the calculated value of the final distillate composition is within 5% of the given value %Error ¼

0:85  0:860  100 0:85

¼ 1:2% Thus, the initial guess for xW,final may be assumed acceptable.

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133

Should the reader so desire, this example may be solved in the same manner as Illustrative Example 9.3, i.e., by regressing the appropriate Raleigh equilibrium plot, and numerically integrating. With only a fourth-order polynomial fit to the above data, a more accurate answer may be obtained upon performing the iterative guess and check procedure. This is left as an exercise for the reader. B

In some situations, the degree of separation achieved in a single equilibrium stage (such as a flash distillation column) or in a batch still is often not large enough to obtain the desired distillate and/or bottoms purities. To improve the recovery of the desired product, a multi-staged distillation column may be employed. The analysis of such multi-stage columns will be the focus of the remainder of this chapter.

CONTINUOUS DISTILLATION WITH REFLUX Equipment and Operation For large-scale operations, continuous distillation is almost always more economical than batch, especially when a steady supply of feed is available. One of the disadvantages of both batch and flash distillation is the multiplicity of sequential distillations that are often necessary to achieve something approaching “complete” separation of the components. Moreover, it is possible to produce a very pure product by batch distillation. However, in order to obtain a high recovery, the liquid residue must be redistilled multiple times.(5) A continuous distillation column is analogous to several small flash units in series and effectively circumvents the need for multiple units. Indeed, unless a mixture contains an azeotrope,(2,6) a product stream of any desired purity may be theoretically obtained. In reality, a massive number of trays or an extremely high reflux ratio (to be defined shortly) would be required in the limit of a 100% pure product; these physical constraints on the system design limit the actual recovery possible. In a continuous distillation column, the mixture to be separated is fed into the column at some predetermined feed point between the top and bottom of the column. Vapor flows up the column and liquid flows countercurrently down the column. For the subsequent discussion regarding continuous distillation, a binary (2-component) feed will be assumed. In the case of a binary feed, the more volatile (lower boiling) component will be referred to as the light component, whereas the less volatile (higher boiling) component is referred to as the heavy component. As is standard practice in binary distillation, all mole fractions (i.e., xD, xB, xF) will be representative of the light component only. A photograph of a unit operations scale continuous distillation column is provided in Figure 9.7. The ascending vapor and descending liquid are brought into contact on either trays (plates) or packing. The plates/packing serves as a “widget” which allows the liquid and vapor to experience more intimate contact. As described earlier, the vapor at the top of the column enters a condenser. Part of the condensate is returned to the top of the column to provide reflux. The reflux descends counter to the rising

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Figure 9.7 Photograph of distillation column.

vapors, and in the case of a total condenser, the remainder of the liquid condensate is withdrawn from the condenser as distillate product. Reflux and Boil-up The molar flow ratio of reflux returned to distillate product collected is defined in this text as the reflux ratio, R. As the liquid/reflux stream descends, it is progressively stripped of the light constituent by the rising vapor. As a result of both heat and mass transfer effects, the vapor stream tends to vaporize the low-boiling constituent from the liquid and the liquid stream tends to condense the heavy constituent from the vapor. This liquid stream travels from the top tray to the bottom of the column, gradually increasing in the heavy component at each tray. The liquid is then collected by a pump at the bottom of the column, sent to a reboiler where it is partially vaporized (steam is usually employed as the heating medium) and returned to the column to provide an ascending vapor stream in the section of the column below the feed

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135

plate. The vapor generated by the reboiler is referred to as the boil-up. The other portion of the reboiler liquid is removed as the bottoms product. Analogous to the reflux ratio, the boil-up ratio, RB, is defined as the molar flow ratio of vapor generated by the reboiler to bottoms product. Rectification and Stripping The vapor stream generated in the reboiler passes up through the portion of the column below the feed tray (the tray onto which the feed stream is admitted). The portion of the column below the feed tray (and including the feed tray itself) is known as the stripping section. Upon reaching the top of the stripping section, the rising vapor enters the portion of the column above the feed tray, referred to as the rectification section. As this rising vapor stream contacts the descending liquid stream on each plate, its concentration of the lighter component is increased. As previously indicated, the vapor exits from the top of the column and passes into the condenser. The coolant is normally water; however, other cooling fluids may be utilized when the situation warrants. (Additional information on condenser coolant will be discussed shortly.) As seen in Figure 9.8, which can be compared to the laboratory unit in Figure 9.7, the rectification section is the portion of the column above the feed tray. In this section, there are n-trays; the first tray (tray 1) is at the very top, onto which the reflux is introduced, and the nth tray (tray n) is the tray above the feed tray. In reality, the liquid and

Figure 9.8 Schematic of a trayed distillation column.

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vapor molar flow rates each vary from plate to plate because of inevitable differences in the molar enthalpy of vaporization between the light and heavy components. The vapor and liquid molar flow rates leaving the ith tray in the rectification section are denoted Vi and Li with concentrations yi and xi, respectively (refer to Fig. 9.9). Similarly, the stripping section is the portion of the column below the feed tray. The stripping section is said to have m-trays (including the feed tray). Thus, the total number of trays in the column is N ¼ n þ m. The vapor and liquid molar flow rates in the stripping section are differentiated from those in the rectification section by using an overbar. For instance, the vapor and liquid molar flow rates leaving the jth tray in the stripping section are denoted V j and Lj with concentrations yj and xj, respectively. A column may consist of one or more feeds and may produce two or more product streams. Any product drawn off at various stages between the top and bottom are referred to as side streams. Multiple feeds and product streams do not alter the

Figure 9.9 Qualitative examination of rectification trays.

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137

basic operation of a column, but they do complicate the analysis of the process to some extent. Not all distillation columns contain both a rectification and stripping section, depending on its prospective use. For instance, if the process requirement is to strip a volatile component from a relatively nonvolatile solvent, the rectification section may be omitted; the unit is then referred to as a stripping column. Tray Designs Many tray designs are in use today, but one of the early favorites was the bubble-cap tray shown qualitatively in Figure 9.10. The descending liquid stream arrives on the plate via the downcomer which can be a short piece of pipe welded into position on the plate or simply a fraction of the plate which has been cut away. While several flow patterns are possible, the liquid generally is guided by the downcomer onto one side of the circular bubble-cap tray below, flows across the middle “capped” section of the tray and leaves by the next downcomer. This flow pattern is referred to as single crossflow; the details of various other flow patterns are available in the literature.(7) Referring to Figure 9.10, it is evident that the length of downcomer pipe protruding above the plate fixes the maximum depth of liquid on a tray. It thus acts as an overflow weir, which will be discussed shortly. The exit to each downcomer must be located below the liquid level on the plate below to create a vapor lock, preventing vapor from rising in the downcomer. The contacting device on a bubble-cap tray is the bubble cap, which consists of two parts—the riser and the cap. The riser is a short piece of pipe welded in place

Figure 9.10

Simplified bubble-cap tray.

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Chapter 9 Distillation

over a hole in the plate. The cap, which surrounds the riser, is a bell-shaped piece containing vertical slots around the lower periphery. The liquid level on the tray should be maintained so as to submerge each cap, which also ensures a vapor lock between trays. The vapor flows upward through the riser, makes a 1808 turn, and is forced under the cap and through the slots to produce intimate contact with the liquid. In order for the vapor to ascend properly, the pressure drop over a given plate must be slightly greater than the head of liquid on said plate. Another type of tray is the sieve tray, which simply consists of a perforated metal plate that has about a 10 – 15% portion of its cross sectional area removed in order to provide a downcomer (see Figs. 9.11 –9.12). A short metal “dam” at the end of the tray acts as an overflow weir which maintains the liquid height and guides the flowing liquid off of the plate and into the downcomer. In this arrangement, the vapor rises from the tray below and passes up through small holes in the tray above. The downcoming liquid flows across this plate and is allowed to achieve intimate contact with the rising vapor. The vapor and liquid come together and are engaged vigorously to form froth. This contact allows for increased rates of mass transfer between phases. Once again, the vapor becomes enriched in the more volatile component while the liquid becomes enriched in the less volatile component. Information on the third major type of tray, the valve tray, is available in the literature.(3)

Figure 9.11 Sieve tray column (single crossflow).

Continuous Distillation with Reflux

Figure 9.12

139

Sieve tray column (single crossflow).

Procedures for designing trays involve two major stages: a basic design procedure, and a detailed mechanical layout of the components for the trays. In the first stage, the following design parameters are usually established: 1 Tray diameter and area 2 Type of tray 3 Bubbling area and hole area 4 Hole diameter for sieve trays, key dimensions for bubble caps, and the type of valve in the case of valve trays. In the second stage, factors such as tray layout and mechanical design are set. Pressure Drop and Flooding There are numerous semi-empirical equations that are available for predicting the pressure drop across tray columns. As a preliminary estimate, one may assume the

140

Chapter 9 Distillation

pressure drop is given by the height of liquid supported on the tray. Typically, this liquid height is in the 4 – 6 inch range. Therefore, a reasonable approximation for pressure drop may be 4 – 6 inches of H2O per tray, which is approximately 0.1 –0.2 psi per tray. The lower value applies to smaller diameter columns and the upper value applying to larger diameter units.(8) It is important that the vapor stream has the correct superficial velocity (linear average velocity, calculated as if the column conduit was empty) as it flows upwards in the column. Should the vapor flow too slowly, liquid may pass down through the tray perforations instead of over the weir, a condition known as weeping. However, if the vapor has a velocity that is too high, some liquid may be carried from the froth to the tray above by the rapidly flowing vapor. This condition is known as entrainment. Should the vapor velocity be increased further, entrainment may become excessive such that the liquid level in the downcomer will reach the plate above. At this point, liquid flow from the tray(s) in question becomes hindered, and the column’s entrainment flooding point has been reached. When calculating the allowable superficial velocity of a vapor, certain effects such as the foaming tendency of a distillation mixture are often taken into account, as foaming increases the likelihood of entrainment. Both excessive entrainment and weeping greatly influence tray efficiency, and negatively impact overall column performance. Condensers and Reboilers In some operations where the top product is required as a vapor, the liquid condensed is sufficient only to produce reflux to the column, and the condenser is referred to as a partial condenser. In a partial condenser, the reflux will be in equilibrium with the vapor leaving the condenser, and the condenser is considered to be a theoretical stage (an equilibrium stage) when estimating the column height. However, in actual practice it may be advisable not to rely on the action of a partial condenser as an extra stage, but instead to add extra plates to the column in order to effect the desired separation. In contrast, when the vapor is totally condensed, the liquid returned to the column will have the same composition as the distillate product and the condenser is not considered to be a theoretical stage. A partial reboiler is generally used at the bottom of the column in order to operate the column and produce vapor which flows upwards through the stripping section and into the rectification section. The liquid in the reboiler generally exits as bottoms product. Since both the liquid and vapor are considered to be in thermodynamic equilibrium, the partial reboiler is usually considered a theoretical stage.

Equilibrium Considerations At this point in the development, each tray in the column is assumed to approach equilibrium conditions, so that the liquid and vapor leaving each tray are in perfect thermodynamic equilibrium. In other words, the contact between the liquid and vapor on a tray is assumed sufficient such that the vapor leaving said tray has the

Continuous Distillation with Reflux

141

same composition as vapor in equilibrium with the liquid overflow from the tray. This simplifying design assumption defines each tray as a theoretical (ideal) tray, or a theoretical stage. Referring to Figure 9.9, Vi and Li are in thermodynamic equilibrium such that yi ¼ Kixi. Distillation columns designed with this simplifying assumption in mind serve as a standard for comparison to actual columns. It is then possible to determine the number of actual trays that is equivalent to a theoretical tray, and by such comparisons, apply this information when designing real columns. In the case of a binary mixture, the aforementioned equilibrium curve may be constructed by either of two methods. The first method is more general, but also more rigorous. It consists of either empirically measuring vapor – liquid equilibrium (VLE) data (such as x, y composition data) in the laboratory, or calculating the data based on thermodynamic VLE equations such as Henry’s and Raoult’s law, the Wilson equation, the NRTL equation, etc.(2,9) This method is considered more rigorous because it requires compiling a table of x, y values from the aforementioned VLE equations. The second method is much simpler in application; however, it applies only to ideal mixtures. Note that, while the ideality of a gaseous mixture is determined based on the pressure of the system, a liquid mixture may be considered nearly ideal when each constituent has similar chemical properties (i.e., similar molecular weight, structure, reactivity, etc.) such as isomers or organic species of the same homologous series. This second method employs the relative volatility, defined earlier in Equation (9.1):

aAB ¼

KA yA =xA ¼ KB yB =xB

(9:14)

where yi, xi ¼ vapor and liquid mole fractions of component i at equilibrium. For binary mixtures, subscripts are usually omitted for convenience, and all mole fractions refer to the more volatile component, such that

aAB ¼

y=x (1  y)=(1  x)

(9:15)

where y, x ¼ mole fractions of the lighter component (A) at equilibrium and (1 2 y), (1 2 x) ¼ mole fractions of the heavier component (B) at equilibrium. Equation (9.15) may be rearranged into a more convenient form, which relates y as a function of x: y¼

aAB x 1 þ (aAB  1)x

(9:16)

This simple formula may be used for ideal mixtures in order to quickly plot an equilibrium curve, assuming a is known and relatively constant over the temperature range encountered in the column. When Raoult’s law applies, the relative volatility between two compounds is given by a ratio of the vapor pressures of each component. While

142

Chapter 9 Distillation

vapor pressure is a strong function of temperature, the ratios of vapor pressures do not vary as much with temperature change.

Constant Molal Overflow As previously mentioned, the liquid and vapor molar flow rates do in fact vary from tray to tray in an actual column. However, a major simplifying assumption which is readily employed in continuous distillation calculations is that the liquid and vapor molar flow rates are invariant within the rectification section, and are also invariant (though perhaps of different values) in the stripping section as well. Examining Figure 9.9, the subscripts on Vi, Li, Viþ1, etc., may all be dropped such that the liquid and vapor molar flows for the rectification section are L and V, respectively, while the liquid and vapor molar flow rates in the stripping section are L and V, respectively. This condition is referred to as constant molal overflow, and it is an accurate assumption to make when the components to be separated are chemically similar, with approximately similar molar enthalpies of vaporization. Taking benzene and toluene as an example, benzene’s molar latent heat of vaporization is 7360 cal/gmol, whereas toluene’s molar latent heat of vaporization is 7960 cal/gmol. Thus, for every 1.0 moles of toluene that is condensed from vapor to liquid, 1.08 moles of benzene may be vaporized, maintaining quasi-constant liquid and vapor molar flow rates in each section of the column. Without assuming constant molal overflow, the column’s operating lines would be non-linear, and enthalpy balances would be required in order to perform design calculations. However, constant molal overflow, when combined with other simplifying assumptions, serves as one of the basic principles that allows simple graphical and analytical methods to be employed to determine the number of theoretical stages in a binary column.

Binary Distillation Design: McCabe –Thiele Graphical Method When designing a binary distillation column, laborious tray-by-tray thermodynamic calculations may be circumvented by utilizing the McCabe – Thiele graphical method. The major simplifying assumptions which are generally employed with the McCabe – Thiele method are: 1 each tray, a partial condenser and/or partial reboiler all serve as theoretical stages, 2 constant molal overflow applies, and 3 Raoult’s law is valid. While these assumptions are not necessary for completing the graphical sizing of a column, they vastly simplify quick-sizing procedures for ideal or near-ideal systems.

Continuous Distillation with Reflux

143

Note that assumption (3) is the assumption most subject to question; however, it may be circumvented by obtaining more accurate equilibrium data. The aforementioned assumptions allow one to determine several important design parameters by simply constructing a McCabe– Thiele diagram. These parameters include the number of theoretical stages, the minimum number of theoretical stages, and minimum reflux ratio. This method is perhaps the most easy to learn and apply to binary systems since it does not require iterative calculations, nor does it include enthalpy balances (though it may be modified to include them when constant molal overflow is not assumed). In fact, this approach is fundamentally a convenient graphical representation of more drawn out calculations.

Calculation of Operating Lines As previously discussed, an equilibrium curve may be developed to describe the equilibrium relation between the liquid and vapor components leaving each stage. However, in the graphical design of a staged column, it is necessary to develop an operating line which relates the passing liquid and vapor streams between each stage. In effect, it is the operating line that provides a mathematical relationship describing the operating conditions within the column. The following analysis will develop operating lines for both the rectification section and the stripping section. To accomplish the following analysis, refer to Figure 9.13. An overall material balance is written for envelope I in the rectification section as: V ¼LþD

(9:2)

This equation relates the vapor (V ) flowing up the column, the liquid reflux (L) flowing down the column, and the distillate (D) collected. A componential material balance for the light component, A, is written as Vynþ1 ¼ Lxn þ DxD

(9:17)

This can be rearranged to form the equation of a straight line, y ¼ mx þ b, ynþ1 ¼

L D xn þ xD V V

(9:18)

where L/V ¼ the slope of the operating line [dimensionless]. The value L/V is defined as the internal reflux ratio, which is the ratio of the liquid reflux molar flow (returned to the top of the column) to the vapor molar flow (exiting the top of the column). By substituting Equation (9.2) into Equation (9.18), then dividing both the numerator and denominator of the terms on the right-hand side by the distillate flow, D, and finally substituting in the definition of the reflux ratio,

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Chapter 9 Distillation

Figure 9.13 Material balances around a distillation column.

R ¼ L/D, one obtains ynþ1 ¼

R xD xn þ Rþ1 Rþ1

(9:19)

which is the final result for the operating line of the rectification section. The corresponding operating line for the stripping section may be developed in a similar manner. The overall material balance around envelope II in the stripping section is L¼V þB

(9:20)

A componential balance on the lighter component in the stripping section may be rearranged to form an equation analogous to that of Equation (9.18). Note

Continuous Distillation with Reflux

145

once again that the overbars are reminders that the flow is occurring in the stripping section: L B (9:21) xm þ xB ymþ1 ¼ V V By performing similar substitutions, and by utilizing the definition of the boil-up ratio, RB ¼ V=B, an equation analogous to that of Equation (9.19) may be developed: ymþ1 ¼

RB þ 1 xB xm þ RB RB

(9:22)

which is the final result for the operating line of the stripping section. Thermal Condition of the Feed (q-factor) The thermal condition of the feed has a major impact on how a column operates. It has been found advantageous to represent the effect of the thermal condition of the feed graphically on a McCabe – Thiele diagram.(10) If the feed is subcooled, it will condense some of the vapor rising through the stripping section; if it is superheated, it will evaporate some of the liquid cascading down the rectifying section. In most applications, the feed is either a saturated liquid, saturated vapor, or a saturated two-phase mixture. Therefore, only these three scenarios will receive a detailed treatment in this text. Figure 9.14 illustrates the operation of a standard feed tray with a saturated feed, where FL is the molar flow rate of the liquid portion of the feed and FV is the molar flow

=

=

Figure 9.14

+

Action on a feed tray: saturated feed.

+

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Chapter 9 Distillation

rate of the vapor portion of the feed. For the purposes of this discussion (as noted above), the feed stream may be saturated liquid (FV ¼ 0), saturated vapor (FL ¼ 0), or a saturated two-phase mixture (F ¼ FL þ FV). When the feed is a saturated liquid (FV ¼ 0) and has an enthalpy equal to that of the mixture on the feed plate, the feed will flow completely into the stripping section such that V ¼ V and L ¼ L þ FL. When the feed is a saturated vapor (FL ¼ 0), the feed will completely flash into the rectification section such that L ¼ L and V ¼ V þ FV. When the feed is a saturated two-phase mixture (F ¼ FL þ FV), the resulting liquid and vapor streams leaving the feed tray are given by L ¼ L þ FL and V ¼ V þ FV, respectively. One can now define a factor which may be used to quantitatively represent the thermal condition of the feed on a McCabe– Thiele diagram. An overall material balance around the feed tray is given by, FþV þL¼V þL

(9:23)

F þ (V  V) ¼ (L  L)

(9:24)

Rearranging,

An enthalpy (H ) balance, neglecting heat losses to the surroundings and any enthalpy of mixing effects, yields FHF þ VHV þ LH L ¼ VHV þ LHL

(9:25)

where Hi ¼ molar enthalpy of stream i at the stream’s temperature and pressure about the feed tray. Since HV  HV and HL  HL , Equation (9.25) becomes FHF þ HV (V  V) ¼ HL (L  L)

(9:26)

Substituting (V 2 V ) from Equation (9.24) into Equation (9.26), one may rearrange the resulting equation to define the feed condition factor, q: q¼

L  L HV  HF ¼ HV  HL F

(9:27)

Note that the term q was employed earlier to represent the volumetric flowrate. The factor q is defined as the moles of liquid flowing in the stripping section which resulted from one mole of feed entering the column. For example, when the feed is a saturated liquid, each mole of feed entering the column adds directly to the stripping section’s liquid flow. Hence, the q factor is equal to unity. Conversely, when the feed is a saturated vapor, each mole of feed entering the column adds directly to the vapor flowing up the rectification section. In this case, none (of the one mole) of the feed adds to the stripping section’s liquid flow and the q factor is equal to zero. By substituting L ¼ L þ FL into Equation (9.27), the L terms cancel, leaving the result

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147

that for any type of saturated feed, q ¼ FL/F, which is equivalent to the fraction of the feed that is liquid. In the case of a subcooled liquid feed, the value of q is determined by: q¼1þ

C P,L (Tb  TF ) l

(9:28)

where TF ¼ the temperature of the feed, Tb ¼ the bubble point temperature of the feed, CP,L ¼ average constant pressure heat capacity of the liquid feed, and l ¼ latent enthalpy of vaporization of the feed at TF. Ensure that consistent units are employed throughout. Analogously, the value of q may be calculated for a superheated feed by Equation (9.29), q¼

C P,V (TF  Td ) l

(9:29)

where Td ¼ the dew point temperature of the feed and C P,V ¼ average constant pressure heat capacity of the vapor feed. Note that some practitioners prefer to employ a factor, f, defined as: f ¼1q

(9:30)

Values of q and f are summarized and provided in Table 9.5.

Table 9.5 Values of q and f for the Five General Feed Conditions Feed condition

q

Subcooled liquid Saturated liquid Saturated 2-phase mixture Saturated vapor Superheated vapor

.1 1 .0, ,1 0 ,0

f ,0 0 .0, ,1 1 .1

The term q also defines the point of intersection between the two operating lines on the McCabe – Thiele diagram. The operating line equations at the point of intersection are: Vyn ¼ L xn1 þ DxD

(9:31)

Vym ¼ Lxm1 þ BxB

(9:32)

for the rectification section and

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Chapter 9 Distillation

for the stripping section. Letting ( yn ¼ ym ¼ y), (xn1 ¼ xm1 ¼ x) at the feed point, and subtracting the latter equation from the former gives (omitting subscripts) y(V  V) ¼ x(L  L) þ DxD þ BxB

(9:33)

An overall componential balance on the column gives FxF ¼ DxD þ BxB

(9:34)

y(V  V) ¼ x(L  L) þ FxF

(9:35)

such that,

Since (L 2 L) ¼ 2qF and (V  V) ¼ F(1  q), substituting into Equation (9.35) yields (9:36)

Fy(1  q) ¼ x(qF) þ FxF which may be rearranged to give y¼

q xF x q1 q1

(9:37)

For a given feed condition (xF and q are fixed), Equation (9.37) plots as a straight line on the McCabe – Thiele diagram and is referred to as the q-line. Substituting xF for x in Equation (9.37) results in y ¼ xF, indicating that the q-line crosses the y ¼ x diagonal line at the point (xF, y). The two operating lines and the q-line intersect at a common point which is generally taken to be the feed tray location. Examples of the slope and y-intercept for the five general feed conditions in Table 9.5 are summarized and provided in Table 9.6. Table 9.6 Values of q-line Slope and Intercept for the Five General Feed Conditions Feed condition Subcooled liquid Saturated liquid Saturated 2-phase mixture Saturated vapor Superheated vapor

q-line slope ,1, .1 1 ,0 0 .0, ,1

y-intercept ,0 1 .xF xF .0, ,xF

A q-line for each of the above five general cases is plotted in Figure 9.15. It should be noted that for a given rectification operating line, decreasing the enthalpy of the feed increases q, which will change the number of theoretical stages required in each section. For example, it can be shown that with an increasingly superheated feed, a larger number of trays are required in the rectification section as opposed to the stripping section. In the case of an increasingly sub-cooled liquid feed, more trays are required in the stripping section.

Continuous Distillation with Reflux

Figure 9.15

149

Examples of q-lines.

Graphical Location of the Feed Tray The optimum theoretical feed plate is identified as the step on the McCabe– Thiele diagram which straddles the intersection of the two operating lines. Figure 9.16 represents a typical McCabe – Thiele diagram, the details of which will be explained in the next section. Notice that since the q-line is vertical, the feed is a saturated liquid, and thus the feed must be introduced on the seventh theoretical tray from the top of the column. In any application, the feed should be introduced on the tray where it will cause the least change in concentration of the process streams. This almost always means that the tray which crosses the q-line is designated the feed tray. Failure to observe this rule will result in extra “work” for the column, i.e., extra theoretical stages will be required for the separation. Minimum Number of Theoretical Stages: Total Reflux The notion of total (infinite) reflux is used in order to determine the theoretical minimum number of stages possible for a given separation. At this condition, the reflux ratio, R ¼ L/D is set equal to infinity since the distillate flow approaches zero. The boil-up ratio, RB ¼ V=B also becomes infinity since the bottoms flow approaches zero. As a result, the feed flowrate, F, also equals zero. At total reflux, the number of trays in the column achieves its minimum value. The physical interpretation for this occurrence is that, as the amount of reflux and boilup increases, the ability of the column to perform rectification and stripping is enhanced. This increased rectification and stripping results in needing less trays for the liquid and vapor streams to contact on. In the limit, at total reflux, the number of trays necessary reaches its minimum value.

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Figure 9.16 Typical McCabe–Thiele diagram.

In order to appreciate the graphical interpretation of this phenomena, one should first note that each of the triangles formed between the equilibrium curve and the operating line on a McCabe – Thiele diagram represents one theoretical stage (see Fig. 9.16). As R and RB approach infinity, Equations (9.19) and (9.22) both approach ynþ1 ¼ xn. Hence, the operating lines for both the rectification and stripping section “collapse” to the line y ¼ x. As a result, the q-line disappears, and the theoretical stages are stepped off between the equilibrium curve and the 458 line producing the minimum number of theoretical stages. Because the flow rates of the feed, distillate and bottoms are all zero when operating under total reflux, this minimum number is not used in actual applications; instead, it should be viewed as representing a limiting case. Minimum Reflux Ratio The case of minimum reflux ratio (Rmin) represents the alternative extreme of total reflux. At this condition, the reflux ratio is at its lowest possible value. Hence, the rectification and stripping abilities of the column are diminished to such a point that it would take an infinite number of trays above and below the feed tray in order to effect the desired separation of products. Once again, this analysis is simply a limiting case; however, it is a concept of great importance in distillation column design. Heat Duty In addition to mole balances, one may also perform energy balances across parts of the system. The duty of the condenser can be calculated with the knowledge of the flow

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rate of coolant (usually water) entering the condenser plus the inlet and outlet temperatures. From a design point of view, the duty may be calculated based on the enthalpy change of the condensing vapor, assuming that the temperature at the top of the column, along with the distillate temperature and pressure are fixed. The condenser heat duty for a total condenser (in a column with light component A, and heavy component, B) may be calculated by assuming that the condenser is adiabatic. First note that enthalpy is a state function,(2) and therefore, the change in enthalpy is independent of the actual process path. As a result, if the entering vapor stream at T1 is condensed and cooled to T2, it does not matter thermodynamically whether the condensation took place at T1, T2, or any temperature in between. Therefore, one may assume that the vapor condenses at T1, which is the temperature at the top tray of the column. This saturated liquid is then cooled further to T2 which is the condenser’s exit temperature, TC (look ahead to Fig. 9.19 for a qualitative condenser schematic). By performing an enthalpy balance around the total condenser, the condenser heat duty, QC, is equal to the enthalpy change of the condensing vapor stream. Thus, QC ¼ V[xD (lA þ C P,L(A) (T1  T2 )) þ (1  xD )(lB þ C P,L(B) (T1  T2 ))]

(9:38)

where C P,L(i) ¼ average molar heat capacity of liquid component i (constant P), li ¼ molar latent enthalpy of vaporization of i at T1. Assuming sensible enthalpy changes are negligible, the heat load may be calculated by summing the latent enthalpy contributions of each species: QC ¼ V[xD lA þ (1  xD )lB ]

(9:39)

The molar flow rate of vapor may be found from the material balance: V ¼DþL

(9:40)

Utilizing R ¼ L/D, one may substitute L ¼ RD to obtain, V ¼ D(R þ 1)

(9:41)

Lastly, the distillate flow, D, may be calculated when the distillate and bottoms compositions are specified. Equation (9.42) is a result of combining the total and componential material balance equations on the column: D xF  xB ¼ F xD  xB

(9:42)

The condenser heat duty may now be readily calculated from either Equation (9.38) or Equation (9.39).

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Once the theoretical reboiler temperature is calculated, one may calculate the reboiler duty, the mass flow rate of heating medium (usually steam) entering the reboiler, and the inlet/outlet temperatures of the steam and/or the required heat exchanger area. A more thorough presentation of how to set the column temperatures and pressures is provided in the section on multicomponent distillation. Constructing a McCabe – Thiele Diagram A detailed step-by-step procedure for designing a binary (trayed) distillation column via the McCabe – Thiele graphical method is provided below. In order to construct a McCabe – Thiele diagram, certain simplifying assumptions are generally made: 1 Each tray acts as an ideal stage, from which the exiting vapor and liquid are in thermodynamic vapor – liquid equilibrium. 2 Constant molal overflow applies (to circumvent enthalpy balances and curved operating lines). 3 Raoult’s law is valid (when the binary mixture is ideal). In order to apply the McCabe – Thiele graphical method, one must first construct a y vs x equilibrium plot. Once an x, y plot is constructed, one should follow these instructions: 1 Plot the line y ¼ x (the 458 line) and locate the exit composition points (xD, xD) and (xB , xB ). 2 Plot the q-line. This line is given by q xF x y¼ q1 q1

(9:37)

The q-line is plotted by starting at the point (xF , xF ) on the 458 line, and extending it until it touches the equilibrium curve. 3 Determine Rmin by drawing a straight line between the point (xD, xD) and the point at which the q-line intersects the equilibrium curve. This represents the pinch point and also corresponds to the minimum reflux ratio at the given feed condition and desired distillate composition. At minimum reflux, the number of ideal trays approaches infinity. The value of Rmin may be determined by noting that the y intercept ( yint) of this line [given by Equation (9.19)] is, xD (9:43) yint ¼ Rmin þ 1 which may be rearranged to: Rmin ¼

xD 1 yint

(9:44)

4 Now that Rmin has been determined, the optimum reflux ratio may be calculated with Equation (9.45) R ¼ mRmin

(9:45)

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Table 9.7 Reflux Ratio Optimization Multipliers Type of coolant

Optimization multiplier, m

Water (1008F) Refrigerant (,1008F) Air (hot conditions)

where

1.3 1.1 1.5

m ¼ an optimization multiplier (which, of necessity is always .1.0) R ¼ the optimum reflux ratio

The value of m is generally dependent upon the type of cooling fluid used in the condenser. Three pertinent values are provided in Table 9.7. These values are approximations, which have been based on both economic considerations and experience. More detailed tabulations and methodologies are available in the literature.(6) 5 Having determined the optimum value of R, the rectification section operating line (ROL) may be drawn. The simplest way of graphing this line is to first calculate its y-intercept, given by yint(ROL) ¼

xD Rþ1

(9:46)

A straight line is then drawn between this y-intercept and the point (xD, xD). 6 The last portion of the diagram’s basic “frame” is the stripping section operating line (SOL). The simplest way to construct this line is to draw a straight line between (xB, xB) and the point where the ROL intersects the q-line. 7 The best course of action at this point is to erase/delete all lines and segments of lines which may be discarded. This includes the portions of the ROL and q-line which extend past the locus where the q-line, ROL, and SOL all intersect. One should also delete the minimum reflux operating line that intersects the q-line at the pinch point. 8 With the basic outline prepared, it is simply a matter of drawing the rectification stages, starting at point (xD, xD) and continuing horizontally leftward, until the line reaches the equilibrium curve. At this point, the next line segment should continue straight down (perpendicular to the previous line segment, until it reaches the ROL). This first triangle represents one ideal stage in the binary distillation column. This procedure is repeated (creating a series of triangles between the equilibrium curve and the ROL), until the intersection point of the ROL, SOL, and q-line is reached. Once a tray crosses this locus, the first possible vertical line segment should be drawn down to the SOL line. The tray at which the line segments switch from ROL to SOL is designated as the feed tray. After this point, all subsequent trays are drawn between the SOL and the equilibrium curve. The last theoretical tray should be drawn such that it just passes the point (xB, xB) since it is a rare occurrence that the bottom tray will achieve the exact bottoms composition.

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Note that in the case of a partial reboiler, the last (bottommost) triangle is representative of the reboiler. Should a partial condenser be employed, the first (topmost) triangle is representative of the condenser. Should a total reboiler and/or total condenser be utilized, they are not counted as equilibrium stages, and the corresponding triangles are instead counted as theoretical trays. The McCabe – Thiele method is quite convenient for the sizing of binary distillation columns. It uses vapor– liquid equilibrium curves to determine the theoretical number of stages required to effect the separation of a binary system. Once again, this method generally assumes constant molal overflow, which implies that molar enthalpies of vaporization of the components are approximately the same, such that for every mole of vapor condensed, one mole of liquid is vaporized. For ideal binary mixtures, Raoult’s law is often assumed in order to make vapor – liquid equilibrium calculations easier. Lastly, it is always assumed that each tray, as well as a partial reboiler and/or condenser, acts as a theoretical equilibrium stage in the column. The importance of the McCabe – Thiele graphical method cannot be overstated as a training tool in distillation. Determining the Minimum Number of Theoretical Stages Graphically At total reflux, R approaches infinity, and the slope of both the ROL and SOL approach unity (the 458 line). The graphical method for determining the minimum number of trays is the same as that for constructing any McCabe – Thiele diagram, i.e., starting at xD and stepping off triangular stages between the equilibrium curve and the operating lines, both of which have collapsed to the 458 line. Thus, the q-line is eliminated (because there is no feed to the column at total reflux). Tray Efficiency In order to determine the actual number of trays, the concept of tray efficiency must be introduced. One of the preliminary assumptions in constructing the McCabe – Thiele diagram was that each tray would serve as an equilibrium stage. However, this is generally far from true, since the vapor and liquid contacted on a given tray do not have sufficient contact time to reach thermodynamic equilibrium before leaving the tray. Thus, an overall efficiency for the plates in a column can be defined. There are several types of efficiencies, including the Murphree (individual tray) efficiency, and local (location on an individual tray) efficiency. However, only the overall efficiency will be discussed here. The overall efficiency, denoted E0, generally ranges between 0.4 and 0.8, with an upper limit of unity. Overall efficiency is affected by a host of factors such as the tray type, tray spacing, operating conditions, and so on (see O’Connell correlation, Fig. 9.21 for more details). In order to calculate the actual number of trays, the number of ideal trays is divided by the overall efficiency: N trays,ideal (9:47) Ntrays ¼ E0

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155

Note that, in the case of the McCabe – Thiele graphical method, Ntrays,ideal is the number of theoretical stages (represented as triangles on the diagram) minus the partial reboiler and/or partial condenser stage(s).

ILLUSTRATIVE EXAMPLE 9.5 Determine the number of stages required to separate a mixture of benzene and toluene containing 40 mole% benzene into a distillate containing 99 mole% benzene and a bottoms containing 1 mole% benzene. The feed is a two-phase mixture consisting of 50 mole% liquid. The process is carried out at 1atm, the condenser employs cooling water at 1008F, and the overall column efficiency is approximately 50%. Employ the step-by-step method provided above. Refer to Figure 9.17 for equilibrium data (curve). SOLUTION:

Refer to Figure 9.17 in the development to follow.

1 Plot the line y ¼ x (also known as the 458 line) and locate the exit composition points (xD, xD) and (xB, xB). 2 Plot the q-line y¼

q xF x q1 q1

(9:37)

For this particular construction, the feed is a 2-phase mixture, which is 50 mole% liquid. Thus, q ¼ 0.50, xF ¼0.40, and the q-line equation becomes y ¼ x þ 0:8 Once again, this line is plotted by starting at the point (xF, xF) on the 458 line, and extending it until it touches the equilibrium curve. 3 The value of Rmin may be determined by drawing a straight line between (xD, xD) and the point at which the q-line touches the equilibrium curve. Observing the y intercept of this line: xD yint ¼ (9:43) Rmin þ 1 which may be rearranged to: Rmin ¼ Since yint ¼0.298, Rmin ¼

xD 1 yint

(9:44)

0:99 1 0:298

¼ 2:32 4 Now that Rmin has been determined, the optimum reflux ratio may be calculated from Table 9.7: (9:45) R ¼ 1:3 Rmin R ¼ 1:3(2:32) ¼ 3:02

156

Chapter 9 Distillation 5 The ROL can be drawn using the optimum value of R. The simplest way of graphing this line is to first calculate its y-intercept: xD Rþ1 0:99 ¼ 3:02 þ 1

yint(ROL) ¼ yint(ROL)

(9:46)

¼ 0:246 Then, a straight line is drawn between this y-intercept and the point (xD, xD). 6 Draw the SOL. The simplest way to construct this line is to draw a straight line between (xB, xB) and the point where the ROL intersects the q-line. 7 Erase all lines and segments that can be discarded. 8 Step off the theoretical stages. Number of Theoretical Stages ¼ 18 Trays þ Partial Reboiler Calculate the actual number of trays by dividing the ideal trays by E0 ¼ 0.50: Ntrays ¼

Ntrays,ideal E0

Ntrays ¼

18 0:50

¼ 36

Figure 9.17 McCabe – Thiele diagram—benzene/toluene, 1 atm (xD ¼ 0.99).

(9:47)

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157

The results for this Illustrative Example are summarized below: Benzene/Toluene system P ¼ 1 atm (absolute) xF ¼ 0.40 xD ¼ 0.99 xB ¼ 0.01 R ¼ 1.3Rmin q ¼ 0.50 (q ¼ FL/F, for two-phase mixture) E0 ¼ 0.50 Number of Required Stages ¼ 36 Trays þ Partial Reboiler

B

ILLUSTRATIVE EXAMPLE 9.6 Refer to Illustrative Example 9.5. Calculate the condenser heat duty, neglecting sensible enthalpy effects. The molar latent heat of vaporization for benzene and toluene at the prevailing temperature (at the top of the column) are lB ¼ 13,251 Btu/lbmol and lT ¼ 14,331 Btu/lbmol, respectively. SOLUTION:

The heat duty for the total condenser is calculated by assuming:

1 sensible enthalpy effects are negligible (in cooling the condensate), and 2 the condenser is adiabatic. By performing an enthalpy balance around the condenser, the condenser heat duty may be set equal to the enthalpy change of the condensing stream. QC ¼ V[xD lB þ (1  xD )lT ]

(9:39)

However, the vapor molar flow rate must first be calculated. As a result of the material balance: V ¼DþL

(9:40)

V ¼ D(R þ 1)

(9:41)

Noting that R ¼ L/D,

Assuming a basis of 100 lbmol/h of feed: D xF  xB ¼ F xD  xB   0:40  0:01 D¼ (100 lbmol=h) 0:99  0:01 ¼ 39:8 lbmol=h Therefore, V ¼ 39:8(3:02 þ 1) ¼ 160 lbmol=h

(9:42)

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Chapter 9 Distillation

Plugging the necessary values into Equation (9.39):      lbmol Btu Btu 0:99 13,251 þ 0:01 14,331 QC ¼ 160 h lbmol lbmol Btu h  Btu lbmol feed ¼ 2,122,000 100 h h ¼ 21,220 Btu lbmol feed ¼ 2,122,000

B

ILLUSTRATIVE EXAMPLE 9.7 Once again, refer to Illustrative Example 9.5. Calculate the minimum number of theoretical stages for the given separation. SOLUTION: The minimum number of theoretical stages is determined by designing the column to operate at total reflux. Recall that, for the condition of total reflux, the ROL and SOL both collapse to the line y ¼ x. See Figure 9.18, following a similar procedure to that in Illustrative Example 9.5: Minimum Number of Theoretical Stages ¼ 10 Trays þ Partial Reboiler

Figure 9.18 McCabe – Thiele diagram (total reflux)—benzene/toluene VLE, 1 atm.

B

B

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159

Column Diameter The following discussion regarding column diameter calculation is equally applicable to both binary and multicomponent distillation. Column diameter must generally be calculated before column height since the column diameter affects the choice of tray spacing, which in turn affects column height. The diameter of a distillation column is generally controlled by the vapor velocity, u. Several factors go into determining an appropriate operating vapor velocity. As previously mentioned, when the velocity is too low in a trayed column, weeping occurs. Under these conditions, the liquid passes through the tray perforations instead of frothing and passing over each subsequent overflow weir. Alternatively, should the vapor velocity be too high, the column will experience the aforementioned entrainment of liquid from one tray to the tray above. At excessively high vapor velocities, the entrainment flooding point may be reached. At this point, the vapor is flowing up the column so rapidly that it effectively prevents the liquid from flowing down the column. Hence, the column will begin to flood with liquid, impacting product purity and perhaps posing a danger to those around the column. Several correlations have been developed which aid in determining the operational vapor velocity of a column. One such correlation which has withstood the test of time is the Fair flooding correlation (look ahead to Fig. 9.23). This correlation uses data on the liquid and vapor mass flows, mass densities, and surface tension in the general location which is most likely to flood (based in part on feed condition) to determine the flooding velocity, uF, which may be corrected for foaming considerations. Once corrected, the operating velocity may be calculated. Generally, design engineers may choose a vapor velocity about 60% of the flooding velocity (u  0.6uF). However, the exact value for this fractional approach to flooding velocity is left up to the individual protocols for a specific design. Additional details are provided in the next chapter. The reader should note that, while the values of V and L (which are molar flows) are assumed constant throughout each section of the column, the equivalent mass flow rates of the vapor and liquid streams do in fact vary as their respective molar compositions vary from stage to stage. As molar composition changes, the average molecular weight also changes and hence, the mass flow rate varies even though the molar flow rates are assumed constant. Here, the variation of mass flow rates is important because flooding is a fluid mechanics phenomenon, which is more directly related to mass, not moles. As a result, it is important to evaluate certain chemical and physical characteristics (i.e., average molecular weight, density) of the vapor and liquid streams in pertinent sections of the column (i.e., top tray, bottom tray, below and above the feed tray) in order to calculate what the flooding velocity would be in that section. Once again, flooding correlations such as the Fair flooding correlation are dependent on the liquid and vapor mass flow rates, mass densities and liquid surface tension. Therefore, in calculating the flooding velocity at each section of interest, the local composition, temperature, and pressure must be determined via material balances, and temperature/pressure estimations based on the reboiler and condenser design temperatures. The Fair flooding correlation is demonstrated later in Illustrative Example 9.19 for a multicomponent distillation column.

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Once a suitable vapor velocity has been determined, the column diameter may be calculated from the continuity equation, given below: _ ¼ rV uB SB m

(9:48)

where m ˙ ¼ mass flow rate of vapor, rV ¼ density of the vapor, uB ¼ vapor velocity (based on bubbling area of tray), and SB ¼ the bubbling area of a given tray. However, a different form of Equation (9.48) is usually applied in distillation. Dividing the mass flow rate by the density of the vapor yields units of volume per time, q ¼ uB SB

(9:49)

where q ¼ the volumetric flowrate of vapor (not to be confused with the feed thermal condition factor). The volumetric flowrate of the vapor stream may be determined from the product of molar volume of the vapor (at the local column temperature) and molar flow rate of the vapor, V, q ¼ vV

(9:50)

where v ¼ molar volume (the inverse of the molar density) of the vapor. The molar volume of vapor at a given point in the column may be readily approximated by the commonly employed ideal gas law ratio(2):    Pstd T v ¼ vstd (9:51) P Tstd where Tstd, Pstd ¼ standard temperature and pressure, for which vstd is known vstd ¼ molar volume of vapor at Tstd and Pstd T, P ¼ temperature and pressure at the area most likely to flood first v ¼ molar volume of vapor at T and P A simple form of the column diameter equation may be expressed as Dcolumn

 0:5 2 q ¼ pffiffiffiffi p (1  h)u

(9:52)

where h represents the fraction of the column cross section which is taken up by one downcomer (e.g., for a tray in which the downcomer accounts for 15% of the column cross section, h ¼ 0.15). Note that h only represents the area of one downcomer because the aforementioned Fair flooding correlation, and hence the calculated vapor velocity u, is based on a tray’s net area. Referring to Figure 9.12, the net area is here defined as the bubbling area plus the cross sectional area of one downcomer. With other correlations, the actual bubbling area may be the basis, for which (122h) should be employed in Equation 9.52. The interested reader is referred to the literature for a more detailed analysis.(7,11) As previously mentioned, different sections of a column have liquid and vapor streams of differing densities, average molecular weights, and mass flow rates. As a

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result, one should consider calculating the flooding velocity at different points in the column. The location with the lowest flooding velocity serves as the limiting vapor velocity in the column, and hence, the column diameter should be calculated based on that velocity. However, it is a common practice to design a distillation column of varying diameter based on calculated flooding velocities in each section of the column. In some instances, the diameter of a column changes about the feed tray, such that the rectification and stripping sections may each be operated at their optimal vapor velocities. Column Height Once again, the subsequent discussion of column height applies equally well to binary and multicomponent distillation. Column height is mainly determined by two factors: the number of trays and the spacing between the trays. The number of actual trays necessary may be determined from the overall column efficiency; however, the tray spacing may vary with column diameter as well as the nature of the distillation mixture. Components with a tendency to foam excessively should be allowed greater tray spacing, both to prevent entrainment and, in the case of larger columns, to allow for a technician to climb into the column for maintenance. Typical tray spacings are 18, 24, or 36 inches. Generally, larger tray spacing is employed with increasing column diameter. In certain types of columns such as those which separate liquefied air components at cryogenic temperatures, smaller tray spacings of 4 to 8 inches may be employed. The bulk of the actual column height is determined by multiplying the number of trays times the tray spacing. After this initial height is determined, various considerations are often employed such as designing a surge volume at the bottom of the column which allows the column extra space to collect liquid should a pump malfunction. It is also customary to allow extra space in the column so that more trays may be added if necessary.

Multicomponent Distillation: Fenske –Underwood – Gilliland (FUG) Method Multicomponent distillation is classified as a distillation in which the feed stream contains more than two components. There are several methods of calculation which have been developed for multicomponent distillation. Among these, there are both rigorous and shortcut column design methods. In this text, the Fenske – Underwood – Gilliland (FUG) shortcut method will be developed. It is important for the reader to note that the FUG method consists of three main parts: 1 the Fenske equation, which calculates the minimum number of theoretical stages and the recoveries of the non-key components, 2 the Underwood equations, which calculate the minimum reflux ratio, and 3 the Gilliland correlation, which allows one to determine the number of theoretical stages necessary to produce a desired separation.

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Almost all of the principles which find application in multicomponent distillation may also be applied to the limited case of binary distillation. Hence, the FUG method may be applied to a binary column just as well as it could to a multicomponent mixture. However, the McCabe – Thiele method is often preferred for binary systems due to its simplicity. After the FUG method is explored, additional pertinent correlations are introduced in order to facilitate a complete column quick-sizing. However, before proceeding to the three individual components of the FUG method, some introductory material is discussed. Multicomponent distillation differs from binary in that the entire composition of the product streams may generally not be defined. When a feed consists of three or more components, it is generally impossible to exactly specify the mole fractions of all the components in the distillate and bottoms, as one can in binary distillation. For instance, if a design engineer was to specify all of the mole fractions in the distillate for a ternary (three-component) mixture consisting of A, B, and C, only one mole fraction specification in the distillate could be met exactly by changing the reflux ratio and/or the number of trays. Suppose that the mole fractions for A, B, and C in the distillate were set as 0.60, 0.25, and 0.15, respectively. When the column is designed such that any one of the above mole fractions is obtained (i.e., xD, A ¼ 0.60), the mole fractions in the distillate of components B and C are solely dependent upon the thermodynamics in the column which allowed xD, A ¼ 0.60. Any chance of xD, B and xD,C equaling their specified design values would be purely coincidental. Thus, the mole fractions of B and C are independent of the mole fraction of A. In contrast, a specified distillate composition of A for a binary feed would automatically fix the amount of the second component in the distillate. Moreover, the thermodynamics that govern multicomponent vapor– liquid equilibrium is much more complex and often produces intricate concentration profiles. As a result of such difficulties, the practice of sizing and designing actual multicomponent distillation columns is often performed by computer programs, which are iterative in nature. These programs are designed to perform tedious thermodynamic calculations on a stage-by-stage basis, while attempting to converge at the desired result. The FUG method is utilized as an efficient means of quick-sizing a multicomponent column without such computer programs. Indeed, some of the more robust distillation programs need large amounts of user-specified inputs regarding the number of trays, reflux ratio, condenser and reboiler temperatures and pressures, feed tray location, etc. Hence, the FUG method still serves a purpose for obtaining necessary ballpark figures. Key Components Since not all of the mole fractions in the distillate and bottoms may be specified, the column designer generally specifies two components whose mole fractions or fractional recoveries in the distillate and bottoms streams will serve as the basis for column design. These two components are referred to as the key components, with the more volatile of the two being referred to as the light key (LK) and the less volatile of the two being referred to as the heavy key (HK). With a feed of n components, any two species may be chosen as the key components. However, it is generally useful to

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choose a sharp separation, for which two components of adjacent volatility are labeled as the keys. Here, the volatilities of each component may be compared by observing their respective normal boiling points. If the components are listed in order of increasing boiling point (decreasing volatility), any two sequential pairs of components are of adjacent volatility. By choosing a sharp separation, the column will be able to specifically divide the feed between the two adjacent key components. With a proper design, the LK, along with all components lighter than the LK will proceed almost completely to the distillate. Similarly, the HK, and all components heavier than the HK will proceed almost completely to the bottoms. In reality, minute amounts of the component(s) lighter than the LK do end up in the bottoms stream and minute amounts of the component(s) heavier than the HK do end up in the distillate product. Hence, all components (not just the keys) are technically distributed components, in that they are distributed between the distillate and bottoms. However, the concentrations are often so small for non-keys that they may be considered insignificant and are assumed to be undistributed components. Generally, unless stage-by-stage calculations are employed in which a finite value of each species must be present on each tray, all non-key components may be assumed to be undistributed, assuming that there are no components between the keys (sharp separation). Of course, this is not always the case with multicomponent mixtures, since it is possible that a given component may be more volatile than the average mixture in one part of the column, yet less volatile than the average mixture in another part of the column.(12) Moreover, when non-keys have relative volatilities which are very similar to that of the key components, they may be distributed in some appreciable quantity. Great care should be exercised in making assumptions regarding distribution with non-ideal chemical mixtures. A more detailed example of the undistributed components assumption will be discussed whilst examining the Fenske equation.

ILLUSTRATIVE EXAMPLE 9.8 Assuming a sharp separation is desired, suggest any set of key components for the following feed composition: Component Propane n-Butane i-Butane Hexane Heptane Nonane

Feed Mole Fraction 0.05 0.06 0.06 0.20 0.23 0.40

SOLUTION: The given six feed compounds are already listed in order of increasing boiling point (decreasing volatility). Hence, any two sequential pairs may be chosen as the LK and HK

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in order to effect a sharp separation. Depending on the desired product streams, the five possible pairs are shown below: Pair #

LK

HK

1 2 3 4 5

Propane n-Butane i-Butane Hexane Heptane

n-Butane i-Butane Hexane Heptane Nonane

B

Fractional Recovery As previously mentioned, in multicomponent distillation, either the fractional recoveries or mole fractions of the key components may be specified in the distillate and bottoms streams. In this text, specification of the fractional recoveries will take precedence over mole fractions. The most convenient definition of fractional recovery, r, from a design point of view is expressed in Equations (9.53) and (9.54) for the LK and HK, respectively, DxD,LK (9:53) rLK ¼ FxF,LK rHK ¼

BxB,HK FxF,HK

(9:54)

Note that other authors may define the fractional recovery of the HK, rHK, with respect to the distillate stream, in a similar manner to Equation (9.53). This definition is equally valid; however, the reader should be cautioned that utilizing this alternative definition would require that the Fenske equation (see later section) be rearranged. When defined as in Equations (9.53) and (9.54), a typical range of values for the recoveries of each key might be between 0.9 and 0.999, depending on the circumstances. While there are a wide variety of methods for quick-sizing a multicomponent distillation column, what follows is one particular interpretation of the FUG method. This series of procedures may be used in sequence in order to efficiently quick-size a typical distillation column. Setting the Column Pressure Before proceeding to explain the FUG method, it is of primary importance to understand how a design engineer sets the column operating pressure, since this is generally performed first. The condenser is important because it determines the operating pressure of the column. If one assumes that the liquid distillate product is a saturated liquid at its bubble point, then knowledge of the distillate temperature and composition allows the calculation of the condenser pressure, PC.(2) If cooling water is the condensing medium of choice, as it usually is, there are certain considerations which must be integrated into the condenser design. Cooling water may be available at a variety of temperatures; however, it is important to not allow the exit temperature of the cooling water to increase well beyond 1208F. This requirement

Continuous Distillation with Reflux

Figure 9.19

165

Total condenser schematic.

arises because normal process water may contain considerable amounts of naturally occurring calcium carbonate and calcium sulfate salts, the solubilities of which decrease with increasing water temperature. Hence, allowing the water temperature to be in excess of 1208F poses a serious problem with respect to salt deposition and subsequent condenser maintenance. Figure 9.19 shows a basic line diagram for a condenser where T1 ¼ temperature at the top tray of the column, TC ¼ condenser temperature, and ti, tf ¼ inlet and final cooling water temperatures, respectively. Based on the available cooling water temperature, a cool end approach of approximately 208F may be employed in order to set a reasonable temperature for TC. By setting TC, both the distillate temperature and composition are fixed. Therefore, for a total condenser, by assuming the distillate is a saturated liquid at its bubble point, a bubble point pressure calculation may be performed in order to determine the condenser pressure, PC. Bubble point calculations(2) are carried out with the equation N X

Ki xD,i ¼ 1:0

(9:55)

i¼1

where

N ¼ the number of components in the feed Ki ¼ equilibrium ratio of the ith component ( yi/xi) xD,i ¼ distillate mole fraction of component i

When Raoult’s law is assumed applicable, the K values for each species may be approximated by the ratio of that species’ vapor pressure to the total pressure. This reduces Equation (9.55) to(2) N X p0i (TC ) xD,i ¼ 1:0 PC i¼1 N X i¼1

p0i (TC ) xD,i ¼ PC

(9:56)

(9:57)

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Chapter 9 Distillation

When the Antoine equation is used to calculate vapor pressures, a convenient bubble point pressure equation can be derived: N X

10ai [bi =(ci þTC )] xD,i ¼ PC

(9:58)

i¼1

where PC ¼ condenser pressure and TC ¼ condenser temperature.

ILLUSTRATIVE EXAMPLE 9.9 A total condenser produces saturated liquid condensate containing 96% n-heptane, 3% n-octane, and 1% i-octane (mole%). If the condenser temperature has been set to 3928F, calculate the condenser pressure in psia (assume Raoult’s law is valid for this ideal liquid mixture). SOLUTION: Since Raoult’s law is valid, Equation (9.58) may be employed in order to calculate PC. The Antoine equation coefficients are given in Table 9.8, with T [8C] and P [mm Hg]. Table 9.8 Antoine Equation Coefficients for Illustrative Example 9.9 Antoine coefficient ai bi ci

n-heptane

n-octane

i-octane

6.90253 1267.828 216.823

6.91857 1351.756 209.100

6.88814 1319.529 211.625

Employing Equation (9.58), N X

10ai [bi =(ci þTC )] xD,i

i¼1

Substituting TC ¼ 200 C and xD;1 ¼ 0:96 xD;2 ¼ 0:03 along with the appropriate coefficients provided in Table 9.8, gives PC ¼ 7140:4 mm Hg ¼ 138 psia

B

Use of the Antoine equation is preferred here because Equation (9.58) does not require a trial-and-error solution and the condenser pressure can be calculated at any condenser temperature. Depending on the type of Antoine coefficients used, one should be sure to employ the correct units in Equation (9.58).(2) Upon determining PC, it is up to the designer’s discretion as to whether or not the condenser pressure is too low or too high. In most applications, PC , 30 psia is too low such that the condenser pressure should be reset to a value of 30 psia. At this

Continuous Distillation with Reflux

167

new PC, the condenser temperature TC must then be re-calculated with a bubble point temperature calculation. Alternatively, if PC . 215 psia, the condenser pressure is so high that cooling water may not completely condense the vapor. In this case, switching the condenser design to a partial condenser is recommended. At very high condenser pressures, for example, PC . 365 psia, use of a partial condenser cooled with refrigerant is suggested. In this case, the condenser pressure should be reset to 415 psia. At this new PC, the condenser temperature TC must then be re-calculated by a dew point temperature calculation. In summary, if 30 psia , PC , 215 psia, a total condenser is recommended which operates at the calculated PC. However, for values of PC outside of this range, the aforementioned adjustments should be made. Once a suitable condenser pressure has been calculated, the pressure at the top of the column may be assumed by adding a slight pressure drop across the condenser, denoted DPcond , P1 ¼ PC þ DPcond

(9:59)

where P1 ¼ the pressure at the top tray of the column. A reasonable value for DPcond would be approximately 2 psia. The next step is to approximate the reboiler pressure by assuming a reasonable number of trays and a corresponding pressure drop per tray. In practice, one may first assume 50 trays with a reasonable pressure drop such as 0.1 psi/tray. In this event, the pressure drop across the actual column would be 50 times 0.1 or 5 psi. This pressure drop between the reboiler and top tray may be denoted as DPcolumn. Thus, the reboiler pressure may be approximated by PR ¼ PC þ DPcond þ DPcolumn

(9:60)

This is obviously a gross approximation; however, it is a useful technique in obtaining preliminary values for quick-sizing the column. With an approximate reboiler pressure and known bottoms composition, the temperature of the reboiler, TR, may be calculated using a bubble point temperature calculation(2) analogous to Equation (9.55): N X

Ki xB,i ¼ 1:0

(9:61)

i¼1

Mean Relative Volatilities Before the Fenske equation is utilized, the relative volatilities of all components should be calculated based on the condenser and reboiler temperatures and pressures. When Raoult’s law is valid, a form of Equation (9.1) may be applied. However, in multicomponent distillation, all relative volatilities are generally calculated with respect to the HK, such that Ki p0 ai ¼ ¼ 0i (9:62) KHK pHK where i ¼ any component in the feed.

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Chapter 9 Distillation

Since the temperatures vary from the condenser to the reboiler, the relative volatilities may also vary slightly. Hence, a relative volatility for each component can be calculated at the condenser and reboiler temperature and pressure. In order to average the differences in relative volatility, a geometric mean relative volatility(13) may be calculated for each component, as given by pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (9:63) ai ¼ aD,i aB,i where

ai ¼ geometric mean relative volatility of component i aD,i ¼ relative volatility of component i in the distillate at TC and PC aB,i ¼ relative volatility of component i in the bottoms at TR and PR

The Fenske Equation The Fenske equation is used to calculate the minimum number of theoretical stages when the column is being operated under total reflux. While many forms of the Fenske equation have been presented in the literature, Equation (9.64) is preferred because it is highly useful from a design perspective,    rLK rHK ln 1  rLK 1  rHK (9:64) Nmin ¼ ln aLK where Nmin ¼ minimum number of theoretical stages (including partial reboiler) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi aLK ¼ aD,LK aB,LK , a geometric mean of light key relative volatilities Note that Nmin may also be considered to include a partial condenser if the design engineer decides to count a partial condenser as a theoretical stage. This will be the case for all subsequent discussion when referring to the number of theoretical stages, since “stages” does not refer explicitly to trays. As previously mentioned, this form of the Fenske equation is based on the definitions of fractional recovery given in Equations (9.53) and (9.54). In the case of a binary mixture, Equation (9.64) may be employed where LK refers to the lighter component of the feed and HK refers to the heavier component of the feed. In the event that the non-keys of a multicomponent feed are assumed to be distributed (as when non-keys have boiling points similar to a key component), the Fenske equation takes on another important use: to determine the approximate recoveries of the non-keys. In order to determine the fractional recovery of non-key component i, Nmin is determined in the usual manner with Equation (9.64). Nmin is then substituted back into Equation (9.64), at which time all rLK terms may be replaced by the recovery of any non-key component i, ri, which is unknown. Similarly, the aLK term in the denominator is replaced with ai , which is calculated from Equation (9.63). Hence, the unknown ratio ri/(12 ri) may be determined and the fractional recovery (in the distillate) of non-key i may ultimately be calculated. This procedure is repeated for all non-key components in order to obtain estimates for their respective recoveries in the distillate stream.

Continuous Distillation with Reflux

169

The Underwood Equations While the Fenske equation calculates the minimum number of equilibrium stages required for separation at total reflux, Underwood developed equations that estimate the minimum reflux ratio. The Underwood equations are a set of two mathematical expressions which are generally solved sequentially (unless there are one or more components in between the light and heavy keys, in which case they should be solved simultaneously in order to determine the correct root). These equations are listed below: N X ai xF,i ¼ 1  q; a Q i¼1 i

Rmin ¼

(1 , Q , aLK )

N X ai xD,i 1 a Q i¼1 i

(9:65)

(9:66)

where q ¼ the q-factor, dependent upon the thermal condition of the feed Q ¼ root of the first Underwood equation (9.65) pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ai ¼ aD,i aB,i , mean relative volatility of the ith component Equation (9.65) should first be solved for the correct value of Q, which must lie between the relative volatility of the heavy key (1.0) and the relative volatility of the light key (aLK ). All other roots of Equation (9.65) are erroneous and have no physical interpretation. Once the correct value of Q has been determined, it is substituted into Equation (9.66) and Rmin is calculated directly. As previously mentioned, if the designer does not choose a sharp separation, and there are one or more components in between the keys, Equations (9.65) and (9.66) should be solved in a slightly different manner. Instead of being solved sequentially, Equations (9.65) and (9.66) must be solved simultaneously since there are at least two values of Q which both satisfy Equation (9.65) and are between unity and aLK . Solving simultaneously will ensure that the correct value of Q is calculated. The Gilliland Correlation Now that both the Fenske and Underwood equations have been utilized to determine Nmin and Rmin, respectively, the last step in the FUG procedure is to employ the Gilliland correlation in order to determine the number of theoretical trays. The Gilliland correlation is shown in Figure 9.20.(14) As is evidenced by Figure 9.20, one need only know the minimum reflux ratio and the operating reflux ratio in order to compute the abscissa of the correlation. The corresponding ordinate is then read from the plot. Since the minimum number of theoretical stages is known, the actual number of theoretical stages, Nt, may be calculated. However, it should be noted that the Gilliland correlation was derived for systems with nearly constant relative volatilities throughout the column.(12) Therefore, the Gilliland correlation may not be a suitable short-cut method for non-ideal systems, in which relative volatilities may vary drastically.

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Chapter 9 Distillation

Figure 9.20 Gilliland correlation.

While the Gilliland correlation itself is immensely useful for use as a shortcut method in column quick-sizing, it is inconvenient in that the graph must be read manually. In order to make the Gilliland correlation more applicable to computer programming, several analytical expressions have been developed. One such outstanding correlation, as suggested by Chang,(15) is provided in Equation (9.67) below:   1:805 (9:67) Y ¼ 1  exp 1:490 þ 0:315X  0:1 X where X is the abscissa (x-axis) of this correlation, given by R  Rmin Rþ1

X¼

(9:68)

and Y is the ordinate ( y-axis) of the correlation, Y¼

Nt  Nmin Nt þ 1

(9:69)

Nmin þ Y 1Y

(9:70)

which may be rearranged to: Nt ¼

where Nt ¼ the number of theoretical stages.

Continuous Distillation with Reflux

171

Thus, Gilliland’s graphical correlation may be replaced by using Chang’s convenient mathematical expression. The calculation of the number of theoretical trays is the last step in the FUG procedure. However, the sizing of distillation columns does not end there. To the contrary, as discussed earlier, there are several other considerations which must be taken into account in order to completely quick-size a distillation column, including: theoretical location of the feed tray, calculation of the actual number of trays, calculation of column diameter, calculation of column height, etc. While the last two considerations have already been discussed in detail, the remaining topics will be further developed in the analysis to follow. The Kirkbride Equation The Kirkbride equation may be employed to determine the location of the theoretical feed tray. The Kirkbride equation is given by Equation (9.71) below, "    #0:206 NR B xF,HK xB,LK 2 ¼ NS D xF,LK xD,HK

(9:71)

where NR ¼ number of theoretical stages above the feed tray and NS ¼ number of theoretical stages below the feed tray. Equation (9.71) does not directly give the feed tray location, but rather provides a ratio of theoretical stages above and below the feed tray. In order to find the feed tray location, Equation (9.71) must be solved simultaneously with NR þ NS ¼ Nt  1

(9:72)

where Nt ¼ the number of theoretical stages. Note that the right-hand side of Equation (9.72) contains a minus one in order to account for the feed tray. In practice, Equation (9.72) is solved for either NR or NS and the result is then inserted into the Kirkbride equation. All known values are substituted and the unknown (either NR or NS) is calculated. If on solving the Kirkbride equation for NR one obtains the value n, then there are n theoretical stages above the feed tray in the rectification section. Hence, the theoretical feed tray location would be tray n þ 1 from the top of the column if a total condenser is employed, or tray n from the top of the column if a partial condenser is employed and considered a theoretical stage. The O’Connell Correlation As noted above, three major types of tray efficiency exist: the overall efficiency which is a single value pertaining to the entire column, the Murphee efficiency which pertains to a single tray, and the local efficiency which provides an efficiency at a particular location on a single tray. Equation (9.47) demonstrates how one calculates a column’s actual number of trays given the overall efficiency, E0. The O’Connell correlation, as shown in Figure 9.21,(16) may be used to obtain a column’s overall efficiency, given

172

Chapter 9 Distillation

a

Figure 9.21 Lockhart and Leggett version of the O’Connell correlation for overall tray efficiency. (Adapted from F. J. Lockhart and C. W. Leggett, Advances in Petroleum Chemistry and Refining, Vol. 1, K. A. Kobe and John J. McKetta, Jr., (eds), Interscience Publishers, Inc., New York City, NY, 1958, 323–326.)

the product of two pertinent quantities: the feed’s average viscosity and the average relative volatility of the LK. Note that, the O’Connell correlation applies equally well to a binary mixture where the LK is simply the light component. The viscosity of a liquid or gas is strongly dependent upon temperature. Therefore, if the feed temperature is not specified, an approximate temperature for the entering feed may be calculated by taking the arithmetic mean of the condenser and reboiler temperatures T feed 

TC þ T R 2

(9:73)

While calculating the actual average viscosity of a multicomponent mixture is best left to a thermodynamicist, taking a mole fraction weighted-average of each component’s viscosity can be used as a bare-minimum approximation for quick-sizing purposes:

m feed 

N X

mi (T feed )  xF,i

(9:74)

i¼1

Note that approximation may be subject to large error in some feed mixtures. Once a suitable feed viscosity has been calculated, the abscissa of the O’Connell correlation is found via

m feed aLK

(9:75)

Lastly, the overall efficiency is read off of the ordinate. Ntrays,ideal is found by calculating either Nt – 1, which subtracts out the partial reboiler, or Nt – 2, if both

Continuous Distillation with Reflux

173

a partial reboiler and partial condenser are considered theoretical stages. Equation (9.47) is then readily employed in order to calculate Ntrays, the actual number of trays in the multicomponent distillation column.

ILLUSTRATIVE EXAMPLE 9.10 Calculate the condenser pressure in a multicomponent distillation column intended to perform a sharp separation of n-pentane and i-octane in a saturated liquid feed, described in Table 9.9. Pertinent information is found below: Table 9.9 Feed Flows Fi, lbmol/h

Component (#) n-Butane (1) LK n-Pentane (2) HK i-Octane (3) n-Nonane (4) n-Decane (5)

50 200 150 50 50

Total

† † † †

† † †

500

rLK ¼ 0.99 rHK ¼ 0.99 Cooling water is available at 908F The non-keys are undistributed such that all keys lighter than the LK appear exclusively in the distillate and those heavier than the HK appear exclusively in the bottoms The liquid density is approximately 58 lb/ft3 The column is operated at 75% of its flooding velocity Raoult’s law may be considered valid for calculating the relative volatilities: use the three coefficient Antoine equation to calculate vapor pressures. (Note that differences in the Antoine coefficients employed may lead to noticeable deviation from the answers presented here; be sure to check that the Antoine data utilized is as accurate as possible over the temperature range of interest.)

SOLUTION: Much of the information given above is extraneous, but will be utilized in subsequent Illustrative Examples. The recoveries of the keys are defined as follows: rLK ¼

DxD,LK FxF,LK

(9:53)

rHK ¼

BxB,HK FxF,HK

(9:54)

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Chapter 9 Distillation

When the 0.99 recoveries are employed and the non-keys are assumed undistributed, the resulting distillate and bottoms flows are given in Table 9.10.

Table 9.10 Column Flows Component (#)

Fi, lbmol/h

Di, lbmol/h

Bi, lbmol/h

50 200 150 50 50

50 198 1.5 0 0

0 2 148.5 50 50

500

249.5

250.5

n-Butane (1) LK n-Pentane (2) HK i-Octane (3) n-Nonane (4) n-Decane (5) Total

Component mole fractions need to be calculated in order to be used in subsequent distillation equations. These results are given in Table 9.11. Table 9.11 Mole Fractions Component (#) n-Butane (1) LK n-Pentane (2) HK i-Octane (3) n-Nonane (4) n-Decane (5)

xF,i

xD,i

xB,i

0.1000 0.4000 0.3000 0.1000 0.1000

0.2004 0.7936 0.0060 0.0000 0.0000

0.0000 0.0080 0.5928 0.1996 0.1996

First, one should note that the cooling water is available at 908F. By assuming a 208F approach, the exit distillate temperature is set to 1108F. See Figure 9.22 for the condenser schematic.

Figure 9.22 Condenser schematic.

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Continuous Distillation with Reflux

The condenser temperature is therefore set to 1108F (43.338C). Since the temperature and composition of the liquid distillate are known (which is specified to be a saturated liquid), one may perform a bubble point pressure (BPP) calculation in order to determine the condenser pressure: N X Ki xD,i ¼ 1:0 (9:55) i¼1

Since Raoult’s law is assumed applicable, the K values for each species may be approximated by the ratio of that species’ vapor pressure to the total pressure. When the Antoine equation is used to calculate vapor pressures, one may employ: N X

10 ai [bi =(ci þTC )] xD,i ¼ PC

i¼1

(9:58)



PC ¼ 1389:48 mm Hg

 14:7 psi ¼ 26:9 psia 760 mm Hg

Since this pressure is below 30 psia, the condenser pressure is reset at 30 psia (this is done in order to avoid a too low of a column pressure that would result in a larger vapor volumetric flow rate, which in turn would lead to a larger and less economical column diameter). Now that the total condenser has a pressure of 30 psia, the new condenser temperature must be determined by a trial-and-error bubble point temperature (BPT) calculation, as described in Equation (9.53). This may be performed by hand or with a computer package. The final result is TC ¼ 47:08C at PC ¼ 30 psia

B

ILLUSTRATIVE EXAMPLE 9.11 Refer to Illustrative Example 9.10. Calculate the relative volatilities of each component at the condenser temperature. SOLUTION: When Raoult’s law is valid, the relative volatility of a component may be expressed in terms of vapor pressures:

ai ¼

Ki p0 ¼ 0i KHK pHK

(9:62)

The relative volatilities of each species at the condenser temperature are tabulated in Table 9.12. B

ILLUSTRATIVE EXAMPLE 9.12 Refer to Illustrative Examples 9.10 –9.11. Calculate the reboiler temperature and pressure. SOLUTION: First, it is assumed that the pressure drop across the condenser is 2 psi. For initial design purposes, it may be assumed that approximately 50 trays will be employed in

176

Chapter 9 Distillation

the column, each having a pressure drop of 0.1 psi. The pressure drop across the column (from the reboiler to condenser) is calculated as in Equation (9.60): PR ¼ PC þ 2 psi þ ð0:1 psi=trayÞð50 traysÞ ¼ 30 þ 2 þ 5 ¼ 37 psia which equates to 1912.93 mm Hg. Last, the reboiler temperature is calculated by observing that the liquid in the reboiler is a saturated liquid at its bubble point. Therefore, a BPT calculation is performed to find TR, assuming Raoult’s law. To perform this calculation, Equation (9.58) can be modified for the reboiler, as shown below: N X

10 ai [bi =(ci þTR )] xB,i ¼ PR

(9:76)

i¼1

Upon iteratively solving for TR with the appropriate Antoine coefficients: TR ¼ 165:58C at PR ¼ 37 psia

B

ILLUSTRATIVE EXAMPLE 9.13 Refer to Illustrative Examples 9.10 –9.12. Calculate the relative volatilities of each component at the reboiler temperature. Then, use these results coupled with those of Illustrative Example 9.11 to calculate the geometric mean relative volatilities of each component. SOLUTION: The relative volatilities are now calculated for each component at TR. Once again, these relative volatilities are based on the HK:

ai ¼

Ki p0 ¼ 0i KHK pHK

(9:56)

The distillate, bottoms, and geometric mean relative volatilities are tabulated in Table 9.12. B

Table 9.12 Distillate, Bottoms, and Mean Relative Volatilities Component (#) n-Butane (1) LK n-Pentane (2) HK i-Octane (3) n-Nonane (4) n-Decane (5)

aD,i

aB,i

ai

55.95 17.78 1.00 0.25 0.09

12.80 6.08 1.00 0.45 0.25

26.77 10.40 1.00 0.34 0.15

Continuous Distillation with Reflux

177

ILLUSTRATIVE EXAMPLE 9.14 Refer to Illustrative Examples 9.10 –9.13. Calculate the minimum number of trays. SOLUTION:

Employ the Fenske equation as provided by    rLK rHK ln 1  rLK 1  rHK Nmin ¼ ln aLK

Upon substitution, Nmin

(9:64)

   0:99 0:99 ln 0:01 0:01 ¼ ln(10:40) ¼ 3:9251

Thus, if rounded, the minimum number of theoretical stages is four (three plus a partial reboiler). Note that as a rule of thumb in distillation design, any decimal answer regarding the number of trays or stages is usually rounded up to the nearest whole number, never down. However, with the exception of the Kirkbride equation, unrounded values are employed in these calculations until the actual number of trays (Ntrays) is determined. This is because Ntrays has a physical interpretation as the actual number of trays needed and is, of necessity, an integer. Employing rounded values in previous calculations (e.g., Gilliland and O’Connell correlations) may propagate error in the final result. B

ILLUSTRATIVE EXAMPLE 9.15 Refer to Illustrative Examples 9.10 –9.14. Calculate the minimum reflux ratio and optimum reflux ratio. SOLUTION: Employ the Underwood equations, as indicated in the text. Noting that, for a saturated liquid feed, q ¼ 1, N X ai xF,i ¼ 1  q; (1 , Q , aLK ) a Q i¼1 i

(9:65)

One finds the root, Q, Q ¼ 1:5572 Substituting into Rmin ¼

N X ai xD,i 1 a Q i¼1 i

gives Rmin ¼ 0:1354

(9:66)

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Chapter 9 Distillation

Referring to Table 9.7, since cooling water is used in the condenser, the optimization multiplier is 1.3. Hence, the optimum reflux ratio is R ¼ 1.3Rmin. R ¼ 0:1760

B

ILLUSTRATIVE EXAMPLE 9.16 Refer to Illustrative Examples 9.10 –9.15. Calculate the number of theoretical stages necessary to affect the desired separation. SOLUTION: Using the Chang equation in order to calculate the proper value of the Gilliland correlation abscissa: R  Rmin Rþ1 0:1760  0:1354 X¼ 0:1760 þ 1

X¼

(9:68)

¼ 0:03454 The ordinate is given by   1:805 Y ¼ 1  exp 1:490 þ 0:315X  0:1 X

(9:67)

for which Y ¼ 0:6417 and thus Nmin þ Y 1Y 3:2951 þ 0:6417 Nt ¼ 1  0:6417 Nt ¼

(9:70)

¼ 12:7457 Therefore, when rounded, this column requires 13 theoretical stages (12 theoretical trays plus a reboiler). B

ILLUSTRATIVE EXAMPLE 9.17 Refer to Illustrative Examples 9.10 –9.16. Determine the theoretical feed tray location for this column.

Continuous Distillation with Reflux SOLUTION:

179

Employing the Kirkbride equation [Equation (9.71)] and substituting, "    #0:206 NR B xF;HK xB;LK 2 ¼ D xF;LK xD;HK NS "    #0:206 250:5 0:30 0:0080 2 ¼ 249:5 0:40 0:0060

(9:71)

¼ 1:062 Utilizing the complimentary equation NR þ NS ¼ Nt  1;

Nt ¼ 13

(9:72)

Rearranging and substituting Equation (9.72) into Equation (9.71), NR NR ¼ ¼ 1:062 NS Nt  NR  1 Solving for NR, NR ¼ 6:18 ! 7 theoretical stages Calculating NS, NS ¼ Nt  NR  1 ¼ 13  7  1 ¼ 5 theoretical stages Employing integer values for the number of stages, there should be seven theoretical stages above the feed tray (NR) and five theoretical stages below the feed tray (NS). Since a total condenser is employed, the feed should be admitted on the eighth tray from the top of the column. B

ILLUSTRATIVE EXAMPLE 9.18 Refer to Illustrative Examples 9.10–9.17. Determine the actual number of trays via the O’Connell correlation. SOLUTION: Using Equation (9.73), Tfeed is approximately 106.728C. Assuming that the average viscosity of the feed at this temperature is taken to be approximately 0.17 centipoise (this approximation is subject to improvement at the reader’s discretion), and employing the result given in Table 9.12 gives

m feed aLK ¼ (0:17)(10:40) ¼ 1:7673

(9:75)

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Chapter 9 Distillation

Reading the corresponding ordinate off of Figure 9.21, the fractional efficiency (ordinate) is approximately 48%. The actual number of trays is calculated by dividing the theoretical number of trays by 0.48: Ntrays ¼

Ntrays,ideal E0

Ntrays ¼

12:7457  1 0:48

(9:47)

¼ 24:47 Thus, the final result is that this column requires approximately 25 trays plus a partial reboiler in order to produce the desired separation. B

ILLUSTRATIVE EXAMPLE 9.19 Refer to Illustrative Examples 9.10–9.18. Calculate the flooding velocity, vapor velocity and column diameter. Assume that the trays to be employed have downcorners which each occupy 10% of the cross sectional area. SOLUTION: Note that the details of the more lengthy calculations found in this illustrative example are left as an exercise for the reader.

Figure 9.23 Fair flooding correlation for crossflow trays.

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181

The Fair entrainment flooding correlation (see Fig. 9.23)(17) is one of several flooding correlations that can be used to calculate the diameter of a sieve tray distillation column. The abscissa of this correlation of the flow parameter, Xflood (dimensionless) is

FLV (flooding) ¼ X flood ¼

  L(MWL ) rV 0:5 V(MWV ) rL

(9:77)

where L ¼ liquid molar flow rate (in the section where flooding first occurs), V ¼ vapor molar flow rate (in the section where flooding first occurs), MWL ¼ average molecular weight of liquid (where flooding first occurs), and MWV ¼ average molecular weight of vapor (where flooding first occurs). Since the feed in this example is a saturated liquid, it may be assumed (due to larger flow rates) that flooding will most likely first occur in the stripping section of the column. Hence L and V must be calculated. The average molecular weights may be calculated based on the composition of the liquid and vapor leaving the reboiler. In reality, the liquid composition should be that of the liquid leaving the last tray, which is where flooding may first occur; however, using the bottoms composition as an approximation does not normally lead to significant error. The composition of the boil-up vapor leaving the reboiler is determined by a vapor–liquid equilibrium calculation, with the liquid composition, reboiler temperature, and reboiler pressure all known ( yreb,i ¼ KixB,i). The average molecular weight of the liquid and vapor at the bottom of the column is taken as a mole fraction weighted average MWV ¼ 115:02, MWL ¼ 122:28 The density of the liquid is given as 58 lb/ft3, while that of the vapor may be calculated using the ideal gas law:

rV ¼ (MWV )

PR RTR

¼ (115:02)

(9:78)

(1912:93) (998:970)(165:5 þ 273)

rV ¼ 0:502 lb=ft3 Note that the reboiler temperature here must be expressed in absolute temperature units.(2) The liquid and vapor molar flow rates in the rectification section may be calculated via material balances (assuming constant molal overflow): L ¼ (R)(249:5) þ 500 ¼ 543:9 lbmol=h V ¼ (249:5)(R þ 1) ¼ 293:4 lbmol=h

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Chapter 9 Distillation

The flooding parameter may now be calculated by using Equation (9.77): XFlood ¼

  L(MWL ) rV 0:5 V(MWV ) rL

¼

  (543:9)(122:28) 0:502 0:5 (293:4)(115:02) 58

XFlood ¼ 0:1833 This value is now used in order to read the ordinate of the correlation, CF, which is in ft/s. In order to read the ordinate, a value for tray spacing must be assumed. Initially, choose a 24 inch tray spacing. Treybal(6) presents values of tray spacing per a given range of column diameters, as shown in Table 9.13. Once the column diameter is calculated, it will be compared to the diameter range recommended for 24 inch tray spacing. Table 9.13 Recommended Tray Spacing Column diameter, ft

Tray spacing, inches 20 24 30 36

4 4–10 10–12 12 

Individual practice may call for tray spacing different than those here described.

From the Fair flooding correlation, with a tray spacing of 24 inches: CF ¼ 0:30 ft=s This value is now used to calculate the vapor flooding velocity uF via Equation (9.79):  uF ¼ CF

rL  rV rV

0:5   s 0:2 20

(9:79)

where s ¼ surface tension correction factor (dyne/cm). Hydrocarbon mixtures often have surface tensions of approximately 20 dyne/cm. Therefore, the surface tension correction factor term may be neglected, leading to:

uF ¼ (0:30)

  58  0:502 0:5 0:502

¼ 3:21 ft=s

Continuous Distillation with Reflux

183

Once the flooding velocity is calculated, the actual velocity is determined by multiplying it by the approach to flooding. In this case, use 0.75. u ¼ (0:75)(3:21 ft=s) ¼ 2:41 ft=s Note that this vapor velocity is based on the aforementioned net area of a tray. The column diameter is now calculated based on the vapor velocity, molar flow rate, and tray geometry:  0:5 2 q Dcolumn ¼ pffiffiffiffi p (1  h)u   2 (MWV )V 0:5 ¼ pffiffiffiffi p rV (1  h)u  0:5 2 (115:02)(293:4) ¼ pffiffiffiffi p (3600)(0:502)(1  0:10)(2:41)

(9:52)

¼ 3:31 ft Therefore, the diameter is 3.31 ft in this particular case. Referring to Table 9.13, this diameter is generally considered too small for a tray spacing of 24 inches. Therefore, the calculation is repeated at a tray spacing of 18 inches on the Fair flooding correlation. A new value of CF is determined and a new vapor velocity calculated. The final result for the new column diameter is calculated as 3.70 ft, which is an acceptable diameter for tray spacing of 18 inches. B

ILLUSTRATIVE EXAMPLE 9.20 Refer to Illustrative Examples 9.10–9.19. Calculate the column height, taking into account a surge volume that will allow for five minutes of “extra” operation time plus a 5 ft height safety factor. SOLUTION: These calculations are relatively self-explanatory. Notice that a surge volume is calculated for safety in operation: the factor of (60/5) is to account for a surge volume at the bottom of the column which will take 5 minutes to fill should the bottoms pump fail. Also, an additional 5 ft is added to the column height, which is roughly a 15% height safety factor; this extra height may be used to install more trays for better operation. The three contributions to the total height may now be calculated:    in ft htrays ¼ 18 (25 trays) ¼ 37:5 ft tray 12 in The surge volume is: Vsurge ¼

(MWL )L ¼ 95:56 ft3 (60=5)rL

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Chapter 9 Distillation

Hence, surge height is: hsurge ¼

Vsurge ¼ 8:9 ft (pD2column =4)

Finally, hcolumn ¼ htrays þ hsurge þ 5 ft ¼ 37:5 þ 8:9 þ 5:0 ¼ 51:4 ft The column height should be approximately 52 ft.

B

Packed Column Distillation Packed-bed distillation towers, generally referred to as packed columns, are employed commercially in small-scale applications where low throughput and small column diameter make them cost competitive with tray towers. Packed columns generally have a smaller pressure drop, which also decreases operational costs. The calculation of column diameter and pressure drop across packing are calculated in a similar manner to the Fair flooding correlation. Details are available in the literature,(6,7) but information is provided in the next chapter. The operation of a packed column is similar to that of a trayed column. Indeed, by examining a column from the outside, they may seem virtually identical. However, while trayed columns have discrete stages for separation, the packing in a packed column creates a continuous surface for mass transfer. In designing packed columns, the column height is based on the number of theoretical plates, and the height equivalent to a theoretical plate (HETP). Let the height of a packed zone be denoted by h. HETP is then defined as follows: HETP ¼

h Np

(9:80)

where Np ¼ the number of theoretical stages achieved in the packed zone. Hence, the HETP may be physically interpreted as the height of packing that produces a liquid and vapor stream (exiting this height of packing from the bottom and top, respectively), which are in thermodynamic equilibrium as the liquid and vapor streams would be if they contacted on a theoretical plate. Other approaches have been developed in order to determine the height of a packed column, namely that of the transfer unit, the details of which are available in the literature.(6) The two main classes of packing are random dumped packing or structured packing. There are packings of various shapes and sizes: some of the more typical dumped packing includes Raschig rings, pall rings, Berl saddles and Intalox saddles.(6) The packing (be they structured or dumped) serves the same purpose as trays do in a trayed column, that is, to provide a “widget” for the vapor and liquid to experience more intimate contact and better mass transfer.

References

185

As liquid flows through dumped packing, it may preferentially flow in certain channels towards the wall of the column. As a result, liquid distribution is usually degraded. In order to ensure that liquid is spread evenly over the packing, liquid redistributors can be placed intermittently in the column. When the liquid distribution is well maintained, most of the packing tends to wet, creating a thin film of liquid on the surface of the packing. As the rising vapors contact the liquid film, mass transfer occurs. For a 1.5 or 2 inch size of a common packing, the HETP is generally in the range of 1 to 2 ft. Smaller packing generally gives rise to a smaller HETP. However, it has been noted that structured packing produces a somewhat better separation than random dumped packing.(12) More information regarding the design and operation of packed columns is available in the literature(3,6,12,18) as well as the next chapter.

REFERENCES 1. P. MARNELL, adopted from class notes (with permission), 2009. 2. L. THEODORE, F. RICCI, and T. VAN VLIET, “Thermodynamics for the Practicing Engineer,” John Wiley & Sons, Hoboken, NJ, 2009. 3. B. D. SMITH, “Design of Equilibrium Stage Processes,” McGraw-Hill, New York City, NY, 1963. 4. L. THEODORE and J. BARDEN, “Mass Transfer Operations,” Theodore Tutorials, East Williston, NY, 1995. 5. J. COATES and B. PRESSBURG, “Analyze Material and Heat Balances for Continuous Distillation,” Chem. Eng., New York City, NY, 2/20/61. 6. R. TREYBAL, “Mass Transfer Operations,” 3rd edition, McGraw-Hill, New York City, NY, 1980. 7. T. R. FAIR, “Distillation,” chapter 5, in R. W. ROUSSEAU (ed),“ Handbook of Separation Process Technology,” John Wiley & Sons, Hoboken, NJ, 1987. 8. L. THEODORE, personal notes, 2009. 9. J. M. SMITH, H. C. VAN NESS, and M. M. ABBOTT, “Introduction to Chemical Engineering Thermodynamics,” 7th edition, McGraw-Hill, New York City, NY, 2005. 10. W. MCCABE and E. THIELE, Ind. Eng. Chem., New York City, NY, 17, 605, 1925. 11. Generalized Pressure Drop Correlation, Chart No. GR-109, Rev 4, U.S. Stoneware Co., Akron, OH, 1963. 12. W. L. MCCABE, J. C. SMITH, and P. HARRIOT, “Unit Operations of Chemical Engineering,” 7th edition, McGraw-Hill, New York City, NY, 2005. 13. S. SHAEFER and L. THEODORE, “Probability and Statistics Applications in Environmental Science,” CRC Press/Taylor & Francis Group, Boca Raton, FL, 2007. 14. E. R. GILLILAND, Ind. Eng. Chem., New York City, NY, 32, 1220, 1940. 15. H. Y. CHANG, Hydrocarbon Processing, New York City, NY, 64(3), 48, 1985. 16. H. E. O’CONNELL, Trans. AIChE, New York City, NY, 42, 741, 1946. 17. J. R. FAIR, Petro/Chem. Eng., New York City, NY, 33(10), 45, 1961. 18. L. THEODORE, “Air Pollution Control Equipment Calculations,” John Wiley & Sons, Hoboken, NJ, 2008.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

10

Absorption and Stripping INTRODUCTION The removal of one or more selected components from a gas mixture by absorption is an important operation in engineering. The process of absorption conventionally refers to the intimate contacting of a mixture of gases with a liquid so that part of one or more of the constituents of the gas will dissolve in the liquid. The contact usually takes place in some type of packed or plate column. This chapter will therefore deal exclusively with packed or plate equipment. Only equipment and design procedures are emphasized, as a detailed presentation of the theory, including diffusional process, mass transfer coefficients, equilibrium (lines), operating lines, etc., has already been covered in Chapters 6 – 8. Since gas absorption is concerned with the removal of one or more species from a gas stream by treatment with a liquid, necessary information includes the solubility of these constituents in the absorbing liquid. In gas absorption operations, the equilibrium of interest is that between a nonvolatile absorbing liquid (solvent) and a solute gas. The solute is ordinarily removed from its mixture in a relatively large amount of a carrier gas that does not dissolve in the absorbing liquid. Temperature, pressure, and the concentration of solute in one phase are independently variable. The equilibrium relationship of importance, again, is a plot of x, the mole fraction of solute in the liquid, against y (or y ), the mole fraction in the vapor in equilibrium with x. Thus, for cases which follow Henry’s law (see Chapter 6), Henry’s law constant, m, can be defined by y ¼ mx

(10:1)

or the equivalent y ¼ mx . The engineering design of gas absorption equipment must be based on a sound application of the principles of diffusion, equilibrium, and mass transfer. The main requirement in equipment design is to bring the gas into intimate contact with the liquid, i.e., to provide a large interfacial area and a high intensity of interface renewal, and to minimize resistance and maximize driving force. This contacting of the phases can be achieved in many different types of equipment, the most important of which are Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

187

188

Chapter 10 Absorption and Stripping

either packed or plate columns. The final choice often rests with the various criteria that may have to be met. For example, if the pressure drop through the column is large enough such that horsepower costs become significant, a packed column may be preferable to a plate-type column because of the lower pressure drop. Again, primary emphasis in this section is placed on packed and plate columns. In most processes involving the absorption of gaseous constituents from a gas stream, the gas stream is the process fluid; hence, its inlet conditions (flow rate, composition, and temperature) are usually known. The temperature and composition of the inlet liquid and the composition of the outlet gas are also usually specified. The main objectives, then, in the design of an absorption column, are the determination of the solvent (liquid) flow rate and the calculation of the principal dimensions of the equipment (column diameter and height). These three topics are reviewed sequentially later in this chapter. The general design procedure consists of a number of steps that have to be taken into consideration (details of which follow shortly). These include(1,2): 1 Solvent selection 2 Equilibrium data evaluation 3 Estimation of operating data (usually obtained from a mass and energy balance, where the energy balance determines whether the absorption process can be considered isothermal or adiabatic) 4 Column selection (should the column selection not be obvious or specified, calculations must be carried out for the different types of columns, and the final selection based on economic considerations) 5 Calculation of column diameter (for packed columns this is usually based on flooding conditions, and for plate columns is based on the optimum gas velocity or the liquid handling capacity of the plate) 6 Estimation of the column height or the number of plates (for a packed column, the column height is obtained by multiplying the number of transfer units, obtained from a knowledge of equilibrium and operating data by the height of a transfer unit; for plate columns, the number of theoretical plates, often determined from the plot of equilibrium and operating lines, is divided by the estimated overall efficiency to give the number of actual plates, which in turn allows the column height to be estimated from the plate spacing) 7 Determination of pressure drop through the column (for packed columns, correlations dependent on packing type, column operating data, and physical properties of the constituents involved need to be available to estimate the pressure drop through the packing; for plate columns, the pressure drop per plate is obtained and multiplied by the number of plates) Although detailed absorber calculations will be provided, a brief introduction follows. The usual operating data to be provided or estimated are the flow rates, terminal concentration, and terminal temperature of the phases. The flow rates and concentrations fix the operating line, while the terminal temperatures provide an indication as to what extent the operation can be considered isothermal (i.e., whether the

Description of Equipment

189

equilibrium line needs to be corrected for changes in liquid temperature). The operating line is obtained by a mass balance, and the outlet liquid temperature is evaluated from an energy balance on the column. In applications where relatively small quantities of gaseous constituents are being absorbed, temperature effects are usually negligible. In gas absorption operations, the choice of a particular solvent is also important. Frequently, water is used as it is inexpensive and plentiful, but the following properties must also be considered. 1 Gas solubility—a high gas solubility is desired since this increases the absorption rate and minimizes the quantity of solvent necessary; generally, a solvent of a chemical nature similar to that of the solute to be absorbed will provide good solubility 2 Volatility—a low solvent vapor pressure is desired since the gas leaving an absorption unit is ordinarily saturated with the solvent and much will therefore be lost 3 Corrosiveness 4 Cost (particularly for solvents other than water) 5 Viscosity—low viscosity is preferred for reasons of rapid absorption rates, improved flooding characteristics, lower pressure drops, and good heat transfer characteristics 6 Chemical stability—the solvent should be chemically stable and, if possible, nonflammable 7 Toxicity 8 Low freezing point—if possible, a low freezing point is favored since any solidification of the solvent in the column could prove disastrous Once the solvent is specified, the choice (and design) of the absorption system may be determined.

DESCRIPTION OF EQUIPMENT The principal types of gas absorption equipment may be classified as follows: 1 Packed columns (continuous operation) 2 Plate columns (stage operation) 3 Miscellaneous Of the three categories, the packed column is most commonly used. Note that plate columns received treatment in the previous chapter (Distillation).

Packed Columns Packed columns are usually vertical columns that have been filled with packing or material of large surface area. The liquid is distributed over and trickles down through the packed bed, thus exposing a large surface area to contact the gas. The

190

Chapter 10 Absorption and Stripping

countercurrent packed column (see Fig. 10.1) is the most common unit encountered in gaseous removal or recovery. A photograph of this unit in the Unit Operations Laboratory at Manhattan College is provided in Figure 10.2. The gas stream moves upward through the packed bed against an absorbing or reacting liquor (solvent-scrubbing solution), which is introduced at the top of the packing. This results in the highest possible efficiency. Since the solute concentration in the gas stream decreases as it rises through the column, there is fresh solvent constantly available for contact. This provides the maximum average driving force for the mass transfer process throughout the packed bed. Mist eliminators also play an important role in absorbers. Mist eliminators are used to remove liquid droplets entrained in the gas stream. Ease of separation depends Gas outlet

Liquid inlet

Entrainment separator (demister)

Liquid distributor Packing restrainer

Shell Random packing

Access manway for packing removal Liquid re-distributor

Access manway for packing removal Packing support Gas inlet Overflow Liquid outlet

Figure 10.1 Typical countercurrent packed column.

Description of Equipment

Figure 10.2

191

Absorption column.

on the size of the droplets. Droplets formed from liquids are usually large—up to hundreds of microns in diameter. However, drops formed in condensation or chemical reactions may be less than one micron in size. Entrainment removal (mist separation) is possible by a number of methods including the following: 1 Knitted wire or plastic mesh 2 Swirl vanes or zigzag vanes 3 Cyclones 4 Gravity settling chambers 5 Units in which the gas is forced to make a 1808 turn 6 Additional packing above the packed bed One of the simplest and most efficient means of mist separation is to use a porous blanket of knitted wire or plastic mesh. For most processes, the pressure drop across these mist eliminators range from 0.1 to 1.0 inches of water, depending on vapor and liquid

192

Chapter 10 Absorption and Stripping

flowrates and the size of the eliminators. The efficiency of separation is generally highusually 90% or better. The packing is the heart of this type of equipment. Its proper selection entails an understanding of packing operational characteristics and the effect of performance of the points of significant physical difference between the various types. The main points to be considered in choosing the column packing include: 1 Durability and corrosion resistance (the packing should be chemically inert to the fluids being processed) 2 Free space per unit volume of packed space (this controls the liquor holdup in the column as well as the pressure drop across it; ordinarily, the fractional void volume, or fraction of free space, in the packed bed should be large) 3 Wetted surface area per unit volume of packed space. (This is very important since it determines the interfacial surface between liquid and gas; it is rarely equal to the actual geometric surface since the packing is usually not completely wetted by the fluid.) 4 Resistance to the flow of gas (this effects the pressure drop over the column) 5 Packing stability and structural strength to permit easy handling and installation 6 Weight per unit volume of packed space 7 Cost per unit area of packed space Table 10.1 illustrates some of the various types and applications of the different column packings available. One additional distinction should also be made: the difference between random and stacked (structured) packings. Random packings are those that are simply dumped into the column during installation and allowed to fall at random. It is the most common method of packing installation. During installation prior to pouring the packing into the column, the column may first be filled with water. This prevents breakage of the more fragile packing by reducing the velocity of the fall. The fall should be as gentle as possible since broken packing tightens the bed and increases the pressure drop. Stacked packing, on the other hand, is specially laid out and stacked by hand, making it a tedious operation and rather costly; it is avoided where possible except for the initial layers on supports. Liquid distributed in this latter system usually flows straight down through the packing immediately adjacent to the point of contact. The aforementioned liquid distribution plays an important role in the efficient operation of the packed column. A good packing from a process viewpoint can be reduced in effectiveness by poor liquid distribution across the top of its upper surface. Poor distribution reduces the effective wetted packing area and promotes liquid channeling. The final selection of the mechanism of distributing the liquid across the packing depends on the size of the column, type of packing, tendency of the packing to divert liquid to column walls, and materials of construction for distribution. For stacked packing, the liquid usually has little tendency to cross distribute and thus moves down the column in the cross sectional area that it enters. In the dumped condition, most liquids follow a conical distribution down the column with the apex of the cone at the liquid impingement point. For uniform liquid flow and reduced channeling

Description of Equipment

193

Table 10.1 Some Typical Packings and Applications Packing

Application features

Raschig rings

Originally, the most popular type, usually cheaper per unit cost but sometimes less efficient than others; available in widest variety of materials to fit service: very sound structurally; usually packed by dumping wet or dry, with larger 4- to 6-in sizes sometimes handstacked; wall thickness varies between manufacturers; available surface changes with wall thickness; produce considerable side thrust on tower; usually has more internal liquid channeling and directs more liquid to walls of column.

Berl saddles

More efficient than Raschig rings in most applications, but more costly; packing nests together and creates “tight” spots in bed that promotes channeling but not as much as Raschig rings; do not produce much side thrust and have lower unit pressure drops with higher flooding points than Raschig rings; easier to break in bed than Raschig rings.

Intalox saddles

One of the most efficient packings, but more costly; very little tendency or ability to nest and block areas of bed; higher flooding limits and lower pressure drop than Raschig rings or Berl saddles; easier to break in bed than Raschig rings.

Pall rings

Lower pressure drop (less than half) than Raschig rings; higher flooding limit; good liquid distribution; high capacity; considerable side thrust on column wall; available in metal, plastic, and ceramic.

Spiral rings

Usually installed as stacked, taking advantage of internal whirl of gas –liquid and offering extra contact surface over Raschig rings.

(Continued)

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Chapter 10 Absorption and Stripping

TABLE 10.1 Continued Packing

Application features

Teller rosette (Tellerette)

Available in plastic; lower pressure drops; higher flooding limits than Raschig rings or Berl saddles; very low unit weight; low side thrust; relatively expensive.

Cross-partition rings

Usually stacked as first layers on support grids for smaller packing above; pressure drop relatively low; channeling reduced for comparative stacked packings; no side wall thrust.

Lessing rings

Not much performance data available, but in general slightly better than Raschig rings; pressure drop slightly higher; high side-wall thrust.

Ceramic balls

Tend to fluidize in certain operating ranges, self-cleaning, uniform bed structure, higher pressure drop, and better contact efficiency than Raschig rings; high side thrust; not much commercial data.

Goodloe packing and wire mesh packing

Available in metal only, used in large and small columns for distillation, absorption, scrubbing, liquid extraction; high efficiency; low pressure drop.

Description of Equipment

195

of gas and liquid, the introduction of the liquid onto the packed bed must be as uniform as possible. Any impingement of the liquid on the wall of the column should be redistributed after a bed depth of approximately three column diameters for Raschig rings and five to ten column diameters for saddle packings. As a guide, Raschig rings usually have a maximum 10– 15 ft of packing per section, while saddle packing can use a maximum of 12– 20 ft. As a general rule of thumb, however, the liquid should be redistributed every 10 ft of packed height. The redistribution brings the liquid off the wall and directs it toward the center area of the column for a new start of distribution and contact in the next lower section. Occasionally, cocurrent flow may be used where the gas stream and solvent both enter the top of the column. Initially, there is a very high rate of absorption that constantly decreases until, with an infinitely tall column, the gas and liquid leave in equilibrium and effectively operate as one theoretical stage. In this case, high gas and liquid rates are possible since the pressure drop tends to be rather low. However, these columns are efficient only when large driving forces are available (e.g., with very soluble gases or acid scrubbing in caustic media). The design for this case utilizes minimum column diameter because of the low pressure drop and nonflooding characteristics. Packed columns may also operate in a crossflow mode (see Fig. 10.3) where the air stream moves horizontally through the packed bed and is irrigated by the scrubbing liquid which flows vertically down through the packing. Crossflow designs are characterized by low water consumption and fairly high air flow capacity at a low pressure drop. Where highly soluble gases are to be recovered, the crossflow packed scrubber has several advantages over the countercurrent scrubber. For example, when operating with the same liquid and gas mass flow rates, a crossflow scrubber has a lower pressure drop. Besides reducing water consumption drastically, the crossflow principle also reduces pump and fan motor sizes. Other advantages include less piping, less plugging from solids dropout at the packing support plate, and the possible use of higher gas and Spray nozzles

Baffles Top spray header Packed bed

Liquid inlet

Outlet transition

Solute-laden gas “Treated” gas Packing support grids Inlet transition

Pump suction

Front spray header Drain

Baffles

Overflow

Figure 10.3

Cross flow operation in a packed column.

Sump

196

Chapter 10 Absorption and Stripping

liquid rates because of the extremely low pressure drop. On the other hand, liquid entrainment from these systems is rather high and mist eliminators are usually required downstream. Packed columns are characterized by a number of features to which their widespread popularity may be attributed. 1 Minimum structure—the packed column usually needs only a packing support and liquid distributor approximately every 10 feet along its height 2 Versatility—the packing material can be changed by simply discarding it and replacing it with a type providing better efficiency 3 Corrosive-fluids handling—ceramic packing is used and may be preferable to metal or plastic because of its corrosion resistance. When packing does deteriorate, it is quickly and easily replaced; it is also preferred when handling hot combustion gases 4 Low pressure drop—unless operated at very high liquid rates where the liquid becomes the continuous phase as the flowing films thicken and merge, the pressure drop per lineal foot of packed height is relatively low 5 Range of operation—although efficiency varies with gas and liquid feed rates, the range of operation is relatively broad 6 Low investment—when plastic packings are satisfactory or when the columns are less than about 3 or 4 feet in diameter, cost is relatively low

Plate Columns Plate columns (also commonly referred to as “tray columns”) are essentially vertical cylinders in which the liquid and gas are contacted in stepwise fashion (staged operation) on plates or “traps,” as shown schematically for one type in Figure 10.4. The liquid enters at the top and flows downward via gravity. On the way, it flows across each plate and through a downspout to the plate below. The gas passes upward through openings of one sort or another in the plate, then bubbles through the liquid to form a froth, disengages from the froth, and passes on to the next plate above. The overall effect is a multiple countercurrent contact of gas and liquid. Each plate of the column is a stage since the fluids on the plate are brought into intimate contact, interface diffusion occurs, and the fluids are separated. The number of theoretical plates (or stages) is dependent on the difficulty of the separation to be carried out and is determined solely from material balances and equilibrium considerations. The diameter of the column, on the other hand, depends on the quantities of liquid and gas flowing through the column per unit time. The actual number of plates required for a given separation is greater than the theoretical number because of plate inefficiencies. To achieve high plate efficiencies, the contact time between the gas and liquid on each plate should be high so as to permit mass transfer to occur, the interfacial surface between phases must be as large as possible, and a relatively high degree of turbulence is required to obtain high mass transfer coefficients. In order to increase contact time, the liquid pool height on each plate should be deep so that the gas bubbles will require

Description of Equipment

197

Gas out

Mist eliminator Shell

Liquid in

Tray Downspout

Bubble-cap

Tray support ring Tray stiffener Vapor riser Froth

Side stream withdrawal Intermediate feed

Gas path through cap Gas in

Liquid out

Figure 10.4

Typical bubble-cap plate column.

a relatively long time to rise through the liquid. On the other hand, great depths of liquid on the plates, although leading to high plate efficiencies, results in a higher pressure drop per plate. Relatively high gas velocities are also preferred for high plate efficiencies. This results in the gas being thoroughly dispersed into the liquid and causes froth formation, which provides large interfacial surface areas. High gas velocities, although providing good vapor – liquid contact, may lead to excessive entrainment accompanied by high pressure drop. Hence, the various arrangements and dimensions chosen for the design of plate columns are usually those which experience has proven to provide reasonably good compromises. Additional information on the design of plate columns can be found in Chapter 9 and in a later section. The particular plate selection and its design can materially affect the performance of a given absorption operation. Each plate should be designed so as to provide as efficient a contact between the vapor and liquid as possible, within reasonable economic limits. The principal types of plates encountered are discussed .

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Chapter 10 Absorption and Stripping

In bubble-cap plates (as discussed in the previous chapter), the vapor rises up through a “riser” into the bubble-cap, out through the slots as bubbles, and into the surrounding liquid on the plates. Figure 10.5 demonstrates the liquid–vapor action for a bubble-cap plate. The bubble-cap plate design was once the most favored of plate designs.

Straight downcomer Adjustable outlet weir Less tray action more due to hydraulic gradient Liquid gradient Liquid

Throw over weir, tw Recessed seal pan

Vapor with mist

Froth

Disengaging space Foam, froth, and bubbles

Clear liquid

No seal

Tapered downcomer

Inlet weir

Figure 10.5 Bubble-cap plate schematic-dynamic operation.

Description of Equipment

199

In sieve or perforated plates, the vapor rises through small holes (usually 18 to 1 inch in diameter) in the plate floor and bubbles through the liquid in a fairly uniform manner. The perforated plate is made with or (occasionally) without the downcomer. With the downcomer, the liquid flows across the plate floor and over a weir (if used), then through the downcomer to the plate below. Figure 10.6 shows the operation

Downcomer edge or inlet weir (if used) Holes on 60° triangular pitch Deflector baffles Hole-shell clearance 2 to 3 Downcomer weir

2 to 3 min

3 to 5 min Hole area Active hole area

Clear liquid

Vapor with mist

Top of froth

Figure 10.6

Sieve or perforated plate with downcomers.

tw

200

Chapter 10 Absorption and Stripping

schematically. These plates are generally not suitable for columns operating under variable load. Plate spacing in this case usually averages about 15 inches. At the same time that the vapor rises through the holes, the liquid head forces liquid countercurrent through these holes and onto the plate below. Perforated plates have become the preferred choice in recent years.

DESIGN AND PERFORMANCE EQUATIONS—PACKED COLUMNS Design and performance equations are provided for both packed and plate columns, and some overlap does exist. However, the emphasis is on packed columns since they are often the choice in absorption applications. On the other hand, plate columns, the preferred choice in distillation (see previous chapter), will receive attention in the next section. For most absorption applications, sufficient information is either provided or available to enable one to completely describe the system through simple yet standard calculational procedures. These calculations generally involve the determination of three unknown system variables: the liquid rate, the column height (and corresponding pressure drop), and the column diameter. Each of the topics is treated sequentially below.

Liquid Rate As described earlier, the equilibrium of interest in gas absorption is that between a relatively nonvolatile absorbing liquid (solvent) and a soluble gas (solute). For cases that follow Henry’s law (see Chapter 6), Henry’s law constant, m, is defined by y ¼ mx

(10:1)

The usual operating data to be determined or estimated for isothermal systems are the liquid rates and the terminal concentrations or mole fractions. An operating line, which describes operating conditions in a countercurrent flow column, is obtained by a mass (or a mole basis) balance around the column (as shown in Fig. 10.7). Note that the notation normally employed for the gas rate in absorption calculations is G, not V, as employed for the vapor rate in Chapter 9. The subscript m is often carried if the rate is based on moles. The overall mole balance is: Gm1 þ Lm2 ¼ Gm2 þ Lm1

(10:2)

For component A, the mass (or mole) balance becomes Gm1 yA1 þ Lm2 xA2 ¼ Gm2 yA2 þ Lm1 xA1

(10:3)

Design and Performance Equations—Packed Columns

201

Treated gas Gm2 yA2

Lm2

xA2

Lean solution

Feed gas Gm1 yA1

Rich solution Lm1

Figure 10.7

xA1

Mole balance; countercurrent flow.

Assuming Gm1 ¼ Gm2 ¼ Gm and Lm1 ¼ Lm2 ¼ Lm (reasonable for many applications where solute concentrations are reasonably small), then Gm yA1 þ Lm xA2 ¼ Gm yA2 þ Lm xA1 or, on rearrangement, Lm yA1  yA2 ¼ Gm xA1  xA2

(10:4)

This is the equation of a straight line known as the operating line. It has a slope of Lm/Gm on x, y coordinates and passes through the points (xA1, yA1) and (xA2, yA2) as indicated in Figure 10.8. In the design of most absorption columns, the quantity of gas to be treated, Gm, the concentrations, yA1 and yA2, and the composition of the entering liquid, xA2, are ordinarily fixed by process requirements. However, the quantity of liquid solvent to be used is subject to some choice. This can be resolved by setting or obtaining a minimum liquid– to – gas ratio. With reference to Figure 10.8, the operating line must pass through point A (top of column) and must terminate at the ordinate yA1. If such a quantity of liquid is used to

202

Chapter 10 Absorption and Stripping B

C

D

yA1

E

Slope = (Lm /Gm)act

yA

yA2

Equilibrium curve y * = f (x)

Slope = (Lm /Gm)min

A

xA2

xA1

xA1max

xA

Figure 10.8 Operating and equilibrium lines.

produce operating line AB, the exiting liquid will have the composition xA1. If less liquid is used, the exit liquid composition will clearly be greater, as at point C, but since the driving forces for mass transfer are less, the absorption is more difficult. The time of contact between the gas and liquid must then be greater and the absorber must be correspondingly taller. The minimum liquid that can be used corresponds to the operating line AD, which has the greatest slope for any line touching the equilibrium curve (tangent to the curve at E). At point E, the mass transfer driving force is zero, the required contact time for the concentration change desired is infinite, and an infinitely tall column results. This then represents the minimum liquid – to – gas ratio. The importance of the minimum liquid – to – gas ratio lies in the fact that column operation is frequently specified as some factor of the minimum liquid – to –gas ratio. For example, a typical situation frequently encountered is that the actual operating line, (Lm/Gm)act is 1.5(Lm/Gm)min.

ILLUSTRATIVE EXAMPLE 10.1 Experimental data are provided for an absorption system to be used for scrubbing ammonia (NH3) from air with water. The water rate is 300 lb/min and the gas rate is 250 lb/min at 728F. The applicable equilibrium data for the ammonia– air system is shown in Table 10.2. The air to be scrubbed has 1.5% (mass basis) NH3 at 728F and 1 atm pressure and is to be vented with 95% of the ammonia recovered. The inlet scrubber water is ammonia-free.

Design and Performance Equations—Packed Columns

203

1 Plot the equilibrium data in mole fraction units. 2 Perform the material balance and plot the operating line on the equilibrium plot. Table 10.2 Ammonia Equilibrium Data I NH3 concentration (lb NH3/100 lb H2O)

Equilibrium partial pressure (mm Hg) 3.4 7.4 9.1 12.0 15.3 19.4 23.5

0.5 1.0 1.2 1.6 2.0 2.5 3.0

SOLUTION 1 Employing the data provided in the problem statement, convert the equilibrium partial pressure data and the liquid concentration data to mole fractions as shown in Table 10.3. Plotting the mole fraction values on the graph in Figure 10.9 results in a straight line. The slope of the equilibrium line is approximately 1.0. 2 Convert the liquid and gas rates to lbmol/min, assuming that the molecular weights are approximately that of pure water and air: Lm ¼ L=18 ¼ 300=18 ¼ 16:67 lbmol=min Gm ¼ G=29 ¼ 250=29 ¼ 8:62 lbmol=min Determine the inlet and outlet mole fractions for the gas, y1 and y2, respectively (dropping the subscripts): y1 ¼

1:5=17 1:5=17 þ 98:5=29

¼ 0:0253 Table 10.3 Ammonia Equilibrium Data II Gas mole fraction, y 0.00447 0.00973 0.0120 0.0158 0.0201 0.0255 0.0309

Liquid mole fraction, x 0.0053 0.0106 0.0127 0.0169 0.0212 0.0265 0.0318

204

Chapter 10 Absorption and Stripping 0.035 0.03

Gas mole fraction, y

0.025 0.02 0.015 0.01 0.005

0 0

0.005

0.01

0.015 0.02 0.025 Liquid mole fraction, x

0.03

0.035

Figure 10.9 Ammonia equilibrium curve in water (728F, 1 atm).

y2 ¼

(0:05)(1:5=17) (0:05)(1:5=17) þ (98:5=29)

¼ 0:0013 The inlet liquid mole fraction x2 is given as zero, and the describing equation for x1, the outlet liquid mole fraction, is x1 ¼ (Gm =Lm )( y1  y2 ) þ x2

(10:4)

Substituting gives x1 ¼ (8:62=16:67)(0:0253  0:0013) þ 0 ¼ 0:0124 One may now use the inlet and outlet mole fractions to plot the operating line on the graph in Figure 10.10. The slope of the operating line is Slope ¼ (Lm =Gm )act ¼

0:0253  0:0013 0:0124  0:0

¼ 1:935 The reader is left the exercise of calculation (Lm/Gm)min. (The answer is slightly B greater than 1.0).

Design and Performance Equations—Packed Columns

205

0.035

0.03

Gas mole fraction, y

0.025

0.02

0.015

0.01

0.005

0 0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

Liquid mole fraction, x

Figure 10.10

Operating line for Illustrative Example 10.1.

ILLUSTRATIVE EXAMPLE 10.2 Given the following information for a packed countercurrent gas scrubber, determine the liquid flux in lbmol/h . ft2. Gas flux ¼ 18 lbmol/h . ft2 The mole fractions of the solute in the inlet and outlet gas are 0.08 and 0.002, respectively The mole fractions of the solute in the inlet and outlet liquid are 0.001 and 0.05, respectively. SOLUTION: Applying a componential mole balance (dropping the subscripts once again) on the solute gives an equivalent form of Equation (10.4): Lm (xout  xin ) ¼ Gm ( yin  yout )

(10:4)

Substituting, one obtains Lm (0:05  0:001) ¼ 18(0:08  0:002) Lm ¼ 18(0:078)=0:049 ¼ 28:7 lbmol=h  ft2

B

206

Chapter 10 Absorption and Stripping

ILLUSTRATIVE EXAMPLE 10.3 The EPA has conducted investigations on the exhaust streams at the Buonicore Chemical Company. To Buonicore’s dismay, their hydrocarbon emissions are too high and must be reduced in order to continue operation. At present conditions, the exhaust stream contains 1.0% benzene at a total flow of 40,000 ft3/h. It has been determined that if the exhaust stream is reduced to 0.01% benzene, the company can continue operation. An absorber that is currently out of commission will be used to absorb the benzene, employing a light wash oil as the absorbent. The pump on the liquid feed has a maximum liquid pumping rate of 50 ft3/h. Does this pump have enough capacity to do the job if the outlet benzene concentration in the wash oil cannot exceed 5.0%? Operating data are provided below: Gas temperature ¼ 1008F Liquid temperature ¼ 608F Light oil molecular weight (MW) ¼ 156 lb/lbmol Light oil density ¼ 55.1 lb/ft3 Actual liquid flow rate ¼ (1.5) Lm,min SOLUTION:

Find the gas molar flow rate Gm by applying the ideal gas law: Gm ¼ (40,000 ft3 =hr)(lbmol=379 ft3 )[(460 þ 100)=(460 þ 60)] ¼ 114 lbmol=hr

Calculate the minimum liquid rate Lm,min by applying Equation (10.4): Lm yin  yout ¼ Gm xout  xin   0:01  0:0001 Lm,min ¼ 114 0:05  0 ¼ 22:6 lbmol=h The actual operating liquid molar flow rate is Lm ¼ 1:5(22:6) ¼ 33:9 lbmol=h The required liquid mass flow rate is L ¼ 33:9(156 lb=lbmol) ¼ 5288 lb=h The liquid volumetric flow rate is then qL ¼ 5288=55:1 ¼ 96:0 ft3 =h The present pump does not have enough capacity.

B

Design and Performance Equations—Packed Columns

207

Column Diameter Consider a packed column operating at a given liquid rate and the gas rate is then gradually increased. After a certain point, the gas rate is so high that the drag on the liquid is sufficient to keep the liquid from flowing freely down the column. Liquid begins to accumulate and tends to block the entire cross section for flow (a process referred to as loading). This, of course, increases both the pressure drop and prevents the packing from mixing the gas and liquid effectively, and ultimately some liquid is even carried back up the column. This undesirable condition, known as flooding, occurs fairly abruptly, and the superficial gas velocity at which it occurs is called the flooding velocity. The calculation of column diameter is usually based on flooding considerations, with the usual operating range being taken as 50– 75% of the flooding rate. One of the more commonly used correlations is U.S. Stoneware’s(3) generalized pressure drop correlation, as presented in Figure 10.11. The procedure to determine the column diameter is as follows: 1 Calculate the abscissa, (L/G)(rG/rL)0.5; mass basis for all terms 2 Proceed to the flooding line and read the ordinate (design parameter) 3 Solve the ordinate equation for Gf at flooding 4 Calculate the column cross-sectional area, S, for the fraction of flooding velocity chosen for operation, f, by the equation: W (10:5) fGf _ is the mass flow rate of the gas in lb/s and S is the area in ft2. where W (m) S¼

Parameter of curves in pressure drop in inches of water/foot of packed height

0.60 0.40 Floo

ding

0.20

G2F m0.2 L rG rL gc

0.10 .060 .040

line

0.50

G = lb/ft . s F = packing factor, dimensionless = liquid density/water density m = lb/ft . s

0.25

rG = lb/ft3

0.10

rL = lb/ft3 L = lb/s G = lb/s

1.50 1.00

.020 .010 .006 .004

gc = 32.2 (lb/lbf) (ft/s2) 0.05

.002 .001 .01 .02

.04 .06 0.1

0.2 0.4 0.6 1.0 L rG 0.5 G rL

2.0

4.0

10.0

( )

Figure 10.11

Generalized pressure drop correlation to estimate column diameter.

208

Chapter 10 Absorption and Stripping

5 The diameter of the column is then determined by 

4 D¼ S p

0:5

¼ 1:13S 0:5 ; ft

(10:6)

Note that the proper units, as designated in the correlation, must be used as the plot is not dimensionless. The flooding rate is usually evaluated using total flows of the phases at the bottom of the column where they are at their highest value. The pressure drop may be evaluated directly from Figure 10.11 using a revised ordinate that contains the actual, not flooding, value of G. Chen(4) developed the following equation from which the tower diameter can easily be obtained:  D ¼ 16:28

W fL

0:5

rL rG

0:25 (10:7)

where log10 f ¼ 32:5496  4:1288 log10

 2 0:5 L Av m0:2 L rL2 13

(10:8)

where (employing Chen’s notation) Av is the specific surface area of dry packing (ft2/ ft3 packed column), L is the liquid flux (gal/min . ft2 of superficial tower cross section), W is the mass rate of flow of gas (lb/h), 1 is the void fraction, mL is the liquid viscosity (cP), and the density terms are in lb/ft3.

ILLUSTRATIVE EXAMPLE 10.4 A packed column is used to absorb a toxic pollutant from a gas stream. From the data given below, calculate the height of packing and column diameter. The unit operates at 50% of the flooding gas mass velocity, the actual liquid flow rate is 40% more than the minimum, and 95% of the pollutant is to be collected. Employ the generalized correlation provided in Figure 10.11 to estimate the column diameter. Gas mass flow rate ¼ 3500 lb/h Pollutant concentration in inlet gas stream ¼ 1.1 mol% Scrubbing liquid ¼ pure water Packing type ¼ 1-inch Raschig rings; packing factor F ¼ 160 HOG of the column ¼ 2.5 ft Henry’s law constant m ¼ 0.98 Density of gas (air) ¼ 0.075 lb/ft3 Density of water ¼ 62.4 lb/ft3 Viscosity of water ¼ 1.8 cP

Design and Performance Equations—Packed Columns SOLUTION:

209

First calculate the equilibrium outlet concentration x1 at y1 ¼ 0.011: x1 ¼ y1 =m

(10:1)

Substituting x1 ¼ 0:011=0:98 ¼ 0:0112 Determine y2 for 95% removal: y2 ¼ ¼

(0:05)y1 (1  y1 ) þ (0:05)y1 (0:05)(0:011) (1  0:011) þ (0:05)(0:011)

¼ 5:56  104 The minimum ratio of molar liquid flow rate to molar gas flow rate, (Lm/Gm)min, is determined by a material balance employing the equilibrium value at the top of the column, i.e., x rather than x:   Lm y1  y2 ¼ (10:4) Gm min x1  x2 ¼

0:011  5:56  104 0:0112  0

¼ 0:933 The actual ratio of molar liquid flow rate to molar gas flow rate Lm/Gm is (Lm =Gm )act ¼ Lm =Gm ¼ (1:40)(Lm =Gm )min ¼ (1:40)(0:933) ¼ 1:306 In addition (mGm )=Lm ¼ (0:98)=(1:306) ¼ 0:7504 To determine the diameter of the packed column, the ordinate of Figure 10.11 is first calculated:   0:5    0:5    L rG Lm 18 rG 18 0:075 0:5 ¼ ¼ (1:306) G rL 29 62:4 Gm 29 rL ¼ 0:0281 The value of the abscissa at the flooding line is determined from the same figure: G2 FCm0:2 L ¼ 0:21; rL rG gc The flooding gas mass flux Gf in lb/ft2 . s is  Gf ¼

0:21rL rgc FCm0:2 L

1=2

¼ 0:419 lb=ft2  s

 ¼

 (0:21)(62:4)(0:075)(32:2) 1=2 (160)(1:0)(1:8)0:2

210

Chapter 10 Absorption and Stripping

Apply Equation (10.5) to determine the column cross-sectional area. W S¼ FGF ¼

(3500=3600) ¼ 4:64 ft2 (0:5)(0:419)

Apply Equation (10.6) to calculate the column diameter.  0:5 4 S ¼ 1:13 S 0:5 ¼ (1:13)(4:64)0:5 D¼ p ¼ 2:43 ft

B

Column Height The column height may be estimated from Z ¼ NOG HOG

(10:9)

where NOG is the number of overall transfer units, HOG is the height of a single transfer unit and Z is the height of the column packing. In most design applications, the number of transfer units (NOG) is obtained experimentally or calculated using any of the methods to be explained later in this section. The height of a transfer unit (HOG) is also usually determined experimentally for the system under consideration. Information on many different systems using various types of packings has been compiled by the manufacturers of gas absorption equipment and should be consulted prior to design. The data may be in the form of graphs depicting, for a specific system and packing, the HOG vs the gas mass flux (lb/h . ft2) with the liquid rate as a parameter. The packing height Z is then simply the product of the HOG and the NOG. Although there are many different approaches to determine the column height, the HOG – NOG approach is the simplest and presently the most used, with the HOG usually being obtained from the manufacturer. Details on NOG follow. In many operations, the constituent to be absorbed (e.g., HCl) is in the very dilute range. For this condition yð1

NOG ¼

dy y  y

(10:10)

y2

If the operating line and equilibrium line are both parallel and straight: y1  y2 NOG ¼ y  y

(10:11)

If the operating line and equilibrium line are just straight (and not necessarily parallel) NOG ¼

y1  y2 ( y  y )1  ( y  y )2 ; ( y  y )lm ¼  ( y  y )lm ln[( y  y )1 =( y  y )2 ]

(10:12)

Design and Performance Equations—Packed Columns

211

However, one can show that if Henry’s law applies, the number of transfer units is given by Coburn’s equation     y1  mx2 1 1 þ ln 1 y2  mx2 A A (10:13) NOG ¼ 1 1 A where A¼

Lm mGm

(10:14)

1.0 0.8

0.3

0.6

0.5 0.6

0.4

0.7

0.3

0.8

0.2

0.9

0.1 0.08

0.95

y2 – mx2 (absorption) y1 – mx2

0.06 0.04

1.0

0.03 0.02

1.05 1.1

0.01 0.008 1.2

0.006 0.004 0.003

1.3 1.4

0.002 3.0 A (absorption factor)

0.001 0.0008 0.0006 0.0005

Figure 10.12

1

2

5.0

1.5 2.0 2.5 1.6 1.8

8 10 20 30 3 4 5 6 Number of transfer units, NOG (absorption)

NOG for absorption column with constant absorption factor.

40

50

212

Chapter 10 Absorption and Stripping

and A is defined once again the absorption factor and m is the slope of the equilibrium curve. The solution to this equation can be conveniently found graphically from Figure 10.12. However, note that the flow rates Lm and Gm are based on moles in Equation (10.14). If the gas is highly soluble in the liquid and/or reacts with the liquid, Theodore(5) has shown that   y N OG ¼ ln 1 (10:15) y2 If the operating line and/or equilibrium line are curved, the integral above in Equation (10.10) should be evaluated. Qualitatively, the height of a transfer unit is a measure of the height of a contactor required to affect a standard separation, and it is a function of the gas flow rate, the liquid flow rate, the type of packing, and the chemistry of the system. As indicated above, experimental values for HOG are generally available in the literature or from vendors.(1,5)

ILLUSTRATIVE EXAMPLE 10.5 When a gas is highly soluble, the number of overall gas transfer units NOG in a packed tower is given by  NOG ¼ ln

y1 y2

 (10:15)

Calculate NOG if y1 ¼ 200 ppm and y2 ¼ 0.5 ppm. SOLUTION:

Substituting in Equation (10.15) yields NOG ¼ ln(200=0:5) ¼ ln(400) ¼ 5:99

B

ILLUSTRATIVE EXAMPLE 10.6 A steel pickling operation emits 300 ppm HCl with peak values of 500 ppm, 15% of the time. The air flow is a constant 25,000 acfm at 758F and 1 atm. Only sketchy information was submitted with a scrubber permit application for a spray tower. You are requested, as a regulatory official, to determine if the spray unit is satisfactory. System information: Emission limit ¼ 25 ppm HCl Maximum gas velocity allowed through the tower ¼ 3 ft/s Number of sprays ¼ 6 Diameter of the tower ¼ 14 ft

Design and Performance Equations—Packed Columns

213

The plans show a countercurrent water spray tower. For a very soluble gas (Henry’s law constant is approximately zero), and the number of transfer units (NOG) can be determined by Equation (10.15), i.e.,(5)   y1 NOG ¼ ln y2 where y1 is the concentration of inlet gas and y2 is the concentration of outlet gas. In a spray tower, the number of transfer units (NOG) for the first (or top) spray is about 0.7. Each lower spray will have only about 60% of the NOG of the spray above it. The final spray, if placed in the inlet duct, has a NOG of 0.5. The spray sections of a tower are normally spaced at three foot intervals. The inlet duct spray adds no height to the column. SOLUTION:

Calculate the gas superficial velocity through the tower in ft/s: q v¼ ; q ¼ volumetric flow rate pD2 =4 ¼

25,000 ¼ 162:4 ft=min ¼ 2:7 ft=s p (14)2 =4

The gas velocity meets requirements since it is less than 3 ft/s. Calculate the number of transfer units required to meet the regulation under worst case conditions:   y1 NOG ¼ ln (10:15) y2   500  106 ¼ ln ¼ 3:0 25  106 Determine the total number of transfer units provided by a tower with six spray sections. The result is given in Table 10.4. Note that this value is below the required value of 3.0. The spray unit is therefore not satisfactory. Using the total NOG from Table 10.4, calculate the outlet concentration of gas: y1 ¼ exp(NOG ) ¼ exp(2:114) ¼ 8:28 y2 y2 ¼

500 ¼ 60:4 ppm 8:28

B

Table 10.4 NOG Calculation for Illustrative Example 10.6 Spray section Top 2nd 3rd 4th 5th Inlet Total

NOG 0.7 (0.7)(0.6) ¼ (0.42)(0.6) ¼ (0.252)(0.6) ¼ (0.1512)(0.6) ¼ (Given)

0.7 0.42 0.252 0.1512 0.0907 0.5 2.114

B

214

Chapter 10 Absorption and Stripping

ILLUSTRATIVE EXAMPLE 10.7 Solve the following two problems: 1 Find the number of overall gas transfer units (NOG) in a packed tower required to recover 90% of a gas in an inlet air stream containing 10 mole percent (mol%) solute using pure water at a rate 20% greater than the minimum rate. Assume m ¼ 1.485. 2 How many NOG values would be required if, instead of pure water, water containing 0.1, 0.3, 0.5, and 0.65 mol% (mole percent) of the solute in the gas were used instead? SOLUTION 1 Calculations can be performed on a mole fraction or solute-free mole fraction basis. Since y1 ¼ 0.1, the solute-free mole fraction is Y1 ¼ y1 =(1  y1 ) ¼ 0:1=(1  0:1) ¼ 0:1111 In addition, y2 ¼ (0:1)(0:1111) ¼ 0:0111 and Y2 ¼ 0:0111=(1  0:0111) ¼ 0:0112 For the minimum rate, one obtains X1 ¼ Y1 =1:485 ¼ 0:1111=1:485 ¼ 0:0748 The minimum liquid-to-gas ratio is then   Lm 0:1111  0:0112 ¼ X1  X0 Gm min ¼

0:1111  0:0112 0:0748  0

¼ 1:34 The actual ratio is 

Lm Gm

 ¼ (1:2)(1:34) act

¼ 1:60

Design and Performance Equations—Packed Columns

215

Since A¼

Lm mGm

(10:14)

substituting gives A ¼ 1:60=1:485 ¼ 1:08; 1=A ¼ 0:928 and y2 Y2 ’ y1 Y1 ¼ 0:111;

y1 ¼ 9:01 y2

One may employ either Equation (10.13) or Figure 10.12. From Figure 10.12 NOG ’ 6:0 From Equation (10.13):

NOG

    y1  mx2 1 1 1 þ ln A A y2  mx2 ¼ 1  (1=A)    0:1 (0:072) þ 0:928 ln 0:0111 ¼ 0:072 ¼ 6:32

2 For this condition x2 ¼ X2 ¼ 0:001 Once again 

Lm Gm

 ¼ min

0:1111  0:0111 0:0748  0:001

¼ 1:35 and 

Lm Gm

 ¼ (1:2)(1:35) act

¼ 1:63 If Equation (10.13) is used, then A ¼ 1:63=1:485 ¼ 1:098; 1=A ¼ 0:911

216

Chapter 10 Absorption and Stripping and y2  mx2 0:011  0:001485 ¼ 0:1  0:001485 y1  mx2 ¼ 0:096 The NOG value from Equation (10.13) is NOG ¼ 6:82 The corresponding values for y2 ¼ 0.003, 0.05, and 0.065 (employing the same procedure above) are 7.45, 9.66, and 13.5, respectively. B

ILLUSTRATIVE EXAMPLE 10.8 Determine the packing height of a packed countercurrent absorber required to reduce the Cl2 concentration in a gas by 99% assuming that a dilute NaOH solution is employed. The following information is given: Liquid mass flux ¼ 1000 lb/h . ft2 (essentially water) Gas mass flux ¼ 750 lb/h . ft2 (essentially air) Mole fraction of Cl2 in inlet gas ¼ 0.00043 HOG ¼ 1.63 ft SOLUTION: It can be assumed that Cl2 will react with the dilute NaOH solution. When the solute reacts with the liquid, it can be assumed that m ¼ 0(5) so that Equation (10.15) applies. For a 99% reduction: NOG ¼ ln(100=1) ¼ 4:6 The packing height is therefore Z ¼ HOG NOG ¼ (1:63)(4:6) ¼ 7:5 ft

(10:9) B

ILLUSTRATIVE EXAMPLE 10.9 Determine the packing height (in feet) of a countercurrent scrubber required to reduce an inlet ammonia concentration by 90%, given the following information: Liquid molar flux (water) ¼ 1000 lbmol/h . ft2 Gas molar flux (air) ¼ 700 lbmol/h . ft2 Mole fraction of NH3 in inlet gas ¼ 0.023 Mole fraction of NH3 in outlet liquid ¼ 0.015

Design and Performance Equations—Packed Columns

217

Assume no NH3 in inlet water stream Slope of equilibrium line ¼ 0.93 HOG ¼ 1.5 ft SOLUTION:

First check whether the material balance is satisfied:

1000(0:015  0) ¼ 700(0:023  0:0023); yout ¼ (0:023)(1  0:9) 15 ¼ 14:49 The material balance is satisfied. Now apply Coburn’s equation to calculate NOG:     y1  mx2 1 1 1 þ ln A A y2  mx2 NOG ¼ 1 1 A Lm A¼ mGm

(10:13)

¼ 1000=(0:93)(700) ¼ 1:536 Substituting, one obtains

NOG

    0:015  0 1 1 1 þ ln 0:0015  0 1:536 1:536 ¼ 1 1 1:536 ¼ 4:07

Therefore Z ¼ HOG NOG ; HOG ¼ 1:5 ft ¼ (1:5)(4:07) ¼ 6:12 ft

(10:9) B

ILLUSTRATIVE EXAMPLE 10.10 Doyle Unlimited, a Daniel F. Rodenci Corporation, has submitted design plans to Theodore Consultants for a packed ammonia scrubber on an air stream containing NH3. The operating and design data provided by Doyle Unlimited, Inc. are given below. Theodore Consultants remember reviewing acceptable plans for a nearly identical scrubber for Doyle Unlimited, Inc. in 2005. After consulting old files, the consultants find all the conditions are identical except for the gas flow rate. What recommendation should be made? Tower diameter ¼ 3.57 ft Packed height of column ¼ 8 ft Gas and liquid temperature ¼ 758F inlet

218

Chapter 10 Absorption and Stripping Operating pressure ¼ 1.0 atm Ammonia-free liquid mass flux ¼ 1000 lb/h . ft2 Gas flow rate ¼ 1575 acfm Inlet NH3 gas concentration ¼ 2.0 mol% Air density ¼ 0.0743 lb/ft3 Molecular weight of air ¼ 29 Molecular weight of water ¼ 18 Henry’s law constant m ¼ 0.972 Figure 10.13; packing type A is used Colburn chart (Fig. 10.12) applies Emission regulation ¼ 0.1% NH3 (by mole or volume) Packing height ¼ HOGNOG 4.0 Packing A Packing B

G = 700

3.2

G = Gas flow, lb/h · ft2

G = 500

HOG, ft

2.4 G = 700

1.6

G = 500

0.8

0.0

0

500

1000 Liquid rate, lb/h · ft2

1500

2000

Figure 10.13 HOG values for different types of packing. SOLUTION: Calculate the cross-sectional area of the tower S (note that S is now employed for the area) in ft2: S ¼ pD2 =4 Substituting, S ¼ (p)(3:57)2 =(4) ¼ 10:0 ft2

(10:6)

Design and Performance Equations—Packed Columns

219

Calculate the gas molar flux (molar flow rate per unit cross section) and liquid molar flux in lbmol/(ft2 . h): Gm ¼ qr=S(MW)G ¼ (1575)(0:0743)=[(10:0)(29)] ¼ 0:404 lbmol=(ft2  min) ¼ 24:2 lbmol=(ft2  h) Lm ¼ L=(MW)L ¼ (1000)=(18) ¼ 55:6 lbmol=(ft2  h) The value mGm/Lm is therefore mGm =Lm ¼ (0:972)(24:2=55:6) ¼ 0:423 The absorption factor A is defined as A¼

Lm mGm

A¼

1 0:423

(10:14)

Substituting,

¼ 2:364 The value of ( y1 2 mx2)/( y2 2 mx2) is y1  mx2 0:02  (0:972)(0) ¼ y2  mx2 0:001  (0:972)(0) ¼ 20:0 NOG is calculated from Colburn’s equation chart,

NOG

    y1  mx2 1 1 1 þ ln A A y2  mx2 ¼ 1 1 A

Substituting,   ln (20:0) 1  NOG ¼

¼ 4:30

   1 1 þ 2:364 2:364 1 1 2:364

(10:13)

220

Chapter 10 Absorption and Stripping

To calculate the height of an overall gas transfer unit, HOG, first calculate the gas mass flux G in lb/h . ft2: G ¼ qr=S ¼ (1575)(0:0743)=10:0 ¼ 11:7 lb=(min  ft2 ) ¼ 702 lb=(h  ft2 ) From Figure 10.13, one obtains (for packing A) HOG ¼ 2:2 ft The required packed column height Z, in feet, is Z ¼ NOG HOG ¼ (4:3)(2:2)

(10:9)

¼ 9:46 ft The proposal should be rejected.

B

ILLUSTRATIVE EXAMPLE 10.11 The calculations for an absorber indicate that it will be excessively tall. Thus, three schemes using two shorter absorbers are considered, as shown on the left hand side of Figure 10.14. Make freehand sketches of operating lines, one for each scheme, showing the relation between the operating lines for the two absorbers and the equilibrium curve. Mark the mole fractions on an equilibrium line-operating line diagram. No calculation is required. Assume dilute solutions. It is suggested that the reader attempt to solve this application prior to looking at the solution which appears in Figure 10.14. SOLUTION:

The solution is presented in the right hand side of Figure 10.14.

B

ILLUSTRATIVE EXAMPLE 10.12 Qualitatively outline how one can size (diameter, height) a packed tower to achieve a given degree of separation without any information on the physical and chemical properties of a gas to be absorbed. SOLUTION: To calculate the height, one needs both the height of a gas transfer unit HOG and the number of gas transfer units NOG. Since equilibrium data are not available, assume that m (slope of equilibrium curve) approaches zero. This is not an unreasonable assumption for most solvents that preferentially absorb (or react with) the solute. For this condition:   y1 NOG ¼ ln (10:15) y2

Design and Performance Equations—Packed Columns

yb2

xb2

(xa1, ya1) (xb2, yb2) (xa2, ya2)

b

a

yb1

y

ui lib riu m

xa2

Eq

ya2

lin e

1.

(xb1, yb1)

xb1

ya1 xa1 x 2.

yb2 xb2

lin

e

ya2 xa2

riu lib Eq

ui

(xa2, ya2) (xb1, yb1)

y a

m

(xa1, ya1)

b (xb2, yb2)

yb1

xb1

ya1 xa1 x 3. yb2 (xa1, ya1)

(xa2, ya2) b

a

y

(xb2, yb2) (xb1, yb1)

xa1 ya1

Figure 10.14

yb1 xb1 Solution to Illustrative Example 10.11.

x

Eq ui

lib riu

m

xb2

e

xa2

lin

ya2

221

222

Chapter 10 Absorption and Stripping Table 10.5 Packing Diameter versus HOG Packing diameter, inches 1.0 1.5 2.0 3.0 3.5

Plastic packing HOG, feet

Ceramic packing HOG, feet

1.0 1.25 1.5 2.25 2.75

2.0 2.5 3.0 4.5 5.5

where y1 and y2 represent inlet and outlet concentrations, respectively. Since it is also reasonable to assume the scrubbing medium to be water or a solvent that effectively has the physical and chemical properties of water, HOG can be assigned values usually encountered for water systems. These are given in Table 10.5. For plastic packing, the liquid and gas flow fluxes are both typically in the range of 1500– 2000 lb/(h . ft2 of cross-sectional area). For ceramic packing, the range of flow rates is 500 –1000 lb/h . ft2. For difficult-to-absorb gases, the gas flow rate is usually lower and the liquid flow rate higher. Superficial gas velocities (velocity of the gas if the column is empty) are in the 3–6-ft/s range. The height Z is then calculated from Z ¼ (HOG )(NOG )(SF) where SF is a safety factor, the value of which can range from 1.25– 1.5. Pressure drops can vary from 0.15–0.40 inch H2O/ft packing. Packing size increases with increasing tower diameter. Packing diameters of 1 inch are recommended for tower diameter in the 3 ft range. One should use large packing for larger diameter packing; for smaller towers, smaller packing is usually employed. The reader is left the exercise of verifying the chart in Table 10.6 for plastic packing.(5) (Note: This problem and design procedure were originally developed by one of the authors in 1985 and later published in 1988.) B

Table 10.6 Packing Height, Z (ft), as a Function of Efficiency and Plastic Packing Size Plastic packing size, inches Removal efficiency, %

1.0

1.5

2.0

3.0

3.5

63.2 77.7 86.5 90 95 98 99 99.5 99.9 99.99

1.0 1.5 2.0 2.3 3.0 3.9 4.6 5.3 6.9 9.2

1.25 1.9 2.5 3.0 3.75 4.9 5.75 6.6 8.6 11.5

1.5 2.25 3.0 3.45 4.5 5.9 6.9 8.0 10.4 13.8

2.25 3.4 4.5 5.25 6.75 8.8 10.4 11.9 15.5 20.7

2.75 4.1 5.5 6.25 8.2 10.75 12.7 14.6 19.0 25.3

223

Design and Performance Equations—Packed Columns

ILLUSTRATIVE EXAMPLE 10.13 A 1600 acfs gas stream is to be treated in a packed tower containing ceramic packing. The gas stream contains 100 ppm of a solute that is to be reduced to 1 ppm. Estimate the tower’s cross-sectional area, diameter, height, pressure drop, and packing size. Use the procedure outlined in Illustrative Example 10.12. SOLUTION: Key calculations from Illustrative Example 10.12 are provided in Table 10.7 for ceramic packing.(5) The equation for the cross-sectional area of the tower S in terms of the gas volumetric flow rate q in acfs is (assuming a 4 ft/a superficial velocity) S [ft2 ] ¼ q [acfs]=4:0 An equation to estimate the tower packing pressure drop DP in terms of Z is DP [in H2 O] ¼ (0:2)Z; Z ¼ ft The following packing size(s) is (are) recommended: For D  3 ft, use 1-inch packing For D , 3 ft, use ,1-inch packing For D . 3 ft, use .1-inch packing

As noted earlier, recommended packing size increases with increasing diameter. For the problem at hand S ¼ 1600=4 ¼ 400 ft2 The diameter D is D ¼ (4S=p)0:5

(10:6)

Table 10.7 Packing Height, Z (ft), as a Function of Efficiency and Ceramic Packing Size Ceramic packing size, inches Removal efficiency, %

1.0

1.5

2

3

3.5

63.2 77.7 86.5 90 95 98 99 99.5 99.9 99.99

2.0 3.0 4.0 4.6 6.0 7.8 9.2 10.6 13.8 18.4

2.5 3.7 5.0 5.75 7.5 9.8 11.5 13.25 17.25 23.0

3.0 4.5 6.0 6.9 9.0 11.7 13.8 15.9 20.7 27.6

4.5 6.75 9.0 10.4 13.5 17.6 20.7 23.8 31.1 41.4

5.5 8.25 11.0 12.7 16.5 21.5 25.3 29.1 38.0 50.7

224

Chapter 10 Absorption and Stripping

Substituting, D ¼ [(4)(400)=p]0:5 ¼ 22:6 ft For a tower diameter this large, the 3.5-inch packing should be used. Additional information is available in the literature.(5) B

Pressure Drop The pressure drop through a packed column for any combination of liquid and gas flows in the operable range is an important economic consideration in the design of such columns. For most random packings, the pressure drop incurred by the gas is influenced by the gas and liquid flow rates. At constant gas rate, an increase in liquid throughput—which takes up more room in the packing (increased holdup) and, therefore, leaves less room for the gas (greater restriction)—is accompanied by an increase in pressure drop until the liquid flooding rate is reached. At this point, any slight liquid excess that cannot pass through remains atop the packing, building up a deeper and deeper head (or pressure drop), hypothetically reaching an infinite value. Similarly, at constant liquid downflow, increasing the gas flow is again accompanied by a rising pressure drop until the flooding rate is reached, whereupon the slightest gas increase will cause a decline in permissible liquid throughput. This causes the liquid to again accumulate atop the packing, so that pressure drop again continues to increase. For a particular packing, the most accurate pressure drop data will be those available directly from the manufacturer. However, for the purposes of estimation, Figure 10.11 is simple to use and usually provides reasonable results. Some general “rules of thumb” in the design of packed columns do exist. They are by no means final in that there are other considerations that might have to be taken into account (allowable pressure drop, possible column height restrictions, and so on). The rules must therefore be applied discriminately. For approximation purposes, if the gas rate is greater than about 500 acfm, a nominal packing size smaller than 1 inch would probably not be practical: similarly, at about 2000 acfm, sizes smaller than 2 inches would also likely be impractical. The nominal size of the packing should not exceed about 1/20th of the column diameter. The usual practice is to design so that the operating gas rate is approximately 75% of the rate that would cause flooding. If possible, column dimensions should be in readily available sizes (i.e., diameters to the nearest half foot and heights to the nearest foot). If the column can be purchased “off-theshelf”, as opposed to being specially made, substantial savings can be realized.

ILLUSTRATIVE EXAMPLE 10.14 A packed column is designed to absorb ammonia from a gas stream. The unit operates at 60% of the flooding gas mass velocity, the actual liquid flow rate is 25% more than the minimum, and 90% ammonia to be collected. Given the operating conditions and type of packing below, calculate the height of packing, the column diameter, and the operating pressure drop. Data

Design and Performance Equations—Packed Columns

225

includes: gas mass flow rate ¼ 5000 lb/h; NH3 concentration in inlet gas stream ¼ 2.0 mol%; scrubbing liquid is pure water; packing type ¼ 1 inch Raschig rings; HOG of the column ¼ 2.5 ft; Henry’s law constant, m ¼ 1.20; density of gas (air) ¼ 0.075 lb/ft3; density of water ¼ 62.4 lb/ft3; viscosity of water ¼ 1.8 cP, c ¼ 1 (ratio of liquid to water density); and F ¼ 160 (for 1 inch Raschig rings). SOLUTION: First calculate the equilibrium outlet liquid composition, and the outlet gas composition for 90% removal: y1 x1 ¼ (10:1) m Substituting, x1 ¼

0:02 ¼ 0:0167 1:20

The minimum liquid to gas ratio (molar basis) is obtained by a material balance: y2 ¼

0:1y1 0:1(0:02) ¼ 0:00204 ¼ [(1  y1 ) þ 0:1y1 ] [(1  0:02) þ 0:1(0:02)]

The minimum liquid –gas ratio (molar basis) is obtained by a material balance:   Lm y1  y2 ¼  Gm min x1  x2 Substituting,



Lm Gm

 ¼ min

(10:4)

0:02  0:00204 ¼ 1:08 0:0167  0

The actual ratio is 25% above the minimum. Thus,     Lm Lm ¼ 1:25 ¼ 1:25(1:08) ¼ 1:35 Gm act Gm min Two parameters are needed to use the Colburn chart to evaluate NOG: y1  mx2 0:02  1:2(0) ¼ 9:80 ¼ y2  mx2 0:00204  1:2(0) and mGm 1:2 ¼ 0:889 ¼ 1:35 Lm From Colburn’s Chart (Fig. 10.12) NOG ¼ 6:2 The packing height is then Z ¼ NOG HOG Substituting, Z ¼ 6:2(2:5) ¼ 15:5 ft

(10:9)

226

Chapter 10 Absorption and Stripping

Figure 10.11 is employed to calculate the tower diameter and packing pressure drop,      L r 0:5 18 0:075 0:5 ¼ 1:35 G rL 29 62:4 ¼ 0:0291 From Figure 10.11, G2 FCmL0:2 ¼ 0:19 rL rgc Thus, the flooding mass velocity is sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:19rL rgc 0:19(62:4)(0:075)(32:2) ¼ Gf ¼ FCmL0:2 160(1)(1:8)0:2 ¼ 0:40 lb=ft2  s The actual velocity is G ¼ 0:6G f ¼ 0:6(0:40) ¼ 0:24 lb=ft2  s ¼ 864 lb=ft2  h The tower diameter may now be calculated directly from D ¼ 1:13S 0:5

(10:6)

Substituting  D ¼ 1:13

W fG

0:5

 ¼ 1:13

5000 (1)(864)

0:5

¼ 2:72 ft The operating pressure drop can be estimated from Figure 10.11. At 60% of flooding, the ordinate becomes (0:19)(0:6)2 ¼ 0:068 Employing this ordinate and an abscissa of 0.029 gives an operating pressure drop of approximately DP ¼ 0:065 in H2 O=ft packing Calculating the overall pressure drop across the column is left as an exercise for the reader. B

ILLUSTRATIVE EXAMPLE 10.15 Consider the absorber system shown in Figure 10.15. Corenza Engineers designed the unit to operate with a maximum discharge concentration of 50 ppm. Once the unit was installed and running, the unit operated with a discharge of 60 ppm. Rather than purchase a new unit, what options are available to bring the unit into compliance with the specified design concentration?

Design and Performance Equations—Plate Columns

227

50 ppm (design) 90°F water

1 atm

5 ft (one inch Raschig rings)

1000 acfm 1000 ppm (solute) 90°F

Water

Figure 10.15

Absorber failure to meet design performance.

SOLUTION: This is obviously an open-ended question. On the basis of the material presented earlier and the solutions to several of the problems in this chapter, one may employ any one or a combination of several suggestions recommended by Theodore.(6) The reader is referred to Illustrative Example 20.4 in Chapter 20 for the “solution” to this illustrative example. B

DESIGN AND PERFORMANCE EQUATIONS—PLATE COLUMNS The most important design considerations for plate columns include the calculation of the column diameter, type and number of plates to be used (usually either bubble-cap or sieve plates), actual plate layout and physical design, and plate spacing; these, in turn, determine column height. To consider each of these to any great extent is beyond the scope of this chapter, particularly since it received attention in Chapter 9. The discussion that follows, therefore, will be a relatively concise presentation of some of the general absorber design techniques that will provide satisfactory results for purposes of estimation.(7–9) The column diameter, and consequently its cross section, must be sufficiently large to handle the gas and liquid at velocities that will not cause flooding or excessive entrainment. The superficial gas velocity for a given type of plate at flooding is given

228

Chapter 10 Absorption and Stripping

by the relation VF ¼ CF

  rL  rG 0:5 rG

(10:16)

where VF (the notation usually employed for plate columns) is the gas velocity through the net column cross sectional area for gas flow, ft3/s . ft2, the densities are in lb/ft3, and CF is an empirical coefficient that depends on the type of plate and operating conditions. The net cross section is the difference between the column cross section and the area taken up by downcomers. In actual design, some percent of VF is usually used— for nonfoaming liquids 80 – 85% of VF and 75% or less for foaming liquids. Of course, the value is subject to a check of entrainment and pressure drop characteristics. The calculation of column diameter based on Equation (10.16) assumes that the gas flow rate is the controlling factor in its determination. After a plate layout has been assumed, it is then necessary to check the plate for its liquid handling capacity. If the liquid– to – gas ratio is high and the column diameter large, the check will indicate whether the column will show a tendency toward flooding or gas maldistribution on the plate. If this is the case, then the liquid rate is the controlling factor in estimating the column diameter and a satisfactory assumption for design purposes is a plate-handling capacity of 30 gal/min of liquid per foot of diameter.(9) However, well-designed single-pass cross-flow plate can ordinary be expected to handle up to 60 gal/min of liquid per foot of diameter without an excessive liquid gradient. It should also be noted that low gas rates can lead to weeping, a condition where the liquid flows down through the holes in the plate rather than across the plate. The column height is determined from the product of the number of actual plates (theoretical plates divided by the overall plate efficiency) and the plate spacing chosen. The theoretical plate (or stage) is the theoretical unit of separation in plate column calculations. It is defined as a plate in which two dissimilar phases are brought into intimate contact with each other and then are physically separated. During the contact, various diffusing components of the mixture redistribute themselves between the phases. In an equilibrium stage, the two phases are well mixed for a sufficient time to allow establishment of equilibrium between the phases leaving the stage. In effect, no further net change of composition of the phases is possible at equilibrium for a given set of operating conditions. The number of theoretical plates can be determined graphically from the operating diagram composed of an operating line and equilibrium curve. In the above discussion of equilibrium stages, it was assumed that the phases leaving the stage were in equilibrium. In actual countercurrent multistage equipment, it is not practical to provide the combination of residence time and intimacy of contact required to accomplish equilibrium. Hence, the concentration change for a given stage is less than that predicted by equilibrium considerations. Stage efficiencies are employed to characterize this condition. The efficiency term frequently used is the overall stage (plate efficiency), given by the ratio of theoretical contacts required for a given separation to the actual number of contacts required for the same operation.

Design and Performance Equations—Plate Columns

229

While reliable information on such an efficiency is most desirable and convenient to use, so many variables come into play that really reliable values for the overall stage efficiency are difficult to come by. This value is generally obtained by experiment or field test data, or may be specified by the vendor. The number of theoretical plates may be determined directly without recourse to graphical techniques for cases where both the operating line and the equilibrium curve may be considered straight (dilute solutions). This will frequently be the case for a relatively dilute gas (as usually encountered in air pollution control) and liquid solutions where, more often than not, Henry’s law is usually applicable. Since the quantity of gas absorbed is small, the total flows of liquid and gas entering and leaving the column again essentially remain constant. Hence, the operating line will be substantially straight. For such cases, the Kremser – Brown – Sounders(10,11) equation applies for determining the number of theoretical plates, Np:  Np ¼ log

   yNpþ1  mx0 1 1 þ 1 y1  mx0 A A log A

(10:17)

Note that ln may be employed rather than log in both the numerator and denominator. Here mx0 is the gas composition in equilibrium with the entering liquid (m is Henry’s law constant ¼ slope of the equilibrium curve). If the entering liquid contains no solute gas, then x0 ¼ 0 and Equation (10.17) can be simplified further. The solute concentrations in the gas stream, yNpþ1 and y1 represent inlet and outlet conditions, and L and V (that appear in A) the total mole rates of liquid and gas flow per unit time per unit column cross-sectional area. Small variations in L and V may be roughly compensated for by using the geometric average value of each taken at the top and bottom of the column. Equation (10.17) has been plotted in Figure 10.16 for convenience and may be used for the solution to this equation. Chen(12) derived a simplified algebraic equation that could be used to estimate the theoretical plates, n, in either an absorber or stripper. The final equation took the form (retaining Chen’s notation): yb þ f yn þ f

(10:18)

yt  A(B þ mxt ) A1

(10:19)

An ¼ where

f¼

and A is the absorption factor, yt is the top plate gas mole fraction, yb is the bottom plate gas mole fraction and yn ¼ B þ mxt (equilibrium line) is the gas mole fraction at plate n. The number of actual trays, which is based on the tray efficiency, is determined by the mechanical design and conditions of operation. For the case where the equilibrium curve and operating lines are straight, the overall tray efficiency E0 can be computed

230

Chapter 10 Absorption and Stripping 1.0 0.3 0.6

0.5 0.6

0.4

0.7

0.3

0.8

0.2

0.9

0.1 0.08

0.95

y1 – mx0 yNP +1 – mx0

0.06 0.04

1.0

0.03 0.02

1.05 0.01 0.008 1.1

0.006 0.004 0.003

1.2

0.002 5.0 0.001 0.0008 0.0006 0.0005

1.3 2.5 1.8 1.5 3.0 1.6 1.4 2.0 4.0

(Absorption factor, L /mV) 1

2

3

4 5 6

8

10

20

30 40

50

Number of theoretical trays, Np

Figure 10.16 Number of theoretical stages for countercurrent absorption columns.

and the number of actual trays determined analytically: E0 ¼ ¼

equilibrium trays actual trays log(1 þ EMGE )(1=A  1) log(1=A)

(10:20)

where EMGE ¼ Murphree efficiency, as noted in Chapter 9, corrected for entrainment (values available in the literature). Empirical data for standard tray designs within standard ranges of liquid and gas rates are available. These data, as shown in Figure 10.17, are accurate for bubble-cap trays and can be used as rough estimates for sieve and valve

231

Design and Performance Equations—Plate Columns

Fractional overall tray efficiency, E0

1.0 0.6 0.4 0.2 0.1 = Commercial hydrocarbon absorbers = Laboratory hydrocarbon absorbers = Laboratory absorption, CO2 in water and glycerol = Laboratory absorption, NH3 in water m = y */x mL = Liquid viscosity, kg/m . s

0.06 0.04 0.02 0.01

ML = Liquid mol wt rL = Liquid density, kg/m3

0.006 0.004

4 6 10–5

Figure 10.17

2

4 6 10–4 2

4 6 10–3 2 mMWL mL rL

4 6 10–2 2

4 6 10–1

2

4

Overall tray efficiencies of bubble-cap tray absorbers.

trays. After the overall efficiency of the tower is determined, the number of actual trays is calculated using: Nact ¼

N Np ¼ E0 E0

(10:21)

The general procedure to follow in sizing a plate tower is given below.(13) 1 Calculate the number of theoretical stages, N, using Figure 10.16 or Equation (10.17). 2 Estimate the efficiency of separation, E. This may be determined at the local (across plate), plate (between plates), or overall (across column) level. The overall efficiency, E0, is generally employed. 3 Calculate the actual number of plates: Nact ¼

N E0

(10:21)

4 Obtain the height between plates, h. This is usually in the 12- to 36-inch range. Many towers use a 24-inch plate spacing. 5 The tower height, Z, is then Z ¼ Nact h 6 The diameter may be calculated directly from Equation (10.16).

(10:22)

232

Chapter 10 Absorption and Stripping

7 The plate or overall pressure drop is difficult to quantify accurately. It is usually in the 2- to 6-inch H2O per plate range for most columns with the lower and upper values applying to small and large diameters, respectively.

ILLUSTRATIVE EXAMPLE 10.16 Refer to Illustrative Example 10.7. Repeat the calculations as they apply to a plate column. In effect, determine the number of theoretical plates N. Employ Equation (10.17). SOLUTION:

Equation (10.17) applies for a plate column: log N¼

    yNþ1  mx0 1 1 1 þ A A y1  mx0 log A

(10:17)

once again m ¼ 1:485 (Lm =Gm )act ¼ 1:2(Lm =Gm )min Results for Np are provided in Table 10.8.

B

Table 10.8 Number of Plates for Illustrative Example 10.16 X2 0 0.001 0.003 0.005 0.0065

(Lm/Gm )min

(Lm/Gm )act

A

Np

1.384 1.4031 1.4438 1.4872 1.5215

1.661 1.6837 1.7326 1.7846 1.8258

1.119 1.1339 1.1667 1.2018 1.2295

5.48 5.87 6.93 8.82 12.21

ILLUSTRATIVE EXAMPLE 10.17 In an attempt to quantify the effect of enthalpy of solution effects on the absorption of HCl into scrubbing water in an absorber, Pallechi Consultants reviewed the literature(6) and obtained the following rough estimates of this effect. The data provided temperature increases as a function of increasing HCl concentration (mass percent basis) in water: 0–1:5% ¼ 108C 0–3:0% ¼ 158C 0–5:0% ¼ 208C

Design and Performance Equations—Plate Columns

233

Apply the above data and estimate the discharge temperature increase for the following two HCl scenarios: Scenario 1: 0.0% inlet to 1.5% outlet (mass percent) Scenario 2: 0.5% inlet to 3.0% outlet (mass percent) SOLUTION: Since enthalpy is a point function, it is reasonable to assume that the temperature effects are additive. Therefore, the temperature increases are Scenario 1: DT ¼ DT1:5  DT0:0 ¼ 10  0 ¼ 108C Scenario 2: DT ¼ DT3:0  DT1:5 ¼ 15  10 ¼ 58C The reader should note that this is an effect that often should be reflected in engineering applications since any increase in the temperature of the scrubbing liquid adversely affects the equilibrium, reducing the equilibrium capacity of the liquid. Note that it is the temperature of the liquid, not the gas, that affects equilibrium. B

ILLUSTRATIVE EXAMPLE 10.18 You are requested to outline a procedure to calculate the height of a plate tower required to absorb R% of organics (A and B) in an organic –air mixture. The following assumptions can be made: 1 2 3 4 5 6 7 8

Ideal gas and liquid solutions Isothermal operation Liquid-to-gas molar flow rate ratio (Lm/Gm) through the tower is constant Henry’s law applies; dimensionless constants available Plate efficiency, E0, is given Height between stages, h, is also given Absorbing liquid contains no A and B Inlet gas concentrations of A and B are given

Calculate the height of the tower for a methylamine (A)/dimethylamine –(B) air–water system if: Lm =Gm ¼ 1:0

234

Chapter 10 Absorption and Stripping

The absorption factors are, Ai ¼ Lm/miGm; AA ¼ 0.85, AB ¼ 0.75. In addition, E 0 ¼ 0:5 h ¼ 2:0 ft yA,i ¼ 0:01 yB,i ¼ 0:008 The required recovery efficiency of the tower, R ¼ 77.78% (total). SOLUTION: The Kremser –Souter–Brown (KSB) equation, referred to by some as the Kremser equation, applies individually to both components A and B, and may be written in the following form: yNþ1  y1 ANþ1  A ¼ Nþ1 yNþ1  mx0 A 1

(10:23)

where A is the absorption factor, yNþ1 is the inlet gas concentration, y1 is the outlet concentration, x0 is the inlet absorbing liquid concentration, and N is the number of theoretical plates required. Generally, one also assumes no condensation, no mixing (heat) effects, no chemical reaction, etc. Since the concentration is usually dilute, the liquid and gas rates are also assumed to be constant. Thus, the absorption factor A is also constant. The key to the multicomponent calculation suggested here is to assume: 1 no interaction effects between the various components, and 2 the absorption of each component occurs as if the other components are not present (i.e., treat each component separately); the KSB equation is then employed for each species present in the gas mixture. Note also that the assumption of an ideal solution and isothermal operation is valid for many operations. For component A: 0:01  yA1 0:85Nþ1  0:85 ¼ 0:85Nþ1  1 0:01  0

(1)

0:008  yB1 0:75Nþ1  0:75 ¼ 0:75Nþ1  1 0:008  0

(2)

For component B:

Calculate the total outlet concentration of the gas mixture: yA,1 þ yBþ1 ¼ (1  R)(yA,Nþ1 þ yB,Nþ1 ) ¼ (1  0:7778)(0:01 þ 0:008) ¼ 0:0040

(3)

Solve the three equations obtained above for N þ 1, yA1, and yB1. The result via a trial-and-error calculation is N þ 1 ¼ 10 yA1 ¼ 0:00187 yB1 ¼ 0:00212

Stripping

235

Note that a simplified equation is available for calculating the total outlet concentration or loading for a given unit when more than two components are absorbed: n X

yi,1 ¼

n X (1  Ai )( yi,in )

i

i

1  ANþ1 i

(10:24)

where i is the component in question and n is the number of components. Using the above results, calculate the number of theoretical plates required and the height of the plate tower, Z: N ¼ (N þ 1)  1 ¼ 9 Z¼ ¼

Nh E0 9(2) ¼ 36 ft 0:5

(10:21) B

The reader should note that there presently exists little to no information available in the literature for a simple treatment of multicomponent absorption. The chemical engineering literature does provide a “shortcut” method where variations in flow rates and temperature are taken into account. However, the shortcut method requires a double trial-and-error calculation. A rigorous technique—involving tray to tray calculations—is also available. The method presented here, assuming ideal conditions, can also be solved graphically. Furthermore, it can be set up to solve for either the outlet concentration or the required liquid flow to achieve a given separation in a particular tower. Finally, this simplified technique can also be used to design or predict the performance of packed towers. Here Z ¼ (HOG )A (NOG )A ¼ (HOG )B (NOG )B ¼    where it has been assumed (HOG )A ¼ (HOG )B .

STRIPPING Quite often, an absorption column is followed by a liquid absorption process in which the gas solute is removed from the absorbing medium by contact with an insoluble gas. This operation is called “stripping” and is utilized to regenerate the solute “rich” solvent so that it can be recycled back to the absorption unit. The rich solution enters the stripping unit and the volatile solute is stripped from solution by either reducing the pressure, increasing the temperature, using a stripping gas to remove the vapor solute dissolved in the solvent, or any combination of these process options. While the concept of stripping is opposite to that of absorption, it is treated in the same manner. The operating line developed for absorption (see Illustrative Example 10.1) can be applied to a stripping unit (see Fig. 10.18 for component A). As was developed in Equation (10.4), the operating line for a stripping unit is also given by Lm yA1  yA2 ¼ Gm xA1  xA2

(10:4)

236

Chapter 10 Absorption and Stripping Rich gas Gm2

yA2

Lm2 Rich solution

xA2

Stripping gas Gm1 yA1 Lean solution Lm1 xA1

Figure 10.18 Stripping unit.

However, since this process is the opposite of absorption, solute is transferred from the liquid to the gas, and thus the operating line lies below the equilibrium curve. When absorption was addressed, a minimum liquid to gas ratio, (Lm/Gm )min, could be set in order to determine limits on the design. However, for stripping operations, a minimum gas to liquid ratio, (Gm/Lm)min, is now used and corresponds to the minimum gas rate required to achieve the desired separation. The minimum gas to liquid ratio can be found by the following procedure. 1 As shown in Figure 10.19, a line is drawn from the point ( yA1, xA1), which represents of the mole fractions of solute in the stripping gas feed and the stripped liquid stream, respectively, to the intersection of x ¼ xA2 (which represents the mole fraction of the liquid stream to be stripped) and the equilibrium line. If a plot of the equilibrium data results in a curve, then a tangent is drawn on the curve at the point corresponding to the value of the inlet rich solution concentration (xA2). 2 Obtain yA2 by reading or calculating the corresponding value on the y-axis. 3 Rearrange Equation (10.5) in terms of (Gm/Lm)min and insert the known quantities. 4 Multiply the results of (3) by the liquid flow rate to obtain the minimum stripping gas rate. 5 Typically, an actual value of 1.3(Gm/Lm)min is employed to assure an efficient separation.

Stripping

237

Equilibrium curve yA2

(G m/L m) min

yact

y (G m/L m) act yA1

x

xA2

xA1

Figure 10.19

Minimum gas to liquid ratio.

Other calculations essentially remain the same for both packed and plate towers, except that the height of a packed tower is given by Z ¼ HOL NOL

(10:25)

where  ln NOL ¼

  x2  ( y1 =m) (1  A) þ A x1  ( y1 =m) 1A

(10:26)

ILLUSTRATIVE EXAMPLE 10.19 Following the absorption of ethylene oxide (EO) from a process stream with water, the water stream is stripped of the EO as part of the regeneration step using steam. For a feed EO concentration of 0.4 mol%, determine the actual amount of steam required for stripping EO to a concentration of 0.02 mol% if the liquid flow rate is 500 lbmol/h and 30% excess EO-free steam is required for the separation in a packed column with 1 inch Raschig rings. Also, determine the diameter of the column. Assuming that Henry’s law applies, use the equation y ¼ 20x as the EO equilibrium data for this system. The system is at 30 psia and employs saturated steam at system conditions. SOLUTION: Generate the EO equilibrium data for x ¼ 0 to 0.008 if y ¼ 20x (see Table 10.9). The plot of the equilibrium line is provided in Figure 10.20. Determine the minimum gas to liquid ratio. Refer to Figure 10.20. yA2 ¼ 0:08

238

Chapter 10 Absorption and Stripping Table 10.9 Ethylene-Oxide Equilibrium Data y

x

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12

0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045 0.005 0.0055 0.006

Since Gm xA1  xA2 ¼ Lm yA1  yA2

(10:5)

Substituting gives 

Gm Lm

 ¼ min

0:0005  0:004 0  0:08

¼ 0:04375 0.14

EO gas mole fraction

0.12 0.1 yA2

0.08 0.06 0.04 0.02 0 0

(yA1, xA1)

0.001

0.002

0.003 0.004 0.005 EO liquid mole fraction

Figure 10.20 Stripping of EO-water system.

0.006

0.007

0.008

Stripping

239

The actual steam rate is therefore   Gm ¼ (1:3)0:04375 Lm act ¼ 0:0569 and _ st ¼ 0:0569(500)(18) m ¼ 511:8 lb steam=h  0:5 L rG Determine the abscissa, from Figure 10.11: G rL      0:5 L rG 0:5 Lm 18 rG ¼ G rL Gm 18 rL The density of steam can be found from the ideal gas law:

r¼

P(MW) (30)(18) ¼ RT (10:73)(250:34 þ 460) ¼ 0:07085 lb=ft3

Note that the temperature used is the saturated steam condition at 30 psia. Since rL ¼ 62.4 lb/ft3,   1 18 0:07085 0:5 ¼ 0:592 0:0569 18 62:4 From Figure 10.11, the ordinate is 0.048. Solve for the flooding gas mass velocity, Gf, in lb/ft2 . s. 0:048 ¼

Gf2 FCmL0:2 rL rgc

Substituting C ¼ 1 (ratio of water to liquid density) mL ¼ 0.19 cP F ¼ 160 (for 1 inch Raschig rings) gc ¼ 32.2 ft . lb/lbf . s2 leads to Gf ¼ 0:0595 lb=s  ft2 The actual gas mass velocity, Gact, is 60% of the flooding velocity: Gact ¼ 0:6Gf ¼ 0:6(0:0595) ¼ 0:0357 lb=s  ft2

240

Chapter 10 Absorption and Stripping

Finally, calculate the diameter of the column in ft: S¼

_ st m 511:8 ¼ Gact 0:0375(3600) ¼ 3:98 ft2 D ¼ 1:13S 0:5 ¼ 2:25 ft

(10:6)

B

ILLUSTRATIVE EXAMPLE 10.20 An atmospheric packed tower air stripper is used to clean contaminated ground-water with a concentration of 100 ppm trichloroethylene (TCE). The stripper was designed such that the packing height is 13 ft, the diameter is 5 ft, and the height of a transfer unit (HTU) is 3.25 ft. Assume that Henry’s law applies with a constant (H ) of 324 atm at 688F. Also, at these conditions, the molar density of water is 3.47 lbmol/ft3 and the air–water mole ratio (Gm/Lm) is related to the air –water volume ratio (G00 /L00 ) through G00 /L00 ¼ 130 Gm/Lm, where the units of G00 and L00 are ft3/s . ft2. If the stripping factor (R) used in the design is 5.0, what is the removal efficiency? In addition, the following equation has been developed for the calculation of the number of transfer units (NTUs) for an air –water stripping system and is based on the stripping factor R and the inlet/outlet concentrations:     R (Cin =Cout )(R  1) þ 1 NTU ¼ ln (10:27) R1 R where

Cin ¼ inlet contaminant concentration, ppm Cout ¼ outlet contaminant concentration, ppm R ¼ stripping factor

SOLUTION:

As described earlier, the height of packed tower can be calculated by Z ¼ (NOG )(HOG ) ffi (NTU)(HTU)

Rearranging Equation (10.27), one obtains Cout ¼ ¼

Cin (R  1) ; R exp½(NTU)(R  1)=R  1

NTU ¼ 13=3:25

(100)(5:0  1) (5:0) exp[(13=3:25)(5:0  1)=5:0]  1

¼ 3:3 The removal efficiency (RE) is then RE ¼ [(Cin  Cout )=Cin ]100% ¼ [(100  3:3)=100]100% ¼ 96:7%

B

Packed vs Plate Tower Comparison

241

PACKED VS PLATE TOWER COMPARISON Of the various types of gas absorption devices, packed columns and plate columns are the most commonly used in practice. Although packed columns are used more often, both have their special areas of usefulness, and the relative advantages and disadvantages of each are worth considering. In general: 1 The pressure drop of the gas passing through the packed column is smaller. 2 The plate column can stand an arbitrarily low liquid feed and permits a higher gas feed than the packed column. It can also be designed to handle liquid rates that would ordinarily flood the packed column. 3 If the liquid deposits a sediment, the plate column is more advisable. By fitting the column with manholes, the plate column can be cleaned of accumulated sediment that would clog many packing materials and warrant necessary costly removal and refilling of the column. Packed columns are also susceptible to plugging if the gas contains particulate contaminate(s). 4 In mass transfer processes accompanied by considerable heat effects, cooling or heating the liquid is much easier in the plate column. A system of pipes immersed in the liquid can be placed on the plates between the caps, and heat can be removed or supplied through the pipe wall directly to the area in which the process is taking place. The solution of the same problem for a packed column leads to the division of this process into a number of sections, with the cooling or heating of the liquid taking place between these sections. 5 The total weight of the plate column is usually less than the packed column designed for the same capacity. 6 A well-installed plate column avoids serious channeling difficulties insuring good, continuous contact between the gas and liquid throughout the column. 7 In highly corrosive atmospheres, the packed column is simpler and cheaper to construct. 8 The liquid holdup in the packed column is considerably less than in the plate column. 9 Temperature changes are apt to do more damage to the packed column than to the plate column. 10 Plate columns are advantageous for absorption processes with an accompanying chemical reaction (particularly when it is not very rapid). The process is favored by a long residence time of the liquid in the column and by easier control of the reaction. 11 Packed columns are preferred for liquids with high foaming tendencies. 12 The relative merits of the plate column and packed column for a specified purpose are normally determined only by comparison of the actual cost figures resulting from a detailed design analysis for each type. Most conditions

242

Chapter 10 Absorption and Stripping

being equal, packed columns in the smaller sizes (diameters up to 2 or 3 ft) are on the average less expensive. In the large sizes, plate columns tend to be the more economical.

SUMMARY OF KEY EQUATIONS The key equations for absorption and stripping calculations for tower height, including a summary of earlier material, are presented below. For packed tower absorption     y1  mx2 1 1 1 þ ln y2  mx2 A A (10:13) NOG ¼ 1 1 A For stripping

 ln NOL ¼

  x2  (y1 =m) (1  A) þ A x1  (y1 =m) 1A

(10:26)

where the subscripts 1 and 2 refer to bottom and top conditions, respectively. In addition, A ¼ Lm/mGm and S ¼ 1.0/A. To use Figure 10.12 for stripping calculations, replace the y coordinate, x coordinate, and parameter by [x1 2 ( y1/m)]/[x2 2 ( y2/m)], NOL, and S, respectively, where S ¼ 1.0/A ¼ mGm/Lm. For plate tower absorption     yNþ1  mx0 1 1 þ log 1 y1  mx0 A A N¼ (10:17) log A Note: The term ln, rather than log, may also be employed in both the numerator and denominator. If A approaches unity, Equation (10.17) becomes N¼

yNþ1  y1 y1  mx0

(10:28)

or yNþ1  y1 N ¼ yNþ1  mx0 N þ 1

(10:23)

Note that the subscripts 1 and N refer to the top and bottom of the column, respectively. For stripping in plate towers     x0  (yNþ1 =m) 1 1 1 þ log xN  (yNþ1 =m) S S (10:29) N¼ log S

References

243

or x0  xN SNþ1  S ¼ Nþ1 x0  ( yNþ1 =m) S 1

(10:30)

If S is approximately 1.0, one may use either of the following equations: N¼

x0  xN xN  ( yNþ1 =m)

x0  xN N ¼ x0  ( yNþ1 =m) N þ 1

(10:31) (10:32)

To use Figure 10.16 for stripping, replace the y coordinate and the parameter A by [xN 2 ( yNþ1/m)]/[x0 2 ( yNþ1/m)] and S, respectively.

REFERENCES 1. L. THEODORE and A. BUONICORE, “Control of Gaseous Emissions,” USEPA Training Manual, Research Training Pack, NC, 1982. 2. L. THEODORE, personal notes, 1979. 3. Generalized Pressure Drop Correlation, Chart No. GR-109, Rev. 4, U.S. Stoneware Co., Akron, OH, 1963. 4. N. CHEN, “New Equation Gives Tower Diameter,” Chem. Eng. New York City, NY, May 2, 1962. 5. L. THEODORE, “Engineering Calculations: Sizing Packed-Tower Absorbers without Data,” Chem. Eng. Progress, New York City, NY, pp. 18–19, May, 2005. 6. L. THEODORE, personal notes, 1986. 7. R. TREYBAL, “Mass Transfer Operations,” 2nd edition, McGraw-Hill, New York City, NY, 1967. 8. E. HENLEY and H. STAFFIN, “Stage Process Design,” John Wiley & Sons, Hoboken, NY, 1963. 9. H. SAWISTOWSKI and W. SMITH, “Mass Transfer Process Calculations,” Interscience, New York City, NY, 1963. 10. A. KREMSER, Nat’l Petrol. News, Washington, DC, 22(21), 42, 1930. 11. M. SOUDERS and G. BROWN, Ind. Eng. Chem., New York City, NY, 24, 519, 1932. 12. N. CHEN, “Calculating Theoretical Plates on Absorbers or Strippers,” Chem. Eng. New York City, NY, May 11, 1964. 13. L. THEODORE and J. BARDEN, “Mass Transfer Operations,” A Theodore Tutorial, East Williston, NY, 1995.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

11

Adsorption(1,2) INTRODUCTION It is already well established that the molecular forces at the surface of a liquid are in a state of imbalance or unsaturation. The same is true of the surface of a solid, where the molecules or ions on the surface may not have all their forces satisfied by union with other particles. As a result of this unsaturation, solid and liquid surfaces tend to satisfy their residual forces by attracting and retaining onto their surfaces gases or dissolved substances with which they come in contact. This phenomenon of the concentration of a substance on the surface of a solid (or liquid) is called adsorption. Thus, the substance attracted to a surface is said to be the adsorbed phase or adsorbate, while the substance to which it is attached is the adsorbent. Adsorption should be carefully distinguished from absorption, the later process being characterized by a substance not only being retained on a surface, but also passing through the surface to become distributed throughout the phase. Where doubt exists as to whether a process is true adsorption or absorption, the noncommittal term “sorption” is sometimes employed. The study of adsorption of various gases (or vapors) on solid surfaces has revealed that the forces operative in adsorption are not the same in all cases. Two types of adsorption are generally recognized: “physical” or van der Waals adsorption and “chemical” or activated adsorption. Physical adsorption (physisorption) is the result of intermolecular forces of attraction between molecules of the solid and the substance adsorbed. When, for example, the intermolecular attractive forces between a solid and a gas (or vapor) are greater than those existing between molecules of the gas itself, the gas will condense upon the surface of the solid even though its pressure may be lower than the vapor pressure corresponding to the prevailing temperature. The adsorbed substance does not penetrate within the crystal lattice of the solid and does not dissolve in it but remains entirely upon the surface. Should the solid, however, be highly porous, containing many fine capillaries, the adsorbed substance will penetrate these interstices if it “wets” the solid. The partial pressure of the adsorbed substance at equilibrium equals that of the contacting gas phase, and by lowering the pressure of the gas phase, or by

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

245

246

Chapter 11 Adsorption

raising the temperature, the adsorbed gas is readily removed or desorbed in unchanged form. Physical adsorption is characterized by low heats of adsorption (approximately 40 Btu/lbmol of adsorbate) and by the fact that the adsorption equilibrium is both reversible and established rapidly. This latter point allows one to simplify calculations in most real-world adsorption applications. Chemisorption, or activated adsorption, on the other hand, is the result of chemical interaction between the solid and the adsorbed substance. The strength of the chemical bond may vary considerably and identifiable chemical compounds in the usual sense may not form. Nevertheless, the adhesive forces are generally much greater than that found in physical adsorption. Chemisorption is also accompanied by much higher enthalpy changes (ranging from 80 to as high as 400 Btu/lbmol) with the heat liberated being on the order of the enthalpy of chemical reaction. The process is frequently irreversible, and, on desorption, the original substance will often have undergone a chemical change. Although it is probable that all solids adsorb gases (or vapors) to some extent, adsorption, as a rule, is not very pronounced unless an adsorbent possesses a large surface area for a given mass. For this reason, such adsorbents as silica gel, charcoals, and molecular sieves are particularly effective as adsorbing agents. These substances have a very porous structure and, with their large exposed surface, can take up appreciable volumes of various gases. The extent of adsorption can be increased further by “activating” the adsorbents in various ways. For example, wood charcoal can be “activated” by heating between 350 and 10008C in a vacuum or in air, steam, and certain other gases to a point where the adsorption of carbon tetrachloride, (e.g., at 248C) can be increased from 0.011 g/g of charcoal to 1.48. The activation apparently involves a distilling out of hydrocarbon impurities and thereby leads to exposure of a larger free surface for possible adsorption. The amount of gas adsorbed by a solid depends on a host of factors, including the surface area of the adsorbent, the nature of the adsorbent and gas being adsorbed, the temperature, and the pressure of the gas. Since the adsorbent surface area cannot always be readily determined, common practice is to employ the adsorbent mass as a measure of the surface available and to express the amount of adsorption per unit mass of adsorbing agent used. A concept, which becomes especially important in determining adsorbent capacity, is that of “available” surface, i.e., surface area accessible to the adsorbate molecule. It is apparent from pore size distribution data that the major contribution to surface area is located in pores of molecular dimensions. It seems logical to assume that a molecule, because of steric effects, will not readily penetrate into a pore smaller than a certain minimum diameter—hence, the concept that molecules are screened out. This minimum diameter is the so-called critical diameter and is a characteristic of the adsorbate and related to molecular size. Thus, for any molecule, the effective surface area for adsorption can exist only in pores that the molecule can enter. Figure 11.1 attempts to illustrate this concept for the case in which two adsorbate molecules in a solvent (not shown) compete with each other for an adsorbent surface site.

Adsorption Classification

Figure 11.1

247

˚ ). Concept of molecular screening in micropores (diameter range ¼ 10 to 1000 A

ADSORPTION CLASSIFICATION Four important adsorbents widely used industrially will be briefly considered, namely, activated carbon, activated alumina, silica gel, and molecular sieves. The first three of these are amorphous adsorbents with a nonuniform internal structure. Molecular sieves, however, are crystalline and therefore have an internal structure of regularly spaced cavities with interconnecting pores of definite size. Details of the properties peculiar to the various materials are best obtained directly from the manufacturer. The following is a brief description of these principal adsorbents.

248

Chapter 11 Adsorption Table 11.1 Properties of Activated Carbon Bulk density Heat capacity Pore volume Surface area Average pore diameter Regeneration temperature (steaming) Maximum allowable temperature

22–34 lb/ft3 0.27–0.368F Btu . lb 0.56–1.20 cm3/g 600– 1600 m2/g ˚ 15–25 A 100– 1408C 1508C

Activated Carbon Charcoal, an inefficient form of activated carbon, is obtained by the carbonization of wood. Various raw materials have been used in the preparation of adsorbent chars, resulting in the development of active carbon, a much more adsorbent form of charcoal. Industrial manufacture of activated carbon is obtained from shells or coal that is subjected to heat treatment in the absence of air, in the case of coal, followed by steam activation at high temperature. Other substances of a carbonaceous nature also used in the manufacture of active carbons include wood, coconut shells, peat, and fruit pits. Zinc chloride, magnesium chloride, calcium chloride, and phosphoric acid have also been used in place of steam as activating agents. Some approximate properties of typical granular gas adsorbent carbons are given in Table 11.1. Gas adsorbent carbons find primary application in gas purification, solvent recovery (hydrocarbon vapor emissions), and pollution control.

Activated Alumina Activated alumina (hydrated aluminum oxide) is produced by special heat treatment of the precipitate of native alumina or bauxite. It is available in either granular or pellet form with typical properties given in Table 11.2. Activated alumina is primarily used for the drying of gases and is particularly useful for the drying of gases under pressure. Table 11.2 Properties of Activated Alumina Density in bulk granules Density in pellets Heat capacity Pore volume Surface area Average pore diameter Regeneration temperature Stable up to

38 –42 lb/ft3 54 –58 lb/ft3 0.21 –0.25 Btu/lb . 8F 0.29 –0.37 cm3/g 210–360 m2/g ˚ 18 –48 A 200–2508C 5008C

Adsorption Classification Table 11.3

249

Properties of Silica Gel

Bulk density Heat capacity Pore volume Surface area Average pore diameter Regeneration temperature Stable up to

44 –46 lb/ft3 0.22 –0.26 Btu/lb . 8F 0.37 cm3/g 750 m2/g ˚ 22 A 120–2508C 4008C

Silica Gel The manufacture of silica gel consists of the neutralization of sodium silicate by mixing with dilute mineral acid, washing the gel formed free from salts produced during the neutralization reaction, followed by drying, roasting, and grading processes. It is generally used in granular form, although bead forms are available (see Table 11.3). Silica gel also finds primary use in gas drying and also finds application in gas desulfurization and purification.

Molecular Sieves Unlike the amorphous adsorbents (i.e., activated carbon, activated alumina, and silica gel), molecular sieves are crystalline, being essentially dehydrated zeolites (i.e., alumino-silicates in which atoms are arranged in a definite pattern). The complex structural units of molecular sieves have cavities at their centers to which access is by pores or “windows”. For certain types of crystalline zeolites, these pores are precisely uniform in diameter. Due to the crystalline porous structure and the near precise uniformity of the small pores, the adsorption phenomena only takes place with molecules that are of small enough size and of suitable shape to enter the cavities through the pores. The fundamental building block is a tetrahedron of four oxygen

Table 11.4 Properties of Molecular Sieves

Type Density in bulk (lb/ft3) Heat capacity (Btu/lb . 8F) Effective diameter of ˚) pores (A Regeneration temperature Stable up to (short period)

Anhydrous sodium aluminosilicate

Anhydrous calcium aluminosilicate

4A 44 0.19 4

5A 44 0.19 5

200 –3008C 6008C

200–3008C 6008C

Anhydrous aluminosilicate 13X 38 – 13 200–3008C 6008C

250

Chapter 11 Adsorption

anions surrounding a smaller silicon or aluminum cation. The sodium ions or other cations serve to make up the positive charge deficit in the alumina tetrahedra. Each of the four oxygen anions is shared, in turn, with another silica or alumina tetrahedron to extend the crystal lattice in three dimensions. The resulting crystal is unusual in that it is honeycombed with relatively large cavities, each cavity connected with six adjacent ones through apertures or pores. The sieves are manufactured by hydrothermal crystal growth from aluminosilicate gels followed by heat treatment to effect dehydration. Typical properties are given in Table 11.4.

ADSORPTION EQUILIBRIA The adsorption process involves three necessary steps. The fluid carrying the adsorbate must first come in contact with the adsorbent, at which time the adsorbate is preferentially, or selectively, adsorbed on the adsorbent. Next, the unadsorbed fluid must

Figure 11.2 Vapor–solid adsorption equilibrium isotherms of some hydrocarbons on silica gel.

Adsorption Equilibria

251

be separated from the adsorbent-adsorbate, and, finally, the adsorbent must be regenerated by removing the adsorbate or discarding the used adsorbent and replacing it with fresh material. Regeneration is performed in a variety of ways, depending on the nature of the adsorbate complex. Gases or vapors are usually desorbed by either varying the temperature (thermal cycle) or reducing the pressure (pressure cycle) of the adsorbent-adsorbate. The more popular thermal cycle is accomplished by passing hot gas through the adsorption bed in the opposite direction to the flow during the adsorption cycle. This ensures that the gas passing through the unit during the adsorption cycle always meets the most active adsorbent last and that the adsorbate concentration in the adsorbent at the outlet end of the unit is always maintained at a minimum. In the first step mentioned above where the molecules of the fluid come in contact with the adsorbent, an equilibrium is established between the adsorbed fluid and that remaining in the fluid phase. Although adsorption equilibrium and equilibrium relationships were reviewed in Chapter 6, material associated specifically with the adsorption process was not included. Because of the unique nature of adsorption equilibrium and the method of representation, this topic is included in the next section. Figures 11.2– 11.4 show experimental equilibrium adsorption isotherms.

Figure 11.3

Vapor–solid equilibrium isotherms of some hydrocarbons on activated carbon.

252

Chapter 11 Adsorption

Figure 11.4 Vapor–solid equilibrium isotherms on molecular sieves.

Consider Figure 11.2, where the concentration of adsorbed gas on the solid is plotted against the (equilibrium) partial pressure, p, of the adsorbate vapor or gas at constant temperature. At 408C, for example, pure propane vapor at a pressure of 550 mm Hg is in equilibrium with an adsorbate concentration at point P of 0.04 lb adsorbed propane/lb silica gel. Increasing the pressure of the propane will cause more propane to be adsorbed, while decreasing the pressure of the system at P will cause propane to be desorbed from the carbon. As described earlier, adsorption is an exothermic process; hence, the concentration of adsorbed gas decreases with increased temperature at a given equilibrium pressure. This is evident from the behavior of the isotherm curves. The process of a gas being brought into contact with an evacuated porous solid, and part of it being taken up by the solid, is always accompanied by the liberation of heat. The extent to which the process is exothermic depends on the type of sorption and the particular system. For physical adsorption, the amount of heat liberated is usually equal to the latent enthalpy of condensation of the adsorbate plus the heat of wetting of the solid by the adsorbate. The heat of wetting is usually only a small fraction of the heat of adsorption. On the other hand, in chemisorption, the heat evolved approximates the enthalpy of chemical reaction. As discussed in Chapter 6, the relation between the amount of substance adsorbed by an adsorbent and the equilibrium pressure or concentration at constant temperature is called the adsorption isotherm. The adsorption isotherm is the most important and by far the most often used of the various equilibria data that can be measured. Five general types of isotherms have been observed in the adsorption of gases on solids. These are shown in Figure 11.5. In cases of chemisorption, only isotherms of Type I are encountered, while in physical adsorption, all five types occur.

Adsorption Equilibria

Figure 11.5

253

Types of adsorption isotherms. P 0 represents the saturation pressure.

Freundlich Equation In isotherms of Type I, the amount of gas adsorbed per given quantity of adsorbent increases relatively rapidly with pressure and then much more slowly as the surface becomes covered with gas molecules. To represent the variation of the amount of adsorption per unit area or unit mass with pressure, Freundlich proposed the following equation: Y ¼ kp1=n

(11:1)

where Y is the weight or volume of gas adsorbed per unit area or unit mass of adsorbent, p is the equilibrium partial pressure and k and n are empirical constants dependent on the nature of solid and adsorbate, and on the temperature. Equation 11.1 may be rewritten as follows. Taking logarithms of both sides, 1 log(Y) ¼ log(k) þ log( p) n

(11:2)

If the log(Y ) is now plotted against log(p), a straight line should result with the slope equal to (1/n) and an ordinate intercept equal to log(k). Although the requirements of the equation are met satisfactorily at lower pressures, the experimental points curve away from the straight line at higher pressures, indicating that this equation does not have general applicability in reproducing adsorption of gases by solids.

Langmuir Isotherms A much better equation for Type I isotherms was deduced by Langmuir from theoretical considerations. Langmuir postulated that gases, on being adsorbed by a solid surface, cannot form a layer more than a single molecule in depth. Further, he visualized the adsorption process as consisting of two opposing actions, a condensation of molecules from the gas phase onto the surface and an evaporation of molecules from the surface back into the body of the gas. When adsorption first begins, every molecule colliding with the surface may condense on it. However, as adsorption proceeds, only those molecules that occupy a part of the surface not already covered by adsorbed molecules may be expected to be adsorbed. The result is that the initial rate of

254

Chapter 11 Adsorption

condensation of molecules on a surface is high and then falls off as the surface area available for adsorption is decreased. On the other hand, a molecule adsorbed on a surface may, by thermal agitation, become detached from the surface and escape into the gas. The rate at which desorption will occur will depend, in turn, on the amount of surface covered by molecules and will increase as the surface becomes more fully saturated. These two rates, condensation (adsorption) and evaporation (desorption), will eventually become equal and when this happens, an adsorption equilibrium will be established. If u is the fraction of the total surface covered by adsorbed molecules at any instant, then the fraction of bare surface available for adsorption is (1 2 u). According to kinetic theory, since the rate at which molecules strike a unit area of a surface is proportional to the pressure of the gas, the rate of condensation of molecules should be determined both by the pressure and the fraction of bare surface, or k1(1 2 u)p, where k1 is a proportionality constant. If k2 is the rate at which molecules evaporate from a unit surface when the surface is fully covered, then for a fraction u of a fully covered surface, the rate of evaporation will be k2u. For adsorption equilibrium, these rates must be equal. Therefore, k1 (1  u)p ¼ k2 u or

u¼

k1 p bp ¼ k2 þ k1 p 1 þ bp

(11:3)

where b ¼ k1/k2. Now, the amount of gas adsorbed per unit area or per unit mass of adsorbent, Y, must obviously be proportional to the fraction of surface covered; hence, Y ¼ ku ¼

kbp ap ¼ 1 þ bp 1 þ bp

(11:4)

where the constant a has been written for the product kb. Equation (11.4) is the Langmuir adsorption isotherm. The constants a and b are characteristic of the system under consideration and are evaluated from experimental data. Their magnitude also depends on the temperature. The validity of the Langmuir adsorption equation at any one temperature can be verified most conveniently by first dividing both sides of Equation (11.4) by p and then taking reciprocals. The result is p 1 b ¼ þ p Y a a

(11:5)

Since a and b are constants, a plot of ( p/Y ) vs p should yield a straight line with slope equal to (b/a) and an ordinate intercept equal to (1/a). ILLUSTRATIVE EXAMPLE 11.1 Of the following vapors, which one would be most readily adsorbed using activated carbon? 1 H2O at 1408F 2 CH4 at 708F

255

Adsorption Equilibria 3 C4H10 at 1408F 4 C4H10 at 708F

SOLUTION: The general rule of thumb is that organics are more easily adsorbed on activated carbon. Furthermore, the higher the molecular weight, the more easily it can be captured (because of the larger molecular diameter). Finally, increasing the temperature decreases the adsorption capability. The correct answer is therefore (4). B

ILLUSTRATIVE EXAMPLE 11.2 The carbon dioxide adsorption isotherms for Columbia (“Columbia” is a registered trademark of Union Carbide Corporation) activated carbon are presented in Figure 11.6 for temperatures of 30, 50, and 958C. Determine the constants of the Freundlich equation at a temperature of 508C. SOLUTION:

For the Freundlich equation Y ¼ kp1=n

(11:1)

The data/calculations presented in Table 11.5 are obtained/generated from Figure 11.6.

Figure 11.6

CO2 adsorption isotherms on activated carbon.

256

Chapter 11 Adsorption Table 11.5 Y (cm3/g) 30 51 67 81 93 104

Calculations for the Freundlich Equation at 508C p (atm)

log Y

log p

1 2 3 4 5 6

1.477 1.708 1.826 1.909 1.969 2.017

0.000 0.301 0.477 0.602 0.699 0.778

Figure 11.7 Adsorption isotherms from Illustrative Example 11.2. For the Freundlich equation (1), the plot of (log Y ) vs (log p) in Figure 11.7 leads to the equation: Y ¼ 30p 0:7

ILLUSTRATIVE EXAMPLE 11.3 Refer to Illustrative Example 11.1. Determine the constants of the Langmuir equation.

B

Description of Equipment Table 11.6

SOLUTION:

257

Calculations for the Langmuir Equation at 508C p/Y

p

0.033 0.039 0.045 0.049 0.054 0.058

1 2 3 4 5 6

For the Langmuir equation, employ p 1 b ¼ þ p Y a a

(11:5)

and the information generated in Table 11.6 from Figure 11.6. From Figure 11.7 (see insert), a plot of ( p/Y ) vs p leads to the equation Y¼

35:7p 1 þ 0:186p

B

ILLUSTRATIVE EXAMPLE 11.4 Which equation provides a better fit to the experimental data discussed in the two previous illustrative examples. SOLUTION: As is evident from the lines drawn in Figure 11.7, the Langmuir equation provides the better fit. Strictly speaking, a regression coefficient should be calculated for both results to provide a better basis for comparison. B

DESCRIPTION OF EQUIPMENT Because of the high cost of maintenance of air recovery and purification systems for applications with high concentrations of organic vapors, scientists and engineers have been forced into researching and designing systems for solvent recovery. The result has been the development of three types of systems, differentiated by the manner in which the adsorbent bed is maintained or handled during both phases of the adsorption-regeneration cycle: (1) fixed or stationary bed, (2) moving bed, and (3) fluidized bed.(3) Figure 11.8 presents a flow diagram of a dual stationary-bed solvent recovery system with auxiliaries for collecting the vapor – air mixture from various point sources, then transporting through the particulate filter and into the on-stream carbon adsorber—in this case, bed 1. The effluent air, which is virtually free of vapors, is usually vented outdoors. The lower carbon adsorber (number 2) is regenerated during the service time of bed 1. A steam generator or other source of steam is

258

Chapter 11 Adsorption

Figure 11.8 Stationary-bed carbon system with auxiliaries for vapor collection and solvent separation from steam condensate.

required. The effluent steam – solvent mixture from the adsorber is directed through the condenser and the liquified mixture then passes into the decanter and/or distillation column for separation of the solvent from the steam condensate.(3) Figures 11.9 and 11.10 show two designs of a stationary-bed solvent recovery system.(3) The type shown employs vertical cylindrical bends wherein the solventladen air flows axially down through the bed. This particular design is advertised for use in the recovery of solvents used in degreasing and dry cleaning, although equally well-suited for recovery of solvents from other industries. Solvents mentioned are trichloroethylene, tetrachloroethylene, toluene, freon TF, and dichloromethane. Regeneration is accomplished with steam flowing upward through the bed and, since the above-mentioned solvents are essentially immiscible in water, decantation is used to separate the condensed solvent from the steam condensate. The valves

259

Figure 11.9

a

(a) Adsorption cycle; (b) desorption cycle.

b

260

Chapter 11 Adsorption

Figure 11.10 Flow diagram of a solvent recovery system.

are of disk type, opened and closed by air-driven pistons. Water is used as coolant in the condenser. Steam, electric power to drive the blower, and cooling water are three operating-cost items. The cost of steam and cooling water increases with frequency of regeneration. Figure 11.9 shows the features and arrangement of the component parts of a twoadsorber system. This particular unit is a Vic Model 572 with two adsorber beds used alternately, i.e., while one unit is on-stream adsorbing, the other is regenerating.(4) Dual adsorber systems can also be operated with both beds on-stream simultaneously, especially when solvent concentrations are low. In this situation, regeneration is less frequent than a full work shift and may be accomplished during off-work hours. Operation in parallel almost doubles the air-handling capacity of the adsorption unit and may be an advantage in terms of operating cost. The concept of the moving-bed system is illustrated in Figure 11.11. The rotary component of the system consists of four coaxial cylinders. The outer cylinder is impervious to gas flow except at the slots near the left end. They serve as air inlet ports where they are shown uncovered and as steam – vapor outlet ports as shown at the lower left end of the rotary. The carbon bed is retained between two cylinders made of screen or perforated metal and also segmented by partitions placed radially between the two cylinders. The inner cylinder is again impervious to gas flow except at the slots near the right end of the rotary. These slots serve as outlet ports for the discharge air and as inlet ports for the regenerating steam. On each rotation of the rotary, each segment of the bed undergoes adsorption and regeneration. The desorbed solvent can then be separated from the steam by decantation or distillation.(3) Because of the continuous regeneration capability of the rotary bed, more efficient utilization of the carbon is possible than with stationary-bed systems. In most solvent

Description of Equipment

261

a

b

Figure 11.11

Continuous rotary bed: (a) cross-sectional view; (b) horizontal exterior view.

recovery operations, the adsorption zone and the saturated bed behind it are idle but add to the bulk of the system and increase airflow resistance through the bed. In deep beds of 12 to 36 inches, as required in the stationary-bed systems, a large portion of the bed is idle at any one time.(3) By continuous regeneration, the regeneration time for each segment of the bed is shortened, and thereby shorter bed lengths can be used. This leads to two advantages: a more compact system and a reduced pressure drop. The disadvantages are those associated with wear on moving parts and maintaining seals in contact with moving parts. The use of shorter beds also decreases the steam utilization efficiency. Figure 11.12 shows a flow diagram for a fluidized-bed solvent recovery system.(3) The carbon is recirculated continuously through the adsorption– regeneration cycle. The spent carbon, saturated with solvent, is elevated to the surge bin and then passed down into the regeneration bed where it is contacted with an upward flow of steam. The regenerated carbon is then metered into the adsorber, where the carbon

262

Chapter 11 Adsorption

Figure 11.12 Fluidized-bed solvent recovery system.

traverses nine beds while fluidized with the upward flow of the vapor-laden air. An air velocity of approximately 240 ft/min is required to cause fluidization of the bed. In both the regeneration and adsorption phases, the carbon is moved countercurrent to the gas or vapor. The countercurrent movement increases the efficiency of regeneration (i.e., the steam : solvent ratio is less than for a stationary bed under otherwise comparable conditions). In addition to the beneficial effects of the countercurrent movement, the bed length can be increased to further improve steam utilization.

Description of Equipment

263

Countercurrent flow also increases the effective use of the carbon; more solvent can be recovered with less carbon than with stationary- or rotary-bed systems. By adjustments or balance of the carbon and solvent input rates, the total carbon in the adsorber can be made part of the adsorption zone reaching saturation in the lowest bed just before it is discharged into the elevator. Very little of the carbon is then idle; hence, maximum utilization is made of the carbon. The fact that the carbon has reached saturation when delivered to the regeneration phase is another factor in the reduced steam requirement; in this respect, the fluidizedbed system is most favorable. In rotary-bed operations, the carbon moved into the regeneration phase has reached the lowest state of saturation in the three systems. When large air volumes are treated and available space for the installation is at a premium, the smaller size and lower initial cost are definite advantages over the stationary-bed system. A serious disadvantage is that of high attrition losses of the carbon caused by the fluidization of the beds. Because of the attrition, filtration of the effluent airstream may be required. The influent airstream need not be highly filtered because plugging of the fluidized beds with particulate matter is minimal. The major manufacturer of these units has given consideration to discontinue operations. No adsorber system could operate alone without sufficient auxiliary equipment and the components required to collect, transport, and filter vapor-laden airstreams being delivered to the adsorber. Proper design of these components is necessary to provide proper service to the adsorber. The ducts and piping need to be sized properly for required air velocities to optimize the efficiency of the adsorber. If the air velocity is too high, the stationary-bed adsorber may become a fluidized-bed adsorber, or low flows may create severe channeling through the beds. The fan is the catalyst for forcing the gas stream in and out of the unit, so it is important that careful attention to design and sizing is given to this equipment.(5) Because there are several adsorber configurations, the location of a filter can be varied: before the inlet airstream (in a stationary bed) to reduce possible contamination to the adsorbent, or after the fluidized-bed adsorber to reduce particulate emissions. Monitoring of the filter efficiency can be accomplished by measuring the pressure drop across the filter using either a manometer or pressure gauge. Compressed-air systems are necessary in some adsorber systems for valve and damper operation. For best results, the air supply should be kept contaminant-free through the use of a filter installed close to the adsorber. The compressed air supply should be equipped with an in-line filter, pressure regulator, and lubricator. Several devices are installed in series following adsorption for recovery of the solvent after regeneration. Condensers and separators are examples of recovery devices. The condenser is installed just after the system for removal of the heat from the vapors. There are two basic types of condensers: surface condensers and contact condensers. In a surface condenser, the coolant does not contact the vapors or condensate. Most surface condensers are of shell-and-tube configurations. Water flows through the tubes and vapors condense on the shell side. In contact condensers, the coolant vapors and condensate are intimately mixed. These condensers are more flexible, simpler, and considerably less expensive to install. Sizing of this condenser is also more straightforward than the design for surface condensers.

264

Chapter 11 Adsorption

Figure 11.13 Water separator.

Separators (decanters) are installed following the condenser to separate the contaminant from the water. Figure 11.13 shows a typical separator used with activated carbon bed adsorbers; note, however, that water is generally the heavier phase. Separators work on the principle of gravitational forces where the heavier material to be separated is removed from the bottom of the canister and the lighter material is removed through a line located at the top of the canister. Water separators are more effective with single solvent applications, and only when the solvent is immiscible in water.

DESIGN AND PERFORMANCE EQUATIONS The movement of vapors through an adsorbent bed is often referred to as a dynamic process. The term “dynamic” refers to motion, both in the movement of gas through

Design and Performance Equations

Figure 11.14

265

Adsorption wavefront.

the adsorbent bed and the change in vapor concentration(s) as it moves through the bed. There are a variety of configurations in which the solute-ladened air stream and adsorbent are brought into contact. The most common configuration is to pass the air stream down through a fixed volume or bed of adsorbent. Figure 11.14 illustrates how adsorption (mass transfer) occurs as a binary solution containing an adsorbed solute passes down through the bed. The gas stream containing the solute at an initial concentration C0 is passed down through a (deep) bed of adsorbent material that is free of any solute. Most of the solute is readily adsorbed by the top portion of the bed. The small amount of solute that is left is easily adsorbed in the remaining section of the bed. The effluent from the bottom of the bed is essentially solute-free, denoted as C1. After a period of time, the top layer of the adsorbent bed becomes saturated with solute. The majority of adsorption (approximately 95%) now occurs in a narrow portion of the bed directly below this saturation section. The narrow zone of adsorption is referred to as the mass transfer zone (MTZ). As additional solute vapors pass through the bed, the saturated section of the bed becomes larger and the MTZ moves further down the length of the adsorbent. The actual length of the MTZ remains fairly constant as it travels through the adsorbent bed. Additional adsorption occurs as the vapors pass through the “unused” portion of the bed. The effluent concentration at C2 remains essentially zero, since there is still an unsaturated section of the bed.

266

Chapter 11 Adsorption

Finally, when the lower portion of the MTZ reaches the bottom of the bed, the concentration of solute in the effluent suddenly begins to rise. This is referred to as the breakthrough point (or breakpoint)—where untreated vapors are being exhausted from the adsorber. If the air stream is not switched to a fresh bed, the concentration of the solute in the outlet will quickly rise until it approaches the initial concentration, illustrated at point C4. It should be noted that in most air pollution control applications even trace amounts of contaminants in the effluent stream are undesirable.(1) To achieve continuous operation, adsorbers must be either replaced or cycled from adsorption to desorption before breakthrough occurs. In desorption or regeneration, the adsorbers solute vapors are removed from the used bed in preparation for the next cycle. Most commercial adsorption systems are the regenerable type. In regard to regenerable adsorption systems, three important terms are used to describe the equilibrium capacity (CAP) of the adsorbent bed. These capacity terms are measured in mass of vapor per mass of adsorbent. First, the breakthrough capacity (BC) is defined as the capacity of the bed at which unreacted vapors begin to be exhausted. The saturation capacity (SAT) is the maximum amount of vapor that can be adsorbed per unit mass of carbon (this is the capacity read from the adsorption isotherm) and thus is equal to CAP. The working capacity or working charge (WC) is the actual amount adsorbed in the adsorber. Thus, the working capacity is a certain fraction of the saturation capacity. Working capacities can range from 0.1 to 0.7 times the saturation capacity. (Note: A smaller capacity increases the amount of carbon required.) This fraction is set by the designer for individual systems by often balancing the cost of carbon and adsorber operation vs preventing breakthrough while allowing for an adequate cycle time. Another factor in determining the working capacity is that it is uneconomical to desorb all the vapors from the adsorber bed. The small amount of residual vapors left on the bed is referred to as the HEEL. This HEEL accounts for a large portion of the difference between the saturation capacity and the working capacity. In some cases, the working capacity can be estimated by assuming it to be equal to the saturation capacity minus the HEEL. The following equations and procedures may be used to estimate some of the terms described above for an adsorber bed of height Z: BC ¼ [(0:5)(CAP)(MTZ) þ (CAP)(Z  MTZ)]=Z

(11:6)

WC ¼ [(0:5)(CAP)(MTZ)=Z] þ [(CAP)(Z  MTZ)]=Z  HEEL ¼ BC  HEEL

(11:7)

As discussed, the working charge can sometimes be taken to be some fraction of the saturated (equilibrium) capacity of the adsorbent (CAP): WC ¼ ( f )(CAP)

(11:8)

0  f  1:0

(11:9)

with

Design and Performance Equations

267

ILLUSTRATIVE EXAMPLE 11.5 Calculate the working capacity of an adsorption bed given the saturation (equilibrium) capacity (CAP), mass transfer zone (MTZ), and HEEL provided. The depth of the adsorption bed is 3 ft, the saturation capacity is 39%, the MTZ is 4 in and the HEEL is 2.5%. SOLUTION:

Calculate the breakthrough capacity, BC. BC ¼

0:5(CAP)(MTZ) þ (CAP)(Z  MTZ) Z

(11:6)

Substituting, BC ¼

0:5(0:39)(4) þ (0:39)(36  4) ¼ 0:368 ¼ 36:8% 36

Calculate the working capacity, WC. WC ¼ BC  HEEL  PF; PF ¼ packing factor

(11:7)

Substituting, WC ¼ 36:8  2:5  0 ¼ 34:3%

B

For multicomponent adsorption, the working charge may be calculated from(6) WC ¼ Xn i¼1

1:0 (wi =CAP)

(11:10)

where n is the number of components, wi is the mass fraction of i in n components (not including the carrier gas) and CAPi is the equilibrium capacity of component i. For a two component system (A, B), the above equation reduces to WC ¼

(CAPA )(CAPB ) wA (CAPB ) þ wB (CAPA )

(11:11)

ILLUSTRATIVE EXAMPLE 11.6 Organic vapors in an air stream are to be recovered with an adsorber using activated carbon as the adsorbent. The organic vapor concentrations in the air stream are provided in Table 11.7. Calculate the “theoretical” working capacity and the “actual” capacity given a HEEL of 0.025 (fractional basis) and a fractional packing factor of 0.03 (negative). SOLUTION:

The working capacity is calculated from WC ¼ Xn i¼1

1:0 (wi =CAP)

(11:10)

268

Chapter 11 Adsorption Table 11.7

Equilibrium Data for Organic Vapor

wi (air-free basis)

Mass fraction, wi (air-free basis)

Equilibrium capacity, lb/lb

Methane Toluene Propane Diphenyl Benzyl alcohol

0.67 0.05 0.25 0.02 0.01

0.39 0.08 0.40 0.11 0.16

Substituting, one obtains WC ¼ ¼

1:0 (0:67=0:39) þ (0:05=0:08) þ (0:25=0:40) þ (0:02=0:11) þ (0:01=0:16) 1:0 1:718 þ 0:625 þ 0:625 þ 0:182 þ 0:063

¼ 0:3113 ¼ 31:13% This represents the “theoretical” or maximum working charge. The actual WC, including the HEEL effect, is WC ¼ 0:3113  0:025 ¼ 0:2863 ¼ 28:63% Including a packing factor leads to WC ¼ 0:3113  0:025  0:03 ¼ 0:2563 ¼ 25:63%

B

After determining the service time and/or working charge necessary for a particular application, there are several other possible factors to be considered: 1 The adsorbent particle size 2 The physical adsorbent bed depth 3 The gas velocity 4 The temperature of the inlet gas stream and the adsorbent 5 The solute concentration to be adsorbed 6 The solute concentration(s) not to be adsorbed, including moisture 7 The removal efficiency 8 Possible decomposition or polymerization on the adsorbent 9 The frequency of operation 10 Regeneration conditions 11 The system pressure Temperature has an inversely proportional effect on adsorption capacity, such that when temperature increases, the adsorption capacity decreases. The adsorption

Design and Performance Equations

269

Table 11.8 Adsorbent Heat Capacity Values (Ambient Conditions Btu/ft3 . 8F) Activated carbon Alumina Molecular sieve

0.25 0.21 0.25

process is exothermic. As the adsorption activity moves through the bed, a temperature front follows and heat is transferred to the gas stream. When the gas leaves the area of adsorption activity, the heat exchange reverses (gas will transfer heat to the bed). The temperature differential during the adiabatic operation of the adsorber can be estimated as follows: DT ¼

(Cp =Ci ) 

6:1 þ 0:51(Cp =Cst )

105

(11:12)

where DT is the temperature rise, 8F, Cst is the saturation capacity of bed at T þ DT, Ci is the inlet concentration (ppm), and Cp is the volumetric heat capacity of the adsorbent, Btu/ft3 . 8F (see Table 11.8). The design of fixed-bed adsorption systems also requires the need and capability of estimating the pressure drop through the bed. Ergun(7) derived a correlation to estimate the pressure drop for the flow of a fluid through a bed of packed solids when it alone fills the voids in the bed. This correlation is given by the relationship DPgc dp 13 75(1  1) þ 0:875 ¼ 2Z r v2 (1  1) Re

(11:13)

where DP is the pressure drop of gas in psf, Z is the depth of packing (ft), gc ¼ 4.18  108 ft . lb/lbf . h2, dp is the effective particle diameter (ft), 1 is the fractional void volume in dry-packed bed, r is the gas density (lb/ft3), v is the superficial velocity of the gas through the bed (ft/h) and Re is the Reynolds number. Information on different mesh carbon sizes is presented in Figure 11.15. This data is used in pressure drop calculations. There is a simpler Ergun equation provided by Union Carbide for molecular sieves:(8) DP fT CT G2 ¼ r dp Z

(11:14)

where CT is the pressure drop coefficient (ft . h2/in2), fT is the friction factor, G is the superficial mass velocity (lb/h . ft2) and the pressure drop is given in psi. The friction factor, fT, is determined from Figure 11.16 as a function of the modified Reynolds number. The pressure drop coefficient, CT, is also determined from the same figure, which has CT plotted as a function of 1. For molecular-sieve pellets, the effective particle diameter can be obtained from dp ¼ 2 3

þ

dc 1 3 (dc =lc )

(11:15)

270

Chapter 11 Adsorption

Figure 11.15 Activated carbon pressure drop curves (EPA chart). e

e

m

Figure 11.16 Pressure drop information for molecular sieves.

Design and Performance Equations

271

where dc is the particle diameter (ft) and lc is the particle length (ft). Suggested values for 1 and dp for various sizes of molecular sieves are available in the literature.(1) This section is concluded with a rather simplified overall design procedure for a system adsorbing an organic that consists of two horizontal units (one on/one off) that are regenerated with steam.(6,9) 1 Select adsorbent type and size. 2 Select cycle time; estimate regeneration time; set adsorption time equal to regeneration time; set cycle time to twice the regeneration time; generally, try to minimize regeneration time. 3 Set velocity; v is usually 80 ft/min, but may be increased to 100 ft/min. 4 Set steam/solvent ratio. 5 Calculate (or obtain) WC for above. 6 Calculate amount of solvent adsorbed during 12 cycle time (tads)ms. ms ¼ qcs t ads ;

cs ¼ inlet solvent concentration

(11:16)

7 Calculate adsorbent mass required, MAC, M AC ¼ ms =WC

(11:17)

8 Calculate adsorbent volume requirement, V AC ¼ M AC =rB ; rB ¼ adsorbent bulk density

(11:18)

9 Calculate the face area of the bed: AAC ¼ q=VAC

(11:19)

Z ¼ H ¼ (V=A)AC

(11:20)

10 Calculate bed height:

11 Set L/D (length-to-diameter) ratio. 12 Calculate L and D from A ¼ LD; constraints: L , 30 ft, D , 10 ft; L/D of 3 to 4 acceptable if v , 30 ft/min. 13 Design (structurally) to handle if filled with water. 14 Design vertically if q , 2500 actual cubic feet per minute (acfm).

ILLUSTRATIVE EXAMPLE 11.7 The R&D group at a local adsorbent manufacturer has recently developed a new granulated activated column (GAC) adsorbent, JB26, for the removal of common water pollutants. Some data have been collected on the adsorption isotherm for a few solutes, but extensive tests have not

272

Chapter 11 Adsorption

been conducted as of yet. However, a major client is very interested in the new adsorbent and would like to know approximately how long one of its 65 ft3 units could operate with JB26 before breakthrough would occur. The following information was given to estimate how many days a 56,000 gal/day unit could run until breakthrough. From limited testing, the isotherm of interest is approximated by YT ¼ 0:002C3:11 where YT is the ratio of lb of adsorbate to lb of adsorbent and C is the adsorbate concentration, mg/L. The bulk density of JB26 is 42 lb/ft3 and it will treat a stream with an inlet concentration (Ci) of 3.5 mg/L. In addition, the breakthrough concentration has been set at 0.5 mg/L. The breakthrough adsorption capacity typically ranges between 25 –50% (assume 50%) of the theoretical capacity, YT, which is determined from the proposed adsorption isotherm, evaluated at the initial solute concentration in solution, Ci. The time to breakthrough is then given by the following equation: tB ¼

YB MAC 8:34q[Ci  (CB =2)]

where tB is the time to breakthrough (days), YB is the adsorption capacity at breakthrough, MAC is the mass of carbon in the column (lb), q is the volumetric flow rate of solution (MMgal/day), and CB is the adsorbate concentration at breakthrough (mg/L). Note: 8.34 lb ¼ 1 MMgal . L/mg. SOLUTION:

The theoretical adsorption capacity YT is YT ¼ 0:002C 3:11 ¼ 0:002(3:5)3:11 ¼ 0:09842 lb adsorbate=lb adsorbent

Assume the actual value is 50% of the theoretical value (see comment above). Thus, YB ¼ (0:5)(0:09842) ¼ 0:04921 lb adsorbate=lb adsorbent The mass of carbon in the unit is MAC ¼ (65 ft3 )(42 lb=ft3 ) ¼ 2730 lb carbon The breakthrough time can now be calculated from the breakthrough equation provided above. tB ¼

(0:04921)(2730) (8:34)(0:56)[3:5  (0:5=2)]

¼ 8:85 days

B

ILLUSTRATIVE EXAMPLE 11.8 A degreaser ventilation stream containing trichloroethylene (TCE) is treated with a horizontal carbon bed adsorber. The adsorber is normally designed to operate at a gas flow of 8000 scfm (608F, 1 atm), and the concentration of TCE at the adsorber inlet is 1500 ppmv.

Design and Performance Equations

273

The capture efficiency of the adsorber is 99% (Ec ¼ 0.99) under normal design conditions. Design parameters are as follows: Actual conditions: 25 psia, 908F SAT ¼ 35% Z ¼ depth of bed ¼ 2.5 ft L ¼ length of adsorber ¼ 2.5 ft D ¼ diameter of adsorber ¼ 8 ft MTZ ¼ 5 in HEEL ¼ 2.0% Bulk density of carbon bed ¼ 35 lb/ft3 Determine the time before breakthrough occurs. SOLUTION:

The calculations are provided below:    14:7 90 þ 460 q ¼ 8000 25 60 þ 460 ¼ 4975 acfm qTCE ¼ (1500  106 )(4975) ¼ 7:46 acfm _ TCE ¼ m

P(MW)qTCE (25)(131:5)(7:46) ¼ (10:73)(90 þ 460) RT

¼ 4:16 lb=min 0:5(Cs )(MTZ) þ (Cs )(Z  MTZ) Z      5 5 þ (0:35) 2:5  0:5(0:35) 12 12 ¼ 2:5

BC ¼

(11:6)

¼ 0:32 WC ¼ BC  HEEL + SF (safety factor) ¼ 0:32  0:02  0 ¼ 0:30 ¼ 30% VAC ¼ (25)(8)(2:5) MAC

¼ 500 ft3 ¼ V AC rB ¼ (500)(35) ¼ 17,500 lb

t (to saturation) ¼ ¼

(11:7)

(11:8)

(WC)(MAC ) _ TCE Ec m (0:30)(17,500) (4:16)(0:99)

¼ 1275 min  21 hr

B

274

Chapter 11 Adsorption

ILLUSTRATIVE EXAMPLE 11.9 Refer to Illustrative Example 11.8. Recalculate the time before breakthrough occurs, based on the following transient condition. The adsorber system is on line for one hour at the previous (normal) design conditions when the inlet concentration of TCE rises to an average value of 2500 ppmv because of a malfunction in the degreaser process. The efficiency also drops to 97.5% during this time. Assume that the SAT remains the same. SOLUTION:

For these transient conditions mTCE (in carbon) ¼ (0:30)(500)(35) ¼ (0:30)(17,500) mTCE

¼ 5250 lbTCE , maximum (flow, first hour) ¼ (4:16)(60)(0:99) ¼ 247:1 lbTCE captured

The remaining capacity of the bed is now available for adsorption after the first hour, mTCE (after first hour) ¼ 5250  247:1 ¼ 5003 lbTCE t (to transient saturation) ¼ tts ¼

5003 _ TCEtransient (0:975) m

_ TCEtransient ¼ (4:16)(2500=1500) m ¼ 6:93 lb= min tts ¼

5003 (6:93)(0:975)

¼ 740 min ¼ 12:34 h The time to breakthrough, following the transient period, is tB ¼ t þ tts ¼ 60 þ 740:1 ¼ 800:1 min ¼ 13:34 h Thus, the time to breakthrough has been reduced from 21 to 13.3 h.

B

ILLUSTRATIVE EXAMPLE 11.10 Theodore Consultants has been selected by Flynn Chemicals Inc. to design an adsorber that treats a degreaser ventilation stream containing trichloroethylene (TCE). Flynn Chemicals has provided basic operating data to Theodore Consultants and requires the use of activated carbon as an adsorbent. Determine the volume of activated carbon required to treat the gas and the height of the adsorption column. Operating data is provided as follows: TCE MW ¼ 131.5 g/gmol Standard conditions ¼ 1 atm, 658F

Design and Performance Equations

275

Operating conditions ¼ 25 psia and 758F Bulk density or activated carbon ¼ 38 lb/ft3 Working capacity of activated carbon ¼ 25 lb TCE per 100 lb carbon Flow rate of air stream ¼ 12,000 scfm Inlet concentration of TCE ¼ 2000 ppm The adsorption column cycle is set at 5 hr in the adsorption mode, 2 hr in the heating and desorbing mode, 2 hr in cooling, and 1 hr in standby. The adsorber recovers 96% of TCE by weight. A horizontal unit with a cross sectional area of 5 ft by 20 ft is recommended. SOLUTION:

Convert the standard flow rate to actual flow rate using the ideal gas law: 

qA ¼ qS

TA TS



PS PA



   75 þ 460 14:7 ¼ 12,000 ¼ 7190:4 acfm ¼ 4:3  105 acfh 65 þ 460 25 The flow rate of TCE is therefore: qTCE ¼ yTCE qA ¼

2000 (4:3  105 ) ¼ 860 acfh 106

_ TCE ¼ qr ¼ 860[25(131:5)=(10:73)(535)] ¼ 492:5 lb=h m _ TCE,adsorbed ¼ (492:5)(0:96)(5) ¼ 2364 lb m VAC ¼ Z¼

_ TCE,adsorbed m 2364 ¼ 248:8 ft3 ¼ 0:25(38) WC(rB ) VAC 248:8 ¼ 2:49  2:50 ft ¼ AAC 5(20)

B

ILLUSTRATIVE EXAMPLE 11.11 Gas emissions are being collected from a landfill and must be treated before being released into the environment. There are several options for treatment. As one of the project engineers, you have been given the task to look at the use of a horizontal activated carbon adsorber to collect the methane in the gas stream (assume 95% removal). Perform the following calculations: 1 Mass of CH4 collected per operating period 2 Mass of activated carbon needed 3 Depth of AC bed The following data are provided: Flow rate ¼ 11,000 acfh Operating pressure of the adsorber ¼ 1 atm Operating temperature of the adsorber ¼ 708F

276

Chapter 11 Adsorption Time between regeneration ¼ 24 h Gas stream contains (by mole fraction): N2 0.10 CH4 0.50 CO2 0.40 Bulk density of activated carbon ¼ 30 lb/ft3 Width of AC bed ¼ 15 ft Length of AC bed ¼ 20 ft CAP ¼ 0.39 lb CH4/lb AC HEEL ¼ 0.05 lb CH4/lb AC

SOLUTION:

The mass flow rate of the methane is (applying the ideal gas law) _ M ¼ (0:5)(11,000)(14:7)(16)=(10:73)(530) m ¼ 227 lb=h

For the mass of methane collected in one day, mM ¼ 0:95(227)(24) ¼ 5176 lb A key assumption that should be made here is that only the methane contributes to the WC: WC ¼ 0:39  0:05 ¼ 0:34 The mass of carbon required is M AC ¼ mM =WC ¼ 5176=0:34 ¼ 15,224 lb

(11:17)

VAC ¼ M AC =rB ¼ 15,224=30

(11:18)

To complete the calculations,

¼ 508 ft3 Z ¼ V AC =AAC ¼ 508=(15)(20) ¼ 1:7 ft ¼ 20:4 in

(11:20)

B

ILLUSTRATIVE EXAMPLE 11.12 A vertical 10-ft-diameter vessel is used to adsorb 40 ppm benzene and 30 ppm pyridine from an air stream at 1 atm and 708F. The superficial velocity through the vessel is 64 fpm. The

Design and Performance Equations

277

adsorbent is 4  6 mesh activated carbon having a bulk density 30 lb/ft3. Inlet concentrations (lb/lb) indicate that the equilibrium capacities for benzene and pyridine are 0.12 and 0.19, respectively. The working charge can be assumed equal to 80% of the ideal value. Determine the pressure drop and the amount of adsorbent in the bed if the bed is in operation for 5 days a week, 24 hr a day. SOLUTION:

Preliminary calculations are provided below: AAC ¼ (p=4)D2 ¼ (0:7854)(10)2 ¼ 78:54 ft2 q ¼ (64)(78:54) ¼ 5027 ft3 =min

Applying the ideal gas law yields

rB ¼

(1)(78:11) ; (530)(0:7302)

MWB ¼ 78:11

¼ 0:20 lb=ft3

rP ¼ 0:20 lb=ft3 ; MWP ¼ 79:1 The volume and mass of each organic are now calculated: VB ¼ (5027)(60)(24)(5)(40=106 ) ¼ 1448 ft3 mB ¼ (1448)(0:2) ¼ 290 lb VP ¼ (5027)(60)(24)(5)(30=106 ) ¼ 1086 ft3 mP ¼ (1086)(0:2) ¼ 217 lb The total mass adsorbed is mT ¼ mB þ mP ¼ 290 þ 217 ¼ 507 lb Calculate the two mass fractions: wB ¼ 290=507 ¼ 0:57 wP ¼ 217=507 ¼ 0:43

278

Chapter 11 Adsorption

Noting that CAPB ¼ 0.12 and CAPP ¼ 0.19 and applying Equation (11.11), one obtains WCI ¼ ¼

(CAPB )(CAPP ) (wB )(CAPP ) þ (wP )(CAPB ) (0:12)(0:19) (0:571)(0:19) þ (0:428)(0:12)

¼ 0:143 The actual WC is WC ¼ (0:143)(0:8) ¼ 0:114 The mass and volume of carbon are therefore MAC ¼ 507=0:114 ¼ 4447 lb VAC ¼ 4447=30 ¼ 148 ft3 The height is given by Z ¼

VAC AAC

Substituting gives Z ¼ 148=78:54 ¼ 1:89 ft ¼ 22:6 in  23 in

B

ILLUSTRATIVE EXAMPLE 11.13 A printing company must reduce and recover the amount of toluene it emits from its Rotograve printing operation. The company submits some preliminary information on installing a carbon adsorption system. You, the primary consultant, are given the following information: Airflow ¼ 20,000 acfm (778F, 1 atm) Adsorption capacity for toluene ¼ 0.175 lb toluene/lb activated carbon Operation occurs at 10% of LEL (lower explosivity limit) for toluene in the exit air from printer LEL for toluene ¼ 1.2% Toluene molecular weight ¼ 92.1 Carbon bulk density (4  6 mesh) ¼ 30 lb/ft3 Working charge ¼ 60% of saturation capacity Regeneration just under one hour; assume 1.0 hr Maximum velocity through adsorber ¼ 100 fpm (ft/min) Determine the minimum size of adsorber you would recommend for a 1  1 system. Calculations should include the pertinent dimensions of the adsorber, the amount of carbon,

Design and Performance Equations

279

the depth of the bed and an estimate of the pressure drop. Also calculate the fan horsepower if the blower/motor efficiency is 58%. SOLUTION: Initially, base the calculations on a 1 hr regeneration time so that 1 hr of adsorption is available. Key calculations and results are provided below for the toluene (TOL) and activated carbon (AC). Design for operation at 10% of LEL. qTOL ¼ (20,000)(0:10)(0:012) ¼ 24 acfm _ TOL ¼ m

(24)(492=537)(92:1)(60) 359

¼ 338 lb=h pTOL ¼ (24=20,000)(14:7) ¼ (0:0012)(14:7) ¼ 0:01764 psia SAT ¼ 17:5% ¼ 0:175 lbTOL =lbAC WC ¼ (0:175)(0:60) ¼ 0:105 lbTOL =lbAC ¼ 10:5 lbTOL =100 lbAC MAC ¼ (338=0:105)(1:0) ¼ 3220 lbAC for one bed ¼ 6440 lbAC for both beds VAC ¼ 3220=30 ¼ 107 ft3 AAC ¼ 20,000=100 ¼ 200 ft2 Z ¼ 107=200 ¼ 0:535 ft ¼ 6:4 in Suggest a horizontal 10-ft diameter  20-ft-long design. Refer to Figure 11.15. At 100 ft/min, DP ¼ 0.625 in H2O/in bed. Therefore DPtotal ¼ (0:625)(6:4) ¼ 4:0 in H2 O HP ¼ qDP=E f ; E f ¼ blower fractional efficiency ¼

(20,000)(4:0)(5:2) (0:58)(33,000)

¼ 22 HP Note: This represents a marginal design since H is only slightly higher than 0.5 ft.

B

ILLUSTRATIVE EXAMPLE 11.14 You have been asked to design a system to recover a 1.3% by volume acetone mixture in air. The air stream flow rate is 4.32  107 acfd (actual ft3/day) at 208C and 1 atm. Your boss has given

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you plenty of latitude but suggests working within the operating conditions and design procedure suggested by the eminent Dr. L. Theodore. The most immediate adsorbent equilibrium data available indicates that the equilibrium capacity for acetone in the 4.0– 8.0% relative saturation range (0.04 , pi/pi0 , 0.08) is 0.35. It has been further suggested to employ 4  6 mesh carbon as the adsorbent and to operate with a 2-hr regeneration period. The average particle diameter can be assumed to be 0.0091 ft. The apparent and bulk density for all types of carbon particles available are 45 lb/ft3 and 26 lb/ft3, respectively. Assume v ¼ 80 fpm and WC ¼ 0.8 (CAP). Also provide horsepower (Ef ¼ 0.65) requirements (use the EPA chart, Fig. 11.15). SOLUTION:

First, calculate the relative saturation (RS) of the acetone on the gas stream: RS ¼ yA P=p0A ¼ pA =p0A ¼ (0:013)(760)=170 ¼ 0:0581 ¼ 5:81%

The equilibrium capacity is therefore 0.35, i.e., CAP ¼ 0:35 Following the usual design procedure q ¼ 4:32  107 =1440 ¼ 30,000 acfm For v ¼ 80 ft/min AAC ¼ 30,000=80 ¼ 375 ft2 _ A ¼ (0:013)(4:32  107 )(50)(273=293)=359 m ¼ 72,878 lb=day In a 2-hr period mA ¼ (72,878)(2=24) ¼ 6073 lb The working charge is WC ¼ (0:35)(0:8) ¼ 0:28 Therefore MAC ¼ 6073=0:28 ¼ 21,689 lb VAC ¼ 21,689=26 ¼ 834 ft3 Z ¼ VAC =AAC ¼ 834=375 ¼ 2:22 ft ¼ 27 in

Design and Performance Equations

281

From Figure 11.15 DP=Z ¼ 0:44 in H2 O=in bed DP ¼ (0:44)(27) ¼ 11:9 in H2 O Finally, HP ¼ (30,000)(11:9)=(33,000)(0:65) ¼ 16:6

B

ILLUSTRATIVE EXAMPLE 11.15 Small volatile organic compound (VOC) emission sources often use activated carbon that is available in small drums often referred to as canisters. An example of this is a modified form of Carbtrol model G-1, which is suitable for low airflow rates. The drum is not regenerated on site; it is returned to the manufacturer and a new drum delivered. A small pilot scale reactor uses this modified model G-1 adsorber to capture methylene chloride emissions in a 50 acfm nitrogen purge source. The following operating and design data are provided: Volumetric flow rate of nitrogen purge ¼ 50 acfm Molecular weight of methylene chloride ¼ 85 Operating temperature ¼ 708F Operating pressure ¼ 1.0 atm Saturation capacity ¼ 30 lb CH2Cl2/100 lb C Methylene chloride concentration ¼ 500 ppm Weight of carbon in drum ¼ 200 lb Height of adsorbent in drum ¼ 24 in Adsorption time ¼ 6 h Mass transfer zone (MTZ) ¼ 2 in HEEL ¼ negligible Based on the above data and information, estimate the number of purge streams that this G-1 model adsorber canister can treat to breakthrough. SOLUTION:

Calculate the working charge, WC, of the carbon drum (canister): WC ¼ [0:5(CAP)(MTZ) þ (CAP)(Z  MTZ)]=Z ¼ [0:5(0:3)(2) þ (0:3)(24  2)]=24 ¼ 0:2875

The drum capacity in lb of methylene chloride is therefore DC ¼ WC(WC ) ¼ 0:2875(200) ¼ 57:5 lb

(11:7)

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Determine the mole fraction of methylene chloride in the purge stream and the volumetric flow rate of methylene chloride in acfh: yMEC ¼

500 ¼ 0:0005 106

qMEC ¼ yMEC q ¼ 0:0005(50)(60) ¼ 1:5 acfh Calculate the density of the methylene chloride vapor in lb/ft3 using the ideal gas law:

rMEC ¼

P(MW) 1(85) ¼ ¼ 0:220 lb=ft3 RT 0:73(70 þ 460)

Calculate the weight of methylene chloride emitted per batch: W ¼ qMEC rMEC (6 hr=batch) ¼ 1:5(0:220)(6) ¼ 1:98 lb=batch Determine the maximum number of purges N per canister: N ¼ DC=W ¼ 57:5=1:98 ¼ 29

B

ILLUSTRATIVE EXAMPLE 11.16 Consider the adsorber system in Figure 11.17. It is designed to operate with a maximum discharge concentration of 50 ppm. Once the unit is installed and running, it operates with a

Figure 11.17 Adsorber not in compliance.

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283

discharge of 60 ppm. Rather than purchase a new unit, what options are available to bring the unit into compliance with the specified design concentration? SOLUTION: This is obviously an open-ended problem.(6) The suggestions and options (if possible) recommended for adsorbers can be found in an Illustrative Example 20.5 in Chapter 20. B

REGENERATION Adsorption processes in practice use various techniques to accomplish regeneration or desorption. The adsorption – desorption cycles are usually classified into four types, used separately or in combination. This important topic is now considered in more detail. 1 Thermal swing cycles using either direct heat transfer by contacting the bed with a hot fluid or indirect transfer through a surface, and reactivating the adsorbent by raising the temperature. A temperature between 300 and 6008F is usually reached, the bed is flushed with a dry purge gas or reduced in pressure, and then it is returned to adsorption conditions. High design loadings on the adsorbent can usually be obtained, but a cooling step is needed. 2 Pressure swing cycles use either a lower pressure or vacuum to desorb the bed. The cycle can be operated at nearly isothermal conditions with no heating or cooling steps. The advantages include fast cycling with reduced adsorber dimensions and adsorbent inventory, direct production of a high purity product, and the ability to utilize gas compression as the main source of energy. 3 Purge gas stripping cycles use an essentially nonadsorbent purge gas to desorb the bed by reducing the partial pressure of the adsorbed component. Such stripping is more efficient at higher operating temperatures and lower operating pressures. The use of a condensable purge gas has the advantages of reduced power requirements, which are gained by using a liquid pump instead of a blower, and an effluent stream that can be condensed to separate the desorbed material by simple distillation. 4 Displacement cycles use an adsorbable purge to displace the previously adsorbed material on the bed. The stronger the adsorption of the purge, the more completely the bed is desorbed using lesser amounts of purge, but the more difficult it becomes subsequently to remove the adsorbed purge itself from the bed. When deciding whether to employ a regenerative system, several factors should be considered. The principal consideration is that of economics. It is important to establish if recovery of the adsorbate will be cost-effective or if regeneration of the adsorbent is the prime consideration. If solvent recovery is the main objective, the design should be based on past experimental data to establish the ratio of sorbent fluid to recoverable adsorbent at different working capacities. Most systems today employ steam as the regenerating medium, but some of the new systems use a hot inert gas such as nitrogen.

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A short discussion on steam systems is now provided. A well-designed system will have a steam consumption in the range of 1 to 4 lb of steam per pound of recovered solvent. The steam entering the bed not only introduces heat but creates adsorption and capillary action of the moisture, which supplies additional heat for the desorption process. Certain parameters should be considered in the design of this stripping process: 1 Minimize the time required for regeneration. If continuous adsorption and recovery are required, multiple systems have to be installed. 2 The short regeneration time requires a higher steaming rate, thus increasing the heat duty of the condenser system. 3 The steam flow should be in a direction opposite to that of adsorption to prevent the possible accumulation of polymerizable substances. 4 To enable fast stripping and efficient heat transfer, it is necessary to sweep out the carrier gas from the adsorber and condenser system as fast as possible. 5 A larger fraction of the heat content of the steam is used to heat the adsorber vessel and the adsorbent; thus, it is essential that the steam condenses quickly in the bed. The steam should contain only a slight super-heat to allow condensation. 6 It is advantageous to use a low-retentivity carbon to enable the adsorbate to be stripped out easily. When empirical data is not available, the following heat requirements have to be taken into consideration: heat to the adsorbent and vessel, heat of adsorption and heat capacity of adsorbate leaving the adsorbent, latent and heat capacity of water vapor accompanying the adsorbate, heat in condensate steam, and, radiation and convection heat loss. During the desorbing cycle, condensation and adsorption will take place in the adsorbent bed, increasing the moisture content of the adsorbent. Also, a certain portion of the solute will remain; this is referred to as the aforementioned “HEEL.” To achieve a minimum efficiency drop from successive adsorption cycles, a drying and cooling cycle should occur before returning to the “adsorb” mode. When using high adsorbate concentrations, it may be desirable to leave some moisture; a moisture-free bed is desired with other solutes. In air pollution control applications where the pollutant concentrations are low, and, in addition, the pollutants may have no recovery value, steam regeneration may not be the best regenerating agent. Because of the adsorption at low concentrations, the vapors may be held tightly by the adsorbent, and, relative to the amounts of pollutant adsorbed, the amount of steam required would be large. Under these conditions, even slightly soluble organic compounds would be completely dissolved in the steam condensate. For these circumstances, the more economical approach is to regenerate using an inert noncondensable gas such as nitrogen or air, or if there is a danger of explosion, regenerate with an inert noncondensable gas such as flue gas. For example, miscible pollutants such as the solvents 4-methyl-2-pentanone and propanone would completely dissolve in large amounts of steam condensate. Their recovery by distillation would not be economical, thus creating a disposal problem if water pollution is be avoided. In these cases, the solution to the problem is to

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285

regenerate the adsorbent with a noncondensable gas and burn the released vapors immediately in a small thermal incinerator.

ILLUSTRATIVE EXAMPLE 11.17 Use the accompanying graph in Figure 11.18 to solve the following problem. Estimate the mass of CO2 that can be adsorbed by 100 lb of Davison 4A Molecular Sieve from a discharge gas mixture at 778F and 40 psia containing 10,000 ppmv (by volume) CO2. SOLUTION:

Calculate the mole fraction of CO2 in the discharge gas mixture: yCO2 ¼ ppm=106 ¼ 10,000=106 ¼ 0:01

Also determine the partial pressure of CO2 in psia and mm Hg: pCO2 ¼ yCO2 P ¼ (0:01)(40) ¼ 0:4 psia ¼ (0:4)(760=14:7) ¼ 20:7 mm Hg Estimate the adsorbent capacity, SAT, at 778F. Refer to Figure 11.18. SAT ¼ 9:8 lb CO2 =100 lb sieve (from Fig: 11:18):

Figure 11.18

Vapor–solid equilibrium isotherms.

B

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Chapter 11 Adsorption

ILLUSTRATIVE EXAMPLE 11.18 Refer to Illustrative Example 11.17. What percentage of this adsorbed vapor would be recovered by passing superheated steam at a temperature of 3928F through the adsorbent until the partial pressure of the CO2 in the stream leaving is reduced to 1.0 mm Hg? SOLUTION: Estimate the adsorbent capacity at 3928F and 1.0 mm Hg. Note that this represents the HEEL: HEEL ¼ 0:8 lb CO2 =100 lb sieve The amount of CO2 recovered is therefore CO2 recovered ¼ 9:8  0:8 ¼ 9:0 lb CO2 =100 lb sieve The percent recovery is % recovery ¼ (9:0=9:8)100 ¼ 91:8% Note that this represents the percent recovery relative to the HEEL.

B

ILLUSTRATIVE EXAMPLE 11.19 Refer to Illustrative Example 11.17. What is the residual CO2 partial pressure in a gas mixture at 778F in contact with the freshly stripped sieve in the previous example? SOLUTION: Estimate the partial pressure of CO2 in mm Hg in equilibrium at 778F with sieve containing 0.8 lb CO2/100 lb sieve (the HEEL): pCO2  0:05 mm Hg The equilibrium CO2 partial pressure may be converted to ppm: ppm ¼ ( pCO2 =P)106 ¼ (0:05)(14:7=760)(106 )=40 ¼ 24:1 This represents the discharge ppm following regeneration. The reader is left the exercise of calculating the percent recovery based on inlet and outlet concentrations (ppm). The data in the isotherm diagram were extrapolated in order to solve parts of this and the two previous examples. This is poor practice. In actual applications, data should be obtained at the operating condition for both the adsorption and regeneration phases. Also note that the last calculation can be extremely critical for some applications. It provides information on the concentration of the adsorbate in the clean gas discharge stream. Note that this can rarely be predicted a priori. B

Regeneration

287

As noted earlier, in order to maintain the concentration of the clean gas discharge stream at its lowest value, the regeneration step is almost always conducted in a direction opposite that for adsorption. A residual concentration gradient exists in the bed after regeneration, with the minimum value located at the inlet side of the regenerating stream. If one adsorbs down and regenerates up—as is typically done—the bottom of the bed will contain a smaller HEEL, thus providing more efficient recovery. Adsorbing down provides an additional advantage. Fine particles, impurities, polymeric materials, and high molecular weight hydrocarbons will be deposited/captured at the inlet or top side of the bed. If the performance of the bed is reduced because of this, one need only replace a small portion of the bed (an easy skimming operation) rather than replace the entire bed.

ILLUSTRATIVE EXAMPLE 11.20 A two-bed carbon adsorption system is used to control the odors being emitted from a drumfilling operation. The material being drummed is a high-purity grade of pyridine (C5H5N) that has a human odor detection level of 100 ppm. It has been reported that odor has been detected from outside the drumming area when the equipment is in service, i.e., when drums are being filled. Discussions with operating personnel have indicated that the adsorption system is the source of the odor. You are requested to determine if the adsorption equipment/emission is in fact the source of the odor or if the equipment is capable of containing/ controlling the pyridine emission. Design and actual operating data are provided below: 1 The adsorption units are twin horizontal units with face dimensions for flow of 5  12 ft. Each unit contains new 4  6 mesh activated carbon B that was installed one month ago. The measured bed height is 12 inches. 2 The carbon manufacturer maintains that the breakthrough capacity of the carbon is 0.49 lb pyridine/lb carbon and that the carbon has a bulk density of 25 lb/ft3. 3 Laboratory tests performed by plant personnel indicate that the carbon contains a HEEL of approximately 0.03 lb pyridine/lb carbon when regenerated with 4.0 lb of steam/lb pyridine at 10 psig. 4 The ventilation blower for the drum-filling station has a flow of 5000 acfm at 258C and 14.7 psia and contains a pyridine concentration of 2000 ppm (toxicology data from plant hygienist). 5 The drum-filling operation operates on a 24-h/day basis and the adsorption units are operated on an 8-h adsorption, 5-h regeneration cycle, with 3 h for cooling and stand-by. The steam used during the 5-h regeneration cycle was determined to be 2725 lb (mass flowmeter). 6 The adsorption unit was designed based on a pressure drop through the bed following the relationship DP ¼ 0:37Z(v=100)1:56 where Z is the bed depth in inches, v is the velocity in ft/min, and the pressure drop is in inches of water. The measured operating pressure drop is 3.3 inches of water. 7 The fractional fan efficiency is 0.58.

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To evaluate the adsorber’s performance, please determine the following: The mass of pyridine to be captured in the adsorption period The working capacity of the carbon B The mass and volume of carbon that should be used in each unit The required bed height The design pressure drop through the bed using the required bed height for full capture 6 The horsepower requirement for this process 7 The required steam to regenerate the bed to the HEEL level of 0.03 lb pyridine/ lb carbon 1 2 3 4 5

SOLUTION:

Calculate the mole fraction of pyridine (P) in the gas stream: yP ¼ 2000=106 ¼ 0:0020

Calculate the volumeteric flow rate of P in acfm: qP ¼ yP q ¼ (0:0020)(5000) ¼ 10:0 acfm Determine the density of the P vapor at the operating conditions:

rP ¼

P(MW) (14:7)(79) ¼ RT (10:73)(537)

¼ 0:2015 lb=ft3 The mass of P collected during the adsorption period is then mP ¼ (10)(0:2015)(8)(60) ¼ 967:2 lb Estimate the working capacity (WC) of carbon B for this system: WC ¼ BC  HEEL ¼ 0:49  0:03 ¼ 0:46 lb P=lb carbon B Calculate the mass of carbon that should be used for each unit: MAC ¼ mP =WC ¼ 967:2=0:46 ¼ 2103 lb carbon B

Regeneration

289

The volume of activated carbon VAC is VAC ¼ MAC =rB ¼ 2103=25 ¼ 84:1 ft3 The cross-sectional area AAC of the carbon that is presently available for flow is AAC ¼ (5)(12) ¼ 60 ft2 Since the bed height is 12 inches or 1.0 ft, the actual volume currently employed is 60 ft3. Because this is below the required 84 ft3, the odor problem exists. Thus, the equipment is not capable of controlling the emission to eliminate the odor. The “required” height of the adsorbent in the unit may now be calculated Z ¼ VAC =AAC ¼ 84:1=60 ¼ 1:40 ft ¼ 16:8 inches  17:0 inches Estimate the pressure drop across the required adsorbent in inches of H2O using the recommended pressure drop equation: DP ¼ 0:37Z

   v 1:56 q=AAC 1:56 ¼ 0:37Z 100 100   5000=60 1:56 ¼ (0:37)(17) 100 ¼ 4:73 in H2 O

The total pressure drop across the bed in lbf/ft2 is then DPtotal ¼ (4:73)(5:2) ¼ 24:6 lbf =ft2 while the HP requirement is HP ¼

(24:6)(5000) (60)(550)(0:58)

¼ 6:43 Finally, the steam requirement for regeneration is msteam ¼ (4:0)(967:2) ¼ 3869 lb steam during regeneration Note that the actual operating steam rate (2725 lb for 5 hr) is also below the required value. B

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Chapter 11 Adsorption

ILLUSTRATIVE EXAMPLE 11.21 Benzene is to be recovered from a dilute mixture with air by adsorption on a 6  10 mesh activated carbon (rB ¼ 30 lb/ft3). The gas enters the adsorber at a rate of 6500 ft3/h (608F, 1.0 atm) and contains 3.8% by volume benzene. A two-bed unit (one on, one off) is used, adsorbing at 808F and 1.0 atm, where approximately 95% of the benzene is removed. Experience indicates that a superficial gas velocity through the adsorber of 25 ft/min is satisfactory. The adsorption time is 4 h. Regeneration is to be accomplished with 150 psia saturated steam and 15% excess carbon is used in the beds. During operation, the activated carbon will retain 0.3 lb benzene per lb carbon at 808F. Determine the following: 1 2 3 4 5

The amount of carbon needed in each bed. The gas flow area of each bed. The dimensions of the adsorber. The pressure drop across the bed (refer to Fig. 11.15). The heat required to heat the vessel to the 150 psia steam temperature is 90,000 Btu and the heat of desorption (with sensible heat) is 40 Btu/lb benzene. The heat capacity of carbon is approximately 0.25 Btu/lb . 8F. How much steam must be theoretically supplied for regeneration of the bed?

SOLUTION 1 Key calculations are provided below: q ¼ 6500(540=520) ¼ 6750 acfh qB ¼ (0:038)(6750) ¼ 256:5 ft3 =h _ B ¼ (256:5)(78=379)(520=540) m ¼ 50:83 lb=h m _ B =4 h ¼ mB ¼ (50:83)(0:95)(4) ¼ 193:2 lb B MAC ¼ (193:2=0:3)1:15 ¼ 740:5 lb AC

2 The face area of the bed may now be calculated: VAC ¼ 740:5=30 ¼ 24:68 ft3 AAC ¼ 6750=(25)(60) ¼ 4:5 ft2

References

291

3 The dimensions of the bed are (assuming a vertical column) Z ¼ 24:68=4:5 ¼ 5:5 ft D ¼ (4:5=0:785)1=2 ¼ 2:4 ft

4 From Figure 11.15, one obtains DP ¼ 0:145 in H2 O=in bed DPtotal ¼ (0:145)(5:5)(12) ¼ 9:57 in H2 O

5 To heat the bed (b) Qb ¼ (0:25)(740:5)(359  80) ¼ 51,650 Btu To desorb (d) Qd ¼ (40)(193:2) ¼ 7727 Btu Therefore, Qtotal ¼ 90,000 þ 51,650 þ 7727 ¼ 149,400 Btu The steam requirement SR is SR ¼ Qtotal =DHvap ¼ 149,400=862 ¼ 173 lb steam for each 4 hour cycle. In actual practice, more steam would probably be required. B

REFERENCES 1. L. THEODORE and A. BUONICORE, adapted from “Control of Gaseous Emissions,” USEPA Training Manual, Research Training Pack, NC, 1982. 2. L. THEODORE, personal notes, 1979. 3. MSA RESEARCH CORP., “Package Sorption Device System Study,” EPA, Washington DC, April 1973.

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4. VIC Manufacturing Co., “Installation, Operation and Maintenance for VIC Air Pollution Control Systems.” 5. F. CROSS and H. HESKETH, “Handbook for the Operation and Maintenance of Air Pollution Control Equipment,” Technomic Publishing, Westport, CT, 1975. 6. L. THEODORE, personal notes, 1986. 7. S. ERGUN, Chem. Eng. Progr. 48, 89, 1952; Ind. Eng. Chem., New York City, NY, 41, 1179, 1949. 8. Union Carbide Corp., Linde Division, Molecular Sieve Department, New York City, NY, Bulletin F-34, F-34-1, and F-34-2. 9. L. THEODORE, “Engineering Calculation: Adsorber Sizing Made Easy,” CEP, New York City, NY, March 2005.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

12

Liquid – Liquid and Solid – Liquid Extraction Contributing Author: PHIL LOGRIPPO

INTRODUCTION Extraction is a term that is used for any operation in which a constituent of a liquid or a solid is transferred to another liquid (the solvent). The term liquid – liquid extraction describes the processes in which both phases in the mass transfer process are liquids. The term solid – liquid extraction is restricted to those situations in which a solid phase is present and includes those operations frequently referred to as leaching, lixiviation, and washing. These terms are used interchangeably below. Extraction involves the following two steps: contact of the solvent with the liquid or solid to be treated so as to transfer the soluble component (solute) to the solvent, and separation or washing of the resulting solution. The complete process may also include a separate recovery procedure involving the solute and solvent; this is normally accomplished by another operation such as evaporation, distillation, or stripping. Thus, the streams leaving the extraction system usually undergo a series of further operations before the finished product is obtained; either one or both solutions may contain the desired material. In addition to the recovery of the desired product or products, recovery of the solvent for recycling is also often an important consideration. In practice, the manner and the equipment in which these operations are carried out is based on the difference in physical states. Because solids are more difficult to handle and do not readily lend themselves to continuous processing, leaching is commonly accomplished in a batch-wise fashion by agitating the crude mixture with the leaching agent and then separating the residual insolubles from the resultant solution. Liquid – liquid extraction may also be carried out in a batch operation. The ease of moving liquids, however, makes liquid extraction more amenable to continuous flow in various types of columns and/or stages. For design calculations or analysis of operations, one can apply either data on the equilibrium attained between the phases or the rate of mass transfer between Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

293

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Chapter 12 Liquid –Liquid and Solid – Liquid Extraction

phases described in earlier chapters. The usual design approach is often through the theoretical-stage concept, as discussed in the absorption and distillation chapters, and also discussed below. Two sections follow. The first is concerned with liquid – liquid extraction and the second with solid – liquid extraction. Equilibrium considerations, equipment and simple design procedures/predicative methods, although different for both processes, are included for both topics. The notation employed is that typically employed in industry.

LIQUID– LIQUID EXTRACTION Liquid – liquid extraction is used for the removal and recovery of primarily organic solutes from aqueous and nonaqueous streams. Concentrations of solute in these streams range from either a few hundred parts per million to several mole/mass percent. Most organic solutes may be removed by this process. Extraction has been specifically used in removal and recovery of phenols, oils, and acetic acid from aqueous streams, and in removing and recovering freons and chlorinated hydrocarbons from organic streams.

The Extraction Process Treybal(1) has described the liquid– liquid extraction process in the following manner. If an aqueous solution of acetic acid is agitated with a liquid such as ethyl acetate, some of the acid but relatively little water will enter the ester phase. Since the densities of the aqueous and ester layers are different at equilibrium, they will settle on cessation of agitation and may be decanted from each other. Since the ratio of acid to water in the ester layer is now different from that in the original solution and also different from that in the residual water solution, a certain degree of separation has occurred. This is an example of stage-wise contact and it may be carried out either in a batch or continuous fashion. The residual water may be repeatedly extracted with more ester to additionally reduce the acid content. As will be discussed shortly, one may arrange a countercurrent cascade of stages to accomplish the separation. Another possibility is to use some sort of countercurrent or crosscurrent continuous-contact device where discrete stages are not involved. More complicated processes may use two solvents to separate the components of a feed. For example, a mixture of para- and ortho-nitrobenzoic acids may be separated by distributing them between the insoluble liquids chloroform and water. The chloroform preferentially dissolves the para isomer and the water the ortho isomer. This is called double-solvent or fractional extraction.(1) The liquid – liquid extraction described above is a process for separating a solute from a solution based on the combination of the concentration and solubility driving force between two immiscible liquid phases. Thus, liquid extraction effectively involves the transfer of solute from one liquid phase into a second immiscible

Liquid– Liquid Extraction

295

liquid phase. The simplest example involves the transfer of one component from a binary mixture into a second immiscible phase such as is the case for the extraction of an impurity from wastewater into an organic solvent. Liquid extraction is usually selected when distillation or stripping is impractical or too costly (e.g., the relative volatility for the two components falls between 1.0 and 1.2). Recovery of the solute and solvent from the product stream is often carried out by stripping or distillation. The recovered solute may be either treated, reused, resold, or disposed of. Capital investment in this type of process primarily depends on the particular feed stream to be processed. The solution whose components are to be separated is the feed to the process. The feed is composed of a dilutent and solute. The liquid contacting the feed for purposes of extraction is referred to as the solvent. If the solvent consists primarily of one substance (aside from small amounts of residual feed material that may be present in a recycled or recovered solvent), it is called a single solvent. A solvent consisting of a solution of one or more substances chosen to provide special properties is a mixed solvent. The solvent-lean, residual feed solution, with one or more constituents removed by extraction, is referred to as the raffinate. The solvent-rich solution containing the extracted solute(s) is the extract. The degree of separation that arises because of the aforementioned solubility difference of the solute in the two phases may be obtained by providing multiple-stage countercurrent contacting and subsequent separation of the phases, similar to a distillation operation. In distillation, large density differences between the gas – liquid phases are sufficient to permit adequate dispersion of one fluid in the other as each phase moves through the column. However, in liquid extraction, the density differences are significantly smaller and mechanical agitation of the liquids is frequently employed at each stage to increase contact and to increase the mass transfer rates. The minimum requirement of a liquid extraction unit is to provide intimate contact between two relatively immiscible liquids for the purposes of mass transfer of constituents from one liquid phase to the other, followed by the aforementioned physical separation of the two immiscible liquids. Any device or combination of devices that accomplishes this is defined in this text as a stage. If the effluent liquids are in equilibrium, so that no further change in concentration would have occurred within them after longer contact time, the stage is considered a theoretical or ideal stage. The approach to equilibrium attained is a measure of the stage efficiency. Thus, a theoretical or equilibrium stage provides a mechanism by which two immiscible phases intimately mix until equilibrium concentrations are reached and then physically separated into clear layers. A multi-stage cascade is a group of stages usually arranged in a countercurrent flow between stages for the purpose of enhancing the extent of separation.

Equipment There are two major categories of equipment for liquid extraction. The first is singlestage units, which provide one stage of contact in a single device or combination of

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Chapter 12 Liquid –Liquid and Solid – Liquid Extraction

Figure 12.1 Multistage extractors.

devices. In such equipment, the liquids are mixed, extraction occurs, and the insoluble liquids are allowed to separate as a result of their density differences. Several separate stages may be used in an application. Second, there are multistage devices, where many stages may be incorporated into a single unit. This type is normally employed in practice. There are also two categories of operation: batch or continuous. In addition to cocurrent flow (rarely employed) provided in Figure 12.1 for a three-stage system, crosscurrent extraction is a series of stages in which the raffinate from one extraction stage is contacted with additional fresh solvent in a subsequent stage. Crosscurrent extraction is usually not economically appealing for large commercial processes because of the high solvent usage and low solute concentration in the extract. Figure 12.1 also illustrates countercurrent extraction in which the extraction solvent enters a stage at the opposite end from where the feed enters and the two phases pass each other countercurrently. A photograph of a bench scale version of this unit (Unit Operations Laboratory—Manhattan College) is provided in Figure 12.2. It can be shown that multistage cocurrent operation only increases the residence time and therefore will not increase the separation above that obtained in a single stage, provided equilibrium is established in a single stage. Crosscurrent contact, in which fresh solvent is added at each stage, will increase the separation beyond that obtainable in a single stage. However, it can also be shown that the degree of

Liquid– Liquid Extraction

Figure 12.2

297

Liquid extraction experiment (Manhattan College).

separation enhancement is not as great as can be obtained by countercurrent operation with a given amount of solvent (see Chapter 8). The maximum separation that can be achieved between two solutes in a single equilibrium stage of the two phases is governed by equilibrium factors and the relative amounts of the two phases used, i.e., the phase ratio. A combination of the overall and component mass balances with the equilibrium data allows the compositions of

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Chapter 12 Liquid –Liquid and Solid – Liquid Extraction

the phases at equilibrium to be computed. If the separation achieved is inadequate, it can be increased by either changing the phase ratio or by the addition of more contacting stages.

Solvent Selection There are several principles that can be used as a guide when choosing a solvent for a liquid extraction process. These are typically conflicting and certainly no single substance would ordinarily possess every desirable characteristic for a process. Compromises are inevitable, and in what follows, an attempt will be made to indicate the relative importance of the various factors that must be considered. Selectivity receives preferential treatment but 10 other factors are also reviewed. Selectivity is the first and most important property ordinarily considered in deciding on the applicability of a solvent; selectivity refers to the ability of a solvent to extract one component of a solution in preference to another. The most desirable solvent from a solubility aspect would be one that would dissolve a maximum of one component and a minimum of the other. As in the case of vapor– liquid equilibrium, numerical data that quantify selectivity can be measured or determined. The numerical values of the selectivity, normally designated as b, are available in literature.(1) Note that there are numerous possible selectivities for a three component system. For example, it could be defined as [(xCB/xAB)/(xCA/xAA)] where xCB is the concentration of solute C in the B rich solution, xAB is the concentration of the third component in B, xCA is the concentration of C in the A rich solution, while xAA is the concentration of A in the A rich solution. Like relative volatility, b has been shown to be approximately constant for a few systems. However, in most cases, b varies widely with concentrations. The importance of “good” selectivity for extraction processes parallels that of relative volatility for distillation. Practical processes require that b exceeds unity and the more so the better. Selectivities close to unity will result in a large extraction unit, a large number of extraction stages, and in general, a more costly investment and operation. As one might suppose, if b ¼ 1, the separation is impossible. Furthermore, in all liquid– liquid extraction processes, it is necessary to remove the extracting solvent from the two products resulting from the separation. This is important not only to avoid contamination of the products with the solvent but also to permit reuse of the solvent in order to reduce the cost of operation. In practically every instance, the recovery process is ultimately carried out with fractional distillation, and the relative volatility of the solvent and substance to be separated must be high in order to ensure that distillation (see Chapter 9 for more details) may be carried out inexpensively. In most extraction processes, the quantity of solvent used is greater than that of the desired products. If in the recovery by distillation, the solvent is the more volatile component, large quantities will need to be vaporized and the process will be costly. Therefore, it is preferable in such cases that the solvent be the less volatile component; distillation will involve vaporization of the desired products that are

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present in smaller amounts. If the solute in the solution is nonvolatile, then distillation will become difficult and it may be necessary to recover the solvent by evaporation. A difference in densities of the contacted phases is also essential and should be as great as possible. Not only is the rate of disengaging of the immiscible layers thereby enhanced but also the capacity of the contacting equipment is increased. It is insufficient to examine merely the relative density of the solution to be extracted and the pure extracting solvent since after mixing the mutual solubility of the two will alter the densities. For continuous contacting equipment, it is important to be certain that a satisfactory density difference for the contacted phases exists throughout the process. The interfacial tension between immiscible phases, which must be settled or disengaged, should preferably be high for continuous processes. However, too high an interfacial tension may lead to difficulties in achieving adequate dispersion of one liquid into the other, while too low a value may lead to the formation of stable emulsions. Chemical reactions between the solvent and components of the solution yielding products extraneous to the process are naturally undesirable. Polymerization, condensation, or decomposition of the solvent at any temperature achieved during the process, including the solvent recovery stage, is not desirable. The extracting solvent and solution to be extracted should be highly immiscible. Solvent recovery in highly insoluble systems is simpler and, for a given distribution coefficient, the selectivity will be greater. In order to reduce the cost of equipment, the solvent should cause no severe corrosion difficulties with common materials of equipment construction. Expensive alloys and other unusual materials should not be required. Low power requirements for pumping, high heat-transfer rates, high rates of extraction, and general ease of handling are all corollaries of low viscosity. Hence, this is a desirable property of solvents employed in extraction processes. The vapor pressure of the solvent should be sufficiently low so that storage and extraction operations are possible at atmospheric or only moderately high pressures. This may conflict with the requirement of high relative volatility with the solution being extracted and a compromise may be necessary. Low flammability is, of course, desirable for safety reasons (see Part III, Chapter 22). Regarding toxicity, highly poisonous materials are difficult to handle industrially. Unless elaborate plant safety devices are planned with frequent medical inspection of personnel, the more toxic substances must be avoided (once again, see Part III, Chapter 22). Low cost and ready availability in adequate quantities usually parallel each other and are of course desirable solvent attributes. While it is true that solvents are recovered from product solutions, make-up solvent to replace inevitable process losses must be expected. Of all the desirable properties described, favorable selectivity, recoverability, interfacial tension, density and chemical reactivity are essential for the process to be carried out. The remaining properties, while not always important from a technical standpoint, must be given consideration in good engineering work and in cost estimation of a process.

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Equilibrium Liquid – liquid extraction processes can involve three components (ternary system). In order to better understand the equilibria associated with a three component system, it is first necessary to become acquainted with the standard method of representing such systems. Two-dimensional equilibrium diagrams for ternary systems can best be plotted on an equilateral triangle, each of whose apexes represents 100% concentration of each component (see Fig. 12.3). A series of grid lines representing fractional concentrations of a particular component is drawn parallel to the base opposite that apex which represents the component. Thus, every point on the diagram corresponds to a certain percentage composition of each of the three components. It is important to note that an apex signifies a single component; any point on one of the sides describes a binary system. There are various types/ classifications of these three component systems. Only one that exhibits a single lion’s share with extraction of pure components will be reviewed in the presentation to follow. Consider the case where acetic acid is the solute in water and n-butanol is used as an “extractant” to extract the acetic acid out of the water phase. This three-component system consists of three liquid components that exhibit partial miscibility. For this system, acetic acid and n-butanol form one pair of partially miscible liquids. Their partial miscibility may be interpreted by means of a ternary diagram seen in Figure 12.4. For simplicity, each of the components is denoted by a letter corresponding to its name as shown.(2) At a given temperature and pressure, water and n-butanol are partially soluble in each other. But, if the mutual solubility limits are exceeded, two layers are formed: one consists of a solution of n-butanol in water, the other of water in n-butanol. Suppose acetic acid, which is completely miscible with both n-butanol and water, is now added to the system. Obviously, acetic acid will distribute itself between the two liquid phases. The two layers will disappear to form a solution if sufficient acetic acid is introduced. Points a and b designate the compositions of two liquid layers resulting from mixing water and n-butanol in some arbitrary overall proportion such as c. The addition of acetic acid to the solution will change the compositions of the two layers from a and b to a1 to b1, respectively. The line a1b1 that passes through c1 connects the compositions of the two layers in equilibrium and is called a tie line. Different tie lines can be constructed through the continuous addition of C until a single solution is obtained. Complete miscibility occurs at the plait point,

Figure 12.3 Triangular concentration diagrams for a ternary system (A, B, C).

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Figure 12.4

301

Ternary equilibrium diagram with tie lines.

at which condition the two solutions coalesce into a single liquid phase of constant composition, i.e., the tie lines converge on the plait point.(2) In the equilateral triangle shown in Figure 12.4, the sum of the perpendicular distances from any point to the three sides of the triangle is equal to the altitude of the triangle; this makes the plot particularly useful for correlating ternary data. As described above, solutions that are A-rich yield the left-hand portion of the bimodal curve up to the plait point and those that are B-rich are found in the right-hand portion. And, as noted, the tie lines are straight lines that connect the concentrations of phases in equilibrium, and the two solutions or phases become identical at the plait point when the tie lines converge to a single point. Once again, there is obviously an infinite number of tie lines within the bimodal curve. A similar graph can be drawn on right triangle coordinates. ILLUSTRATIVE EXAMPLE 12.1 Using the equilibrium data provided in Tables 12.1 and 12.2, plot the equilibrium curve for the n-butanol– acetic acid –water ternary system at 308C. SOLUTION: The plotting of the data is left as an exercise for the reader, but the bottom portion of the curve for an equilateral triangle plot is provided in Figure 12.5. Note that the plait point is usually represented in the literature by “ þ ” in a circle and the (equilibrium) tie line points by “W”. Plotting the tie lines is left as an exercise for the reader as well. B

Graphical Procedures The calculation of the number of equilibrium stages required to achieve a given degree of separation by countercurrent contact requires simultaneous solution of the mass balance equations with equilibrium data from stage to stage. For purposes of calculation,

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Table 12.1 Equilibrium Data for n-Butanol – Acetic Acid –Water System at 308C n-Butanol (wt%)

Acetic acid (wt%)

Water (wt%)

7.30 7.13 8.04 14.25 15.20 30.90 49.35 56.46 69.82 79.50

0.00 3.15 5.46 12.75 13.10 15.10 14.45 12.14 6.02 0.00

92.70 90.00 86.50 73.00 71.70 54.00 36.50 31.70 24.46 20.50

Estimated plait point 27.50

14.90

57.50

Table 12.2 Tie Line Data for n-Butanol –Acetic Acid– Water System at 308C Acetic acid in n-butanol layer (wt%) 1.42 2.50 3.68 4.90 6.04 8.49 10.70 15.00

Acetic acid in water layer (wt%) 0.88 1.82 2.63 3.64 4.37 6.35 8.20 12.00

Figure 12.5 Ternary equilibrium curve for n-butanol, acetic acid, and water.

Liquid– Liquid Extraction

Figure 12.6

303

Graphical construction for calculating the number of theoretical stages (immiscible phases).

it is important to distinguish two classes of system: those in which the two phases are completely immiscible or in which the relative miscibility of the two phases is constant and independent of the solute concentration, and those in which the relative miscibility of the two phases varies with the solute concentrations. In the former case above, the solute-free flow rates of the two phases may be assumed constant throughout a multistage countercurrent contractor (the actual phase flow rates will vary as a result of solute transfer, but not the flow rate of the solvent). The number of equilibrium stages required to effect a given separation can then be obtained conveniently from an X– Y (solute free basis) plot by stepping off stages between the operating and equilibrium lines, just as in a gas absorption or distillation problem. The constant flow rates result in a straight operating line. The technique is illustrated in Figure 12.6 for a simple system. The situation for partially miscible systems is different. Consider the acetic acid, n-butanol, water system discussed above. The acidic nature and polarity of the acetic acid makes it soluble in water while its hydrophobic part makes it soluble in n-butanol. Thus, the two phases have different solubility traits and this leads to a separation of substances according to the solubility of each chemical in other substances. It is for this reason that liquid – liquid extraction is used as a substitute for distillation and evaporation, particularly when the substances to be separated are chemically different. Extraction utilizes differences in the solubilities of the components rather than differences in their volatilities. Since solubility depends on chemical properties, extraction exploits chemical differences instead of vapor –pressure differences. ILLUSTRATIVE EXAMPLE 12.2 Outline how to determine the number of theoretical stages for a given separation.

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SOLUTION: The determination of the number of theoretical stages required for a given separation can be performed graphically as illustrated by the equilateral triangle coordinate plot in Figure 12.5. The known ternary data are plotted on these coordinates and the bimodal curve defined. Points F, S, E1, and RN represent the acid feed solution, solvent feed (n-butanol), final extract and final raffinate, respectively (see X points). Points M and O are arrived at by construction. By using lines extended from point O through the bimodal curve as well as existing and interpolated tie line data, the construction lines drawn will yield the number of theoretical stages required. Treybal provides a more detailed presentation.(1) B

ILLUSTRATIVE EXAMPLE 12.3 Outline how to calculate the overall stage efficiency for a process. SOLUTION:

The overall stage efficiency, E, is calculated from E¼

Ntheo  100 Nact

(12:1) B

ILLUSTRATIVE EXAMPLE 12.4 The actual number of stages in the extraction unit at Manhattan College is 12. If the ideal stages are determined to be 6.2, calculate the overall stage efficiency. SOLUTION: Apply Equation (12.1), incorporating the given values for the actual and theoretical number of stages. This yields, E¼

6:2  100 12

¼ 51:7%

B

Analytical Procedures Fortunately, simple analytical procedures are available to perform many of the key extraction calculations. These remove the need for any graphical solutions (as outlined in Illustrative Example 12.2), while providing fairly accurate results. This development follows. Earlier, the equilibrium constant, K, was defined as the ratio of the mole fraction of a solute in the gas to the mole fraction of the solute in the liquid phase. A distribution coefficient, k, can also be defined as the ratio of the weight fraction of solute in the extract phase, y, to the weight fraction of solute in the raffinate phase, x, i.e., y (12:2) k¼ x For shortcut methods, a distribution coefficient k0 (or m) is represented as the ratio of the weight ratio of solute to the extracting solvent in the extract phase, Y, to the

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weight ratio of solute to feed solvent in the raffinate phase, X. In effect, Y and X are weight fractions on a solute-free basis. k0 ¼ m ¼

Y X

(12:3)

Consider first the crosscurrent extraction process in Figure 12.7. This may be viewed as a laboratory unit since the extract and raffinate phases can be analyzed after each stage to generate equilibrium data as well as to achieve solute removal. If the distribution coefficient, as well as the ratio of extraction solvent to feed solvent (S0 /F0 ) are constant, and the fresh extraction solvent is pure, then the number of crosscurrent stages (N ) required to achieve a specified raffinate composition can be estimated from: N¼

log(XF =XR )  0 0  kS log þ1 F0

(12:4)

Here, XF is the weight fraction of solute in feed, XR is the weight fraction of solute in raffinate, S0 is the mass flow rate of solute free extraction solvent to each stage, and F0 is the mass flow rate of the solute free feed solvent. Once again, XF and XR are weight fractions on a solute-free basis. The term k0 (S0 /F0 ) will later be defined as the extraction factor, 1—analogous to the absorption factor A discussed in Chapter 10. As noted earlier, most liquid– liquid extraction systems can be treated as having either: 1 immiscible (mutually non-dissolving) solvents, 2 partially miscible solvents with a low solute concentration in the extract, or 3 partially miscible solvents with a high solute concentration in the extract. Only the first case is addressed below. The reader is referred to the literature for further information on the second and third cases.(1)

Figure 12.7

Three-stage crosscurrent extraction.

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For the first case where the solvents are immiscible, the rate of solvent in the feed stream (F0 ) is the same as the rate of feed solvent in the raffinate stream (R0 ). Also, the rate of extraction solvent (S0 ) entering the unit is the same as the extraction solvent leaving the unit in the extract phase (E0 ). However, the total flow rates entering and leaving the unit will be different since the extraction solvent is removing solute from the feed. Thus, the ratio of extraction-solvent to feed-solvent flow rates (S0 /F0 ) is equivalent to (E0 /R0 ). ILLUSTRATIVE EXAMPLE 12.5 With reference to Figure 12.7, derive an expression for the feed concentration leaving a single ideal (equilibrium) stage. Assume the extraction solvent is pure. SOLUTION:

Assume both the solvent feed and extraction solvent S to be constant, i.e., F 0 ¼ R1 ¼ constant S0 ¼ E1 ¼ S1 ¼ constant

A material balance around stage 1 gives F 0 XF þ S0 (0) ¼ F 0 X1 þ S0 Y1 Assume this is an equilibrium stage so that m¼

Y1 X1

Substitution into the above equation gives 

 F0 XF X1 ¼ F 0 þ S0 m

B

ILLUSTRATIVE EXAMPLE 12.6 Refer to the three theoretical stage crosscurrent flow systems pictured in Figure 12.7. If the same amount of fresh solvent S0 is fed to each stage of the three equilibrium batch extraction stages, verify that Equation 12.4 is correct. SOLUTION:

For equilibrium stage 1 

 F0 X1 ¼ XF F 0 þ S0 m For equilibrium stage 2 

 F0 X1 F 0 þ S0 m  2 F0 ¼ XF F 0 þ S0 m

X2 ¼

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307

For equilibrium stage 3 

 F0 X2 F 0 þ S0 m  2 F0 ¼ X1 F 0 þ S0 m  3 F0 ¼ XF F 0 þ S0 m

X3 ¼

For N equilibrium stages  XR ¼ XN ¼

F0 F 0 þ S0 m

N XF

Rearranging the above equation gives  0  XF F þ S0 m0 N ¼ XN F0 or  N XF F0 ¼ XN F 0 þ S0 m Taking the log of both sides gives     XF S0 m ¼ N log 1 þ 0 log F XN or N¼

log(XF =XR )  0  mS þ 1 log F0

(12:4) B

ILLUSTRATIVE EXAMPLE 12.7 Calculate the number of theoretical stages required for a crossflow system employing the same quantity of fresh solvent for each stage. Pertinent data include: S0 ¼ 10 lb/min F 0 ¼ 10 lb/min XF ¼ 0.51 XR ¼ 0.01 (design requirement) m ¼ 0.72 SOLUTION:

Employ Equation (12.4). N¼

log(XF =XR )  0  mS log þ 1 F0

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Substitution gives N¼

log(0:51=0:10)   (0:72)(10) þ1 log (10)

¼ 0:7076=0:2355 ¼ 3:01 The reader is left the exercise of calculating the discharge solvent concentration from the first stage. B

ILLUSTRATIVE EXAMPLE 12.8 Calculate the discharge solution concentration XR for a crossflow system with nine actual stages. Employ the same quantity of fresh solvent for each stage. Assume the information provided in the previous example applies. In addition, the overall stage efficiency is 67%. SOLUTION:

Since the overall efficiency is 67%, the number of theoretical stages Ntheo is N theo ¼ (0:67)(9) ¼ 6

Equation (12.4) may be rearranged to solve for XR.  XR ¼

 1 6 0:51 1:72

¼ (0:0386)(0:51) ¼ 0:0199  2%

B

By writing an overall material balance around the countercurrent unit illustrated in Figure 12.8 (a similar unit in the Unit Operations Laboratory at Manhattan College is provided in Fig. 12.2), the describing material balance equation can be rearranged into a McCabe – Thiele type of operating line with a slope F0/S0 (F0 ¼ F ¼ R; S0 ¼ S ¼ E) FXF þ SYS ¼ RS XR þ E1 YE YE ¼

Figure 12.8 Countercurrent extraction.

F (XF  XR ) þ YS S

(12:5)

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where YE is the weight ratio of solute removed to the extraction solvent and YS is the weight ratio of solute to be removed to the extraction solvent. If the equilibrium line is straight, its intercept is zero, and if the operating line is straight, the number of theoretical stages can be calculated with one of the following equations, which are forms of the Kremser equation. When the intercept of the equilibrium line is greater than zero, YS/kS0 should be used instead of YS/m, where kS0 is the distribution coefficient at YS. Also, these equations contain an extraction factor (1), which is calculated by dividing the slope of the equilibrium line, m, by the slope of the operating line, F0 /S0 , i.e.,(3) 1¼

m  S0 F0

(12:6)

If the equilibrium line is not straight, a geometric mean value of m should be used. This quantity is determined by the following equation,(4) m¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m1  m2

(12:7)

where subscripts 1 and 2 denote the feed stage and the raffinate stage, respectively. When 1 = 1.0,  ln N¼

   XF  (YS =m) 1 1 1 þ (XR  YS )=m 1 1 ln 1

(12:8)

When 1 ¼ 1.0,  N¼

 XF  (YS =m) 1 XR  (YS =m)

(12:9)

ILLUSTRATIVE EXAMPLE 12.9 Referring to Figure 12.8, use the following data to determine the degree to which an overall material balance is satisfied: F ¼ 0.13 lb/min S ¼ 0.433 lb/min R ¼ 0.0945 lb/min E ¼ 0.0878 lb/min SOLUTION:

Apply an overall material balance to the unit: FþS¼RþE

(12:10)

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Substitute the data provided into the above equation: ?

0:13 lb=min þ 0:0433 lb=min ¼ 0:0945 lb=min þ 0:0878 lb=min ?

0:1733 lb=min ¼ 0:1823 lb=min

The percent deviation (based on the inlet stream) is: 

 0:1823  0:1733 Error ¼  100 0:1823 ¼ 4:9%

B

ILLUSTRATIVE EXAMPLE 12.10 Referring to the previous example, if the solute free basis mass fraction of the solute (e.g., acetic acid) in the feed, solvent, extract, and raffinate are 0.1339, 0.0, 0.1058, and 0.00183, respectively (lb/lb), determine the degree to which a solute (e.g., acetic acid) balance is satisfied. SOLUTION:

Apply a componential balance to the unit: (F)(XF ) þ (S)(YS ) ¼ (E)(YE ) þ (R)(XR )

(12:11)

Substitute the data provided along with data from the previous example into the above equation ?

ð0:13 lb=minÞ(0:1369) þ ð0:0433 lb=minÞ(0:0) ¼ ð0:0945 lb=minÞ(0:01058) þ ð0:0878 lb=minÞ(0:00183) ?

0:01742 lb=min ¼ 0:009998 lb=min þ 0:00016 lb=min ?

0:01742 lb=min ¼ 0:01015 lb=min The percent deviation (based on the inlet stream) is, Error ¼

0:01742 lb=min  0:01015 lb=min  100 0:01742 lb=min

¼ 41:7% The deviation here is much larger than the previous example, suggesting some inconsistency in the data. B

ILLUSTRATIVE EXAMPLE 12.11 A 200 lb/h process stream containing 20 wt% acetic acid (A) in water (W ) is to be extracted down to 1 wt% with 400 lb/h of a methyl isobutyl ketone (MIBK) ternary mixture containing 0.05 wt% acetic acid and 0.005 wt% water. Determine F0, S0 , XF, XR, YS, and YE.

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SOLUTION: Calculate the feed solvent flow rate (F0 ) and the extraction solvent flow rate (S0 ) in lb/h. Note that the weight percents are not on a solute free basis. F 0 ¼ F  (1  xA ) ¼ 200 lb=h(1  0:2) ¼ 160 lb=h S0 ¼ S  (1  yA  yW ) ¼ 400  (1  0:00005  0:0005) ¼ 400  (1  0:00055) ¼ 399:8 lb=h Calculate the solute-free weight ratios of A in the feed (XF), raffinate (XR), and extraction solvent (YS): XF ¼

0:2  200 lb=h ¼ 0:25 160 lb=h

1 ¼ 0:0101 100  1 0:0005 ¼ 0:0005 YS ¼ 1  0:0005

XR ¼

Calculate the weight ratio of A in the extract (YE) using a modified form of Equation (12.5): YE ¼

F 0 XF þ S0 YS  R0 XF E0

Substituting gives, YE ¼

ð0:25  160Þ þ ð0:0005  399:8Þ  ð0:0101  160Þ ¼ 0:0965 ¼ 9:65% 399:8

B

ILLUSTRATIVE EXAMPLE 12.12 Using the values from Illustrative Example 12.11, determine the number of theoretical stages required to achieve the desired acetic acid removal. The equilibrium data for the MIBK and acetic acid system can be represented by Y ¼ 1.23(X )1.1 for acetic acid weight ratios between 0.01 and 0.25. SOLUTION: Determine the weight ratio of the liquid leaving the first stage (X1), which is in equilibrium with the liquid leaving the same stage (YE):   YE 0:0965 1=1:1 ¼ ¼ 0:0988 X1 ¼ 1:23 1:23  X 1:1 Next, obtain an expression for the slope of the equilibrium line, noting that the slope of a line is the first derivative of the function for the line Y ¼ 1:23  X 1:1 dY ¼ 1:1  (1:23  X 0:1 ) ¼ 1:353  X 0:1 dX Calculate the slope of the equilibrium line at the feed stage (m1), i.e., for X1 ¼ 0.0988, m1 ¼

dY ¼ 1:353(X)0:1 ¼ 1:353  (0:0988)0:1 ¼ 1:073 dX

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Calculate the slope of the equilibrium line at the feed stage (mR), i.e., for XR ¼ 0.0101, mR ¼

dY ¼ 1:353(XR )0:1 ¼ 1:353  (0:0101)0:1 ¼ 0:8545 dX

Determine the geometric mean equilibrium slope (m) from Equation (12.7): pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi m ¼ 1:073  0:8545 ¼ 0:958 Also calculate the extraction factor (1) from Equation (12.6):   399:8 lb=h ¼ 2:39 1 ¼ 0:958 160 lb=h Finally, calculate the number of theoretical stages from Equation (12.8):     XF  (YS =m) 1 1 þ ln 1 (XR  YS )=m 1 1 N¼ ln 1 Substituting,

    0:25  (0:0005=1:074) 1 1 1 þ ln (0:0101  0:0005)=1:074 2:39 2:39 N¼ ln 2:39 ¼ 3:23

Since integer values of actual stages are generally employed, four theoretical stages may be required for the desired separation. B

SOLID –LIQUID EXTRACTION (LEACHING)(4) There are three key unit operations that involve the “mass transfer” between solids and liquids: 1 Crystallization 2 Solid – liquid phase separation 3 Solid – liquid extraction Crystallization receives treatment in Chapter 14 while the physical separation of solids and liquids (2) appears as a section in Chapter 16. This last section in this chapter addresses the important topic of (3) solid – liquid extraction.(4) Solid – liquid extraction involves the preferential removal of one or more components from a solid by contact with a liquid solvent. The soluble constituent may be solid or liquid, and it may be chemically or mechanically held in the pore structure of the insoluble solid material. The insoluble solid material is often particulate in nature, porous, cellular with selectively permeable cell walls, or surface-activated. In engineering practice, solid– liquid extraction is also referred to by several other names such as chemical extraction, washing extraction, diffusional extraction, lixiviation, percolation, infusion, and decantation-settling. The simplest example of a

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313

leaching process is in the preparation of a cup of tea. Water is the solvent used to extract or leach, tannins and other substances from the tea leaf. A brief description/ definition of terms adopted by some is provided below.(5) The entire field of liquid – solid extraction may be subdivided in a number of ways. The authors have chosen to subdivide the field in the following fashion: 1 Leaching—The contacting of a liquid and a solid, e.g., with the potential of imposing a chemical reaction upon one or more substances in the solid matrix so as to render them soluble. 2 Chemical extraction—This is similar to leaching but it applies to removing substances from solids other than ores. The recovery of gelatin from animal bones in the presence of alkali is typical. 3 Washing extraction—The solid is crushed to break the cell walls, permitting the valuable soluble product to be washed from the solid matrix. Sugar recovery from cane is a prime example. 4 Diffusional extraction—The soluble product diffuses across the denatured cell walls (no crushing involved) and is washed out of the solid. The recovery of beet sugar is an excellent case in point.

Process Variables In the design of solid– liquid systems, the rate of extraction is affected by a number of independent variables. These are: 1 Temperature 2 Concentration of solvent 3 Particle size 4 Porosity and pore-size distribution 5 Agitation 6 Solvent selection 7 Terminal stream composition and quantities 8 Materials of construction Details on each of the above design/system variables are provided below: 1 As with most rate phenomena (e.g., chemical reaction), extraction is enhanced by an increase in temperature. The maximum temperature that can be used for a particular system is determined by either the boiling point of the solvent, degradation of the product or solvent, economics, or all of the above. Processes that depend on a chemical reaction are significantly enhanced by a rise in temperature. However, many solid – liquid extraction systems are controlled by diffusional processes and the improvement is less dramatic in these systems. The process temperature for leaching also varies, depending on the raw material

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and desired final product. For instance, in tea production, the feed water temperature enters the extraction unit between 130– 1908F, whereas in coffee production, the extract water temperature enters the unit at 3608F. Temperature affects the solute solubility, solvent vapor pressure, and selectivity, as well as the quality of the final product. 2 The concentration of solvent is also an important factor, particularly in the case of aqueous solutions in which a chemical reaction plays a part. In oil seed extraction, the concentration of the solvent is of minor consequence because the rate of extraction is limited by the diffusion of the oil from the seed. 3 Particle size is significant in most cases since it is a direct function of the total surface area that will be available for either reaction or diffusion. It is probably of greatest importance in extracting cellular materials because a reduction in particle size also results in an increase in the number of cells ruptured. Particle size is of lesser importance in ores since porosity and pore-size distribution often take on greater significance. 4 In ores, porosity and pore-size distribution can affect the rate of extraction because the leaching solution must flow or diffuse in and out of the pores and, in many instances, the movement of the solute through the pores to the surface of the particle is by diffusion. A reduction in particle size usually results in a decrease in the average time of passage of a solute molecule from the interior of the ore particle to the surface of the particle. 5 In nonagitated systems, the solute molecules must not only diffuse to the surface of the particle but must also diffuse to the main stream in order to be carried to a collection point. Agitation tends to reduce resistance to mass transfer and to cancel the effects that restrict efficiency. 6 A solvent needs to provide for a high selectivity of the solute to be extracted from a solid, as well as the capability to produce a high quality extract (see previous section for additional details). Other items to be considered include chemical stability at process conditions, low viscosity, low vapor pressure to reduce losses, low toxicity and flammability, low density, low surface tension, ease and economy of solute recovery from the extract stream, and cost. Solvent cost and potential losses (fugitive or otherwise) must also be taken into consideration. 7 The terminal stream quantity and composition are the primary variables that control the economics of any extraction process. In order for a plant to be profitable, it must be able to meet certain production goals within the constraints of a plant’s capabilities. The plant engineer is responsible for stretching those capabilities to their fullest extent to meet production quotas. 8 The corrosive properties of the solvent and its solutions are definite factors in setting equipment costs, particularly where metal construction is required. Selection of the proper materials of construction can insure a long life of the equipment. It can avoid loss of not only product quality but also value as a result of contamination.

Solid –Liquid Extraction (Leaching)

315

The choices for the above variables are made based on the specific process under evaluation with typical needs often determined through experimentation.

Equipment and Operation The methods of operation of a leaching system can be specified by four categories: operating cycle (batch, continuous, or multibatch intermittent), direction of streams (cocurrent, countercurrent, or hybrid flow), staging (single-stage, multistage, or differential-stage), and method of contacting. The following is a list of typical leaching systems: 1 Horizontal-basket design 2 Endless-belt percolator 3 Kennedy extractor 4 Dispersed solids leaching 5 Batch stirred tanks 6 Continuous dispersed solids leaching 7 Screw conveyor extraction As one might suppose, details on leaching equipment and operation is similar to that for liquid– liquid extraction systems. The simplest method of operation for a solid– liquid extraction or washing of a solid is to bring all the material to be treated and all the solvent to be used into intimate contact once and then to separate the resulting solution from the undissolved solids. The single-contact or single-stage batch operation is encountered in the laboratory and in small-scale operations but rarely in industrial operations because of the low recovery efficiency of soluble material and the relatively dilute solutions produced. If the total quantities of solvent to be used is divided into portions and the solid extracted successively with each portion of fresh solvent after draining the solids between each addition of solvent, the operation is called multiple-contact or multistage. Although recovery of the soluble constituents is improved by this method, it has the disadvantage that the solutions obtained are still relatively dilute. This procedure may be used in small-scale operations where the soluble constituent need not be recovered. If the solid and solvent are mixed continuously and the mixture fed continuously to a separating device, a continuous single-contact operation is obtained. High recovery of solute with a highly concentrated product solution can be obtained only by using countercurrent operation with a number of stages. In countercurrent operation, the product solution is last in contact with fresh solid feed and the extracted solids are last in contact with fresh solvent. Details on the above three methods of operation are provided below.(6) The single stage operation is shown in Figure 12.9 and represents both the complete operation of contacting the solids feed and fresh solvent and the subsequent mechanical (or equivalent) separation.

316

Chapter 12 Liquid –Liquid and Solid – Liquid Extraction

Figure 12.9 Single-stage leaching unit.

The second type is the multistage system shown in Figure 12.10 with the flow direction termed cross-flow. Fresh solvent and solid feeds are mixed and separated in the first stage. Underflow from the first stage is sent to the second stage where more fresh solvent is added. This is repeated in all the subsequent stages. The third type of flow is the multistage countercurrent system shown in Figure 12.11. The underflow and overflow streams flow countercurrent to each other.(3) Figure 12.12 shows a material balance for a continuous countercurrent process.(3) The stages are numbered in the direction of flow of the solid (e.g., sand). The light phase is the liquid that overflows from stage to stage in a direction opposite to that of the flow of the solid, dissolving solute as it moves from stage N to stage 1. The heavy phase is the solid flowing from stage 1 to stage N. Exhausted solids leave stage N, while concentrated solution overflows leave from stage 1. For purposes of analysis, it is customary to assume that the solute free solid is insoluble in the solvent so that the flow rate of this solid is constant throughout the process unit.

Figure 12.10 Multistage cross-flow leaching unit.

Figure 12.11 Multistage countercurrent leaching unit.

Solid –Liquid Extraction (Leaching)

Figure 12.12

317

Material balance-countercurrent process.

Design and Predictive Equations Design and predictive equations for leaching operations can be more involved than those for liquid extraction. As before, the solute/solvent equilibrium and process throughput determine the cross-sectional area and the number of theoretical and/or actual stages required to achieve the desired separation. And, as with many of the previous unit operations discussed so far, the number of equilibrium stages and stage efficiencies can be determined under somewhat similar conditions for the countercurrent units discussed earlier. As in distillation (see Chapter 9) and absorption (see Chapter 10), the quantitative performance of a countercurrent system can be analyzed by utilizing an equilibrium line and an operating line, and, as before, the method to be employed depends on whether these lines are straight or curved. Provided sufficient solvent is present to dissolve all the solute in the entering solid and there is no adsorption of solvent by the solid, equilibrium is attained when the solid is completely “saturated” and the concentration of the solution (as formed) is uniform. Assuming these requirements are met, the concentration of the liquid retained by the solid leaving any stage is the same as that of the liquid overflow from the same stage. Therefore, an equilibrium relationship exists for this (theoretical) stage in question. The equation for the operating line is obtained by writing a material balance. From Figure 12.12, VNþ1 þ Lo ¼ V1 þ LN

(total solution, including solute)

VNþ1 yNþ1 þ Lo xo ¼ V1 y1 þ LN xN

(solute)

Eliminating VNþ1, and solving for yNþ1, gives 2 3 1 V y  Lo x o   xN þ 1 1 yNþ1 ¼ 6 V1  L o 7 LN þ V1  Lo 4 5 1þ LN

(12:12) (12:13)

(12:14)

If the density and viscosity of the solution change considerably with solute concentration, the solids from the lower stages might retain more liquid than those in the higher stages. The slope of the operating line then varies from stage to stage.

318

Chapter 12 Liquid –Liquid and Solid – Liquid Extraction

If, however, the mass of the solution retained by the solid is independent of concentration, LN is constant, and the operating line is straight. The two above mentioned conditions describe variable and constant overflow, respectively. It is usually assumed that the inerts are constant from stage to stage and insoluble in the solvent. Since no inerts are usually present in the extract (overflow) solution and the solution retained by the inerts is approximately constant, both the underflow LN and overflow VN are constant, and the equation for the operating line approaches a straight line. Since the equilibrium line is also straight, the number of stages can be shown to be (with reference to Fig. 12.12)   yNþ1  xN log y  x1  1  (12:15) N¼ yNþ1  y1 log xN  x1 The above equation should not be used for the entire extraction cascade if Lo differs from L1, L2, . . . , LN (i.e., the underflows vary within the system). For this case, the compositions of all the streams entering and leaving the first stage should first be calculated before applying this equation to the remaining cascade.(3,7) ILLUSTRATIVE EXAMPLE 12.13 Calculate the grams of water that need to be added to 40 g of sand containing 9.1 g salt to obtain a 17% salt solution. SOLUTION:

Set V to be the grams of water required. The describing equation is 9:1 ¼ 0:17 9:1 þ V

Solving for V, V ¼ 44:4 g

B

ILLUSTRATIVE EXAMPLE 12.14 Refer to the previous example. If the salt solution is to be reduced to 0.015, calculate the amount of salt that must be removed (“leached”) from the solution. SOLUTION: Let Z be equal to the final amount of salt in the sand–water– salt solution. The describing equation is Z ¼ 0:15 40 þ Z Solving for Z, Z ¼ 0:61 g salt The amount of salt removed is therefore 9.120.61 ¼ 8.49 g.

B

Solid –Liquid Extraction (Leaching)

319

ILLUSTRATIVE EXAMPLE 12.15 A sand–salt mixture containing 20.4% salt enters a solid –liquid extraction at a rate of 2500 lb/h. Calculate the hourly rate of fresh water that must be added for 99% of the salt to be “leached” from the sand–salt mixture if the discharge salt–water solution contains 0.153 (mass) fraction salt. SOLUTION: balance is

Let W equal to the hourly rate of water. The describing equation from a mass (0:204)(2500) ¼ 0:153 (0:204)(2500) þ W

Solving for W, W ¼ 2820 lb=h Also note that the feed consists of 510 lb salt and 1990 lb sand. On discharge, the sand contains only 5.1 lb salt. The discharge water solution consists of the 2820 lb water plus 504.9 lb salt. B

ILLUSTRATIVE EXAMPLE 12.16 A countercurrent leaching system is to treat 100 kg/h of crushed sugar stalks with impurity-free water as the solvent. Analysis of the stalks is as follows: Water ¼ 38% (by mass) Sugar ¼ 10% Pulp ¼ 52% If 95% sugar is to be recovered and the extract phase leaving the system is to contain 12% sugar, determine the number of theoretical stages required if each kilogram of dry pulp retains 2.5 kg of solution. SOLUTION:

For a basis of 100 kg (one hour of operation) of sugar stalks,

Water ¼ 38 kg Sugar ¼ 10 kg Pulp ¼ 52 kg For 95% sugar recovery, the extracted solution contains

0:95(10) ¼ 9:5 kg sugar and

  1  0:12 (9:5) ¼ 69:7 kg water 0:12

320

Chapter 12 Liquid –Liquid and Solid – Liquid Extraction

The total extract solution is then

V1 ¼ 9:5 þ 69:7 ¼ 79:2 kg The underflow solution is

L1 ¼ L2 ¼    ¼ LN ¼ (2:5)(52) ¼ 130 kg Since L0 ¼ 10 þ 38 ¼ 48 kg, a material balance on the initial stage must be performed:

L0 þ V2 ¼ L1 þ V1 ¼ 48 þ V2 ¼ 79:2 þ 130 V2 ¼ 161:2 kg Applying a componential solute (sugar) balance across the first stage,

161:2y2 þ 10 ¼ 9:5 þ 130(0:12) y2 ¼ 0:0937 As indicated above, the remaining (N21) stages operate with the underflow and overflow solutions relatively constant. For this part of the system, and subject to this assumption,

  yNþ1  xN log y  x1  1  N1¼ yNþ1  y1 log xN  x1

(12:16)

For this equation, yNþ1 ¼ 0:0 and xN ¼

0:05(10) ¼ 0:00385 130

Substitution of the values into the above equation gives   0  0:00385 log 0:0973  0:12   ¼ 8:26 N1¼ 0  0:0937 log 0:00385  0:12 N ¼ 9:26 stages

B

ILLUSTRATIVE EXAMPLE 12.17 Refer to Illustrative Example 12.16. Calculate the actual stages required if the overall stage efficiency is 85%.

Solid –Liquid Extraction (Leaching)

321

SOLUTION: The 9.26 stages represent the theoretical number of stages. The actual number of stages, Nact, is Nact ¼

9:26 ¼ 10:89 stages 0:85

Eleven stages are suggested.

B

ILLUSTRATIVE EXAMPLE 12.18 Assume that salt (NaCl) is to be recovered by leaching a salt mixture containing insoluble impurities. The salt content of the mixture is 20 wt% and is to be reduced to 1.0 wt%. Pure water at 808F is the leaching agent. Determine how much water is required per 100 lb of solids for a single-stage operation. Each pound of insoluble matter leached in a stage retains 1.5 lb of solution under the operating conditions. SOLUTION: Assume 100 lb of feed (Lo) as a basis. The feed contains 80 lb of insolubles and 20 lb of NaCl. The solid phase leaving the stage contains the 80 lb of insolubles that represents 99% of the dry solid phase. Therefore, the solids are 80 lb ¼ 80:8 lb (total) 0:99 This represents 80 lb of the insoluble matter plus 0.8 lb of salt. The solution retained by insoluble matter is 1.5 (80 lb) or 120 lb, and includes the 0.8 lb of NaCl. The solution concentration in the insoluble matter is therefore 0:8 lb ¼ 0:00667 120 lb Leaching liquor leaving the stage contains (2020.8) lb, or 19.2 lb of salt. For an equilibrium stage, this is at the same solution concentration as that in the solid phase. Therefore, the mass of this phase is 19:2 lb ¼ 2880 lb 0:00667 The solvent (salt–water solution) required can be determined by an overall material balance. V2 þ Lo ¼ V1 þ L1 V2 ¼ 2880 þ 120  20 V2 ¼ 2980 lb

B

ILLUSTRATIVE EXAMPLE 12.19 Refer to the previous example. Perform the calculation for a three-staged continuous countercurrent system.

322

Chapter 12 Liquid –Liquid and Solid – Liquid Extraction

SOLUTION: This requires a trial-and-error solution. Stream V4, rather then V2, is the unknown. For a three-stage countercurrent operation, assume 450 lb for V4. The solution retained by the solids leaving stage 3 now contains 0.8 lb salt (s). Therefore, x3 ¼

0:8 ¼ 0:00667 120

A material balance on the third stage shows 450 lb leaching agent entering (V4), plus 120 lb solution L2 (L is constant) from the preceding stage. Stream V3 must still be 450 lb and the salt content of stream V3 is (450)(0.00667) ¼ 3.0015 lb salt. The salt content in L2 is (23.001520.8) ¼ 2.2015. Considering the second stage in the same manner, one finds that x2 must be x2 ¼

2:2015 ¼ 0:0183 120

The salt content of stream V2 is (450)(0.0183) ¼ 8.2556 lb and of stream L1 is 6.054. For the first stage to supply the 6.054 lb in stream L1, x1 must be x1 ¼

6:054 ¼ 0:05045 120

The salt content in stream V1 is, V1 ¼ (0:05045)(450) ¼ 22:70 lb The salt content in stream Lowill therefore be (22.70 2 0.8) ¼ 21.90 lb as compared to its “actual” value of 20 lb. The assumed value for V is reasonable. Another trial yields a value for V4 of approximately 440 lb. B

The Baker equation(8) is useful for calculating concentrations when the number of ideal stages is known, or vice versa. If the fresh solvent contains no solute, Baker provided the following equation: n X 1 ¼ 1 þ an an¼1 f 1

(12:17)

where f ¼ ratio of solute in the underflow from first stage to solute in the underflow fed to the last stage a ¼ ratio of overflow solution leaving stage N21 to solution in the underflow leaving the first stage an ¼ ratio of overflow solute leaving the last stage to the solution in the underflow leaving the last stage a0 ¼ ratio of solvent in overflow leaving stage N21 to solvent in underflow leaving first stage 0 an ¼ ratio of solvent in overflow leaving the last stage to solvent in the underflow leaving the last stage In the case of constant solvent-to-inerts ratio, the quantity a0 replaces a and a0n replaces an.

Solid –Liquid Extraction (Leaching)

323

ILLUSTRATIVE EXAMPLE 12.20 One hundred tons/day of ore containing 15% solubles and 5% moisture by mass is to be leached with 100 tons/day of water in a continuous countercurrent system consisting of three ideal stages. The underflow from each stage contains approximately 0.3 lb solution/lb inerts. Determine the percentage of solubles recovered. SOLUTION: Take 1 day of operation as a basis. The underflow from each stage contains 10021525 ¼ 80 tons of inerts and (0.3)(80) ¼ 24 tons of solution. Therefore, the total underflow (assumed constant) from each stage is equal to (80 þ 24) ¼ 104 tons. Applying an overall material balance, the weight of solution leaving the third stage, VNþ1, is 100 þ 1002104 or 96 tons. Material balances around stages 1 and 2 dictates overflows of 100 tons of solution from each of these stages. See also Figure 12.12. Since the fresh solvent contains no solute, Equation (12.17) applies. Therefore, a¼

V1 100 ¼ 4:167 ¼ 24 L1

an ¼

VNþ1 96 ¼ 4:00 ¼ 24 LN

For the case where N ¼ 3, applying Equation (12.17) gives 1 ¼ 1 þ an þ an a þ an a2 f Substituting, 1 ¼ 1 þ 4:0 þ 4:0(4:167) þ 4:0(4:167)2 ¼ 91:1 f f ¼

1 ¼ 0:0110 91:1

Percent solubles recovered ¼ 100(120.0110) ¼ 98.9%.

B

ILLUSTRATIVE EXAMPLE 12.21 Refer to Illustrative Example 12.20. Calculate the composition of the overflow discharge stream. SOLUTION: The solubles in overflow discharge ¼ (0.989)(15) ¼ 14.84 tons. The percent solubles in overflow discharge ¼ (14.84/96) ¼ 15.4%. By difference, the percent water in the overflow discharge is 84.6%. B

For the case of constant underflow, another equation is available for calculating the number of theoretical stages, N, in a continuous multistage countercurrent leaching operation. It is valid for both constant solution-to-inerts or solvent-to-inerts ratios. The

324

Chapter 12 Liquid –Liquid and Solid – Liquid Extraction

equation, developed by Chen,(9) is:    xo  y1 log 1 þ (r  1) xN  y1 N¼ log(r)

(12:18)

where r ¼ s/IF x ¼ weight fraction of solute in underflow solution y ¼ weight fraction of solute in overflow solution s ¼ flow rate of fresh solvent or clear liquid overflow from each stage I ¼ flow rate of inert solids F ¼ liquid retained in solid inerts ILLUSTRATIVE EXAMPLE 12.22 One hundred tons of underflow feed containing 20 tons of solute, 2 tons of water and 78 tons of inert material, I, is to be leached with water to give an overflow effluent concentration of 15% solute and a 95% recovery of solute. The underflow from each stage carries 0.5 lb of solution per lb of inerts. Calculate the number of ideal stages required.(10) SOLUTION:

For a 95% recovery of solute, Recovered solute ¼ (0:95)(20 tons) ¼ 19 tons

For a 15% solute weight fraction in the effluent 19 ¼ 0:15 s þ 19 where s ¼ tons of water. Solving for s, s ¼ 107:67 tons Based on the problem statement: y1 ¼ 0 xo ¼ 0.20 I ¼ 78 tons F ¼ 0.5 lb solution/lb inerts In addition, xN ¼

1:0 1 ¼ (0:5)(78) 39

xN ¼ 0:0256 To employ Equation (12.18), first calculate r: r¼

s 107:67 ¼ 2:736 IF (78)(0:5)

References

325

Substituting into Equation (12.18),    xo  y1 log 1 þ (r  1) xN  y1 N¼ log(r)    0:2 log 1 þ (2:736  1) 0:0256 N¼ log(2:736) N¼

1:163 ¼ 2:66 0:437

B

REFERENCES 1. R. TREYBAL, “Mass Transfer Operations,” McGraw-Hill, New York City, NY, 1955. 2. Author unknown, report submitted to L. THEODORE, Manhattan College, New York City, NY, date unknown. 3. J. REYNOLDS, J. JERIS, and L. THEODORE, “Handbook of Chemical and Environmental Engineering Calculations,” John Wiley and Sons, Hoboken, NJ, 2004. 4. R. TREYBAL, “Liquid Extraction,” McGraw-Hill, New York City, NY, 1951. 5. R. RICKLES, adapted from “Liquid-Solid Extraction,” Chem. Eng., New York City, NY, March 15, 1965. 6. F. TAGIAFERRO, adapted from report submitted to L. THEODORE, Manhattan College, Bronx, NY, Dec. 1964. 7. L. THEODORE and J. BARDEN, “Mass Transfer Operation,” A Theodore Tutorial, East Williston, NY, 1995. 8. E. BAKER, Chem. & Mech. Eng., 42, 699, New York City, NY, 1935. 9. N. CHEN, Chem. Eng., 125–128, New York City, NY, Nov. 23, 1964. 10. Source Unknown.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

13

Humidification and Drying INTRODUCTION In many unit operations, it is necessary to perform calculations involving the properties of mixtures of air and water vapor. Such calculations often require knowledge of: 1 the amount of water vapor carried by air under various conditions, 2 the thermal properties of such mixtures, and 3 the changes in enthalpy content and moisture content as air containing some moisture is brought into contact with water or wet solids, and other similar processes. This chapter will review the properties of mixtures of air and water vapor, the mechanisms of humidification and drying processes, and the equipment in which these processes are carried out. The remainder of this chapter will focus on three topics: 1 Psychrometry and the Psychrometric Chart (including some key definitions) 2 Humidification 3 Drying

PSYCHROMETRY AND THE PSYCHROMETRIC CHART Some key (and important) terms are introduced before proceeding to the general subject of psychrometry. In the discussion of the physical properties of mixtures of air and water vapor, certain terms need to be defined. The definitions of several key terms follow.(1) 1 The humidity, or absolute humidity, YA, is defined as the mass of water (designated with subscript A) carried by one unit of mass of bone dry air (BDA). The molal humidity, Y A0 , can also be defined, but most of the literature on

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

327

328

Chapter 13 Humidification and Drying

humidification and drying is in terms of YA. The two quantities are related by: YA ¼

18 0 Y 29 A

(13:1)

The humidity at 1.0 atm is then related to the mole fraction, yA, and the partial pressure, pA, by:     18 yA 18 pA ¼ (13:2) YA ¼ 29 1:0  yA 29 1:0  pA 2 Air is said to be saturated with water vapor at a given temperature and pressure when it contains the maximum amount of water vapor possible at that condition. This is achieved when the air is in equilibrium with liquid water. The saturation humidity, YAs, is the value of YA corresponding to a partial pressure, pA, equal to the vapor pressure of water, pA0 , at the given temperature. The relative humidity, YR, is defined by:   pA YR ¼ 100 0 (13:3) pA 3 The percent humidity, or percent saturation, PS, is given by:     YA pA 1:0  p0A PS ¼ 100 ¼ 0  100 YAs pA 1:0  pA

(13:4)

where pA0 and pA must be expressed in atmospheres. 4 The humid heat capacity, CpH, is defined as the energy required to raise the temperature of 1 lb of the carrier gas (air, B) and its accompanying vapor (water, A) by 18F. Thus, if CpB and CpA are the heat capacities of the carrier gas and vapor, respectively, CpH is given by: C pH ¼ C pB þ YA C pA

(13:5)

For the air –water system, the above equation becomes: C pH ¼ 0:24 þ 0:46YA ;

[Btu= F  lb BDA]

(13:6)

Note that heat capacities on a mass basis are designated with a capital C in this field. The molal humid heat is based on moles of BDA. 5 The humid volume, VH, is defined as the volume, in ft3, of “moist” gas per unit mass of bone dry gas where VH ¼ VR þ YA VA ; [ft3 =lb BDA]

(13:7)

The terms YR and VA represent the volume, in ft3, of BDA/lb BDA and volume of A/lb A, respectively.

Psychrometry and the Psychrometric Chart

329

6 The enthalpy of moist air, Hy, at Ty is defined by Equation (13.8) in units consistent with the present discussion: Hy ¼ C pH (Ty  To ) þ lo YA ; [Btu=lb BDA]

(13:8)

The term lo is the heat of vaporization at some reference temperature, To, which is normally assumed to be either 08F or 328F. This effectively represents the sensible enthalpy of the air – water mixture relative to To plus the enthalpy of vaporization at To. 7 The adiabatic saturation temperature, Tas, of moist air is the temperature that air reaches when it is saturated adiabatically (i.e., at constant enthalpy). 8 The wet-bulb temperature, Twb, is the temperature attained by a small reservoir of water in contact with a large amount of air flowing past it. Normally, one can assume that Twb and Tas are the same for the water – air system at temperatures low enough to form a dilute gas-phase solution. For other liquid – vapor mixtures, the adiabatic saturation and wet-bulb temperatures are normally not equal. 9 The dew point temperature, Tdp, is the temperature at which a given sample of moist air becomes saturated as it is cooled at constant pressure and absolute humidity. Thus, the dew-point pressure is the (total) pressure to which moist air must be compressed at constant temperature and humidity to bring it to saturation. As an example, suppose the air in a room at 708F is at 50% relative humidity. From steam tables (see Appendix), the vapor pressure of water at 708F is 0.3631 psi, which means that the air at 50% humidity holds water vapor with a partial pressure of (0.50)(0.3631) or 0.1812 psi. If the temperature is dropped at constant pressure to the point where 0.1812 psi equals the water vapor pressure (around 528F), the air becomes saturated with water and any further drop in temperature will cause condensation. The dew point of this air mixture is then 528F. Obviously, if the air were already saturated at 708F (i.e., 100% relative humidity), then the dew point would also be 708F. A summary of the above definitions is provided in Table 13.1. One vapor – liquid phase equilibrium example containing air – water raw data is the psychrometric or humidity chart. A humidity chart is used to determine the properties of moist air and calculate the moisture content in air. The ordinate of the chart is the absolute humidity, Y (with the subscript A dropped for convenience), which was defined earlier as the mass of water vapor per mass of bone dry air. (Note that some charts base the ordinate on moles instead of mass). The previously defined concepts and definitions are normally presented graphically on the aforementioned psychrometric or humidity chart (see Figs 13.1 and 13.2). There are also charts that apply to other single noncondensible gases and single condensable components at a fixed pressure (usually 1 atm). Curves showing the relative humidity (ratio of the partial pressure of the water vapor in the air to the vapor pressure of water at the system temperature) of humid air also appear on the charts. The curve for 100% relative humidity is also referred

330

Chapter 13 Humidification and Drying

Table 13.1 Definitions of Psychrometric Terms Term 1. Absolute humidity 2. Molal humidity 3. Relative saturation or relative humidity 4. Percent saturation or percent humidity 5. Humid volume 6. Humid heat capacity 7. Adiabaticsaturation temperature 8. Wet-bulb temperature 9. Dew-point temperature

Definition Vapor content of a gas Vapor content of a gas Ratio of partial pressure of vapor to partial pressure of vapor at saturation Ratio of concentration of vapor to the concentration of vapor at saturation with concentrations expressed as mole ratios Volume occupied by 1 lbmol of dry gas plus its associated vapor Energy required to raise the temperature of 1 lbmol of bone dry gas plus its associated vapor 18F Temperature that would be attained if the gas were saturated in an adiabatic process Steady-state temperature attained by a wet-bulb thermometer under standardized conditions Temperature at which vapor begins to condense when the gas phase is cooled at constant pressure

Units lb vapor/lb noncondensible gas (BDA) moles vapor/mole noncondensible gas atm/atm, or mole fraction/ mole fraction, often expressed as a percent mole ratio/mole ratio, often expressed as a percent

ft3/lbmole of BDA Btu/lbmol of BDA . 8F

8F

8F

8F

to as the saturation curve. The abscissa of the humidity chart is the air temperature, also known as the dry-bulb temperature, Tdb. The wet-bulb temperature, which has also already been defined (see Table 13.1), is another measure of the humidity. As described earlier, it is the temperature at which a thermometer stabilizes when it has a wet wick wrapped around the bulb. As water evaporates from the wick to the ambient air, the bulb is cooled; the rate of cooling depends on how humid the air is. No evaporation occurs if the air is saturated with water; hence, Twb and Tdb are the same and the lower the humidity, the greater the difference between these two temperatures. On the psychrometric chart, constant wet-bulb temperature lines are straight with negative slopes. The value of Twb corresponds to the value of the abscissa at the point of intersection of this line with the saturation curve. The humid volume, also defined earlier, is the volume of wet air per mass of BDA and is linearly related to the humidity. This quantity is used as an alternate ordinate in Figure 13.2. Note the straight parallel lines labeled with units of cubic feet. The humid enthalpy is the enthalpy of the moist air on a bone-dry basis. Since enthalpy is a measure of the energy content of the mixture, the enthalpy for saturated air can be read from the chart by extending the approximate wet-bulb temperature line upwards to the diagonal scale labeled enthalpy of saturation.

331

Figure 13.1

20

25

25

30

20%

40%

80% 60%

30

18 .

35

+0

tu

45

4B

+0.

tu

3B

+0.

Btu

40

2 +0.

tu 1B

40

45

0

50

tu .5 B

50

55

55

3

65

70

75

50

%

%

60

Dry Bulb Temperature, ∞F

60

39

Psychrometric chart—low temperatures. Barometric pressure, 29.92 in Hg.

Dry Bulb

Wet Bulb, Dew Point or Saturation Temperature

8

7

12

11

10

9

14

13

12

15 16

17 35

ft cu 12.5

22 21 19 20

26 25 24 23

28 27 ft cu

Bt 02

31

35 u

34 65

23 3

36 B

41 %

37 38 70 tu 04

29 30 60 13.0

tu

40 u

ive

%

90

Bt 06

5 44 34 42 4 75 lat re

% 80

hu ft

70

08 B

49 74 8 46 4 ty mi di cu 13.5

85

90

90

95

95

Humid Volume, ft 3/lbm Dry Air

80

10%

20%

% 30

%

80

40

85

ft cu 14.0

100

100

105 0

.0

.001

.002

.003

.004

.005

.006

.007

.008

.009

.010

.011

.012

.013

.014

.015

.016

.017

.018

.019

.020

.021

.022

.023

.024

.025

Pounds of moisture per pound of dry air

110

0

10

20

30

40

50

60

70

80

90

100

110

120

130

140

150

160

170

180

Grains of moisture per pound of dry air

105

Enthalpy deviatio n Btu per pound of dry air –0.3 Btu

Enthalpy at saturation, Btu per pound of dry air

r . pe ir cu ft a 14.5 of dry nd pou

60

80

100

120

140 160 Dry-bulb Temperature, °F

180

200

1%

f Dr

0

y Air

A 15ir 5

3.0 Btu

20

20 t

0 16

0

30 16

0

F Cu.

En 12 tha 5 lpy 13 of S 0 a 13 tura 5 tion 14 –Btu 0 P 14 oun 5 do 15 f D 0 ry

0 5

17 0 17

5

13

0

220 230 240 700

220

Figure 13.2 Psychrometric chart—high temperatures. Barometric pressure, 29.92 in Hg. 400

350

240

0

0.05

300 0.04

250

200 0.03

150 0.02

100

50

0.01

0

Pounds of Water per Pound of Dry Air

Fe

do oun er P et P

12

210

Grains of Moisture Per Pound of Dry Air

1

bic 9 Cu

Air

5

200

–2.5 Btu

s

2%

Dry Per Pound of deviation – Btu

re

–2.0 Enthalpy

ra tu

190

3%

pe

Te m

u.Ft

11

125

180

18 C

0

11

100

4%

6%

idity

hum

lb

Bu

170

5%

17

tive

120

160

t Cu.F

W et

Rela

20%

40%

50%

50 600 0

150

15%

30% 25%

–0.5 Btu

95

G ra in pe s o rI fm b. of ois dr tur y e air

Enthalpy Deviation of Rejected of Added Water

tu –1.5 B

10 % 9% 8% 7%

90

tha lp 80 y at

Sa 85 tura 65 tio 70 90 n–B tu P 75 95 er P ou Wet B 10 nd ulb an 0 of d Dew 100 D Point 10 r y A 105 or Satura 5 ir tion Te m 11 90% 110 perature 0 s 80% 115 70% 60%

75

0 40 50 60 70 80 90 100 110 120 Temperature of Water, °F

140

u.Ft 16 C –1.0 Btu

50

60

130

–0.8 Btu

85

55

Enthalpy of water. Btu Per Pound of Dry Air 1

–0.6 Btu

45

En 2

t

80

40

50

40

–0.4 Btu

Btu

75

35

3

–0.2

70

30

4

.Ft

65

5

u.F 15 C

14 Cu

25

60

332 Chapter 13 Humidification and Drying 010

650 0.09

600

550

0.08

500 0.07

450

0.06

Psychrometry and the Psychrometric Chart

333

A common experimental method for determining the humidity of air is to determine the wet-bulb and dry-bulb temperatures simultaneously. The operation described above can be accomplished by rapidly passing a stream of air over two thermometers, the bulb of one which is dry. The bulb of the other is kept wet by means of a cloth sack either dipped in water or supplied with water. In a sling psychrometer, the two thermometers are fastened in a metal frame that may be whirled about a handle. The psychrometer is whirled for some seconds and the reading of the wet-bulb thermometer is observed as quickly as possible. The operation is repeated until successive readings of the wet-bulb thermometer show that it has reached its minimum temperature. Thus, a psychrometer consists mainly of two thermometers for wet- and dry-bulb readings. The wet-bulb temperature is the equilibrium temperature obtained by the aforementioned action of an unsaturated vapor– gas mixture flowing past a wetted wick completely covering the bulb of the thermometer. At this equilibrium temperature, the sensible heat given to the water from the air is balanced by the loss of heat from the water by evaporation. An expression for the absolute humidity Y can be obtained from both a heat and material balance in terms of the dry-bulb temperature, Tdb, and the wet-bulb temperature, Twb: Y ¼ Yw 

h (Tdb  Twb ) k lw

(13:9)

where Yw ¼ the saturated absolute humidity at the wet-bulb temperature

lw ¼ the latent heat of vaporization at Twb h ¼ the gas film coefficient for heat transfer by conduction and convection k ¼ the mass transfer coefficient The ratio of h to k (English units) has been experimentally determined to equal 0.236 for an air – water system. It is assumed that radiation effects are negligible in Equation (13.9). As mentioned earlier, the relative humidity, YR, is the ratio of the partial pressure of the water vapor, pA, in the mixture to the vapor pressure of liquid water, pA0 , at the system temperature. This was expressed in Equation (13.3) on a percentage basis. In addition, the dew point is the temperature to which a vapor– gas mixture must be cooled at constant total pressure and humidity to become saturated. This can be determined by calculating pA in the formula for YR [Equation (13.3)], and looking up the temperature that corresponds to this vapor pressure of water. In an adiabatic “saturator”, evaporation cools the water to Tas, the adiabatic saturation temperature. If the contact time between the air and water is long enough, the air becomes saturated and leaves the apparatus at the steady-state temperature Tas. The following equation can be derived from an enthalpy balance, where the subscript as refers to conditions at Tas: T  Tas ¼

lo (Yas  Y) C pH

(13:10)

334

Chapter 13 Humidification and Drying

For the air – water system at ordinary temperature ranges, the heat capacity may be assumed as constant and one may employ Equation (13.6) for CpH. Equations (13.9) and (13.10) for absolute humidity can be compared in the following form: h (T  Tw ); for wet-dry bulb data (13:11) Yw  Y ¼ k lw Yas  Y ¼

C pH (T  Tas ); ls

for adiabatic saturation data

(13:12)

where h and k again represent the heat and mass transfer coefficients, respectively. Both equations would be identical if CpH ¼ h/k. This is almost always true for the air– water system. Hence, the adiabatic cooling line on the psychrometric chart may be used for wet-bulb problems under ordinary conditions. Figures 13.1 and 13.2 contain the psychrometric charts from thermodynamic properties drawn from the literature. A General Electric chart is provided in Figure 13.3.(2) The following are some helpful points on the use of psychrometric charts: 1 Heating or cooling at temperatures above the dew point (temperature at which the vapor begins to condense) corresponds to a horizontal movement on the chart. As long as no condensation occurs, the absolute humidity remains constant. 2 If the air is cooled, the system follows the appropriate horizontal line to the left until it reaches the saturation curve and follows it thereafter. 3 In problems involving use of the humidity chart, it is convenient to choose the mass of air on a dry basis since the chart uses this basis.

ILLUSTRATIVE EXAMPLE 13.1 For air with a dry- and wet-bulb temperature of 808F and 658F, respectively, calculate the following: 1 2 3 4 5 6

Relative humidity Moisture content Dew point Total heat Specific volume Vapor pressure

SOLUTION: The solution is obtained by carefully reading the charts in Figures 13.1 and 13.2. Answers are provided below. 1 2 3 4

Relative humidity ¼ 45% Moisture content ¼ 0.151 lb/lb BDA Dew point ¼ 568F Total heat ¼ 30.0 Btu/lb BDA

Psychrometry and the Psychrometric Chart 5 Specific volume ¼ 13.81 ft3/lb BDA 6 Vapor pressure ¼ 0.228 psi

335

B

ILLUSTRATIVE EXAMPLE 13.2 Qualitatively explain the effect of the wet-bulb temperature on how a sling psychrometer works. SOLUTION: As described in this section, the wet-bulb temperature (Twb) is the temperature at which a thermometer with a wet wick wrapped around the bulb stabilizes. As water evaporates from the wick to the ambient air, the bulb is cooled; the rate of cooling depends on how humid the air is. No evaporation occurs if the air is saturated with water; hence, the wet-bulb temperature and the dry-bulb temperature are the same when the air has a 100% relative humidity. B

ILLUSTRATIVE EXAMPLE 13.3 Determine the key properties for humid air at a dry-bulb temperature of 1608F and a wet-bulb temperature of 1008F. SOLUTION: Refer to the psychrometric chart in Figure 13.1. If the air were to be cooled until moisture just begins to condense, the dew point would be reached. This is represented by a horizontal line at constant absolute humidity intersecting the saturation curve at a dew point of approximately 87.58F. The relative humidity is approximately 14% (interpolating between the 10 and 15% relative humidity lines). The absolute humidity is the horizontal line extended to the right; it intersects the ordinate at a humidity of 0.0285 lb H2O/lb BDA. The humid volume is approximately 16.3 ft3 moist air/lb BDA (interpolating between 16 and 17 ft3 moist air volume). The enthalpy for saturated air at a Twb of 1008F is 71.8 Btu/lb BDA. For the unsaturated air, the enthalpy deviation is 21.0 Btu/lb BDA; therefore, the actual enthalpy for the moist air at a Twb of 1008F and a Tdb of 1608F is 70.8 Btu/lb dry air. B

ILLUSTRATIVE EXAMPLE 13.4 A stream of moist air is cooled and humidified adiabatically from Tdb of 1008F and Twb of 708F to a Tdb of 808F. How much more moisture is added per pound of BDA? SOLUTION: Once again, refer to the psychrometric chart in Figure 13.1. Adiabatic cooling follows the wet-bulb temperature line upwards (toward the saturation curve). The difference in the final and initial humidities is the moisture added: State

Y

Initial Final

0.0090 0.0133

336

Chapter 13 Humidification and Drying (b)

10

10

0%

0%

(a)

Surface temp.

Heating Cooling

Coo

(d)

Mi

xin

g

Humidifying

10 0%

10 0%

(c)

g

idifyin

ehum

nd d ling a

(f)

10

10

0%

0%

(e)

A Che

mic

Ev ap

ora tive

coo

ling

al d

ryin

B′ B

g

B″

Figure 13.3 General Electric psychrometric chart.(2) (a) Sensible heating and cooling of air are represented on the psychrometric chart by a straight horizontal line between the dry-bulb temperature limits of the process. These processes are distinguished by a change in dry-bulb temperature, relative humidity, wet-bulb temperature, total heat, specific volume, and by no change in moisture content, dew-point temperature, and vapor pressure of the air. (b) Cooling and dehumidifying of air are represented on the psychrometric chart by a straight line drawn between the initial condition of the air and the point on the 100 per cent line corresponding to the temperature of the cooling surface. This applies only when the surface temperature is below the initial dew point. The final condition of the air will depend on the total heat extracted from the air. This process is distinguished by a change in all properties of the air. (c) Humidifying of air, with no temperature changes, is represented by a straight vertical line along the dry-bulb temperature line of the air between the moisture content limits of the process. This process is distinguished by an increase in relative humidity, wet-bulb temperature, total heat, specific volume, moisture content, dew-point temperature and vapor pressure of the air. (d) Mixing of air at one condition with air at some other condition is represented by a straight line drawn between the points representing the two air conditions. The condition of the resultant mixture will fall on this line at a point determined by the relative masses (lever rule ) of air being mixed. (e) Evaporative cooling of air, by bringing it in contact with water at a temperature equal to the wet-bulb temperature of the air, is represented by a straight line drawn along the wet-bulb

Psychrometry and the Psychrometric Chart

337

Therefore DY ¼ 0:0133  0:0090 ¼ 0:0043 ¼ 4:3  103 lb H2 O=lb BDA

B

ILLUSTRATIVE EXAMPLE 13.5 A gas is discharged at 1408F from an HCl absorber. If 9000 lb/h (MW ¼ 30) of gas enters the absorber essentially dry (negligible water) at 5608F, calculate the moisture content, the mass flow rate, and the volumetric flow rate of the discharge gas. The discharge gas from the absorber may safely be assumed to be saturated with water vapor. SOLUTION:

From Figure 13.1, the discharge humidity of the gas is approximately Y out ¼ 0:0814 lb H2 O=lb BDA

This represents the moisture content of the gas at outlet conditions. If the gas is assumed to have the properties of air (i.e., BDA), the discharge water vapor rate is: m˙ H2 O ¼ (0:0814)(9000) ¼ 733 lb=h The total flow rate leaving the absorber is: m˙ T ¼ 733 þ 9000 ¼ 9733 lb=h The volumetric (or molar) flow rate can only be calculated if the molecular weight of the gas is known. The average molecular weight of the discharge gas must first be calculated from the mole fraction of the gas (g) and water vapor (w): yg ¼

9000=30 ¼ 0:88 (9000=30) þ (733=18)

yw ¼

733=18 ¼ 0:12 (733=18) þ (9000=30)

MW ¼ (0:88)(30) þ (0:12)(18) ¼ 28:6

Figure 13.3 (Continued) temperature line of the air between the limits of the process. In this process, the total heat of the air remains unchanged because the sensible heat extracted from the air is returned as latent heat by an increase of moisture content. This process is distinguished by a change in dry bulb temperature, relative humidity, specific volume, moisture content, dew-point temperature, vapor pressure, and by no change in wet-bulb temperature. (f) Chemical drying of air is represented by a straight line along the wet-bulb temperature between the limits of the process (AB) only in case the drying is purely by adsorption (the drying agent does not dissolve in the water extracted from the air) and only in case the drying agent does not retain an appreciable amount of the heat of vaporization liberated when the water is condensed on the surface of the adsorbent. In case an appreciable amount of this heat is retained by the adsorbent, the process takes place on a line below the wet-bulb temperature (AB0 ). If the drying agent is soluble in water (such as calcium chloride) the drying process is above (AB00 ) or below (AB0 ) the wet-bulb temperature, depending on whether heat is liberated or absorbed when the agent is dissolved in water.

338

Chapter 13 Humidification and Drying

The ideal gas law is employed to calculate the volumetric flow rate, m˙ T RT (MW)   9733 (0:73)(460 þ 140) q¼ 28:6 1:0

Pq ¼

B

q ¼ 1:49  105 ft3 =h

ILLUSTRATIVE EXAMPLE 13.6 A dehumidification unit in a large factory draws in moist air at 500 lb/h. Moist air enters the unit at a temperature of 82.58F and a wet-bulb temperature of 758F. The exit stream of the unit is 60.88F and saturated with water. Determine the amount of water (in lb/h) condensed in the dehumidifier. SOLUTION: This problem is meant to serve as an example of a basic material balance. As with all material balance problems, it is first recommended that a picture be drawn as follows:

500 lb/h yBDA

mE DEHUMIDIFIER

yBDA

yW

yW

T = 82.4° F P = 1 atm Inlet stream

T = 60.8° F P = 1 atm Air saturated with water

mW xW = 1.0 Liquid water

In order to solve this problem, one must make use of the aforementioned relative humidity (YR). As noted earlier, the relative humidity is the ratio of the partial pressure of water in the gas phase to the vapor pressure of water at temperature T. Since the outlet stream is saturated with water, its relative humidity is unity. The relative humidity of the inlet stream, however, can be determined via Figure 13.1. Locating the appropriate intersection of T and Twb for the inlet conditions, YR ¼ 0:73 ¼ 73% Note that, YR ¼

yW P p0w

Humidification

339

At the inlet temperature, the vapor pressure of water is approximately 0.0373 atm (see steam tables in the Appendix). Therefore, at the inlet: YR p0w P (0:73)(0:0373) ¼ (1)

yw ¼

¼ 0:027 Also note yBDA ¼ 1 2 yw. At the exit conditions, the vapor pressure of water is approximately 0.0179 atm. Therefore, at the exit: YR p0w P (1)(0:0179) ¼ (1)

yw ¼

¼ 0:0179 The total material balance equation for the system may be written 500 ¼ m˙ W þ m˙ E Similarly, a componential balance for water may be written around the dehumidifier, as (0:027)(500) ¼ ( m˙ W ) þ (0:0179)( m˙ E ) The details of solving the above two equations simultaneously is left as an exercise for the reader. The final result is, m˙ W ¼ 4:6 lb=h

B

HUMIDIFICATION The evaporation of water into air for the purpose of increasing the air humidity is known as humidification. Closely allied to this is the evaporation of water into air for the purpose of cooling the water. Dehumidification consists of condensing water from air to decrease the air humidity. All these processes are of considerable industrial importance and involve the contacting of air and water accompanied by heat and mass transfer. These processes are discussed in this section. In terms of industrial applications of humidification, it is often necessary to employ air at a known temperature and a known humidity. This can be accomplished by bringing the air into contact with water under such conditions that a desired humidity is reached. This mixing process can occur in any of the gas – liquid contacting devices discussed early in this Part. If conditions in a humidifier are such that the air reaches complete saturation, the humidity is fixed. However, if the equipment is operated in such a manner that the exit air is not saturated, then process conditions are somewhat indeterminate. Many of

340

Chapter 13 Humidification and Drying

these applicable calculations can be obtained directly from the psychrometric chart provided in the previous section, or from an equivalent chart. Direct contact of a condensable vapor – noncondensable gas mixture with a liquid can produce any of several results, including humidifying of the gas or cooling of the liquid. The direction of liquid transfer (either humidification or dehumidification) depends on the difference in humidity of the bulk of the gas and that at the liquid surface. If the liquid is normally a pure substance, no concentration gradient exists within it and the resistance to mass transfer lies entirely within the gas. Since evaporation and condensation of vapor simultaneously involves a latent enthalpy of vaporization or condensation, there will always be a transfer of latent heat in the direction of mass transfer. The temperature differences existing within the system additionally control the direction of any sensible heat transfer that may occur. Furthermore, since temperature gradients may reside within the liquid, within the gas, or within both, the sensible heat transfer resistance may include effects in either or both phases. The effects of latent and sensible heat transfer may be simultaneously considered in terms of the enthalpy changes which occur. The operation of an adiabatic humidifier normally involves make-up water entering the unit at the adiabatic saturation temperature. Under these conditions, the temperature of the water in the system is assumed constant at the adiabatic saturation temperature, and both the air temperature and humidity water remain constant. In addition, all the heat required to vaporize the water is supplied from the sensible heat of the air. When water is present in air, it is possible to extract the water by cooling the air– water mixture below the mixture’s dew point temperature. The dew point was defined earlier as the temperature at which the air can no longer absorb more water (i.e., it is the 100% relative humidity point) and if the temperature is then reduced further, water is forced out of the air – water mixture as condensation (dew) is formed. This process is described as dehumidification. The amount of water that is removed from a mixture as a result of cooling can be determined by drawing a line on a psychrometric chart (see Fig. 13.1) from the mixture’s initial conditions (e.g., dry-bulb temperature and relative humidity), horizontally to the left (i.e., the cooling direction) until the 100% relative humidity line is encountered. The dehumidification process can also be accomplished by bringing moist air into contact with a spray of water—the temperature of which is lower than the dew point of the entering air. An example is passing the air through sprays. Furthermore, dehumidification of air may also be accomplished by passing a cold fluid through the inside of (finned) tubes arranged in banks through which the air is blown. The outside surface of the metal tubes must be below the dew point of the air so that water will condense out of the air.(3) Cooling towers also find application in industry. The same operation that is used to humidify air may also be used to cool water. There are many cases in practice in which warm water is discharged from condensers or other equipment and where the value of this water is such that it is more economical to cool it and reuse it than to discard it. Water shortages and thermal pollution have made the cooling tower a vital part of many plants in the chemical process industry. Cooling towers are normally

Humidification

341

employed for this purpose and they may be destined to have an increasingly important role in almost all phases of industry. Modern (newer) power-generating stations remain under construction or in the planning stage, and both water shortages and thermal pollution are serious problems that must be dealt with. The cooling of water in a cooling tower is accomplished by bringing the water into contact with unsaturated air under such conditions that the air is humidified and the water is brought approximately to the wet-bulb temperature. This method is applicable only in those cases where the wet-bulb temperature of the air is below the desired temperature of the exit water.(3) Qualitatively speaking, water is cooled in cooling towers by the exchange of sensible heat, latent heat, and water vapor with a stream of relatively cool dry air. The basic relationships developed for dehumidifiers also apply to cooling towers although the transfer is in the opposite direction since the unit acts as a humidifier rather than as a dehumidifier of air.(4) Brown and Associates(5) have provided empirical correlations from the litera(6,7) for estimating (roughly) sizes and capacities of conventional towers. ture

Equipment Any of the absorption contact equipment discussed in Chapter 10 are applicable to humidification, dehumidification, and water-cooling applications. Packed towers and plate towers are particularly effective but there are other types of specialized units. These are briefly discussed below. As mentioned above, warm water flows down cooling towers countercurrent to rising unsaturated air. Water is cooled by furnishing part of the latent heat required to vaporize some of the water into the air stream. The air is thus humidified as it rises. The calculation of the tower height for the vaporization process can be accomplished by the stepwise procedure detailed for distillation in Chapter 9. The operation of a cooling tower is indicated diagrammatically in Figure 13.4. As described above, the hot water at TL2 is introduced at the top of the tower and leaves the bottom at TL1. The air flows countercurrent to the water. It enters at the bottom (point 1) and leaves the top (point 2). The temperature of the air – water interface tends to approach the adiabatic saturation or wet-bulb temperature of the air. At the top of the tower, heat is being transferred from the inlet (hot) water to the air, since the temperature of the water is higher than that of the interface, and the interface temperature is usually higher than that of the air. This sensible heat removed from the water appears as sensible and latent heat of the air – water mixture. At the bottom of the tower, the temperature of the water and of the interface may both be lower than that of the air, with sensible heat being transferred both from the liquid and air to the interface, resulting in the vaporization of the water. Water may thus be cooled by air at a higher temperature, provided that a humidity difference driving force (which produces evaporation) is maintained.(5) Cooling towers were originally primarily constructed of redwood, a material which is very durable when in continuous contact with water; however, there are also

342

Chapter 13 Humidification and Drying

L2 = liquid mass velocity, lb/h · ft2 TL2 = temperature of liquid HL2 = enthalpy of liquid

G2 T2 H2 Y2

dZ = differential height

L1 TL1 HL1

G1 = dry gas mass velocity, lb/h · ft2 T1 = temperature (dry-bulb) of air mixture H1 = enthalpy of air (mixture) Y1 = humidity of air (mixture)

Figure 13.4 Adiabatic humidification.

moisture resistant polymer composite materials that are now available. The internal packing is usually in the form of horizontal wooden slats. The void volume is usually greater than 90%, leading to a pressure drop that is extremely low. The air – water interfacial surface includes not only the liquid films that wet the slats but also the surface of droplets, which settle as a “rain” from each tier of slats to the next. Natural-circulation towers consist of two types: atmospheric and natural-draft. In atmospheric towers, air circulation is dependent solely on the prevailing winds, which essentially produces a crossflow of the air and water. The towers are generally long with narrow horizontal cross-sectional areas. This leads to adequate penetration of the air into the central portions of the unit. Louvers on the sides of the tower help reduce the losses of water entrained in the gas stream. Natural-draft towers provide a free convection effect. This ensures air movement even in calm weather; it is similar to a stack or a chimney. Large cross-sectional areas are usually required in order to maintain a low air velocity and, consequently, a low pressure drop. (Both naturalcirculation and natural-draft towers must be relatively tall.) A pump is required for the water, but there are no fans or the accompanying power cost associated with moving the air. The emphasis in recent years has been to employ mechanical-draft equipment. These towers may be of the forced-draft type where the air is blown into the tower by a fan at the bottom, or of the induced-draft type where the air is drawn upward by a fan at the top. Since the forced-draft tower ensures the recirculation of the hot, humid discharged air, the effectiveness of the tower is somewhat compromised.

Humidification

343

However, the induced-draft tower discharges the air at a higher velocity and can lead to a more uniform air distribution in the packing. Based on these considerations, this unit is often preferred despite the fact that the fan power is higher since the air density is lower. Water is usually distributed over the packing by weirs or spray nozzles. Spray eliminators at, or near, the top of the unit can reduce water carryover. All these units incur losses defined as blowdown. In addition, make-up water is required for evaporation and entrainment losses as well as the blowdown. Spray columns (see Chapter 10) are either forced or induced-draft towers without any internal packing. Contact relies entirely on the inlet water sprays at the top to provide interfacial surface for mass transfer. Low gas pressure drop is normal for this unit. Spray chambers are essentially horizontal spray columns. They too are frequently used for adiabatic humidification – cooling operations. Dehumidification is possible by cooling the water prior to spraying or by inserting refrigerating coils in the side spray chamber. Generally, three banks of sprays in series will bring the gas to substantial equilibrium with the incoming spray liquid. Spray ponds are occasionally used for water cooling where a close approach to the air wet-bulb temperature is not required. These units essentially contain fountains from which the water is sprayed vertically upward into the air and allowed to settle by gravity into a collection basin. They are obviously subject to water loss due to any prevailing winds. Cooling ponds or reservoirs are used for removing a relatively small amount of heat from water over a small temperature range. The pond required may simply be estimated from any relationship that provides the rate of evaporation of water into still air. The heat is then calculated from the latent enthalpy (of vaporization) of water. As one might expect, any wind will increase the capacity of a pond. The cooling capacity can be further increased if a system of spray nozzles can be installed above the surface of a cooling pond. Other methods of dehumidification can also include adsorption, such as those methods discussed in Chapter 11. Adsorbents employed include silica gel or alumina (commonly referred to as desiccants). In addition, treating/washing the gas with liquids can also reduce the water content of the gas.

Describing Equations For cooling or humidification, the operating line lies below the equilibrium line. As noted earlier, an operating line refers to the actual vapor– liquid relationship of a key component, in contrast to the true equilibrium relationship. There is, therefore, a minimum air rate that can be used to accomplish a specified water-cooling duty. In dehumidification, cool water is used to reduce both the humidity and the temperature of the air introduced at the bottom of the tower. The operating line for this case lies above the equilibrium line and there exists a minimum amount of water that can be used to dehumidify a given quantity of air. The basic relationships available for humidifiers and cooling towers apply to dehumidifiers, although the transfer occurs in the opposite direction; in effect, the tower is a dehumidifier rather than a humidifier of air.

344

Chapter 13 Humidification and Drying

An adiabatic, gas – liquid, continuous, countercurrent contact system may now be examined. The diagram in Figure 13.4 shows such a system and the corresponding notation of Equations 13.13 and 13.14. For any process involving humidification, a moisture mass balance and an enthalpy balance can be written. The mass balance is: L2 þ G1 Y1 ¼ L1 þ G2 Y2

(13:13)

In addition, the enthalpy balance is: _ ¼ L1 HL1 þ G2 H2 L2 HL2 þ G1 H1 þ Q

(13:14)

_ ¼ 0 if operated adiabatically). ˙ is the heat transferred to the units (Q where Q One can ultimately show that for the differential element dZ the equation describing the rate of sensible heat transfer on the liquid side (assuming L and G are relatively constant) takes the form: LCL dTL ¼ hL a(TL  Ti ) dZ;

CL ¼ CPB

(13:15)

GCG dT ¼ hG a(Ti  T) dZ; CG ¼ CPA

(13:16)

Equation (13.16) applies on the gas side:

The rate of mass transfer can be written as: G dY ¼ kL a(Yi  Y) dZ

(13:17)

where a is the interfacial surface area per unit volume and the subscript i refers to interfacial values. If the interfacial values are constant through column height Z, one may integrate Equations (13.16) and (13.17) to give:   Ti  T1 hG aZ (13:18) ¼ ln Ti  T2 GCG   Yi  Y1 kL aZ (13:19) ¼ ln Yi  Y2 G In the above derivations, it was assumed that Ti and Yi are constant throughout length Z; strictly speaking, this is not necessarily true. However, by taking an average Ti (assumed to be the average liquid temperature of the unit) and corresponding average Yi for the length of the column, satisfactory answers can be obtained once heat and mass transfer coefficient values are provided.(8,9)

ILLUSTRATIVE EXAMPLE 13.7 A flue gas (MW ¼ 30, dry basis) is being discharged from a scrubber at 1808F (dry-bulb) and 1258F (wet-bulb). The gas flow rate on a dry basis is 10,000 lb/h. The absolute humidity at the

Humidification

345

dry-bulb temperature of 1808F and wet-bulb temperature of 1258F is 0.0805 lb H2O/lb dry gas. What is the mass flow rate of the wet gas?(8) SOLUTION:

Calculate the flow rate of water (w) in the air and the total (T ) flow rate: m˙ w ¼ (0:0805)(10,000) ¼ 805 lb=h m˙ T ¼ 10,000 þ 805 ¼ 10,805 lb=h

B

ILLUSTRATIVE EXAMPLE 13.8 Refer to Illustrative Example 13.7. What is the molar and volumetric flow rate of the wet gas? SOLUTION: To determine the actual volumetric flow rate, first calculate the molar rate of water and dry gas (dg): 10,000 ¼ 333:3 lbmol=h 30 805 ¼ 44:7 lbmol=h n˙ w ¼ 18

n˙ dg ¼

Calculate the mole fraction of water vapor: yw ¼

44:7 ¼ 0:12 44:7 þ 333:3

Calculate the average molecular weight of the mixture: MW ¼ (1  0:12)(30) þ (0:12)(18) ¼ 28:6 lb=lbmol Determine the molar flow rate of the wet gas (wg): n˙ wg ¼

10,805 ¼ 378 lbmol=h 28:6

Calculate the volumetric flow rate of the wet gas by employing the ideal gas law. qwg ¼

n˙ wg RT (378)(0:73)(460 þ 180) ¼ ¼ 1:77  105 ft3 =h P 1:0

B

ILLUSTRATIVE EXAMPLE 13.9 Calculate the height of a packed tower operated in a countercurrent mode that is required to cool a utility water discharge from 908F to 808F. Ambient air at a wet-bulb and dry-bulb temperature of 478F and 568F, respectively, is to be employed. The water and gas rates have been set at 500 and 1000 lb/h . ft2, respectively. Based on a review of the literature, one may assume that: hG a ¼ (7:5  103 )(G0 )0:7 (L0 )0:6

346 where

Chapter 13 Humidification and Drying hGa ¼ heat transfer coefficient, Btu/h . ft3 . 8F G0 ¼ air mass flux, lb/h . ft2 L0 ¼ water mass flux, lb/h . ft2

Since the air at discharge conditions is essentially saturated, assume the humid heat capacity to be 0.245 Btu/lb BDA . 8F. Also, assume the adiabatic saturation temperature to be 688F. SOLUTION:

First calculate the heat transfer coefficient: hG a ¼ (7:5  103 )(500)0:7 (1000)0:6 hG a ¼ 36:68 Btu=h  ft3  F

Use Equation (13.18) to solve for Z:   Ti  T1 hG aZ ¼ ln GCG Ti  T2 In order to do this, the unknown values in the above equation must be determined; these are T1, T2, Ti, and CG. T1 is the dry-bulb temperature of the air stream, which is given in the problem statement as 568F. The heat capacity of air may be assumed equal to approximately 0.245 Btu/lb BDA . 8F. Ti is to be assumed constant throughout the height, Z, of the column. This value is therefore assumed to be the average liquid temperature of the unit. This is determined via: Ti ¼

90 þ 80 ¼ 85 F 2

_ 0 of the packed The value of T2 is determined in the following manner. The heat duty Q column is calculated using the information given for the water stream: _ 0 ¼ L0 CPL DTw Q   ¼ 500 lb=h  ft2 ð1:0 Btu=lb   FÞ(90  80) F ¼ 5000 Btu=h  ft2 Next, the duty of the water stream must be equal to the heat duty of the gas stream. Therefore: _ 0 ¼ G0 CPG DTG Q   5000 Btu=h  ft2 ¼ 1000 lb=h  ft2 ð0:245 Btu=lb   FÞ(56  T2 ) F T2 ¼ 76:4 F Substituting the appropriate values in Equation (13.18) yields the height of the packed column, which is calculated below:   Ti  T1 hG aZ ¼ ln GCG Ti  T2     (1000)(0:245) 85  56 ln Z¼ 36:68 85  76:4 ¼ 8:1 ft

B

Drying

347

DRYING

Mass H2O in Solid

In the drying process, a liquid (usually water) is separated from a wet solid by use of a hot dry gas (usually air). The drying of solids to remove moisture involves the simultaneous processes of heat and mass transfer. Heat is transferred from the gas to the solid (and liquid) in order to evaporate the liquid contained in the solid. Mass is transferred as either a liquid or vapor within the solid and then as a vapor from the surface of the solid. Additional details on this process are provided later in this section. The energy required to vaporize the liquid in a solid is almost always furnished by a hot inert carrier gas that enters the drier. In some driers, the solid may be in contact with heated metal surfaces where the required heat of vaporization flows to the solid by conduction. In vacuum drying (where there is essentially no carrier gas), the heat of vaporization is furnished by conduction or radiation; here, the capacity of the drier is largely influenced by the heat-transfer surface available within the dryer. The curve provided in Figure 13.5 is obtained when a substance saturated with water is dried. During the drying process, a thin film of water exists on the surface of the material where water is supplied from the solid fast enough to keep the surface entirely wet. As this water is evaporated, water from the interior of the sample rises to the surface essentially by capillary action with the solid temperature approximately given by the wet-bulb temperature of the air. After drying has proceeded for some time, the surface film begins to disappear and the rate of drying decreases. This critical moisture content leads to dry patches on the surface, and—as noted later in Figure 13.6—the drying rate begins to fall. This corresponds to the curved portion of the graph in Figure 13.5. Ultimately, water ceases to evaporate and a final equilibrium moisture content (denoted by the dashed line) is achieved. The curve in Figure 13.5 can also be plotted as shown in Figure 13.6. In this plot, the horizontal portion corresponds to the constant rate period. The value of the

Time Figure 13.5

Typical drying process: moisture content vs time.

348

Chapter 13 Humidification and Drying

Rc

R=

mass evaporation h · ft2

Ro Xo

Xc

X1

mass H2O X= mass dry solid Figure 13.6 Typical drying process: drying rate vs moisture content.

mass rate of water evaporated per unit time, R [mass/time . area], corresponding to the horizontal portion is defined as the critical drying rate, Rc, while the value of X (mass H2O evaporated/mass solid) corresponding to Rc is called the critical moisture content, Xc. Note that the drying rate may reach zero before the solid is completely dry. The resulting equilibrium moisture content, Xe, for the solid is a function of the drying rate. The reader is reminded that the term X is a concentration term based on a water free basis. An equation for the total time of drying can be derived. First note that: dm ; dm ¼ RA dt (13:20) R¼ A dt The term m is the mass of water, A is the drying surface, and dt is the time of drying. In addition, (13:21) dm ¼ 2sArs dX where s is specified as one-half of the thickness of the material being dried, rs is the density of the dry material and X is the moisture content on a water-free basis. Combining Equations (13.20) and (13.21), while eliminating dm and solving for R, gives: R¼

[2sArs ] dX dX ¼ 2srs A dt dt

(13:22)

The total drying time, tt, is then found by integration, ðtt

Xð1

dt ¼ tt ¼ 2srs 0

dX R

X2

where X1 and X2 are the initial and final moisture contents, respectively.

(13:23)

Drying

349

During the constant rate period, R is constant and may be set equal to Rc. One may therefore integrate Equation (13.23) to give: 

 2srs tc ¼ (X1  X2 ) Rc

(13:24)

In the falling rate period, an assumption can be made that there is a linear relationship, that is, y ¼ mx þ b (as shown in Fig. 13.6), between falling rate, R, and moisture content, X. This effectively assumes that the curve approximates a straight line from the critical moisture content to the origin during the falling rate period. Thus: R ¼ mX þ b

(13:25)

dR ¼ m dX; dR=m ¼ dX

(13:26)

In addition:

For this falling rate period, one may replace dX in Equation (13.23) by Equation (13.26) to give: X ðc

tf ¼ 2srs

dX 2srs ¼ m R

Xo

R ðc

dR R

(13:27)

Ro

where Xo refers to the final moisture content. Upon integration, the time of the falling rate period, tf , becomes:   2srs Rc ln tf ¼ m Ro

(13:28)

If the y-intercept in Figure 13.6 is assumed to be zero, i.e., Ro ¼ Xo ¼ 0, then m¼

Rc  Ro Rc ¼ Xc  Xo Xc

(13:29)

Substituting this value of m into Equation (13.28) gives:     Xc Rc ln tf ¼ 2srs Rc Ro

(13:30)

The total time for drying is the sum of the times for the constant rate period and the falling rate period, i.e., tt ¼ tc þ tf ¼

   2srs Rc (X1  X2 ) þ Xc ln Rc Ro

(13:31)

350

Chapter 13 Humidification and Drying

If both rate periods are present, X2 ¼ Xc so that the above is given by    2srs Rc tt ¼ (X1  Xc ) þ Xc ln Rc Ro

(13:32)

Another term that has received attention in this field is the free moisture content, F. It is defined as the difference between the total moisture content X and the equilibrium moisture content Xe (the content of water in the solid that cannot be removed by the air), expressed as mass of water per mass of dry solid: F ¼ X  Xe

(13:33)

The free moisture content is a function of the same variables as the equilibrium moisture content. As one might surmise, F is of major interest in drying calculations. It should be noted that drying is one method of separating a liquid from a solid. Technically, it is the aforementioned vaporization process in which the heat rate and mass transfer rates control equipment design. In most dryers, heat is transferred by convection from a gaseous drying medium to the surface of the wet solid. (In some designs, radiation from the walls of the dryer to the wet material supplements convection.) This heat vaporizes the liquid, which is usually water. The vapor that is thus formed must diffuse into the gas phase. In so doing, it passes through the same gasphase convective resistance through which the heat passed. However, the transfer of heat is in the opposite direction to the transfer of mass. Depending on operating conditions, a small portion of the heat can act as sensible heat to raise the temperature of the wet solid. In some respects, drying is identical to humidification in that water is provided from a pure liquid while in drying it comes from liquid dispersed in a solid. Thus, drying requires that a new resistance be considered. This resistance can be thought of opposing the movement of the liquid through the solid to the gas – solid interface. Although the next two subsections will key on rotary dryers and spray dryers, the significance of the above resistance becomes apparent if one analyzes typical drying data obtained in a batch dryer. Summarizing, one notes that the rate at which the liquid is vaporized is constant at first. This constant rate holds until the moisture content of the wet stock reaches a critical value, at which point the rate begins to decrease. It continues to decrease thereafter, ultimately falling to or approaching zero when the moisture content has been reduced to an equilibrium value that is the lowest value that it can reach with the drying conditions employed.(4)

ILLUSTRATIVE EXAMPLE 13.10 A granular solid weighing 5000 lb (bone dry) is to be dried under constant drying conditions from a moisture content of 0.20 lb/lb to a final moisture content of 0.02 lb/lb. The material

Drying

351

Table 13.2 Drying Rate Data, Illustrative Example 13.10 Moisture content (lb/lb)

Drying rate, R (lb/ft2-h)

0.300 0.200 0.140 0.114 0.096 0.056 0.042 0.026 0.016

0.35 0.35 0.35 0.30 0.265 0.180 0.150 0.110 0.075

has an effective area of 0.25 ft2/lb. The drying rates shown in Table 13.2 were previously obtained under the same conditions. Calculate the time required for drying. SOLUTION: Based on the problem statement, A ¼ (0.25 ft2/lb) (5000 lb) ¼ 1250 ft2, m ¼ 5000 lb, and the critical moisture content, from Table 13.2, is approximately 0.14 lb/lb. For the constant rate period, tc ¼

2srs (X1  X2 ); Rc

X1 ¼ 0:20;

X2 ¼ Xc ¼ 0:14

(13:24)

with 2srs ¼

m A

This may be rewritten as tc ¼

m (X1  X2 ) ARc

Substituting, tc ¼

5000 (0:20  0:14) ¼ 0:69 h (1250)(0:35)

The rate decreases following this constant rate period. This is treated in the next illustrative example. B

ILLUSTRATIVE EXAMPLE 13.11 Refer to the previous example. If the falling rate period is approximately described by R ¼ 2:17X þ 0:047 calculate the total drying time.

352

Chapter 13 Humidification and Drying

SOLUTION: (13.30):

An integration is required for the falling time calculation. Apply Equation     Xc Rc tf ¼ 2srs ln Rc Ro

This may be rewritten as tf ¼

    m Xc Rc ln A Rc Ro

Substituting,  tf ¼

    5000 0:14 0:35 ln ¼ 2:17 h 1250 0:35 2:17(0:02) þ 0:047

The required total time is then tt ¼ tc þ tf ¼ 0:69 þ 2:17 ¼ 2:86 h

B

Rotary Dryers The rotary dryer is a popular device suitable for the drying of free-flowing materials that can be tumbled without concern for breaking. Moist solid is continuously fed into one end of a rotating cylinder with the simultaneous introduction of heated air. The cylinder is installed at a slight angle and with internal lifting flights so that the solid is showered through the hot air as it traverses the dryer. A bench scale unit, located in the Unit Operations Laboratory at Manhattan College, is provided in Figure 13.7. In the lab’s experiment, a liquid (water) is separated from a wet solid (cornmeal) by use of a hot dry gas (air). The rotary dryer is operated in the cocurrent mode. The hot air is cooled as it is humidified; at the same time, the solid is heated and dried by contact with the hot air. A typical industrial rotary dryer consists of a cylinder, rotated upon suitable bearings and usually slightly inclined to the horizontal. The length of the cylinder may range from four to more than ten feet. Solids fed into one end of the cylinder progress through it by virtue of rotation, head effect, the slope of the cylinder, and discharge as finished product at the other end. Gases flowing through the cylinder may retard or increase the rate of solids flow, depending upon whether the gas flow is countercurrent or cocurrent with respect to solid flow. Rotary dryers have been classified as direct, indirect-direct, indirect, and special types. These terms refer to the method of heat transfer: direct when heat is added to or removed from the solids by direct exchange between flowing gas and solids, and indirect when the heating medium is separated from physical contact with the solids by a metal wall or tube. Rotating equipment is applicable to batch or continuous processing solids, which are relatively free-flowing and granular when discharged as product. Materials that are not completely free-flowing in their feed condition are handled in a special

Drying

Figure 13.7

353

Rotary dryer: Manhattan College’s Unit Operations Laboratory.

manner—either by recycling a portion of the final product and premixing with the feed in an external mixer to form a uniform granular feed to the process, or by maintaining a bed of free-flowing product in the cylinder at the feed end and, in essence, performing a premixing in the cylinder itself. The method of feeding rotating equipment depends upon material characteristics and the location and type of upstream processing equipment. When the feed comes from above, a chute extending into the cylinder is usually employed. For sealing purposes or if gravity feed is not convenient, a screw feeder is normally used. On cocurrent direct-heat units, cold-water jacketing of the feed chute or conveyer may be desirable if it is contacted by the inlet hot gas stream. This will prevent overheating of the metal wall with resultant scaling or overheating of heat sensitive feed materials. One method of feeding direct cocurrent drying equipment utilizes dryer exhaust gases to convey, mix, and pre-dry wet feed. The latter is added to the exhaust gases from the dryer at a high velocity. The wet feed, mixed with dust entrained from the dryer, separates from the exhaust gases in a cyclone (typically) and usually drops into the feed end of the cylinder. This technique combines pneumatic and rotary drying. The dust entrained in the exit-gas stream is customarily removed in the cyclone collector(s). This dust may be returned into the process or separately collected. For expensive materials or extremely fine particles, bag collectors (a baghouse) may follow a cyclone collector, assuming fabric temperature stability is not limiting and there are assurances of no temperature excursions. Rotating equipment, with the exception of brick-lined vessels operated above ambient temperatures, are usually insulated to reduce heat losses. Other exceptions are direct-heat units of bare metal construction operating at high temperatures where heat losses from the shell are necessary to prevent overheating of the metal.

354

Chapter 13 Humidification and Drying

Insulation is the rule with cocurrent direct-heat units. It is not unusual for product cooling or condensation on the shell to occur in the last 10 to 50 percent of the cylinder length if it is not sufficiently insulated. The direct-heat rotary dryer is usually equipped with flights (discussed above) on the interior for lifting and showering the solids through the gas stream during passage through the cylinder. These flights are usually offset every 2 – 6 feet to ensure more continuous and uniform curtains of solids in the gas. The shape of the flights depends upon the handling characteristics of the solids. For free-flowing materials, a radial flight with a 90-degree lip is employed. For sticky materials, a flat radial flight without any lip is used. When materials change their characteristics during drying, the flight design is usually changed along the dryer length. Many standard dryer designs employ flat flights with no lips in the first one-third of the dryer measured from the feed end, flights with 45-degree lips in the middle one-third, and flights with 90-degree lips in the final one-third of the cylinder. Spiral flights are usually provided in the first few feet at the feed end to accelerate forward flow from under the feed chute or the conveyer to prevent leakage over the feed-end retainer. When cocurrent gas –solid flow is used, flights may be left out of the final few feet at the exit end to reduce entrainment of dry product in the exit gas. Showering of wet feed at the feed end of a countercurrent dryer will, on the other hand, frequently serve as an effective means for scrubbing dry entrained solids from the gas stream before it leaves the cylinder. Some dryers are provided with sawtooth flights to obtain uniform showering while others use lengths of chain attached to the underside of the flights to scrape over and knock the walls of the cylinder, thereby removing sticky solids that might normally adhere to it. Solids sticking on flights and walls are usually removed more efficiently by external shell knockers. In dryers of large cross-section, internal elements or partitions are sometimes used to increase the effectiveness of material distribution and reduce dusting and impact grinding. Use of internal members increases the difficulty of cleaning and maintenance unless sufficient free area is left between partitions for the easy access of an individual. Some examples of the more common flight arrangements are shown in Figure 13.8. Countercurrent flow of gas and solids gives greater heat-transfer efficiency with a given inlet gas temperature. However, cocurrent flow is used more frequently to dry

90-degree Lip Flights

Radial Lip Flights

Figure 13.8 Rotary dryer flight arrangements.

45-degree Lip Flights

Shell with Diaphragm

Drying

355

heat-sensitive materials at higher inlet gas temperatures because of the rapid cooling of the gas during the initial evaporation of surface moisture. A major design variable of the rotary dryer is the drying rate. Since the solid (cornmeal in the Manhattan College experiment) loses moisture while the air stream is gaining moisture, the drying rate can be calculated in either of two ways. Data taken from a dryer can provide the drying rate based either on the moisture lost by the solid and/or the humidity gained by the air stream between the inlet and exit.(10) As noted above, the mechanism of drying involves the transfer of heat by convection for vaporizing the material. Mass is transferred as a result of a moisture concentration gradient at a rate dependent on the characteristics of the solid. The dynamic equilibrium prevailing between the rate of heat transfer to the material and rate of vapor removal from the surface during the constant drying rate period also provides a means for calculating the heat and mass transfer coefficients. The drying rate, dm/dt can be measured by the two ways presented above, i.e., based on the loss of water by the solids or on the increase in water content of the air stream. Dry-bulb temperatures and saturated (wet-bulb) temperatures at each end of the dryer are normally employed to calculate the log mean temperature difference. This enables one to determine the heat transfer coefficient, h, based on the air and water data as follows:   dm hV(T  Ts )mean ¼ (13:34) m˙ ¼ l dt with ( m˙ water )solid ¼ ( m˙ water )air The above terms are defined as follows: m˙ ¼ dm/dt ¼ drying rate (lb water evaporated/h) h ¼ heat transfer coefficient based on dryer volume (Btu/h . ft2 . 8F) V ¼ dryer volume (ft3) T ¼ temperature of drying medium, dry-bulb temperature (8F), determined at inlet and outlet conditions Ts ¼ surface temperature of solid (8F), wet-bulb temperature, determined at inlet and outlet conditions

l ¼ latent heat of vaporization (Btu/lb) Similarly, humidities can be obtained at the inlet and outlet (including saturated humidities) and the log mean humidity difference is used to calculate the mass transfer coefficient, k, based on either the air and water data as follows since _ ( m˙ water )solid ¼ ( m˙ water )air ¼ m: m˙ ¼

dm ¼ kV(Yas  Y)mean dt

(13:35)

356

Chapter 13 Humidification and Drying

where k ¼ mass transfer coefficient based on dryer volume (lb/h . ft3 . DY ) Y ¼ absolute humidity of drying medium at dry-bulb temperature (lb water/lb BDA), determined at inlet and outlet conditions Yas ¼ absolute saturated humidity of surface at wet-bulb temperature (lb water/lb BDA), determined at inlet and outlet conditions For comparison purposes, heat transfer coefficients can also be calculated from an empirical equation. Friedman and Marshall(11) expressed the heat transfer coefficient as an empirical equation in terms of air mass flux, G (lb/ft2 . s): aGm h¼ (13:36) D where a and m are empirical constants and D is the diameter of the dryer in feet. The value of m varies from 0.16 –0.67 depending on the individual unit. The value of a may be approximated by

a¼

0:05(Nf  1) 0:786

(13:37)

where Nf is the number of flights in the dryer. Another important parameter in a dryer operation is the holdup F, which refers to the fraction of dryer volume occupied by the solid at any instant. The describing equation is: F¼

uLs rs Z

(13:38)

where F ¼ holdup fraction (solid volume/dryer volume) u ¼ retention time of solids passing through dryer (h) Ls ¼ rate of flow of solids per dryer cross-sectional area (lb/h . ft2) Z ¼ length of dryer (ft)

rs ¼ density of solid material (lb/ft3) ILLUSTRATIVE EXAMPLE 13.12 A rotary dryer is operating at 3208F with a gas flow rate of 600 scfm (608F). The dryer is 4 ft in diameter and 16 ft long. The minimum residence time required for operation is 0.4 min. Calculate (1) the actual gas flow rate, (2) the volume required for minimum residence time and, (3) the maximum residence time. SOLUTION: 1 The actual gas flow rate is calculated by Charles’ law: qa ¼ qs

  Ta Ts

Drying

357

where Ta and Ts are in absolute units  qa ¼ (600 scfm)

 320 þ 460 60 þ 460

¼ 900 acfm 2 The volume required for minimum residence time is Vm ¼ qa tmin ¼ (900 acfm)(0:4 min) ¼ 360 ft3 3 The maximum residence time, tmax, takes into account the actual volume, Va, in the equation: tmax ¼

Va qa

where Va ¼ (p=4)D2 L ¼ (0:785)(4)2 (16)

tmax

¼ 201 ft3 ¼ 210=900 ¼ 0:223 min ¼ 13:4 s

The minimum residence time required is therefore not satisfied. Note that the gas residence time, calculated by dividing the volume of the dryer by the gas flow rate, is only an approximate value since fully developed flow does not necessarily exist. Thus, a residence time distribution exists with real systems. B

ILLUSTRATIVE EXAMPLE 13.13 An engineer involved in the analysis of a rotary dryer unit suspects that the dryer is operating below the desired residence time of 1 hour. The available operating data are Dryer length ¼ 25 ft Dryer diameter ¼ 10 ft Slope of dryer ¼ 0:01 ft=ft of length The present dryer rotation velocity is 0.9 rpm. 1 Is this velocity providing the necessary residence time? 2 What maximum kiln rotation velocity would meet the desired residence time?

358

Chapter 13 Humidification and Drying

The following equation may be assumed to apply: t¼ where

0:19L NDS

t ¼ retention time (min) L ¼ dryer length (ft) N ¼ dryer rotational velocity (rpm) D ¼ dryer diameter (ft) S ¼ dryer slope (ft/ft of length)

SOLUTION:

Calculate the residence time using the equation provided. t¼

(0:19)(25) (0:9)(10)(0:01)

¼ 53 min Therefore, the necessary residence time of 60 min is not being met. Set the residence time of this equation to 60 min to find the minimum dryer rotation rate. Rearrange this equation and solve for N: 60 min ¼

(0:19)(25) (N)(10)(0:01)

N¼

(0:19)(25) (60)(10)(0:01)

¼ 0:8 rpm Therefore, a maximum rotational velocity of 0.8 rpm is required to achieve the desired residence time of 1 hour in this rotary dryer. Note: The lead coefficient of 0.19 in the above equation has been estimated from limited experimental data. Other values appear in the literature. Therefore, the residence time calculated through the use of this coefficient is, at the very best, a rough estimate. B

ILLUSTRATIVE EXAMPLE 13.14 The following data was obtained during a rotary dryer experiment at Manhattan College: Barometric pressure – 29.92 in Hg Air flow rate – 210 ft/min Mass of feed – 29.8 g Feed time – 5.11 min Dryer diameter – 4.0 in Dryer length – 18.75 in Calculate the drying rate, dm/dt.

Inlet air temperature (dry-bulb) – 748F Inlet air temperature (wet-bulb) – 618F Exit air temperature (dry-bulb) – 908F Exit air temperature (wet-bulb) – 678F Inlet solids temperature – 1828F Exit solids temperature – 1238F

Drying

359

SOLUTION: The drying rate equals the mass flow rate of BDA, m˙ (BDA) times the change in humidity (Y ) of the air from inlet to outlet (lb H2O/lb BDA). As a first step, calculate the area available for air flow: cross-sectional area of dryer ¼ (p=4)(D2 ) ¼ (0:785)(4=12)2 ¼ 0:0872 ft2 The density of moist air at these conditions may be assumed to be 0.0739 lb/ft3. Thus, m˙ tot ¼ (210 ft=min)(60 min=h)(0:0872 ft2 )(0:0739 lb=ft3 ) ¼ 81:20 lb=h This represents the mass of air plus water in air in the feed. To convert to a dry air basis, the following relation is used: lb (H2 O) m˙ w ¼ lb (BDA) m˙ tot  m˙ w The lb (H2O)/lb (BDA) is known from the wet-bulb temperature of the inlet air. It is merely the humidity, Y, which is 0.0087. Therefore, Y¼ 0:0087

m˙ w m˙ tot  m˙ w

lb (H2 O) m˙ w ¼ lb (BDA) 81:20  m˙ w m˙ w ¼ 0:70

lb (H2 O) h

_ air ¼ 80:50 lb=h: Therefore, by difference, m Knowing the wet-bulb temperature at the outlet gives Y at the outlet: Y ¼ 0:0095 lb (H2 O)=lb (BDA) The drying rate, dm/dt, can now be solved for in the following manner: m(BDA) ˙ ¼

  dm lb (H2 O) lb (air) ¼ (0:0095  0:0087) 80:50 dt lb (air) h ¼ 0:0644

lb (H2 O) h

B

ILLUSTRATIVE EXAMPLE 13.15 Refer to Illustrative Example 13.14. Calculate the log mean temperature difference (T 2 Ts)lm.

360

Chapter 13 Humidification and Drying

SOLUTION: Since rotary dryers are cocurrent devices, the describing equation for the logarithmic mean temperature difference is given by:

(T  Ts )lm ¼

(T  Ts )exit  (T  Ts )inlet   (T  Ts )exit ln (T  Ts )inlet

Since this is a measure of the temperature difference driving force down the length of the dryer, the inlet and exit solids temperatures are used for T. The corresponding air inlet and exit air temperatures are given by Ts (exit) ¼ 90 F Ts (inlet) ¼ 74 F Thus, (T  Ts )lm ¼

(123  90) F  (182  74) F   (123  90) F ln (182  74) F

(T  Ts )lm ¼ 63:258 F

B

ILLUSTRATIVE EXAMPLE 13.16 Refer to Illustrative Example 13.14. Calculate the heat transfer coefficient, h. SOLUTION:

The dryer volume and the latent heat of vaporization must first be found: V ¼ (length)(cross-sectional area) ¼ (p=4)(D2 )(L) ¼ 0:1363 ft3

The heat of vaporization l is approximately 910 Btu/lb. Thus, the heat transfer coefficient can be calculated from Equation (13.35): h¼

  dm l ; dt (T  Ts )lm V

dm ¼ m(BDA) ˙ dt

Substituting:  h ¼ 0:0644

 910 (66:28)(0:1363)

¼ 6:487 Btu=ft2  h   F

B

Drying

361

ILLUSTRATIVE EXAMPLE 13.17 Refer to Illustrative Example 13.14. Determine the mass transfer coefficient, k. SOLUTION: The drying rate remains the same. Calculations are now required for (Yas 2 Y )lm and then k. The calculation of the log mean humidity difference (Yas 2 Y )lm is as follows: (Yas  Y)lm ¼

(Yas  Y)exit  (Yas  Y)inlet   (Yas  Y)exit ln (Yas  Y)inlet

As given, Y (inlet) ¼ 0.0087 lb (H2O)/lb (BDA) and Y (exit) ¼ 0.0095 lb (H2O)/lb (BDA). Since this is the humidity difference along the length of the drum, Yas is taken at the saturation temperatures listed. Therefore, Yas (inlet) ¼ 0.0115 lb (H2O)/lb (BDA) and Yas (exit) ¼ 0.0144 lb (H2O)/lb (BDA). Substituting into the above equation gives (Yas  Y)lm ¼

¼

(0:0115  0:0095) lb=lb  (0:0144  0:0087) lb=lb   (0:0115  0:0095) ln lb=lb (0:0144  0:0087) 0:002  0:0057   0:002 ln 0:0057

¼ 0:00353 lb=lb (BDA) The calculation of the mass transfer coefficient, k, is determined by: 

  dm 1 dt V(Yas  Y)lm   1 ¼ (0:0644 lb=h) (0:1363 ft3 )(0:00353 lb=lb)

k¼

¼ 133:8 lb(BDA)=h  ft3

(13:35)

B

Spray Dryers Spray dryers (SD) are often utilized in the chemical processing industry to obtain a dry product in a granular or powder form. Various spray-dried products include coffee, detergents, and instant beverages. Spray drying is a drying technique that involves the drying of a solid in solution via atomization of the solution. The atomized drops are then contacted with a hot air stream. The dry product is normally collected by a cyclone (or another particulate control device) at or near the bottom of the unit. Spray dryers may be operated cocurrently or countercurrently. The spray dryer in the Unit Operations Laboratory of Manhattan College is shown in Figure 13.9.

362

Chapter 13 Humidification and Drying

Figure 13.9 Spray dryer: Manhattan College’s Unit Operations Laboratory.

The method of operation of the spray dryer is relatively simple, requiring only two major equipment items—a spray dryer similar to those used in the chemical food-processing and mineral preparation industries and a cyclone or fabric filter (baghouse) or electrostatic precipitator (ESP) to collect the fly ash and entrained solids. In the spray dryer, the solution (or slurry) is atomized into the incoming hot gas stream to increase the liquid – gas interface and to promote the mass transfer of the gas to the slurry droplets, where it is absorbed. The slurry solution may be mechanically atomized by either a rotary atomizer or by spray nozzles and injected into the gas stream. The action of the rotary atomizer results in smaller droplet size and size distribution, and is less subject to plugging and wear than the spray nozzle; however, it is higher in cost. This atomizer propels the droplets radially outward and perpendicular to the gas flow. The droplets decelerate rapidly owing (in part) to the drag forces of the downward moving gas and eventually attain the velocity of the gas. The radial distance between the atomizer and the dryer wall must be sufficient to allow for adequate drying of the largest droplets.

Drying

363

The length to diameter (L/D) ratio of the cylindrical section of the SD is typically 0.8 : 1. In a SD with a dual-fluid pneumatic nozzle, atomization is in the direction of the gas flow and the L/D ratio is typically 2 : 1. Optimum SD performance is achieved through the proper choice of L/D, droplet size, and residence time. Spray dryers have several advantages over other types of dryers. The drying time is very short, permitting drying of highly heat sensitive materials and the creation of a solid or a hollow spherical product. Some products, such as food and detergents, require certain appearances, consistencies, and bulk densities; other drying methods may fail to produce these valuable properties. Another desirable characteristic of spray dryers is their ability to produce a dry product ready for packaging from a feed solution. This may greatly simplify the manufacturing process. Drying a solid means removing small amounts of liquid from a solid material to reduce the liquid content of the solid to an acceptably low value. For example, a solid dissolved in a solution can be dispersed into a stream of hot gas (usually air) in the form of fine droplets. Moisture vaporizes from the droplets, leaving residual particles of dry solid behind. The inlet dry gas gains the moisture that evaporates from the droplets. A dry solids material balance requires the subtraction of the water from the product in performing the balance. This is accomplished by determining the moisture content of the feed and the product. After determining the amount of water in the feed and in the product and subtracting the water content for both the feed and the product, one may compare the dry feed and the dry product to determine any experimental error. As atomized droplets are contacted with the hot, dry air in the drying chamber, the water evaporates and enters the air. The evaporation rate is determined from the following water mass balance equation (employing SI units in this development): dm (water) ¼ ( m˙ 1 )(X1 )  ( m˙ 2 )(X2 ) dt where

(13:39)

dm (water) ¼ m˙ w ¼ evaporation rate (kg H2O/s) dt m˙ 1 ¼ dry solids feed rate (kg dry feed/s) X1 ¼ feed moisture content (kg H2O/kg dry feed) m˙ 2 ¼ dry solids product rate (kg dry feed/s) X2 ¼ product moisture content (kg H2O/kg dry feed)

The moisture gain by the drying gas may be similarly calculated by: dm (gas) ¼ ( m˙ 2 )(Y2 )  ( m˙ 1 )(Y1 ) m˙ g ¼ dt where

dm (gas) ¼ m˙ g ¼ moisture gain by drying gas (kg H2O/s) dt m˙ 2 ¼ dry air mass flow from dryer (kg dry air/s)

(13:40)

364

Chapter 13 Humidification and Drying

Y2 ¼ air humidity at dryer outlet (kg H2O/kg dry air) m˙ 1 ¼ dry air mass flow to dryer (kg dry air/s) Y1 ¼ air humidity at dryer inlet (kg H2O/kg dry air) The air humidities in Equation (13.40) can be determined from a psychrometric chart. The moisture gain of the drying gas should agree within 10% of the evaporation rate provided in Equation (13.39). The average drop diameter, dp, of the atomized feed solution may be estimated from(9):     dp G 0:6 m 0:2 srs Lw 0:1 ¼ 0:4 (13:41) r rs nr2 G G2 where dp ¼ average drop diameter (ft) G ¼ spray mass velocity (lb/ft2 . min)

rs ¼ density of the solution (lb/ft3) n ¼ atomizer disk speed (rev/min) r ¼ radius of the centrifugal disk (ft) m ¼ viscosity of the solution (lb/ft . min) s ¼ surface tension (lb/min2) Lw ¼ wetted disk periphery (ft) In order to completely dry the solution, it is customary to base the drying calculations on the maximum drop diameter, dp,max. The maximum drop diameter is often assumed to be three times the average drop diameter, i.e., d p, max ¼ 3dp

(13:42)

The mass flow rate of the feed is necessary to determine the surface area available for heat transfer. It is often assumed that the drops are dried in one second; this approximation is fairly reasonable. The dimension of a drop also needs to be determined. This may be obtained from the equation for a sphere. It is also necessary to determine the number of drops. The number of drops is simply the volume of the feed divided by the volume of one drop. The surface area available for heat transfer is then obtained by multiplying the surface area of one drop by the total number of drops. These calculations are usually performed using the maximum drop diameter. The heat load of the dryer is necessary to determine the heat transfer coefficient. The following equation is suggested when centrifugal disk atomizers are used(9):  r 2 4:19(kf ) R  DT rffiffiffiffiffiffiffiffiffiffiffi ws rt 2 _ (13:43) Q¼ d2p, max rs (r)(n) where

˙ ¼ heat load (Btu/h) Q kf ¼ thermal conductivity at the film temperature (Btu/h . ft . 8F)

Drying

365

R ¼ radius of the drying chamber (ft) r ¼ radius of the disk (ft) DT ¼ temperature driving force for heat transfer (8F) dp,max ¼ maximum drop diameter (ft)

rs ¼ density of the liquid solution (lb/ft3) ws ¼ feed flow rate (lb/h) rt ¼ feed density (lb/ft3) n ¼ disk speed, rev/h Interestingly, the driving force for heat transfer is the difference between the drying gas outlet temperature and outlet gas wet-bulb temperature. The wet-bulb temperature may be found on a psychrometric chart from the outlet temperature and relative humidity of the outlet gas. A film temperature is also necessary to determine the physical properties of the solution; it is calculated by averaging the outlet air temperature and the outlet wet-bulb temperature. An experimental heat transfer coefficient is then found by: hexp ¼

_ Q ADT

(13:44)

where hexp ¼ experimental heat transfer coefficient (Btu/h . ft2 . 8F) _ ¼ heat load (Btu/h) Q A ¼ surface area available for heat transfer (ft2) DT ¼ temperature driving force for heat transfer (8F) This experimental heat transfer coefficient may be compared to a literature heat transfer coefficient. This coefficient is normally based on the assumption that the Nusselt number is equal to 2.0, i.e., hdp, max ¼ 2:0 kf

(13:45)

Mass transfer coefficients may also be determined for a spray dryer. The experimental mass transfer coefficient is found by: kexp ¼

NA Dp

(13:46)

where kexp ¼ experimental mass transfer coefficient (lbmol/s . ft2 . atm) NA ¼ molar flux (lbmol A/s . ft2) Dp ¼ partial pressure driving force for mass transfer (atm) The molar flux represents the number of moles diffusing per time per surface area. It is obtained by dividing the mass flow rate of the feed by the total surface area available

366

Chapter 13 Humidification and Drying

for mass transfer. The driving force for mass transfer is the difference between the interface partial pressure and the bulk partial pressure. To determine the interfacial partial pressure, one must first determine the mole fraction of water in the dry air at the interface. It is customary to assume that the interface temperature is the outlet wet-bulb temperature. The bulk partial pressure may be found by determining the mole fraction of water in the outlet dry gas. To determine the literature mass transfer coefficient, the Sherwood number is assumed equal to 2.0, which is valid for a relative velocity of the drops through air of zero. Although it is not completely accurate, it provides for a good approximation. For this assumption: klit d p, max Dpf MW ¼ 2:0 DAB r where

(13:47)

klit ¼ literature mass transfer coefficient (lbmol/s . ft2 . atm) Dpf ¼ pressure difference across the mass transfer path (atm) MW ¼ average molecular weight across the mass transfer path (lb/lbmol) DAB ¼ diffusivity of water vapor (A) in air (B) (ft2/s)

r ¼ density of air at exit conditions (lb/ft3) The diffusivity of water vapor in air, DAB, is provided in the literature.(9) At 258C and 1.0 atm, the diffusivity of water vapor in air is 0.202 cm2/s. However, the diffusivity should be estimated at the dryer outlet temperature, which is assumed to be the temperature inside the dryer. A companion spray unit to a spray dryer is the spray tower. This unit can also be employed to cool a hot gas. In one spray tower design, the gas (quench) cooling is accomplished simply by spraying water at the top of the tower as the hot gases travel upwards through the tower. When the unvaporized water reaches the pump at the bottom of the tower, it is recirculated by pumping it back up to the top. Since about 10% of the water stream is usually vaporized during contact with the hot gases, makeup water must be constantly added. For efficient evaporation of the water, the gas velocity should be from 400– 600 fpm and the entire cross section of the gas stream should be covered with a fine spray of water. This necessitates proper location, type, and number of spray nozzles to ensure adequate coverage of the gas-side flow area. Cooling hot gases with a water spray is relatively simple and requires little space. These towers are easy to operate and, with automatic temperature controls, only that amount of water that is needed is used to maintain the desired temperature of the gases at the discharge. Their installation and operating costs are generally considered to be less than that of other cooling methods. Quench towers require careful design review for applications where the gases to be cooled contain a significant amount of acid. Before spray towers can be sized, the water flow rate necessary to bring the hot gas to the desired temperature must first be calculated. This is accomplished by means of

Drying

367

both componential and total energy and material balances. Componential balances, where the mass and enthalpy of each component entering and exiting the tower is accounted for, are necessary because material will be exchanged between the water and gas streams. Design techniques of spray towers are available in the literature.(9)

ILLUSTRATIVE EXAMPLE 13.18 List some advantages of a spray dryer. SOLUTION: 1 2 3 4 5

Among the inherent advantages that the dryer enjoys are:

Low capital costs Low pressure drop losses Reduced auxiliary power Reduced water consumption Continuous, two-stage operation, from liquid feed to dry product.

B

ILLUSTRATIVE EXAMPLE 13.19 Discuss the equipment that is available for collecting the dry product from a spray dryer. SOLUTION: The choice of many individuals in the dryer field is the cyclone. However, fabric filters (baghouses) and electrostatic precipitators (ESPs) may also be employed. Baghouses have an inherent advantage over ESPs in that they have high collection efficiency and simplicity in design. Two disadvantages of using a fabric filter are that the fabric is somewhat sensitive to wetting, so that a margin above the saturation temperature must be kept for bag protection, and there is always the possibility of catastrophic bag failure due to a temperature excursion. However, it is the cyclone that is the preferred choice in the industry.(12,13) Details on these units are provided in Chapter 16. B

ILLUSTRATIVE EXAMPLE 13.20 An air (a) stream at 908F, 14.7 psia, with a relative humidity of 60% is flowing into a water-spray chamber (SC) where enough water (w) is added to reach saturation. The outlet temperature and pressure of the saturated stream are 808F and 12 psia, respectively. After the spray chamber, the stream is heated and compressed (H/C) to 30 psia and has a final relative humidity of 30%. What is the mole ratio of the amount of water added (i.e., absorbed by the air stream) to the amount of incoming moist air in the spray chamber? SOLUTION: Choose a basis of 100 lbmol of feed and draw a flow diagram of the process (see Fig. 13.10). Using Dalton’s law and steam table data, determine the composition of the moist air feed stream, and the stream leaving the compressor/heater unit. Water saturation pressures are

368

Chapter 13 Humidification and Drying Moist air N1 = 100 lbmol

Moist air N3

SC

80°F, 12.0 psia saturated

90°F, 14.7 psia YR = 60%

Moist air N4 = 100 lbmol

H/C

T, 30.0 psia YR = 30%

Water N2

Figure 13.10 Spray chamber/compressor diagram.

obtained from Table C1 in Appendix C.1: pw ¼ (YR )p0w ¼ 0:60(0:69813) ¼ 0:4189 psia; yw ¼

90 F

pw 0:4189 ¼ 0:0285 ¼ 14:7 P

ya ¼ 1  0:0285 ¼ 0:9715 For the stream exiting the spray chamber: pw ¼ (YR )p0w ¼ 1:00(0:50693) ¼ 0:50693 psia yw ¼

pw 0:50693 ¼ 0:0422 ¼ 12:0 P

ya ¼ 1  0:0422 ¼ 0:9578 Since no water is added to or removed from the air stream in the compressor/heater unit, the inlet composition must remain the same as the stream exiting the spray chamber. Perform a material balance on the air solely around the spray chamber unit to find the amount of moist air leaving this unit. For the SC, air in ¼ air out 0:9715(100) ¼ 0:9578N3 N3 ¼ 101:43 lbmol Perform a total material balance around the same unit to find the amount of water added: 100 þ N2 ¼ N3 100 þ N2 ¼ 101:43 N2 ¼ 1:43 lbmol of water added Finally, determine the mole ratio of the amount of water added to the amount of incoming moist air: ratio ¼

N2 1:43 ¼ 0:0143 ¼ 100 N1

B

References

369

ILLUSTRATIVE EXAMPLE 13.21 Refer to Illustrative Example 13.20. What is the dew point temperature of the moist air leaving the compressor/heater unit and what is the final temperature of the moist air in 8F? SOLUTION: Using the steam tables in Appendix C.1, determine the dew point of the moist air leaving the compressor/heater unit. The composition of the inlet and outlet stream to H/C remain the same. Thus, at the outlet conditions, pw ¼ yw P ¼ 0:0422(30) ¼ 1:266 psia From the tables, the temperature at which the air stream would be saturated (dew point) is somewhere between 1008F (0.94924 psia) and 1108F (1.2750 psia). Using linear interpolation: T ¼ 109:7 F  110 F Determine the vapor pressure of the water in the air stream leaving the compressor/heater unit. As before, pw ¼ yw P ¼ 0:0422(30) ¼ 1:266 psia and p0w ¼

pw 1:266 ¼ 4:22 psia ¼ 30 YR

B

REFERENCES 1. C. BENNETT and J. MYERS, “Momentum, Heat, and Mass Transfer,” McGraw-Hill, New York City, NY, 1962. 2. General Electric, Bloomfield, NJ, 1942. 3. Author unknown, New York University, Chemical Engineering Unit Operations Laboratory Report, Bronx, NY, 1960. 4. Author unknown, Manhattan College, Bronx, NY, 1961. 5. G. BROWN and ASSOCIATES, “Unit Operations,” John Wiley & Sons, Chapmann & Hall Limited, Hoboken, London, 1950. 6. FLUOR CORP. LTD., Bulletin T 337, 1939. 7. R. C. KELLY, paper published by the Fluor Corp., presented before the California Natural Gasoline Association, Dec. 3, 1942. 8. L. THEODORE and J. BARDEN, “Mass Transfer Operations,” A Theodore Tutorial, East Williston, NY, 1995. 9. D. GREEN and R. PERRY (editors), “Perry’s Chemical Engineers’ Handbook,” 8th edition, McGraw-Hill, New York City, NY, 2008. 10. J. FAMULARO, et al., “Unit Operations Laboratory Manual,” Manhattan College, Bronx, NY, 1996. 11. G. FRIEDMAN and J. MARSHALL, CEP, 45, No. 9, 573, New York City, NY, 1949. 12. L. THEODORE, Present notes, 1987. 13. L. THEODORE, “Air Pollution Control Equipment Calculations,” John Wiley & Sons, Hoboken, NJ, 2008.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

14

Crystallization CONTRIBUTING AUTHOR: AMANDA WEEDEN

INTRODUCTION Throughout the history of the chemical industry, crystals have been produced via crystallization methods that range between something as simple as allowing vats of hot concentrated solution to cool to those as complex as continuous, carefully controlled, multi-step processes which yield a crystal product of a particular size or size distribution, shape, moisture content, and purity. Today, consumer demands require crystal products meet rigid specifications that can also include color, odor, particle size distribution, and caking characteristics. Crystallization is an important mass transfer operation that is often employed in the preparation of a pure product. In the process, a crystal usually separates out as a substance of definite composition from a solution of varying composition. Any impurities in the liquid (often referred to as the mother liquor) are carried in the crystalline product only to the extent that they adhere to the surface or are occluded (retained) within the crystals that may have grown together during or after the crystallization operation. The separation of a solid from a solution onto a crystal occurs only if there is a state of imbalance involving a mass driving force; namely, a decrease in chemical potential (or concentration) between the bulk of the liquid solution and the crystal interface exists. This effectively means that the solution must be supersaturated. Although crystallization is ordinarily thought of as the deposition of a solid crystalline phase from a liquid phase by cooling, evaporation, or both, the same principles apply to crystal formation by precipitation caused by the addition of a third substance. This other component may either react to form a precipitate or simply decrease the solubility of the precipitated material.

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

371

372

Chapter 14 Crystallization

One may also view crystallization as a phase equilibrium application. These solid crystals are usually formed from a homogenous liquid phase. If one starts with an unsaturated solution formed by dissolving some solid in liquid, more solid can be dissolved until the solution becomes saturated. More solid can still be dissolved in the solution causing it to become supersaturated. At this point, solids begin to deposit out of solution in a process often referred to as nucleation. Crystal growth continues until the solution reaches the previously attained saturation (equilibrium) point. There are several different ways that crystallization can occur. The four most often encountered in practice are: 1 cooling, 2 evaporation, 3 cooling and evaporation (also denoted adiabatic evaporation), and 4 a salting out process. Process (1) is most commonly employed, provided the solubility of the component being crystallized decreases with decreasing temperature. The problems concerning crystallization, which are most frequently encountered by the practicing engineer, are: 1 yield of a given product, 2 purity of the product, 3 energy requirements for cooling, evaporation, etc., 4 shape of the individual crystals, 5 size of the crystals, 6 uniformity or size distribution of the crystals, 7 rate of production of the desired crystals, and 8 caking. It should also be noted that when crystallization begins, heat is evolved (liberated). The amount of heat evolved is defined as the heat (or enthalpy) of crystallization. The following topics receive treatment in this chapter: 1 Phase diagrams. 2 The crystallization process. 3 Crystal characteristics. 4 Equipment. 5 Describing equations. 6 Design considerations. The last two sections are highlighted with several illustrative examples.

Phase Diagrams

373

PHASE DIAGRAMS All two-component solid –liquid equilibrium systems may be classified according to the miscibility of the liquid phases and the nature of the solid phases which crystallize from the solution. As one might suppose, there are various types of these systems. Of these classifications, only the completely miscible liquid systems from which pure components crystallize from the solution will be examined. Two-component systems belonging to the above class have a diagram of the general form shown in Figure 14.1. They are characterized by the fact that components (1) and (2) are completely miscible in the liquid state, and yield only pure (1) or pure (2) as solid phases. In this figure, points E and D are the melting points of pure (1) and pure (2), respectively. Line EG gives the concentrations of the solutions saturated with (1) at temperatures between E and F, or the freezing points of the solutions that yield (1) as a solid phase. Similarly, line DG gives the concentrations of solutions saturated with solid (2) at temperatures between D and F. At G, the solution is saturated with both components (1) and (2), i.e., at G, three phases are in equilibrium. It follows, therefore, that the lines DG and EG represent monovariant twophase equilibria, while G is an invariant point. At this point, the temperature F of the solution must remain constant as long as three phases coexist. The temperature can be lowered below F only when one of the phases has disappeared and, on cooling, this must be the saturated solid solution. In effect, at F, solution G can completely solidify. Temperature F must consequently be the lowest temperature at which a liquid phase may exist in the system for components (1) and (2); below this temperature the system is completely solid. Temperature F is called the eutectic temperature, composition G the eutectic composition, and point G the eutectic point of the system.

Figure 14.1

Two component solid –liquid system.

374

Chapter 14 Crystallization

Above the lines DG and EG is the area in which the unsaturated solution exists. One phase is present in this area only. In order to define any point in this area, both the temperature and composition must be specified. The significance of the remaining portions of the diagram can be made clear by observing the behavior on the cooling of several different mixtures of components (1) and (2). Consider a mixture of overall composition a. If such a mixture is heated to point d, an unsaturated solution is obtained. On cooling this solution, only a decrease in temperature of the liquid phase occurs until point c, corresponding to temperature Tc. At point c, the solution becomes saturated with component (1); c is the freezing point of the solution at temperature Tc. As cooling continues, component (1) continues to separate out, and the composition of the saturated solution changes along the line cG. Thus, at a temperature such as Tb, solid (1) is in equilibrium with a saturated solution of composition e. It is seen, therefore, that for any overall composition falling in the area EFG, solid (1) is in equilibrium with various compositions of solution given by the curve EG at each temperature. However, at temperature F, another solid phase of component (2) appears, and the system becomes invariant. On further extraction of heat from this system, both components (1) and (2) will completely crystallize. Once this process is over, a mixture of solid (1) and solid (2) remains and the system becomes monovariant. When cooling is continued below temperature F into the area 12FGF, there is coexistence of the two solids, (1) and (2). Consider now a three-component (ternary) system. (See also Chapter 12— Equilibrium section, Fig. 12.4) In order to better understand the equilibria associated with a three component system, it is first necessary to become acquainted with the standard method of representing such systems. Two-dimensional equilibrium diagrams for ternary systems can best be plotted on an equilateral triangle, each of whose apexes represents 100% of a particular component [see Fig. 14.2(a)– (d)]. A series of grid lines representing fractional concentrations of a particular component is drawn parallel to the base opposite the apex that represents the component. Thus, every point on the diagram corresponds to a certain percentage composition of each of the three components. It is important to note than an apex signifies a single component; a point on one of the sides describes a binary system. There are various types of classifications of three-component systems. Only one which exhibits a simple liquid phase with crystallization of pure components will be considered in the presentation to follow. The isothermal phase diagram for a threecomponent system such as water (H2O), sodium nitrate (NaNO3), and sodium chloride (NaCl)—where only pure components crystallize from solution—takes the form shown in Figure 14.3. Details on the four phases noted in the figure are given in Table 14.1. (See also Chapter 12 for additional details). a

b

c

Figure 14.2 (a–d) Equilateral triangle; ternary systems.

d

Phase Diagrams

Figure 14.3

375

H2O–NaNO3 – NaCl phase diagram.

As noted in earlier chapters, Gibb’s phase rule states that the number of variables that must be specified to define a system (or degrees of freedom) is determined by both the number of components and the number of phases present. This was represented mathematically by the equation: F ¼CPþ2

(14:1)

where F ¼ degrees of freedom C ¼ number of components present P ¼ number of phases present This may be applied to the above three-component system. Since there are three components, the number of degrees of freedom is: F ¼5P

(14:2)

In order to simplify matters, two variables (temperature and pressure) are generally specified. Thus, the degrees of freedom are reduced to: F ¼3P

(14:3)

Table 14.1 Phase Diagram Details Region I II III IV

# of phases

Type of phase

1 2 2 3

Unsaturated solution Saturated solution þ Pure NaNO3 Saturated solution þ Pure NaCl Saturated solution þ Pure NaNO3 þ Pure NaCl

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Chapter 14 Crystallization

This indicates that when one phase is present, two variables must be specified in order to completely define the system. If two salt phases and one liquid phase are present, the degree of freedom of the saturated solution of NaNO3 and NaCl is zero; this means that the composition of this point is a constant and is known as the isothermal invariant point.

ILLUSTRATIVE EXAMPLE 14.1 Two components are completely miscible in the liquid state and the solid phases consist of pure components. Calculate and discuss the degrees of freedom at the freezing point. SOLUTION: At this temperature, there are two phases, liquid and solid, with the vapor being ignored. Since there are two components, the system has two degrees of freedom according to the phase rule: F ¼CPþ2¼22þ2¼2 The pressure is normally fixed so that the system is univariant; hence, either the temperature or the composition of the liquid phase alone is sufficient to completely define the system. B

With reference to Figure 14.3, point D represents the maximum solubility of NaNO3 in H2O, point E the maximum solubility of NaCl in H2O, and point F a saturated solution of both NaNO3 and NaCl. This last point is not called a eutectic point since it does not represent a temperature but instead is defined as the aforementioned isothermal invariant point. If a dry mixture of NaNO3 and NaCl of composition Q can, by the addition of water, be brought into the region DBF in Figure 14.3, the solid NaNO3 can be separated from the solution above it containing NaCl and NaNO3. This can easily be accomplished by moving along the line QA until point R is reached. To calculate the amount of water to be added, the lever rule may be employed: Mass of salt mixture Q RA ¼ Mass of water to be added RQ

(14:4)

In a similar manner, the mass of NaNO3 that can be recovered may be determined by the following relationship: Mass of NaNO3 captured RS ¼ Mass of liquid phase above NaNO3 BR

(14:5)

Obtaining equilibrium information for this system is demonstrated with the aid of Figure 14.4. Mixtures of various portions of the two solid components in water can be prepared and agitated until equilibrium is established. If the liquid phase is separated from the solid crystals, both can be weighted and analyzed. The composition thus

Phase Diagrams

377

I

III

II

IV

Figure 14.4

H2O–NaNO3 – NaCl phase equilibrium diagram.

obtained for the saturated solution(s) and residue(s) may be plotted on a triangular coordinate diagram as shown in Figure 14.4. It also shows a series of other points arrived at in this manner. If a mixture of composition R1, R2, R3, . . . are prepared, S1, S2, S3, . . ., are respectively, the composition of the saturated solutions that will result in equilibrium with NaNO3. To ascertain the nature of the solid phases in equilibrium with the various solutions, one notes that any mixture concentration point must lie on a straight line joining the composition of the solid phases and that of the saturated solution. Consequently, an equilibrium tie-line drawn between any corresponding pair of R and S points in region II must pass B on extension. Moreover, as several solutions may have the same solid phase, all the tie-lines for such solutions must intersect at a common point, which is the composition of the common solid phase. Therefore, it may be deduced that the solid phase for all solutions between D and F is NaNO3, for those between F and E is NaCl, while area IV is composed of a saturated solution of composition F and saturated with various portions of the solid phases of both NaNO3 and NaCl. The whole diagram may be drawn once various points and tie-lines have been determined in Figure 14.4. An entire separation process can then be devised for a mixture of components NaNO3 and NaCl by using the appropriate amount of water to obtain the maximum yield of each component.

378

Chapter 14 Crystallization

ILLUSTRATIVE EXAMPLE 14.2 The following laboratory data was obtained in an attempt to determine the solubility of NaCl in H2O at 228C: Weight of evaporating dish and saturated solution ¼ 46.73 g Weight of evaporating dish ¼ 34.80 g Weight of evaporating dish and dry salt (NaCl) ¼ 37.90 g SOLUTION:

The mass of saturated solution is: Saturated solution ¼ 46:73 g  34:80 g ¼ 11:93 g

The mass of dry salt (NaCl) in the saturated solution is: NaCl ¼ 37:90 g  34:80 g ¼ 3:10 g The mass percent NaCl in the saturated solution is therefore   3:10 %NaCl ¼ 100 11:93 ¼ 26:0% This compares favorably with the literature value of 26.5% NaCl, 73.5% H2O at 258C.

B

ILLUSTRATIVE EXAMPLE 14.3 Refer to the previous example. Calculate the percent error between the calculated and literature values. SOLUTION:

Base the error on the literature value:   26:5  26:0 %error ¼ 100 26:5 ¼ 1:89%

As noted, there is excellent agreement.

ILLUSTRATIVE EXAMPLE 14.4 The weight of NaNO3, NaCl, and H2O in a saturated solution at 228C is provided below: NaNO3 ¼ 3.94 g

B

The Crystallization Process

379

NaCl ¼ 1.92 g Saturated solution ¼ 13.50 g Is the saturated solution at the invariant point? The literature value for the invariant point at 228C is: 32% NaNO3 13% NaCl 55% H2O SOLUTION:

Determine the mass percent composition of the saturated solution: %NaNO3 ¼

3:94 100 13:50

¼ 29:2% %NaCl ¼

1:92 100 13:50

¼ 14:2% %H2 O ¼ 100  (14:2 þ 29:2) ¼ 56:6% The sample results compare favorably with the literature value for the invariant point.

B

The reader should note that the phase diagrams presented above apply for isothermal conditions, i.e., they represent constant temperature data. A change in temperature will alter these isothermal curves. An increase in solubility generally occurs with rising temperature, i.e., as the temperature increases, the mutual solubilities of the components increase and the curves enclose smaller two-phase and threephases areas.

THE CRYSTALLIZATION PROCESS The crystallization process essentially involves three steps: 1 Formation of crystals. 2 Crystal growth. 3 Separation of the crystals from residual liquid. A simple schematic of this overall process is provided in Figure 14.5. There are various processes that can accomplish the above while meeting the desired design requirements associated with a crystallization operation and/or the rate of crystallization. As with many mass transfer (as well as heat transfer) calculations, the rate is expressed in terms of an overall coefficient which takes into account all individual resistances. This coefficient is almost always determined from experimental data and/or experience. Details on product yield, product purity, crystal

380

Chapter 14 Crystallization

Figure 14.5 Schematic diagram of generalized crystallization process.

size and/or size distribution must also be specified. Each of these topics is discussed below with the latter topic receiving additional treatment in the next section. The rate of crystallization involves two distinct steps: 1 The rate of formation of new crystals, or nucleation. 2 The rate of precipitation on crystals already present, usually referred to as crystal growth. The mechanism of nucleation is essentially unknown. However, the process of crystal growth has been studied extensively, with some researchers suggesting that crystal growth occurs at erratic rates and that the solution concentration at the crystal surface is not uniform. It is common knowledge that the solubility of small crystals is greater than that of larger crystals. Therefore, material may be simultaneously depositing on larger crystals while smaller crystals are dissolving when both are exposed to the same solution. As a result, it is difficult in practice to maintain uniform conditions while growing crystals. Thus, it is usually impossible to avoid locally varying conditions in the solution. Prior to crystallization, it is common practice to inject small crystals, known as seeds, into the solution so that solids will be deposited more easily. Since the solubility of crystals of this size is less than that of submicroscopic crystals, the crystallization unit may be operated in such a manner as to retain the supersaturated solution with respect to the seeds, but not supersaturated with respect to crystals of the size of the aggregations that will inevitably form under the operating conditions. It should be noted that seeding is rarely used in continuous processes; it is instead used often in batch processes. Predetermination of the yield from a crystallizer may be obtained from a material balance. The yield can usually be predicted from the solubility (see Phase Diagrams section of this chapter) of the solid phase being precipitated. It is general practice to assume equilibrium or a saturated condition. In some applications, crystallization occurs slowly and equilibrium is not attained in a short time period. In many cases, controlled precipitation does not occur unless the aforementioned seed crystals are present. If solubilities are obtained from the equilibrium diagrams discussed earlier, the prediction of the equilibrium yield can be made by applying the usual lever arm ratio principle between the phases which are known to be present at the specified operation temperature. The solubilities usually reported in the literature are expressed in terms of mass of “salt” per 100 units of mass of pure solvent. The prediction of a yield of a nonhydrated

The Crystallization Process

381

salt from a solution is simple since the amount of solvent present during crystallization is generally constant and the quantity remaining at the operating temperature is known.(1) Impurities in crystallization operations are introduced primarily from the mother liquor that is not completely removed from the solid product. The extent of occlusion and the completeness of washing are important factors in the determination of the purity of a crystalline product. The agglomeration of crystals into larger crystals, which can occur during the growth of the crystals, makes washing more difficult and accordingly results in a lower purity of product. However, agitation decreases the tendency of the crystals to agglomerate. Purification techniques have become increasingly important in recent years. Mechanisms by which the impurities discussed above can be incorporated into crystalline products include adsorption of impurities on crystal surfaces, solvent entrapment in cracks, crevices and agglomerates, and inclusion of pockets of liquid. The suggested recommendation for producing high purity crystals is to maintain supersaturation at a low level so that large crystals are formed. The characteristics of an impurity can also determine whether it is positioned on the surface or in the interior of the host crystal. Although crystallization is often thought of as solely a separation operation, it is the filtration step at the end of the process that accomplishes the actual crystal separation. The residual liquid (mixed magma underflow) from a crystallizer typically contains approximately 30% mass solids and can be thickened further before filtration by being passed through a gravity settler. Such pre-thickening allows dewatering of the mother liquor, which allows either more time for the final wash step or a shorter filter cycle. The filter cycle almost always includes a washing step at the end of the full cycle. This permits a thorough drain before the crystals are discharged. It should also be noted that the system’s throughput and performance is often limited by the solids –liquid separation step; in effect, this then becomes the critical step in the overall operation.(2) Perhaps the most troubling problem is associated with the final product. Crystalline materials have a tendency to bind together or cake on storage. Most crystalline products are required in a free-flowing form, as with sugar and table salt, or to be capable of being distributed uniformly over surfaces (e.g., fertilizers). It is also important for the crystalline product to remain in a particulate state. Caking may also require a crushing operation before the crystals can be used. Numerous problems can arise during the operation, design, and analysis of a crystallizer. Ten of the most important problems are listed below: 1 Crystal-size distribution (CSD). 2 Purity. 3 Fouling. 4 Vapor release and location. 5 Scale-up. 6 Stability of operation. 7 Liquor – solids separation.

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Chapter 14 Crystallization

8 Space requirements. 9 Capital costs. 10 Operating and maintenance costs. The proper location of the crystallization process is as important to its success as the selection of the equipment to be employed. Careful consideration must be given to the many tangible factors such as labor supply and raw material sources along with a number of intangible factors, which may be more difficult to evaluate. The selection of the site is sometimes based on a detailed study in which all factors are weighed as carefully as possible. For many processes, one or more predominant factors effectively minimize the number of possibilities for site location. Raw material and transportation costs may be such that the process must be located near a source. Thus, only the sites near sources of raw material need to be studied and these may be few in number. These and other factors serve as effective screening agents that can save both time and money. ILLUSTRATIVE EXAMPLE 14.5 Discuss the crystallization separation process relative to the presence (or absence) of a eutectic. SOLUTION: In cases where a complete solid solution occurs with no eutectic, it is possible to separate a binary mixture into its components. However, such systems are the rare exception; eutectic formation is the general rule. In most cases, the degree of separation attainable is therefore limited by the eutectic composition. To accomplish a complete separation, an additional step is required. B

CRYSTAL PHYSICAL CHARACTERISTICS The three principle physical characteristics of crystals are size, shape, and density. Only by knowing these three properties is it possible to determine not only how they will behave in a crystallization process but also information in their later use. Although density is often not treated as an important factor, consider, for example, a ping-pong ball and a golf ball. Under a microscope, they will appear almost equal in size; however, if both were tossed into a moving fluid stream they would behave quite differently. Even though the size and shape are similar, the density is quite different and the behavior of the two objects is far from being similar. This is one of the reasons that many object to physical sizing. A common method of specifying large crystal sizes is to designate the screen mesh that has an aperture corresponding to the crystal diameter. Since various screen scales are in use, confusion may result unless the screen scale involved is specified. The screen mesh generally refers to the number of screen openings per unit length or area. The aperture for a given mesh will depend on the wire size employed. The clear space between the individual wires of the screen is termed the screen aperture. As indicated above, the term mesh is applied to the number of apertures per linear inch (e.g., a 10-inch mesh screen will have 10 openings per inch and the aperture will be 0.1 inch minus the diameter of the wire).

Crystal Physical Characteristics

383

Table 14.2 Tyler and U.S. Standard Screen Scales Tyler mesh 400 325 270 250 200 170 150 100 65 48 35 28 20 14 10 8 6 4 3

Aperture microns

U.S. mesh

Aperture microns

37 43 53 61 74 88 104 147 208 295 417 589 833 1168 1651 2362 3327 4699 6680

400 325 270 230 200 170 140 100 70 50 40 30 20 16 12

37 44 53 62 74 88 105 149 210 297 420 590 840 1190 1680

The Tyler and the U.S. Standard Screen Scales (Table 14.2) are the most widely used in the United States. The screens are generally constructed of wire mesh cloth, with the diameters of the wire and the spacing of the wires being specified. These screens form the bottoms of metal pans about 8 inches in diameter and 2 inches high, the sides of which are fashioned so that the bottom of one sieve nests snugly on top of the next.(3,4)

ILLUSTRATIVE EXAMPLE 14.6 The size of a crystal is specified as Tyler 8-14 screen mesh. Determine its size. SOLUTION: To calculate the size of the crystal in the problem statement, refer to the Tyler screen information in Table 14.2. 8-mesh opening ¼ 2362 mm 14-mesh opening ¼ 1168 mm Since an 8 by 14 mesh size indicates that an object will pass through the 8-mesh screen but not pass (be captured) through the 14-mesh screen, one size cannot be specified for the crystal in question. The crystal is in the size range 1168–2362 mm and the average or “mean” arithmetic size is 1765 mm. B

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Chapter 14 Crystallization

Size distributions are often characterized by a mean diameter. Although numerous “means” have been defined in the literature, the most common are the arithmetic mean and the geometric mean. The arithmetic mean diameter of a number of crystals is simply the sum of each of the diameters divided by the number of diameters measured. The geometric mean diameter is the nth root of the product of the n number of diameters in the sample. In addition to the arithmetic and geometric means, a size distribution may also be characterized by the median diameter. The median diameter is that diameter for which 50% of the crystals are larger in size and 50% are smaller in size. Another important characteristic is the measure of central tendency. It is sometimes referred to as the dispersion or variability. The most common term employed is the standard deviation. These terms are discussed below. One basic way of summarizing data is by the computation of a central value. The most commonly used central value statistic is the aforementioned arithmetic average, or the “mean”. This statistic is particularly useful when applied to a set of crystal size data having a fairly symmetrical distribution. The mean is an important statistic in that it summarizes all the data in the set and because each crystal is taken into account in its computation. The formula for computing the mean is

X¼ where

X1 þ X2 þ X3 þ    þ Xn ¼ n

Pn

i¼1

Xi

n

(14:6)

X ¼ arithmetic mean Xi ¼ any individual measurement n ¼ total number of observations X1, X2, X3, . . . ¼ measurements 1, 2, and 3, respectively.

The most commonly used measure of dispersion, or variability, of sets of data is the standard deviation, s. Its defining formula is given by the expression sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Pn 2 i¼1 (Xi  X) s¼ n1

(14:7)

where s ¼ standard deviation (always positive) Xi ¼ value of the ith data point X ¼ mean of the data sample n ¼ number of observations (Note: The above term s represents the sample standard deviation. In statistical circles, the symbol s represents another, but nearly similar, standard deviation. Unfortunately, s has often been employed to represent the sample standard deviation.(5)) A typical size analysis method of representation employed in the past is provided in Table 14.3. These numbers mean that 40% of the crystals by mass are greater than 5 mm in size, 27% are less than 5 mm but greater than 2.5 mm, 20% are less

Crystal Physical Characteristics Table 14.3

385

Early Crystal Size Distribution (CSD) Representation

— ,5.0 ,2.5 ,1.5

.5.0 .2.5 .1.5 —

mm mm mm mm

40% 27% 20% 13%

—

—

—

100%

than 2.5 mm but greater than 1.5 mm, and the remainder (13%) is less than 1.5 mm. Another form of representing data is provided in Table 14.4. Frequency distribution curves are usually plotted on regular coordinate (linear) paper. The curve describes the amount of material (crystals) contained within each size range. A plot of percent mass versus crystal size (d ) on a linear scale gives a curve with a peak at the preferential size. Such a curve is shown in Figure 14.6. This figure shows a normal probability distribution that is symmetric about the preferential size. This curve is only occasionally encountered for crystal size distribution; however, this curve may occasionally be approached in some applications. Size data can also be plotted as a cumulative plot. The size for each size range is plotted on the ordinate. The cumulative percent by weight (frequency) is plotted on the abscissa. The cumulative percent by weight can be given as a cumulative percent less than stated size (%LTSS) or cumulative percent greater than stated size (%GTSS). The cumulative percent by weight can be plotted on either a linear percentage or a probability percentage scale. The size range (ordinate) is usually a logarithmic scale. Frequently, the cumulative distribution is plotted on special coordinate paper called log-probability paper. The size of each size range is plotted on a logarithmic ordinate. The percent by weight larger than crystal size d is plotted on the probability scale as the abscissa. If the distribution is lognormal, the distribution curve plots out as a straight line. It should be noted that one can just as easily plot percent mass less than size d (%LTSS) on the abscissa. The geometric mean value of a lognormal distribution can be read directly from a plot similar to that represented in Figure 14.7. The geometric mean size is the 50% size on the plot. As discussed previously, the geometric standard deviation is a measure of the dispersion or spread of a distribution. The geometric standard deviation is the root mean square deviation about the mean value and can be read directly from a plot Table 14.4 Size Ranges in Arithmetic Increments Size range, mm 0– 2 2– 4 4– 6 6– 8 8– 10 .10

Percent in size range, % 10 15 30 30 10 5

386

Chapter 14 Crystallization

d

Figure 14.6 Size distribution.

d

Figure 14.7 Lognormal distribution plot.

Crystal Physical Characteristics

387

such as shown in Figure 14.6. For a lognormal distribution (which plots d maximum versus percent mass larger than d ), the geometric standard deviation is given by

s ¼ sgm ¼

50% size 84:13% size

(14:8)

s ¼ sgm ¼

15:87% size 50% size

(14:9)

or

Thus, all one must do is determine the 50% size and the 84.13% size from the plot and divide to determine the geometric standard deviation. It is safe to say that crystals produced via the crystallization process have a distribution of sizes that can vary over large ranges, and the crystal size distribution (CSD) is expressed as a number or mass distribution in terms of the size. Other crystal factors include appearance, purity, the solid – liquid separation process, and other properties involving shape and surface area.

ILLUSTRATIVE EXAMPLE 14.7 The following crystal sizes (in mm) in a sample were recorded from a crystallizer: 22, 10, 8, 15, 13, 18 Find the median, the arithmetic mean, and the geometric mean of these crystal sizes. SOLUTION: The median is simply the middle value of a distribution, or the quantity above which half the data lie and below which the other half lie. If n data points are listed in their order of magnitude, the median is the [(n þ 1)/2]th value. If the number of data is even, then the numerical value of the median is the value midway between the two data nearest the middle. The median, being a positional value, is less influenced by extreme values in a distribution than the mean. However, the median for these data is 14 mm and 15 mm since this data set has an even number of measurements. Another measure of central tendency used in specialized applications is the aforementioned geometric mean XG. The geometric mean can be calculated using the following equation pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d G ¼ n (d1 )(d2 )    (dn ) (14:10) For the particle sizes given above, one obtains dG ¼ [(8)(10)(13)(15)(18)(22)]1=6 ¼ 13:54 mm while the arithmetic mean d is d ¼ (8 þ 10 þ 13 þ 15 þ 18 þ 22)=6 ¼ 14:33 mm

B

388

Chapter 14 Crystallization

ILLUSTRATIVE EXAMPLE 14.8 Refer to the previous example. Calculate the standard deviation of the six crystal sizes. SOLUTION: The following algebraically equivalent formula makes computation much easier (now applied to the crystal diameter d): sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P 2ffi P P n di2  di (di  d)2 s¼ ¼ (14:11) n1 n(n  1) The standard deviation may be calculated for the data at hand: P 2 di ¼ (8)2 þ (10)2 þ (13)2 þ (15)2 þ (18)2 þ (22)2 ¼ 1366 P 2 di ¼ (8 þ 10 þ 13 þ 15 þ 18 þ 22)2 ¼ 7396 Thus, sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 6(1366)  7396 s¼ ¼ 5:16 mm (6)(5)

B

ILLUSTRATIVE EXAMPLE 14.9 A tiny spherical crystal has a diameter of 100 nanometers (nm). Calculate the volume (cm3) and surface area (cm2) of the crystal. SOLUTION:

The volume (V ) of the crystal is V ¼ pD3 =6 ¼ 0:524(100)3 ¼ 0:524  106 nm3

Since there are 107 nm/cm, V ¼ 0:524(106 )(107 )3 ¼ 0:524  1015 cm3 The surface area (A) is given by A ¼ p D2 ¼ 3:14  (100)2 ¼ 3:14  104 nm2 Converting leads to A ¼ 3:14(104 )(107 )2 ¼ 3:14  1010 cm2

B

Crystal Physical Characteristics

389

ILLUSTRATIVE EXAMPLE 14.10 Compare a 10 mm spherical crystal to one with a diameter of 100 mm. Include the ratio of volumes and surface area in the calculation. SOLUTION:

The ratio of diameters (RD) is RD ¼

100 mm (100)(1000) ¼ 10 mm 10

¼ 104 Since the volume and surface area are a function of the (diameter)3 and (diameter)2, respectively, the volume ratio (VR) and surface area ratio (SAR) are VR ¼ (104 )3 ¼ 1012 B

SAR ¼ (104 )2 ¼ 108

With reference to the crystallization process, the average size to which crystals are grown is usually dictated by the manner in which they are to be used. As noted earlier, fine crystals have less opportunity for agglomeration and accordingly occlude less mother liquor; however, fine crystals have a greater surface and offer greater difficulty of complete removal of mother liquor entrained on the crystals. In addition, producing crystalline material of a given size range (a specific upper and lower limit) is difficult to accomplish. It should be noted that the removal rate of crystals of a given size can greatly alter a crystallizer’s ability to continue the production of crystals having that same size and/ or size distribution. This effect is particularly pronounced in a separation unit where the crystals of a desired size – distribution range are preferentially removed and the under-sized particles recycled to the crystallizer. This can shift the distribution to a smaller and narrower size range. ILLUSTRATIVE EXAMPLE 14.11 The crystal size separation efficiency information for a filtration unit is provided in Table 14.5. Calculate the overall separation efficiency for the filter on a 1 number basis 2 mass basis 3 volume basis Table 14.5 Crystal Size–Separation Efficiency Data d, mm 1.0 100

Efficiency, %

Number of crystals

0 100

10,000 1

390

Chapter 14 Crystallization

SOLUTION: 1 On a number basis (10,000)(0) þ (1)(1:0) 10,000 þ 1 1 1 ¼ ¼ 10,000 þ 1 10,001

EN ¼

¼ 0:0001 ¼ 0:01% 2 On a mass basis

E¼ ¼

(1)(100)3 (1:0) þ (10,000)(1:0)3 (0) (1)(100)3 þ (10,000)(1:0)3 106 þ 0 106 þ 104

¼ 0:99 ¼ 99% 3 On a volume basis EV ¼ 0:99 ¼ 99% since volume is proportional to mass.

B

ILLUSTRATIVE EXAMPLE 14.12 Calculate the overall efficiency of the crystallization separation unit using the data provided in Table 14.6.

Table 14.6 Crystal Size Distribution and Separation Information CSR, mm 0–5 5–10 10–15 15–20 20–30 30–50 50–100 100þ

% in PSR

%LTSS

%GTSS

d, mm

Ei, %

2 2 1 4 6 11 13 61

2 4 5 9 15 26 39 –

98 96 95 91 85 74 61 –

2.5 7.5 12.5 17.5 25.0 40.0 75.0 100.0þ

0.6 7.7 19.2 38.3 56.2 79.1 98.9 100

Equipment

391

SOLUTION: Refer to Table 14.6. The overall efficiency is obtained from the cross-product of columns 2 and 6 for each CSR and summing the results:

E¼

n X

[(2)  (6)]=100; percent basis

i¼1

The calculated results are provided in Table 14.7. Table 14.7 Calculation of Overall Efficiency CSR, mm

26

0–5 5–10 10–15 15–20 20–30 30–50 50–100 100þ

1.2 15.4 19.2 153.2 337.2 870.1 1285.7 6100.0 8782.0

From Table 14.7, one obtains E¼

8782:0 ¼ 87:82% 100

B

Finally, two of the goals of the design and practicing engineers who work on the CSD problem are to be able to: 1 understand CSD so as to analyze and modify what a particular operating crystallizer can and cannot do in the way of producing acceptable crystal size and crystal size ranges, and 2 include the effects of CSD on the design of new units, i.e., to design an original process that is not yet operating.(2)

EQUIPMENT Equipment and equipment selection procedures are usually based on maximum throughput capacities for a new crystallization process. The reason for this is to enable the equipment to perform satisfactorily under the most extreme operating conditions. Material and energy balances based on these conditions are also required before the individual equipment is selected. Most importantly, this procedure is also influenced by the manner in which crystallization occurs. As noted in the Introduction, the methods to accomplish this are crystallization via cooling, via

392

Chapter 14 Crystallization

evaporation, cooling and evaporation (adiabatic evaporation), and a salting out process. Details on the equipment that can accomplish these tasks are provided below. Perhaps the simplest type of equipment for crystallization is a tank in which natural cooling is allowed to lower the temperature of a solution. A hot concentrated solution with the solute is simply poured into the tank where it cools by natural convection. However, there are several drawbacks when employing this practice. This type of crystallizer is inefficient relative to the quantity produced per unit space or per unit of time because of the slow cooling rate. It offers no control of size or size distribution of crystals and it also favors the formation of large crystals that are subject to occlusion of the mother liquor since the crystals tend to grow together. This agglomeration phenomena can be reduced with agitation. This class of equipment is employed when precipitating small quantities of a material and where the size range of the product is not critical. Crystallization can also be achieved by cooling a hot concentrated solution on a surface. The cooling, in turn, produces results in solute solubility that favor solid-phase precipitation. This type of equipment is available in several designs. The most popular of these in earlier years was the Swenson – Walker crystallizer. The unit contains a water-jacketed open trough with a semicylindrical bottom. Hot concentrated solution enters at one end of the trough and the crystals that accumulate on the cooling surface are lifted and pushed along by a spiral stirrer to the other end of the crystallizer. Crystals are removed at the end of the crystallizer by an inclined spiral flight conveyor that lifts them onto a draining board or a conveyor which carries the crystals to centrifuges or any other drying and/or separation operation that may be required. This, as well as other units in this category, are usually employed for systems with a steep solubility slope for which a relatively large yield of crystals can be achieved with a modest drop in temperature. The vacuum crystallizer accomplishes cooling via a flashing process. It is a device that cools the solution by the evaporation of a portion of the solvent. The energy for vaporization is obtained from the sensible enthalpy (heat) of the feed. The unit is usually evacuated to a low pressure by steam jet ejectors. The flashing of the feed due to the lower pressure results in both the cooling and concentration of the solution. One of the problems that can arise is that the operating pressure required to attain the desired crystallization is so low that the evolved vapor cannot be condensed by any cooling water available. These units work for systems of intermediate solubility slopes (see Chapter 12 for a further explanation of solubility slopes). It also has the advantage of vaporizing some of the solvent to bring about additional yield. In the evaporative crystallizer, crystallization is caused by evaporating the solvent from the feed, which can either be a weak unsaturated solution or a hot concentrated mixture. These classes of crystallizers are employed when the solute has a small, low slope temperature-solubility curve. The operation of the evaporator seldom allows much flexibility as to the size, size distribution, and shape of crystals produced. Some classifying action does occur in that fine crystals are usually carried in the circulating mother liquor until they attain a size that will result in their precipitation out of the circulating stream into the solid-removal equipment.(1) These units are normally steam-heated although the heating can also be accomplished by passing hot gases

Describing Equations

393

through the solution. If a traditional evaporator is employed, the classic equation for heat transfer may be applied:(4) Q ¼ UADT

(14:12)

where Q is the rate of heat transfer, U is the overall heat transfer coefficient (the resistance to heat transfer), A is the available area for heat exchange, and DT is the temperature difference driving force across the heat transfer area. The salting-out crystallizer uses a third component that does not chemically react to induce supersaturation, but instead displaces the solubility of the solute in the mixture. The term salting-out is employed because sodium chloride (NaCl) is usually added as the third component although any substance that depresses the solubility of the original solute can be added (as the third component). In some applications, the process is referred to as an antisolvent crystallization. Examples include alcohols and ketones. Reactor crystallizers are another category of units, but are rarely employed. In such units, a mass-transfer step (e.g., gas absorption), occurs in the mother liquor. This results in supersaturation and provides additional mass transfer from the liquid to the solid phase in the form of crystallization.(2) Auxiliary equipment for a crystallization process can include slurry pumps, vacuum pumps, mixers, condensers, steam traps, and entrainment separators. These latter units are not different than those described earlier (and required) for other mass transfer operations.(3)

ILLUSTRATIVE EXAMPLE 14.13 Define the terms occlusion and magma. SOLUTION: Occulsion is a word that describes a process that “prevents the passage of.” It is employed with reference to the mother liquor retained within crystals. The word magma is simply “a suspension of precipitated matter in a water substance.” B

DESCRIBING EQUATIONS Overall and componential material balances have already been described in rather extensive detail in Part I. Some of these calculations in the chemical process industry include transient effects that can account for process upsets, startups, shutdowns, and so on. The describing equations for these time-varying (unsteady-state) systems are differential. The equations usually take the form of a first-order derivative with respect to time, where time is the independent variable. However, calculations for most crystallization processes assume steady-state conditions with the ultimate or final design based on the aforementioned worst-case or maximum flow conditions. This greatly simplifies the calculations since the describing equations are no longer differential, but rather algebraic.

394

Chapter 14 Crystallization

The development of mass (and energy to follow) balances follow the same procedure(s) presented in Part I. Consider, as with other mass transfer operations, material and energy balances and any heat transfer equations that can be written for the various crystallization processes. The material balances can provide the process yield, i.e., the mass of crystals formed from a given mass of solution. The effect of evaporation or of cooling must be included if applicable. As an example, the evaporation of solvent from a crystallizer is pictured in Figure 14.8. A solute componential and total material balance equation (on a mole basis) may be written: F ¼ V þ L þ P; total

(14:13)

xF F ¼ V(0) þ xL L þ xC P; componential

(14:14)

where F is the feed, V is the quantity of evaporation, L is liquid (filtrate) withdrawal, P is the crystal production, and x represents the solute content of a stream in units consistent with the flow rate. There is also an equilibrium expression relating xL and xC. Once the material balance is completed, one may then proceed directly to the energy calculations, some of which can play a significant role in crystallizer design. As with material balance calculations, energy calculations for crystallizers almost always are based on steady-state conditions. One of the principal jobs of a practicing engineer involved with crystallizers is to account for the energy that flows into and out of the unit and to determine overall energy requirements. This is accomplished by performing energy balances on the unit.

Figure 14.8 Line diagram of an evaporative crystallizer.

Describing Equations

395

As noted earlier, crystallization operations may be carried out either by cooling a solution or by evaporative concentration, or by both. The cooling of a solution may be accomplished by the removal (transfer) of heat to cooling water or air of both the sensible heat and the heat evolved during crystallization of the product. If the solution is crystallized by evaporation, the required heat may be supplied by the sensible heat of solution as in vacuum crystallizers, or it may be supplied from an external source. In the former case, the heat of vaporization of the solvent from the solution may safely be assumed equal to the heat of vaporization of the pure solvent. In the latter case, the heat of solution (data for which is rarely available) may be assumed equal to the heat of solution at infinite dilution.(1) Obviously, for crystallization problems requiring energy balances, e.g., where evaporation occurs or when the temperature of an adiabatic crystallizer is unknown, enthalpy data must be provided or made available. Referring once again to Figure 14.8, one may also write an energy balance F H^ F þ Q ¼ V H^ V þ LH^ L þ PH^ C

(14:15)

The enthalpies in this equation can be determined, provided the temperatures of the product streams are known. In addition, the operation is usually either adiabatic or may be safely assumed to be adiabatic, so that Q ¼ 0. Five illustrative examples concerned with the describing equations associated with crystallizers follow. Several excellent additional examples are available in the literature.(6) ILLUSTRATIVE EXAMPLE 14.14 Devise a process that will produce 2000 lb/hr of NaNO3 crystals from a 90% (by mass) NaNO3 – NaCl mixture. SOLUTION: This is obviously an open-ended process design problem. One possible flow diagram is provided in Figure 14.9. B

Figure 14.9

Process flow diagram.

396

Chapter 14 Crystallization

Figure 14.10 Process flow diagram with system data.

ILLUSTRATIVE EXAMPLE 14.15 Refer to the previous example. Write pertinent material balance equations around the cooler and crystallizer for the system provided in Figure 14.10. SOLUTION: Overall material balance: A þ B þ C  E  D ¼ 0;

D ¼ 2000

Water balance: A þ 0:487(B)  0:5506(E) ¼ 0 NaCl balance: 0:1(C) þ 0:0999(B)  0:13(E) ¼ 0 NaNO3 balance: 0:9(C) þ 0:4131(B)  0:3194(E)  2000 ¼ 0 Since there are three components—(H2O, NaNO3, and NaCl)—only three independent equations may be employed. B

ILLUSTRATIVE EXAMPLE 14.16 Refer to the previous example. Write pertinent material balance equations around the heater and evaporator in Figure 14.10.

Design Considerations

397

SOLUTION: Overall material balance: EGFB¼0 Water balance: 0:5506(E)  F  0:487(B) ¼ 0 NaCl balance: 0:13(E)  G  0:0999(B) ¼ 0 NaNO3 balance: 0:3194(E)  0:4131(B) ¼ 0 Once again, only three independent equations may be written.

B

ILLUSTRATIVE EXAMPLE 14.17 Obtain the flow rates for each of the eight streams noted in Figure 14.10. SOLUTION: First, examine the material balance equations provided in the three previous examples. Although there are eight equations, one notes that there are a total of only seven independent equations. Since there are seven unknown streams—A, B, C, D, E, F, and G—a unique solution is possible. That unique solution, in lb/hr can be shown to be: A ¼ 733 B ¼ 3254 C ¼ 2222 D ¼ 2000 E ¼ 4209 F ¼ 733 G ¼ 222 H¼0

B

DESIGN CONSIDERATIONS As with any mass transfer device, there are usually five conceptual steps to be considered with the design of equipment. These are: 1 the identification of the parameters that must be specified, 2 the application of fundamentals underlying theoretical equations or concepts, 3 the enumeration, explanation, and application of simplifying assumptions,

398

Chapter 14 Crystallization

4 the possible use of correction factors for “nonideal” behavior, and 5 the identification of other factors that must be considered for adequate equipment specification. Since design calculations are generally based on the maximum throughput capacity for the proposed crystallizer or process, these calculations are never completely accurate. It is usually necessary to apply reasonable safety or “fudge” factors when setting the final design for the crystallizer. Safety factors vary widely, particularly with crystallizers, and are a strong function of the accuracy of the data involved, calculational procedures, and past experience. Nonetheless, simple qualitative design procedures are provided below. Assuming a crystallizer is the mass transfer operation of choice, the class of crystallizer is normally selected first. For high feed rate (e.g., greater that 50,000 lb/hr), a continuous—as opposed to a batch—process is normally selected. Overall process decisions are based on capital, operating and maintenance costs, space, location, production rate, crystal characteristics, size and size distribution of crystals, physical and chemical characteristics of the feed liquor and slurry, corrosion concerns, and so on. Regarding a new design, scale-up problems can be expected when the design is based on pilot-plant or laboratory units. Even after the equipment is finally constructed, significant startup problems are the norm. Finally, it should be noted that the scaling-up of crystallization equipment is probably more difficult than that for any of the other mass transfer operations. If the pilot or small-scale crystallizer produces a satisfactory product, then the design of the larger crystallizer must simulate a number of different conditions obtained in the smaller unit. The four most important conditions are:(7) 1 identical flow characteristics of liquids and solid particles, 2 identical degrees of supersaturation in all equivalent regions of the crystallizer, 3 identical initial seed sizes (if applicable) and magma densities, and 4 identical contact times between growing crystals and supersaturated liquor.

ILLUSTRATIVE EXAMPLE 14.18 Discuss problems associated with the design of crystallization equipment. SOLUTION: As discussed earlier, the design of crystallization equipment can be difficult. Knowledge of the phase equilibrium and physical properties does not allow one to predict the behavior of the actual process. Bench-scale and pilot-plant experiments on the actual stream are usually recommended and some equipment vendors require a pilot-plant test before designing the equipment. Even with such testing, operating adjustments must usually be made on the actual commercial installation before the equipment will operate to yield an acceptable product. B

Design Considerations

399

Marnell(8) has provided a step-by-step procedure for the design of a single solute forced circulation crystallization, as pictured in Figure 14.10. The nine specifications required and listed below are: 1 production rate of crystals, P, 2 feed (liquid) mass fractions, 3 desired CSD and appropriate residence time to achieve it, 4 feed temperature, 5 design temperature and corresponding mother liquor saturation mass fractions, 6 centrifuge wash water ratio, 7 crystal bulk density, rB, 8 magma density, rM, and 9 liquor density, rL. Specifications (5) – (9) are generally drawn from experience. ILLUSTRATIVE EXAMPLE 14.19 Complete a material balance for the process pictured in Figure 14.11. The following information is provided:(8) Production rate: 3125 kg/h Feed composition: 75% urea, 25% water

u u

u

Figure 14.11

Forced circulation crystallizer.

400

Chapter 14 Crystallization Magma density: 1170 kg/m3 Magma crystal concentration: 450 kg crystals/m3 magma Liquor density: 1340 kg/m3 Magma recirculation rate: 567,900 kg/h Centrifuge washwater ratio: 0.12 kg H2O/kg urea

SOLUTION:

First calculate the feed ratio, F: F¼ ¼

P ; xF,U ¼ mass fraction urea xF,U 3125 0:75

¼ 4167 kg=h The wash rate, W, is W ¼ (0:12)P ¼ (0:12)(3125) ¼ 375 kg=h The water vaporization rate, V, is given as V ¼ (F)(xF,W ) þ W; xF,U ¼ mass fraction urea ¼ (4167)(0:25) þ 375 ¼ 1417 kg=h Calculate the magma flow, M, M¼ ¼

(P)(rM ) ; rC,M

rM ¼ magma density, rC,M ¼ crystal density in magma

(3125)(1170) 450

¼ 8120 kg=h The liquor flow, L, is obtained by a total mass balance around the centrifuge: W þM ¼PþL L¼W þMP ¼ 375 þ 8120  3125 ¼ 5370 kg=h The bottoms, B, is B¼MþC ¼ 8120 þ 567,900 ¼ 576,000 kg=h

Design Considerations

401

The recycle, R, is R¼CþL ¼ 567,900 þ 5370 ¼ 573,300 kg=h Finally, the inlet to the crystallizer, I, is I ¼FþR ¼ 4167 þ 573,300 ¼ 577,500 kg=h

B

ILLUSTRATIVE EXAMPLE 14.20 Refer to Example 14.19. Size the crystallizer. Design information is provided below: Average residence time, t (based on M ): 3.38 h Maximum (with safety factor) vapor velocity,  v ¼ Cf

rl  rv rv

0:5 ; Cf ¼ 0:0122 m=s

Crystallizer diameter, D, is based on the maximum vapor velocity, v Crystallizer magma cylinder height, H : H ¼ 1.15D Freeboard height (for droplet disengagement): 2 m Refer to Figure 14.11 for additional details. SOLUTION: The crystallizer area and diameter are to be based on entrainment to ensure vapor droplet disengagement. The vapor velocity is given in:  v ¼ Cf

rl  rv rv

0:5 ; Cf ¼ 0:0122 m=s

For steam at atmospheric pressure, set rv ¼ 0.0563 kg/m3. Therefore,  v ¼ 0:0122

 1340 0:5 0:0543

¼ 1:917 m=s ¼ 6899 m=h

402

Chapter 14 Crystallization

The crystallizer area, A, and diameter, D, are therefore A¼ ¼

V (rv )(v) 1417 (0:0543)(6899)

¼ 3:78 m2 In addition,  0:5 4A p   4(3:78) 0:5 ¼ p

D¼

¼ 2:19 m The magma occupies two sections (of volume) in the crystallizer—the cylinder of height H and the cone below the cylinder. The magma volume MV is therefore MV ¼ cone volume þ cylinder volume The volume of the cone can be shown to be 0.2267D 3.(9) Thus, p  MV ¼ 0:2267D3 þ D2 H; H ¼ 1:15D 4 ¼ (0:2267 þ 0:903)D3 ¼ (1:13)D3 The volume occupied by the magma M t rM   8120 (3:38) ¼ 1170

MV ¼

¼ 23:5 m3 Setting the above two equations equal to each other, 23:5 ¼ (1:13)D3 D ¼ 2:75 m In addition, H ¼ 1:15D ¼ (1:15)(2:75) ¼ 3:16 m

403

Design Considerations The total crystallizer volume, TC, is TC ¼ 3:16 þ 2:0 ¼ 5:16 m

B

ILLUSTRATIVE EXAMPLE 14.21 Quick-size the heat exchanger in Figure 14.11. Assume U ¼ 200 Btu/h . ft2 . 8F. SOLUTION: In order to size the exchanger, first calculate the steam requirement. An energy _ s: balance around the entire process allows one to estimate the heat exchanger steam load, Q _ s ¼ V H^ V  F H^ F  W H^ W þ PH^ P Q Neglecting sensible enthalpy changes associated with F, W, and P reduces the above to _ s ¼ V H^ V ¼ V lw ; lw ¼ enthalpy of vaporization of water Q Thus, _ s ¼ ð1417 kg=hÞð2403 kJ=kgÞ Q ¼ 3,405,000 kJ=h Quick-size the magma exchanger by applying the standard heat transfer equation: _ S ¼ UAS DTLM Q

(14:16)

Employing atmospheric steam at 1008C, DTLM  100  55 ¼ 458C The overall boiling and convective water coefficient (assumed) of 200 Btu/h . ft2 . 8F is converted to 4088 kJ/h . m2 . 8C. Substituting,

AS ¼

_S 3,405,000 kJ=h Q  ¼ UDTLM 4088 kJ=h  m2  8C (458C)

¼ 18:5 m2  ¼ (18:5 m2 )

2 1 ft 0:3048 m

¼ 60:73 ft2 This is a small area. Normally, a double pipe exchanger is used if the area is below 500 ft2. However, to avoid the numerous bends in such a design, a conventional shell and tube exchanger with a single tube and shell pass is recommended. B

404

Chapter 14 Crystallization

ILLUSTRATIVE EXAMPLE 14.22 Outline how to quick-size the overhead condenser. Assume U ¼ 300 Btu/h . ft2 . 8F. SOLUTION:

_ C , is given by The heat load, Q _ C ¼ V lw Q

with (once again), _ C ¼ UAC DTLM Q Therefore, AC ¼

_C Q UDTLM

A vacuum system is often recommended and a steam jet ejector should be adequate for this design. Note that the urea in the solution elevates the operating boiling temperature of water by 55 2 41 ¼ 148C (per Fig. 14.11) . Assume cooling water is available at 308C and employ an approach temperature of 38C. Thus, DTLM ¼

(41  30)  (41  38)   ¼ 6:168C 41  30 ln 41  38

Set U ¼ 300 Btu/h . ft2 . 8F ¼ 6132 kJ/h . m2 . 8C for a water–water condenser. Using the duty found in Illustrative Example 14.21, _ C ¼ 3,405,000 kJ=h Q one may now solve for AC: AC ¼

3,405,000 (6132)(6:168C)

¼ 90:14 m2 ¼ 970 ft2

B

REFERENCES 1. 2. 3. 4.

G. BROWN and ASSOCIATES, “Unit Operations,” John Wiley & Sons, Hoboken, NJ, 1950. A. RANDOLPH, “Crystallization,” Chem. Eng., New York City, NY, May 4, 1970. L. THEODORE, “Air Pollution Control Equipment Calculations,” John Wiley & Sons, Hoboken, NJ, 2008. J. REYNOLDS, J. JERIS, and L. THEODORE, “Handbook of Chemical and Environmental Engineering Calculations,” John Wiley & Sons, Hoboken, NJ, 2004. 5. S. SHAEFER and L. THEODORE, “Probability and Statistics Applications for Environmental Science,” CRC Press/Taylor & Francis Group, Boca Raton, FL, 2007.

References

405

6. N. CHOPEY, “Handbook of Chemical Engineering Calculations,” 2nd edition, McGraw-Hill, New York City, NY, 1994. 7. J. MULLIN, “Crystallization,” Butterworth, London, 1961. 8. P. MARNELL: personal communication to L. Theodore, Manhattan College, Bronx, NY, 2009. 9. P. MARNELL: personal notes, Manhattan College, Bronx, NY, 1982.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

15

Membrane Separation Processes INTRODUCTION Membrane processes are state-of-the-art separation technologies that show continued promise for technical growth and wide-scale commercialization. They are used in many industries for process stream and product concentration, purification, and fractionation. The need for membrane research and development (R&D) is important because of the increasing use of membrane technology in traditional and emerging engineering fields. Membrane processes are increasingly finding their way into the growing engineering areas of biotechnology, green engineering, specialty chemical manufacture, biomedical engineering, as well as the traditional chemical process industry. Membrane technology is also being looked at as either a replacement for or supplement to traditional separations such as distillation (Chapter 9) or extraction (Chapter 12). Membrane processes are generally more efficient and effective since they can simultaneously concentrate and purify, and can perform separations at ambient conditions.(1) Membrane unit operations are often characterized by the following parameters: driving force utilized, membrane type/structure, and species being separated. The following membrane unit operations utilize a pressure driving force to separate a liquid feed into a liquid permeate and retentate: reverse osmosis, nanofiltration, ultrafiltration, and microfiltration. They are listed in ascending order in their ability to separate a liquid feed based on solute size. Reverse osmosis uses non-porous membranes and can separate down to the ionic level as with the example of seawater in the rejection of dissolved salt. Nanofiltration performs separations at the nanometer range. Ultrafiltration uses porous membranes and separates components of molecular weight ranging from the low thousand to several hundred thousand molecules; an example includes components of biochemical processing. Microfiltration uses much more porous membranes and is typically used in the macromolecular range to remove particulate or larger biological matter from a feed stream (e.g., in the range of 0.05 – 2.5 mm).(1,2) Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

407

408

Chapter 15 Membrane Separation Processes

Dialysis membrane processes use a concentration driving force for separation of liquid feeds across a semipermeable membrane with the major application in the medical field of hemodialysis. Electrodialysis separates a liquid feed solution through ion selective membranes by means of an electrical driving force and is widely used in water purification and industrial processing. Gas separation processes can be divided into two categories—gas permeation through non-porous membranes and gas diffusion through porous membranes. Both of these processes utilize a concentration driving force. The gas permeation processes are used extensively in industry to separate air into purified nitrogen and enriched oxygen. Another commercial application is hydrogen recovery in petroleum refineries.(1) Since membrane separation processes are one of the newer (relatively speaking) technologies being applied in practice, the subject matter is and has been introduced into the engineering curriculum. There are four major membrane processes of interest to the practicing engineer: 1 reverse osmosis (hyperfiltration), 2 ultrafiltration, 3 microfiltration, and 4 gas permeation The four processes have their differences. The main difference between reverse osmosis (RO) and ultrafiltration (UF) is that the size/diameter of the particles or molecules in solution to be separated is smaller in RO. In microfiltration (MF), the particles to be separated/concentrated are generally solids or colloids rather than molecules in solution. Gas permeation (GP) is another membrane process that employs a nonporous semipermeable membrane to “fractionate” a gaseous stream. The heart of the membrane process is the membrane itself. A membrane is an ultra-thin semipermeable barrier separating two fluids that permits the transport of certain species through the barrier from one fluid to the other. The membrane is typically made from various polymers such as cellulose acetate or polysulfone, but ceramic and metallic membranes are also used in some applications. The membrane is selective since it permits the transport of certain species while rejecting others. The term semipermeable is frequently used to describe this selective action.(1) The presentation to follow will key on the above four membrane separation processes, particularly RO because of its widespread use in desalinization applications. RO is reviewed first and receives the bulk of the treatment. This is then followed by UF, MF, and GP. The reader should note that the notation and units previously adopted by this industry are primarily employed in the development.

REVERSE OSMOSIS The most widely commercialized membrane process by far is reverse osmosis (RO), which belongs to a family of pressure driven separation operations for liquids that includes reverse osmosis, ultrafiltration, and microfiltration. Care should be exercised

Reverse Osmosis

409

here since some terms are used interchangeably. For example, RO is considered by some as hyperfiltration. Reverse osmosis is an advanced separation technique that may be used whenever low molecular weight solutes such as inorganic salts or small organic molecules (e.g., glucose) are to be separated from a solvent (usually water). In normal (as opposed to reverse) osmosis, water flows from a less concentrated salt solution to a more concentrated salt solution as a result of an inate driving force (the aforementioned chemical potential). As a result of the migration of water, an “osmotic pressure” is created on the side of the membrane to which the water flows. In reverse (as opposed to normal) osmosis, the membrane is permeable to the solvent or water and relatively impermeable to the solute or salt. In order to make water pass through an RO membrane in the desired direction (i.e., away from a concentrated salt solution), a pressure must be applied that is higher than the osmotic pressure. Reverse osmosis is widely utilized today by a host of industries for a surprising number of operations. Aside from the classic example of RO for seawater desalination, it has found a niche in the food industry for concentration of fruit juices, in the galvanic industry for concentration of waste streams, and in the dairy industry for concentration of milk prior to cheese manufacturing.(3–5) A more novel use of reverse osmosis is the production of low-alcohol beer by breweries in Denmark, France, and Germany. Reverse osmosis processes are classified into the following two basic categories: 1 Purification of a solvent such as in desalination where the permeate or purified water is the product. 2 Concentration of the solute such as in concentration of fruit juices where the retentate is the product. The membranes used for RO processes are characterized by a high degree of semipermeability, high water fluxes, mechanical strength, chemical stability, and relatively low operating and capital costs. Early RO membranes were composed of cellulose acetate, but restrictions on process stream pressure, temperature, and organic solute rejection spurred the development of non-cellulosic and composite materials (membrane “sandwiches”). Reverse osmosis membranes may be configured into certain geometries for system operation: plate and frame, tubular, spiral wound (composite), and hollow fiber. These are detailed in the next paragraph.(5) In the plate and frame configuration, flat sheets of membrane are placed between spacers with heights of approximately 0.5– 1.0 mm. These are, in turn, stacked in parallel groups. Tubular units are also used for RO. This is a simpler design in which the feed flows inside of a tube whose walls contain the membrane. These types of membranes are usually produced with inside diameters of 12.5 –25 mm. Also, they are generally made in lengths of 150– 610 cm. There is also the hollow fine fiber (HFF) arrangement. This geometry is used in 70% of worldwide desalination applications. Millions of hollow fibers are oriented in parallel and fixed in epoxy at both ends. The feed stream is sent through a central distributor where it is forced out radially through the fiber bundle. As the pressurized feed contacts the fibers, the permeate is

410

Chapter 15 Membrane Separation Processes

forced into the center of each hollow fiber. The permeate then moves along the hollow bore until it exits the permeator. A spiral wound cartridge is occasionally employed. In this configuration, the solvent is forced inward towards the product tube while the concentrate remains in the spacing between the membranes. A flat film membrane is made into a “leaf.” Each leaf consists of two sheets of membrane with a sheet of polyester tricot in between to act as a collection channel for the product water. Plastic netting is placed between each leaf to serve as a feed channel. Each leaf is then wrapped around the product tube in a spiral fashion. Water covers around 70% of the Earth’s surface but 97.5% of it is unfit for human consumption. With the world facing a growing fresh water shortage, from which the United States will not be spared, one method for producing fresh water that has been around for decades is desalination. Until recently, a thermal process was used to separate water from salt. Saline water was boiled until it evaporated, leaving the salt behind; the salt-free water is then reclaimed when the steam condensed. Unfortunately, it is a very expensive method because it requires significant amounts of energy. The technology of choice today is RO. Essentially, water is pumped under high pressure through membranes that filter off the salt. It too requires energy but not nearly as much as the thermal method. Recent advances in membrane technology and energy recovery methods have made the RO desalination process much more cost-competitive. Thus, the major application of RO is water desalination. Areas of the world that do not have a ready supply of fresh water may choose to desalinate seawater or brackish water using RO to generate potable drinking water. Because no heating or phase change is required, the RO process (often referred to as hyperfiltration) is a relatively low energy water purification process. A typical salt water RO system consists of an intake, a pre-treatment component, a high-pressure pump, membrane apparatus, remineralization and pH adjustment components, as well as a disinfection step. Generally, a pressure of about 1.7– 6.9 MPa is required to overcome the osmotic pressure of salt water.(2) The wine and juice industries have applications for hyperfiltration as well. Flavor tests have shown positive results indicating the potential of membrane processing for improving taste. Using membranes with pore sizes controlled within a specific range has resulted in bitterness and “off-flavors” being removed from finished wine products as well as grapefruit and orange juice. After using a RO process, a taste-testing panel found a low quality Chenin Blanc to have a significantly better taste.(2) However, this is a difficult application as desirable and undesirable taste elements have a similar molecular size along with steric and polar characteristics that are also similar. Another important application of hyperfiltration is the aforementioned dialysis. This technique is used in patients who suffer from kidney failure and can no longer filter waste products (urea) from the blood. In general, RO equipment used for dialysis can reduce ionic contaminants by up to 90%. In this process, the patient’s blood flows in a tubular membrane while a dialysate flows countercurrently on the outside of the feed tube. The concentrations of undesirables such as potassium, calcium and urea are high in the blood and low or absent in the dialysate. Although this treatment successfully mimics the filtration capabilities of the kidney, it cannot replace its endocrine functionality.(1)

Reverse Osmosis

Figure 15.1

411

Seawater desalination by RO.

The membrane operation that incorporates a selective barrier can be reviewed using the process line diagram provided in Figure 15.1 for the purification of seawater. This membrane operation typifies the case where a feed stream (seawater) is separated by a semi-permeable membrane that rejects salt but selectively transports water. A purified water stream (the permeate) is therefore produced while, at the same time a concentrated salt stream (the retentate) is discharged. With reference to Figure 15.1, a simple material balance can be written on the overall process flows and for that of the solute: q ¼ volumetric flow rate qf ¼ qr þ qp ; Cf qf ¼ Cr qr þ Cp qp ; C ¼ solute concentrate

(15:1) (15:2)

Subscripts f, r, and p refer the feed, retentate, and permeate, respectively. ILLUSTRATIVE EXAMPLE 15.1 Verify that the quantities provided in Figure 15.1 satisfy both a componential and overall material balance. SOLUTION:

An overall balance yields (in m3/day) qf ¼ qr þ qp 800 ¼ 600 þ 200 800 ¼ 800

A componential balance on the salt gives (in mg/day) (3500)(800)(1000) ¼ (46,650)(600)(1000) þ (350)(200)(1000); 1000 L ¼ 1 m3 2:80  108 ¼ 2:80  108 Both balances are satisfied.

B

412

Chapter 15 Membrane Separation Processes

Figure 15.2 Before osmosis equilibrium.

Osmosis must first be better explained in order to fully describe RO. As previously mentioned, osmosis occurs when a concentrated solution is partitioned from a pure solute or a relatively lower concentration solution by a semi-permeable membrane. The semi-permeable membrane only allows the solvent to flow through it freely. Equilibrium is reached when the solvent from the lower concentration side ceases to flow through the membrane to the higher concentration side (thus reducing the concentration) because the mass transfer driving force has been “quenched”. This is shown in Figures 15.2 and 15.3. Osmotic pressure is termed as the pressure needed to stop the flux of solvent through the membrane or the force that pushes up on the concentrated side of the membrane (see Fig. 15.4). Applying a pressure on the concentrated side stops the solvent flux. Reverse osmosis (Fig. 15.5) takes place when an applied force (pressure)

Figure 15.3 Osmosis of solvent.

Figure 15.4 Osmotic pressure.

Reverse Osmosis

Figure 15.5

413

Reverse osmosis.

on the concentrated side overcomes the osmotic pressure and forces the solvent from the concentrated side through the membrane and leaves the solute on the concentrated side. As noted above, osmotic pressure occurs when two solutions of different concentrations (or a pure solvent and a solution) are separated by a semi-permeable membrane. In simplest terms, it means that the membrane is permeable to the solvent but not to the solute. The solvent molecules in the dilute phase have a higher chemical potential than the molecules in the concentrated phase. This difference in chemical potential causes a flow of solvent molecules from the dilute phase (high chemical potential) to the concentrated phase (low chemical potential). Flow of solvent molecules into the concentrated solution continues until osmotic equilibrium is reached, i.e., until the chemical potential of the solvent molecules in both phases are equal. Summarizing, Figure 15.6 provides a more detailed pictorial representation of what happens on the surface of a membrane in RO. A concentrated solution enters as the feed and is separated with the assistance of the membrane and the filtered solvent

Figure 15.6

Explanation of what happens in a membrane process.

414

Chapter 15 Membrane Separation Processes

exits as the permeate. The retentate is the solvent that is not cleaned and is more concentrated than the feed solution. A step-by-step explanation (see Fig. 15.6) of this phenomenon follows: 1 The feed enters and the solution is forced to the surface of the membrane. 2 Some of the solvent passes through the membrane (not shown because it is smaller than solute) and some passes on to the retentate side. 3 The solute builds up a layer on the surface of the membrane, causing the flux to decrease (which reduces the quantity of feed being purified). 4 The feed solution that is still flowing comes into contact with the solute buildup on the membrane surface and removes some or even all of the solute on the membrane and re-entrains it in the flow to become the retentate. Downstream, the re-entrained solute can then be re-trapped on the surface of the membrane. 5 The whole process can be repeated with a small film on the membrane (where solute has been trapped already) or a totally clean part (downstream). In the process, the retentate becomes more concentrated with the solute because the solute that is removed and trapped on the membrane becomes re-entrained by the tangential force of the feed solution that is still passing through the unit. The solvent is effectively forced through the membrane to make permeate (pure solvent) by the design of the filter itself.(6) Reverse osmosis is the most selective of the three membrane processes described earlier that are used in industry for (primarily) liquid purification: microfiltration, ultrafiltration, and hyperfiltration. The three processes are shown schematically in Figure 15.7. In microfiltration, the particles to be concentrated are generally solids or colloids rather than molecules in solution. As previously stated, if there is a difference between RO and ultrafiltration; it is that the size of the particles or molecules to be separated in solution is smaller in RO. Various process flow schematics of RO are provided in Figure 15.8.(7)

Describing Equations As noted above, one important characteristic of the RO process is the osmotic pressure of the solvent. Osmotic pressure is related to both the solute concentration and the temperature of the solution as described in the Van’t Hoff equation below:

p ¼ iC s RT where

p ¼ osmotic pressure, psi i ¼ Van’t Hoff factor, dimensionless Cs ¼ solute concentration, mol/L R ¼ Universal gas constant, L . atm/mol . K T ¼ absolute temperature, K

(15:3)

Reverse Osmosis

Figure 15.7

Figure 15.8

415

Membrane separation processes.

Process flow diagram (RO). (Adapted from C.S. Slater and J.D. Paccione, “A reverse osmosis system for an advanced separation process laboratory,” Chemical Engineering Education, 22, New York City, NY, pp. 138– 143, 1987).

416

Chapter 15 Membrane Separation Processes

The Van’t Hoff factor, i, takes into account the number of ions in solution. For example, NaCl separates into two ions, Naþ and Cl2, therefore making the Van’t Hoff factor equal to 2. Upon closer inspection of Equation (15.3), one can readily observe that the Van’t Hoff equation is analogous to the ideal gas law. The change in osmotic pressure across the membrane in this operation must be overcome in order to achieve RO. This is shown by Dp ¼ p f  p p

(15:4)

where Dp ¼ change in osmotic pressure, psi

pf ¼ osmotic pressure in the feed, psi pp ¼ osmotic pressure in the permeate, psi This change in osmotic pressure can also be found using the concentrations of both the feed and the permeate, as well as a coefficient denoted as C. This formula is shown by Dp ¼ C(C f  C p )

(15:5)

where Dp ¼ osmotic pressure change, psi C ¼ constant, L . psi/g Cf ¼ feed concentration, g/L Cp ¼ permeate concentration, g/L The permeate flux is an important characteristic of the RO operation. It is related to the permeate flow as well as the area of the membrane. This can be seen in the following equation qp (15:6) Jp ¼ Am where

Jp ¼ permeate flux, gal/ft2 . day qp ¼ permeate flow, gal/day Am ¼ membrane surface area, ft2

The flux can be determined by measuring each incremental volume of permeate, DV, collected in time period Dt, and dividing by the surface area of the membrane. In water-based processes such as desalination, the permeate consists of mostly water. Therefore, the permeate flux can be considered to be equal to the water flux. Equation (15.7) defines the water flux: J p ¼ J w ¼ Aw (DP  Dp) ¼ where

Ks (DP  Dp) tm

Jw ¼ water flux, gal/ft2 . day Jp ¼ permeate flux, gal/ft2 . day Aw ¼ water permeability coefficient, gal/ft2 . day . psi

(15:7)

Reverse Osmosis

417

DP ¼ pressure drop, psi Dp ¼ osmotic pressure change, psi Ks ¼ permeability constant, gal/ft . day . psi tm ¼ membrane thickness, ft The water permeability coefficient can be experimentally determined by obtaining data on the unit with pure water; this eliminates the change in osmotic pressure since both sides of the membrane contain pure water. Another important factor is the solute flux. This can be determined through the utilization of: Js ¼ where

q pC p m _p ¼ Am Am

(15:8)

Js ¼ solute flux, g/ft2 . min qp ¼ permeate flow rate, L/day Cp ¼ permeate concentration, g/L _ p ¼ permeate flow rate, g/day m Am ¼ membrane surface area, ft2

The solute flux can also be related to the solute concentration by utilizing the solute permeability factor. This relationship is provided by J s ¼ Bs (DC s ) where

Js ¼ solute flux, g/ft

2.

(15:9)

min

Bs ¼ solute permeability coefficient, L/ft2 . min DCs ¼ change in concentration, g/L The selectivity of a membrane to filter out a solute can be expressed as the percent rejection (%R). Percent rejection represents a membrane’s effectiveness and is a measure of the membrane’s ability to selectively allow certain species to permeate and others to be rejected. This is an important characteristic when selecting a membrane for a separation process. The percent rejection represents the percentage of solute that was not allowed to pass into the permeate stream, and is given by   Cf  Cp  100% (15:10) %R ¼ Cf where %R ¼ solute rejection, % Cp ¼ permeate concentration, g/L Cf ¼ feed concentration, g/L Finally, the solvent recovery, Y, is a measure of how much solvent is allowed to pass through the membrane. This is defined as the quotient of the permeate flow

418

Chapter 15 Membrane Separation Processes

divided by the feed flow, as shown by Y¼ where

qp qf

(15:11)

Y ¼ solvent recovery, dimensionless – fractional basis qp ¼ permeate flow, L/min qf ¼ feed flow, L/min

ILLUSTRATIVE EXAMPLE 15.2 With reference to Figure 15.1, calculate the solvent flux and the membrane selectivity. SOLUTION:

For the flux, Js ¼

_s m Am

(15:8)

Substituting, Js ¼

200 ¼ 20 m3 =m2  day 10

For the selectivity,

 %R ¼

 Cf  Cp 100 Cf

(15:10)

Substituting,  %R ¼

 35,000  350 100 ¼ 99:0% 35,000 B

The reader should note the following two basic membrane transport equations. J¼P where

(DF) tm

(15:12)

J ¼ flux P ¼ membrane permeability DF ¼ driving force across the membrane tm ¼ membrane thickness

The driving force can be a pressure, concentration, or electric field. The flux, J, may also be written as J ¼ Aw (DP  Dp) where

Aw ¼ water or solvent permeability coefficient DP ¼ total pressure drop across the membrane Dp ¼ osmotic pressure drop

(15:13)

Reverse Osmosis

419

ILLUSTRATIVE EXAMPLE 15.3 Consider the seawater desalination example discussed earlier. If the applied pressure gradient across the membrane is 500 psi and the membrane thickness is 10 mm, determine the permeability of the membrane in m2/s . Pa. SOLUTION:

Employing Equation (15.12), J¼P

(DF) tm

This equation may be rearranged and solved for P, being careful to employ consistent units, P¼

Jtm (20)(1=86,400)(10)(106 ) ¼ DF 500(1:01325  105 =14:7)

¼ 6:71  1016 m2 =s  Pa

B

ILLUSTRATIVE EXAMPLE 15.4(1) A new membrane material (#CSS-1) is to be evaluated for its solute and solvent permeability. A small test cell is utilized with a 5.0 cm diameter circular membrane. The test solution is 6000 mg/L of NaCl in water at 258C. Assume that the following relationship holds for the osmotic pressure of NaCl in water (0.0114 psi/mg/L). At an operating pressure gradient of 750 psi, the permeate flow rate is 0.0152 cm3/s and the permeate solute concentration is 150 mg/L. Assuming there is no concentration polarization and that the operating conditions remain constant, determine the water flux in g/cm2 . s and the solute flux in g/cm2 . s. SOLUTION:

The membrane surface area is Am ¼

p (5)2 ¼ 19:63 cm2 4

Employing Equation (15.8) for water, Jw ¼

_ p 0:0152 m ¼ 7:74  104 g=cm2  s ¼ 19:63 Am

Noting that Cp Js ¼ Cw Jw Js ¼

Cp Jw 150 ¼ 6 (7:74  104 ) ¼ 1:16  107 g=cm2  s 10 Cw

ILLUSTRATIVE EXAMPLE 15.5(1) Refer to Illustrative Example 15.4. Calculate the percent solute rejection.

B

420

Chapter 15 Membrane Separation Processes

SOLUTION:

The percent rejection is given by Equation (15.10).  %R ¼

   Cf  Cp 6000  150 100 ¼ 100 ¼ 97:5% 6000 Cf

B

ULTRAFILTRATION Ultrafiltration (UF) is a membrane separation process that can be used to concentrate single solutes or mixtures of solutes. Trans-membrane pressure is the main driving force in UF operations and separation is achieved via a “sieving” mechanism. The UF process can be used for the concentration of oily wastewater, for pretreatment of seawater prior to RO, and for the removal of bacterial contamination (pyrogens). In the food industry, UF is used to separate lactose and salt from cheese whey proteins, to clarify apple juice, and to concentrate milk for ice cream and cheese production.(8–10) The most energy intensive step in ice cream production is the concentration of skimmed milk, where membrane processes are more economical for this step than vacuum evaporation.(9) UF processes are also used for the concentrating or dewatering of fermentation products, and for the purification of blood fractions and vaccines. UF membranes are rated in terms of their molecular weight cutoff, thereby separating proteins and other biochemicals according to their molecular weight differences. Thin, mechanically strong, flexible, non-adsorbing, and flat-textured, UF membranes are available in a wide variety of cutoff sizes. For example, the YM-30 membrane (employed in a Manhattan College laboratory experiment) will retain any material with a MW .30,000. Most UF membranes have an asymmetric structure with a thin selective membrane supported by a thicker porous structure, which is the case for the YM-30 membrane. Ultrafiltration may be thought of as a membrane separation technique where a solution is introduced on one side of a membrane barrier while water, salts, and/or other low molecular weight materials pass through the unit under an applied pressure. Membrane separation processes can be used to concentrate single solutes or mixtures of solutes. The variety in the different membrane materials makes a wide temperature/pH processing range possible. The main economic advantage of UF is a reduction in design complexity and energy usage since UF processes can simultaneously concentrate and purify process streams. The fact that no phase change is required leads to highly desirable energy savings. Also, as no catalysts are needed for these separations or chemical reagents required, there are further operational savings. A major disadvantage is the high capital investment that might be required if low flux rates for purification demand a large system design. However, UF processes stand out as economically sound in comparison to other traditional separation techniques. In addition to the application decribed above, ultrafiltration membrane processes are utilized in various commercial applications. They are found in the treatment of industrial effluents and process water, concentration, purification, and separation of macromolecular solutions in the chemical, food, and drug industries; sterilization,

Ultrafiltration

421

clarification, and prefiltering of biological solutions and beverages; and, production of ultra pure water and preheating of sea water in RO processes. The most promising area for the expansion of UF process applications is in the biochemical area. Some of its usage in this area includes purifying vaccines and blood fractions; concentrating gelatins, albumin, and egg solids; and, recovering proteins and starches. Ultrafiltration processes have also been used in the production of leukocyte interferon from white blood cells and fibroblast interferon from cell culture, and for the production of human insulin. Usage of the leukocyte interferon includes the treatment of chronic viral hepatitis in heart transplant recipients. Ultrafiltration is an effective purification and concentration process for enzymes. Processes to concentrate and purify enzymes are becoming increasingly important to the biochemical industry where UF is typically employed in downstream separation and recovery of fermentation products. Ultrafiltration of milk is also an important industrial process with the retentate of thickened milk product used in the manufacture of cheese and other milk products. A polysulfone membrane with a 20,000 MWCO (molecular weight cutoff) is normally employed. At low temperatures and pressures, ultrafiltration separates high molecular weight components from the lower molecular weight ones. It allows the lower molecular weight components to permeate along with water. The molecular weight cutoff of a membrane might be defined, for example, as the molecular weight that is 90% rejected by the membrane which indicates that a 10,000 MWCO membrane will reject 90% of solutes having a MW of more than 10,000 Daltons (1 Da ¼ 1 amu). The rejection of a solute is a function of the size, size distribution, shape, and surface binding characteristics of the hydrated molecule. It is also a function of the poresize distribution of the membrane and therefore molecular weight cutoff values can only be used as a rough guide for membrane selection. The retention efficiency of the solutes depends to a large extent on the proper selection and condition of membranes. Replacement of highly used membranes and regular inspections of the separation units averts many problems that would otherwise occur because of clogging and gel formation.

Describing Equations The general design factors for any membrane system (including UF) are reported by Wankat as:(11) 1 Thin active layer of membrane 2 High permeability for species A and low permeability for species B 3 Stable membrane with long service life 4 Mechanical strength 5 Large surface area of membrane in a small volume 6 Concentration polarization elimination or control

422

Chapter 15 Membrane Separation Processes

7 Ease in cleaning, if necessary 8 Inexpensive to build 9 Low operating costs System performance is usually defined in terms of permeate flux, Jp, with dimensions of volume/area . time. The typical units are L/m2 . h. As with RO, Jp can be obtained experimentally by measuring the incremental volume of the permeate, DV, collected in a time period, Dt. Thus, the permeate flux describing equation is Jp ¼ where

DV=Dt surface area

(15:14)

Jp ¼ permeate flux, L/m2 . h DV ¼ incremental volume of permeate, m3 Dt ¼ collection time period, h Surface area ¼ surface area of the membrane, m2

Other consistent units for the flux may be employed, for example, cm3/cm2 . s. Trans-membrane pressure is the main driving force in UF operations, and separation is achieved through the aforementioned sieving mechanism. Since UF is a pressure-driven separation process, it is appropriate to examine the effects of pressure on flux. Equation (15.15) shows how the flux varies with pressure. It is seen that the flux of a pure solvent through a porous membrane is directly proportional to the applied pressure gradient across the membrane, DP, and inversely proportional to the membrane thickness, tm: Js ¼ where

Ks (DP  Dp) tm

(15:15)

Ks ¼ permeate constant, cm2/psi . s DP ¼ pressure drop across the membrane (trans-membrane pressure), psi Dp ¼ osmotic pressure difference across the membrane, psi tm ¼ membrane thickness, cm

Such factors as the membrane porosity, pore size distribution, and viscosity of the solvent are accounted for by the permeability constant, Ks. When tm is not available or is not known, the water permeability coefficient, Aw, may be used in place of Ks. The water permeability coefficient is a function of the distribution coefficient (solubility), diffusion coefficient, membrane thickness, and temperature. The value of Aw can be determined by conducting ultrapure water-flux experiments at varying operating pressures while the permeate collection occurs at atmospheric pressure. The osmotic pressure is relatively low for macromolecular solutions, which are typically the ones recommended for UF processes, and therefore the Dp term can be neglected in Equation (15.15). This is the case since the molar concentration of the high molecular weight molecules separated by UF is quite low, even when the

Ultrafiltration

423

mass concentrations are high. When the Dp term is neglected and Ks/tm is replaced with Aw, the following equation is obtained: Js ¼ Aw DP

(15:16)

2

where Aw ¼ water permeability coefficient, cm /psi . s. When a solute such as milk solids dissolved in water flows through a typical UF process, some of the solute usually passes through the membrane since real membranes are partially permeable. The apparent rejection on a fraction basis is then once again calculated as follows (see also Eq. 15.10): Rapp ¼

Cr  Cp Cr

(15:17)

where Rapp ¼ apparent rejection, dimensionless – fractional basis Cp ¼ permeate concentration, g/cm3 Cr ¼ retentate concentration, g/cm3 There are three important factors that need to be considered in UF separations: concentration polarization, gel formation, and fouling. A concentration gradient or boundary layer typically forms during a UF process. This concentration gradient appears near the membrane surface and is referred to as concentration polarization. It results from the counteracting effects of the convective flow of solute towards the membrane and diffusion of the solute back toward the bulk fluid. While concentration polarization is regarded as a reversible boundary-layer phenomenon that causes a rapid initial drop in flux to a steady-state value, fouling is considered as an irreversible process that leads to a flux decline over the long term. Gel formation, however, may be reversible or irreversible. When the gel is difficult to remove, the membrane is said to be fouled and thus the gel formation is irreversible. Concentration polarization may occur with or without gel formation. Concentration polarization occurs in many separations, and for large solutes where osmotic pressure can be neglected, concentration polarization without gelling is predicted to have no effect on the flux. Therefore, if a flux decline is observed, it can be attributed to the formation of a gel layer with a concentration Cg. The gel layer, once formed, usually controls mass transfer so that Equation (15.15) is no longer applicable. When this happens, Equation (15.18) can be used to determine the solvent flux: J¼

DP Rm þ Rg

(15:18)

where Rm ¼ resistance to flow through the membrane, psi . s . cm2/cm3 Rg ¼ resistance to flow through the gel, psi . s . cm2/cm3 The value of Rg varies with pressure, bulk concentration, and cross-flow velocity at lower transmembrane pressure, but tends to become pressure independent at higher transmembrane pressures. This value can be, and often is, measured experimentally.

424

Chapter 15 Membrane Separation Processes

When the gel layer controls mass transfer and Cp ¼ 0 or the apparent rejection is unity, the solvent flux can be expressed in terms of a mass transfer coefficient, k, as follows:   Cg (15:19) Js ¼ k ln Cr where Cg ¼ gel layer concentration, g/cm3 k ¼ mass transfer coefficient, cm3/cm2 . s To determine an experimental value for k, data can be measured when R  1, for the flux as a function of the bulk concentration. This information can be graphed using Equation (15.20) which is a rearrangement of Equation (15.19) above. This plot is obtained for a constant temperature and cross-flow velocity. A plot of Js vs ln Cg on arithmetic coordinates has a slope of 2k and the y-axis intercept is the ln (natural log) of Cg, Js ¼ ln Cg  k ln Cb

(15:20)

Empirical equations can be employed in the determination of the mass transfer coefficient, k. Fully developed turbulent flow in UF devices appears to occur at Re2000. The Reynolds number can be calculated using an equivalent diameter as follows:     cross-sectional area 2hw ¼4 (15:21) Deq ¼ 4 wetted perimeter 2w þ 4h where w ¼ width of the channel, mm h ¼ height of the channel, mm (the channel height is usually defined as 2h rather than h) The following equation can be used to determine the mass transfer coefficient for turbulent flow through a channel: k ¼ 0:023

D Re0:83 Sc1=3 Deq

(15:22)

where Re ¼

Deq ub r m

and

Sc ¼

m Dr

(15:23)

where ub ¼ linear velocity through the channel, m/s D ¼ diffusivity, m2/s m ¼ viscosity, kg/m . s

r ¼ density, kg/m3 The above physical properties are based on the average concentration of the fluid on the feed side of the membrane. The density is estimated on a weight percent solids basis.

Ultrafiltration

425

For laminar flow through a channel, the average mass transfer coefficient can be estimated using the following equation:  1=3 u b D2 k ¼ 1:177 hL

(15:24)

where L ¼ length of the flow channel, mm This correlation is used when the concentration polarization is thin, which holds when the axial distance is much less than the entrance length. Energy costs need to be considered. It has been found that the energy costs for the laminar flow system are generally lower, and thus it is normally the desirable mode of operation.

ILLUSTRATIVE EXAMPLE 15.6 Determine the volume of milk solution and water to be mixed in order to produce a total of 400 ml for a 1 : 8 milk to water volume ratio. SOLUTION:



 1 400 ml ¼ 44:4 ml 1þ8   8 Amount of water ¼ 400 ml ¼ 355:6 ml 1þ8 Amount of milk ¼

B

ILLUSTRATIVE EXAMPLE 15.7 During a 12 psi UF run with pure water, the incremental volume of water collected for a 580 s time interval was 50 cm3. If the effective surface area is 40 cm2, calculate the permeate flux. SOLUTION:

Apply Equation (15.14): Jp ¼

DV=Dt Am

The permeate flux is therefore found to be Jp ¼

50 cm3 =580 s ¼ 0:002155cm3 =cm2  s 40 cm2

B

ILLUSTRATIVE EXAMPLE 15.8 The average concentration of the retentate for a UF run is 1.117 g/cm3. If a 19 cm3 permeate sample is placed on a 1.0534 g tray and dried to a weight of 1.1454 g, calculate the apparent rejection.

426

Chapter 15 Membrane Separation Processes

SOLUTION:

The apparent rejection, Rapp is determined as follows: Rapp ¼

Cr  Cp Cr

(15:17)

The permeate concentration is found using the data provided: Cp ¼

(1:1454  1:0534) g ¼ 0:00484 g=cm3 19 cm3

The apparent rejection is therefore Rapp ¼ 1 

0:00484 g=cm3 ¼ 0:9957 1:117 g=cm3

¼ 99:57%

B

ILLUSTRATIVE EXAMPLE 15.9 The mass transfer coefficient can be estimated from empirical equations. The empirical equation used depends on whether the flow is laminar or turbulent. The Reynolds number needs to be calculated to determine the type of flow. Consider the following UF bench scale system and operating conditions. The height and width of the channel available for flow are 0.038 and 0.95 cm, respectively. The average velocity in the channel is 0.116 cm/s and the flowing fluid’s density and viscosity have been estimated to be 1.013 g/cm3 and 0.020 g/cm . s, respectively. Calculate the Reynolds number. SOLUTION:

The equivalent diameter Deq can be estimated from Equation (15.21)     cross-sectional area 2hw ¼4 Deq ¼ 4 wetted perimeter 2w þ 4h

Substituting   2(0:00038)(0:0095) Deq ¼ 4 2(0:0095) þ 4(0:00038) ¼ 0:00146 m The Reynolds number can now be estimated from Equation (15.23).   (0:146 cm)ð0:116 cm=sÞ 1:0127 g=cm3 Re ¼ ¼ 0:8576 ð0:02 g=cm  sÞ The flow is therefore laminar.

B

ILLUSTRATIVE EXAMPLE 15.10 Refer to the previous illustrative example. Calculate the mass transfer coefficient. The length of the channel is 41.4 cm.

Microfiltration

427

SOLUTION: The flow is laminar. For a laminar flow through a channel, the mass transfer coefficient can be calculated using Equation (15.24)  1=3 u b D2 k ¼ 1:177 hL where D is the diffusivity. The calculation for k cannot be completed since D is not specified. If the system is lactose in water, the diffusivity is approximately 4.9  10210m2/s. Substituting into the above equation (maintaining dimensional consistency) "  2 #1=3 (0:00116 m=s) 4:9  1010 m2 =s k ¼ 1:177 (0:00038 m)(0:414 m) ¼ 1:424  106 L=m2  h ¼ 1:424  106 m3 =m2  s ¼ 1:424  103 L=m2  s ¼ 5:126 L=m2  h

B

MICROFILTRATION Microfiltration (MF) is employed in modern biochemical and biological separation processes. For example, in cell harvesting, microfiltration can be used instead of centrifugation or pre-coat rotary vacuum filtration to remove yeast, bacteria, or mycelial organisms from fermentation broth. Both MF and UF are used for cell harvesting. Microfiltration is used to retain cells and colloids, while allowing passage of macromolecules into the permeate stream. Ultrafiltration is used to concentrate macromolecules, cells, and colloidal material, while allowing small organic molecules and inorganic salts to pass into the permeate stream. Pore sizes in microfiltration are around 0.02– 10 mm in diameter as compared with 0.001 – 0.02 mm (300– 300,000 Daltons) for ultrafiltration (ranges vary slightly depending on the source).(12,13) Similar types of equipment are used for MF and UF, but membranes with larger pore sizes are installed in MF applications. MF and UF belong to a group of separation processes that depend on pressure as the driving force for separation. MF processes operate at lower pressures than UF, but at higher pressures than conventional particulate filtration. Ideal membranes possess high porosity, a narrow pore size distribution, and a low binding capacity. When separating microorganisms and cell debris from fermentation broth, a biological cake is formed. Principles of cake filtration apply to MF systems except that the small size of the yeast particles results in a cake with a relatively high resistance to flow and a relatively low filtration rate. In dead-end filtration, feed flow is perpendicular to the membrane surface, and the thickness of the cake layer on the membrane surface increases with filtration time; consequently, the permeation rate decreases. Cross-flow filtration, on the other hand, features feed flow parallel to the membrane surface, which is designed to decrease formation of a cake layer by sweeping previously deposited solids from the membrane surface and returning them to the bulk feed stream. Cross-flow filtration is far superior

428

Chapter 15 Membrane Separation Processes

to dead-end filtration for cell harvesting because the biological cake is highly compressible, which causes the accumulated layer of biomass to rapidly blind the filter surface in dead-end operation. Therefore, MF experiments are often conducted using crossflow filtration because of the advantages that this mode offers.(14,15) In addition to the mode of operation and cross-flow rate, a number of other factors affect system performance including the following: 1 Operating Temperature: Temperature affects the viscosity of the feed suspension, and subsequently affects permeate flux through both the membrane and the biological cake. Viscosity decreases as temperature increases; hence, it is desirable to operate at the highest temperature that can be tolerated by the species being separated, and the membrane material being used. 2 Average Transmembrane Pressure (ATP): ATP is the average pressure on the retentate side of the membrane minus the average pressure on the permeate side. Increased operating pressure increases permeate flux for ultrapure water but tends to compact biomass on the membrane surface in MF processes. For yeast slurries, increased ATP should increase the transient flux but it may increase, decrease or have little effect on final steady-state fluxes depending on cake compressibility. 3 Yeast Concentration in the Feed: Permeate flux is inversely related to feed concentration, i.e., the final steady-state flux values decrease as the yeast concentration increases. Also, transient fluxes decline faster at higher concentrations because of accelerated cake buildup. 4 pH: The pH of the feed suspension affects the binding characteristics of the membrane and solubility of macromolecules, which in turn influences membrane fouling. The flux should vary inversely with pH. 5 Feed Preparation: Redkar and Davis(16) report that steady-state fluxes for slurries of unwashed Fleischmann’s yeast are significantly lower than fluxes observed when the yeast is washed prior to preparing the feed suspension. Differences are attributed to the presence of extracellular proteins and other macromolecules in the unwashed yeast suspensions, factors that tend to increase specific cake resistance.

Describing Equations Separation principles and governing equations for MF are similar to that for RO and UF. This presentation is primarily directed toward the theoretical aspects of MF. System performance is usually defined in terms of permeate flux, Jp, with dimensions of (volume/area . time), i.e., typical units are L/m2 . h. As noted earlier, the flux can be determined by measuring each incremental volume of permeate, DV, collected in time period, Dt, and dividing by surface area of the membrane as follows: Jp ¼

DV=Dt surface area

(15:25)

Microfiltration

429

Since MF is a pressure-driven separation process, it is appropriate to comment on the effects of pressure on flux. Flux of a liquid solution through a porous membrane is directly proportional to the applied pressure gradient across the membrane, DP, and inversely proportional to the solution viscosity, m, and membrane thickness, tm (see Equation 15.26): Ks DP DP ¼ (15:26) Js ¼ mRm tm The hydrodynamic resistance of the membrane, Rm, is inversely related to the solvent permeability constant, Ks. The permeability constant accounts for factors such as membrane porosity, pore size distribution, and viscosity of the liquid. Permeate is normally collected at atmospheric pressure. In membrane separation processes with pure solvent, temperature effects on flux generally follow the Arrhenius relationship where Jo is the flux at 258C, Ea is the activation energy, R is the universal gas constant, and T is absolute temperature. Js ¼ Jo e(Ea =RT) Equation (15.27) may also be written as:

 Ea 1 ln(Js ) ¼ ln(Jo ) R T

(15:27)



(15:28)

Changes in flux with temperature result from changes in solution viscosity. As previously stated, viscosity decreases as temperature increases; thus, water permeability through the membrane subsequently increases. This relationship can be shown to hold for a Newtonian fluid like distilled water. Fermentation broth containing suspended microorganisms is a non-Newtonian fluid; therefore, increased temperatures tend to improve flux but not to the same magnitude as observed with dilute aqueous solutions. It can be seen from Equation (15.26) that the product of (Jsm) should be a constant value in temperature studies on water at constant DP. Substituting Equation (15.26) into (15.27) and taking the logarithm of both sides of the resultant equation leads to the Arrhenius-type relationship similar to Equation (15.28). Thus, if one employs ultrapure water at varying temperatures and constant transmembrane pressure, Ea can be determined from the slope of a graph of ln(1/m) vs 1/T,       1 1 Ea 1 ln (15:29) ¼ ln  m mo R T The primary factor limiting flux in MF processes is cake buildup. Fouling, caused by factors such as proteins being adsorbed on the membrane surface and increased cake resistance because of cell debris, antifoam, precipitates, etc., which fill the void space in the biological cake, may also contribute to the flux decline. As noted above, cross-flow filtration is designed to sweep the membrane surface so as to sweep deposited solids off of the membrane surface. Cross-membrane flow rate can be varied and its effect on flux determined. While the cross-flow mode is a significant improvement over dead-end filtration, the permeate flux still decreases to

430

Chapter 15 Membrane Separation Processes

some steady-state value of limiting flux, J1. The limiting flux can be modeled in terms of the resistances to permeation through the membrane, Rm, the biological cake, Rc, and the gel or fouling layer, Rg, as follows: J1 ¼

DP m(Rm þ Rc þ Rg )

(15:30)

The resistances in Equation (15.30) can be measured experimentally. For example, the value of Rm can be found by initial clean water flux measurements (Equation 15.26). The total resistance (Rm þ Rc þ Rg) is measured from the final steady-state flux through the system after cake buildup, e.g., with yeast slurry as feed. After completing yeast runs, the MF system may be cleaned in two steps: 1 cleaning with water to remove yeast cake and other reversible deposits, and 2 chemical cleaning with a solution (e.g., hypochlorite) to remove fouling deposits. After cleaning with water, the value for (Rm þ Rg) is measured by clean water fluxes. As Rg is usually negligible in microfiltration processes, thorough cleaning with water should result in flux values that are very close to the original water values. Therefore, chemical cleaning should not be required under normal operating conditions but may be required if membranes are to be reused after high pressure or low cross-flow studies. The final concentration of the retentate (e.g., yeast cells), Cr, can be used as an absolute measure of system performance. A relative measurement of performance would be the concentration factor, c, defined as the ratio of the initial feed volume, V0, to the final retentate volume, Vr, i.e., c ¼ V0 =Vr . Initial and final volumes and concentrations can also be used to calculate the recovery, Y, where C0 is the initial cell concentration in the feed. Recovery, Y ¼

(Cr )(Vr )  100% (C0 )(V0 )

(15:31)

Solute rejection, R o, is another parameter that can be used to measure performance in these systems(17) where Cp is the concentration of yeast cells in the permeate. This parameter is given by: Ro ¼ 1  (Cp =C0 )

(15:32)

ILLUSTRATIVE EXAMPLE 15.11 The data corresponding to a pressure run of 1.5 psi in a MF resulted in a flux for the pure water system of 196.7 mL/m2 . s. Calculate the membrane hydrodynamic resistance (Rm).

Microfiltration SOLUTION:

431

When converted to the “proper” units, the value of the flux is J ¼ 708 L=m2  h

Rm is calculated using the following formula: Rm ¼

DP Jp m

(15:26)

For water at 258C:

m ¼ 3:60 kg=m  h ¼ 0:001 kg=m  s Substituting gives Rm ¼

1:5 ¼ 5:89  104 psi  m3  h2 =L  kg (708)(3:60)

The reader is left the exercise of converting the above results to units of m . h2/L. (The answer is 0.173). B

ILLUSTRATIVE EXAMPLE 15.12 Refer to Illustrative Example 15.11. Calculate the cake resistance if fouling is neglected. Assume the steady-state (or limiting) flux value to be approximately half the value calculated in the previous example. SOLUTION:

The cake resistance is calculated using the following formula: J1 ¼

DP ; (Rg ¼ 0:0) m(Rm þ Rc )

(15:30)

When rearranging to solve for the cake resistance, the following formula is obtained: Rc ¼

DP  Rm ; DP ¼ 1:5 psi ¼ 440 kgf =m2 J1 m

Substituting the values in the equation gives: Rc ¼

440  0:173 (3:6)(708=2)

¼ 0:172 m  h2 =L

B

ILLUSTRATIVE EXAMPLE 15.13 Data for a yeast run in a MF system yielded the following concentration–volume data over a 5 min sampling period: Cr ¼ final concentration of yeast cells in retentate ¼ 0.5 g/L Co ¼ initial cell concentration in feed ¼ 1.2 g/L

432

Chapter 15 Membrane Separation Processes Vr ¼ final retentate volume ¼ 290 L Vo ¼ initial feed volume ¼ 150 L

Calculate the recovery. SOLUTION:

The recovery of the system is calculated using the following formula: Y¼

(Cr )(Vr )  100% (Co )(Vo )

(15:31)

Substituting the values gives: Y¼

(0:5)(2:9)  100% (1:2)(1:5)

¼ 80:56% ¼ 0:8056

B

ILLUSTRATIVE EXAMPLE 15.14 Refer to the previous example. Calculate the solute rejection, R o, if the concentration of the yeast cells in the permeate is 0.10 g/L. SOLUTION:

The solute rejection is calculated using the following formula: Ro ¼ 1  (Cp =Co )

(15:32)

Ro ¼ 1  (0:1=1:2) ¼ 0:92

B

Substituting the values gives:

GAS PERMEATION Several different types of membrane separation processes are used in the chemical process industries, including systems for gas separation. These processes are generally considered as new and emerging technologies because they are not included in the traditional chemical engineering curriculum.(18–23) The use of membranes in gas separation processes was commercialized by Monsanto in the early 1970s with the development of the hollow fiber Prism system for which the Monsanto Company won the 1981 Kirkpatrick Chemical Engineering Achievement Award. The hollow fiber membrane allowed, for the first time, the practical use of membranes in large-scale gas separation and purification processes. Several other firms, including UOP, Air Products and Chemicals, Dow, DuPont and Grace produce these gas permeation membrane units.(18) Gas permeation systems have and continue to gain popularity in both traditional and emerging engineering areas. These systems were originally developed for hydrogen recovery. There are presently numerous applications of gas permeation in industry and other potential uses of this technology are in various stages of development. Applications include gas recovery from waste gas streams, landfill gases, and ammonia

Gas Permeation

433

and petrochemical products. Gas permeation membrane systems are also employed in gas generation and purification, including the production of nitrogen and enriched oxygen gases.(18) Gas permeation is the term used to describe a membrane separation process using a non-porous semi-permeable membrane. In this process, a gaseous feed stream is fractionated into permeate and non-permeate streams. The non-permeating stream is typically referred to as the non-permeate in gas separation terminology and defined as the retentate in liquid separation. Transport occurs by a solution diffusion mechanism. Membrane selectivity is based on the relative permeation rates of the components through the membrane. Each gaseous component transported through the membrane has a characteristic permeation rate that depends on its ability to dissolve and diffuse through the membrane material. The mechanism for transport is based on solubilization and diffusion; the two describing relationships upon which the transport are based are Fick’s law (diffusion) and Henry’s law (solubility).(18)

Describing Equations Diffusive flux through the membrane can be expressed by Fick’s law, as related to the membrane system, and given by(18) Ji ¼ where

Di (Cim2  Cim1 ) tm

(15:33)

Ji ¼ flux of component i, mole/m2 . s Di ¼ diffusivity of component i, m2/s tm ¼ thickness of the membrane, m Cim1 ¼ concentration of component i inside membrane wall on feed side, mole/m3 Cim2 ¼ concentration of component i inside membrane wall on permeate side, mole/m3

Henry’s law may be written in the following form(18) Cim ¼ Si pi

(15:34)

where Si ¼ solubility constant for component i in the membrane pi ¼ partial pressure of component i in the gas phase Substituting Equation (15.34) into Equation (15.33) yields(18) Ji ¼

Di (Si pi2  Si pi1 ) tm

(15:35)

The terms pi2 and pi1 are the respective partial pressures of gas i on the feed and permeate side of the membrane. Permeation through the membrane is a function of solubility and diffusivity, as provided by(18) Pi ¼ Di Si

(15:36)

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Chapter 15 Membrane Separation Processes

Substitution of Equation (15.36) into Equation (15.35) provides the relationship for local flux through the membrane(18) Ji ¼

Pi ( pi2  pi1 ) tm

(15:37)

The separation efficiency aij is based on the different rates of permeation of the gas components: aij ¼

Pi Pj

(15:38)

This data is available for some commonly separated gases and the polymer(s) used.(13,23) An experimental separation factor aij is frequently used to quantify the separation of a binary system of components i (oxygen, O2) and j (nitrogen, N2), where Cp and Cr represent molar concentrations in the permeate and retentate (non-permeate) streams, respectively.(22) The separation factor can also be defined in terms of Cp and Ci,j, i.e., concentrations in the permeate and feed streams, respectively.(18,21) These relationships can be written in terms of mole fractions yp, yr, and yf, which is often more convenient since (the oxygen) analyzers measure concentrations in mol%:(17) a0ij ¼

Cip =C jp yip =y jp ¼ Cir =C jr yir =y jr

(15:39)

a00ij ¼

Cip =C jp yip =y jp ¼ Cif =C jf yif =y jf

(15:40)

Recovery is defined by the equations below, where qp, qr, and qf represent the volumetric flow rates of the permeate, retentate (or non-permeate), and feed streams, respectively (m3/s). Volumetric flow rates of the permeate and non-permeate are measured as the difference between final and initial cumulative gas volumes for the permeate and non-permeate DV (m3) measured during time period Dt, i.e., q ¼ DV/Dt.(18) For air, Recovery of O2 ¼

qp CO2 ,p qf CO2 , f

(15:41)

Recovery of N2 ¼

qr CN2 ,r qf CN2 , f

(15:42)

The term “stage cut” is used to define the ratio of permeate flow rate to total flow rate as shown in Equation (15.43). The concentrations and volumetric flow rates are usually measured at atmospheric pressure for both the permeate and the non-permeate streams. If this were not the case, stage cut would be defined as the ratio of molar flows instead of volumetric flows(18) qp Stage cut ¼ (15:43) qp þ qr

Gas Permeation

435

The total flux of a component, Ji, may be calculated from Equation (15.44) Ji ¼

qip r nA

(15:44)

where qip ¼ volumetric flow rate of species i in the permeate (m3/s)

r ¼ density of permeate (gmol/m3) A ¼ area of membrane (m2/module) n ¼ number of modules used If values of the aforementioned terms Pi and t cannot be determined independently from experiment or the literature, an intrinsic permeability Pi is used where Pi has units of lb/ft2 . h . psi: Pi ¼

Pi Ji ¼ tm pi2  pi1

(15:45)

Note that permeate pressure is assumed to be atmospheric (0 psig) in these equations. The operating pressure should be expressed as a pressure differential (usually psi), although some references use absolute pressure on the feed side of the membrane.(22)

ILLUSTRATIVE EXAMPLE 15.15 Figure 15.9 shows a block flow diagram representing mass balances for a gas permeation system which was provided by a recent Manhattan College chemical engineering student. Note that the feed stream is air and the two product streams are a non-permeate nitrogenenriched stream and a permeate oxygen-enriched stream). Comment on Figure 15.9. SOLUTION: Overall:

As noted below, the mass balances are satisfied. F F ¼ F NP þ F P 0:104 ¼ 0:0932 þ 0:011 ¼ 0:1042

Figure 15.9

Gas permeation flow diagram.

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Chapter 15 Membrane Separation Processes

Componential O2: F O2 ,F ¼ F O2 ,NP þ F O2 ,P 0:0159 ¼ 0:0112 þ 0:00472 ¼ 0:01592 Componential N2: F N2 ,F ¼ F N2 ,NP þ F N2 ,P 0:0884 ¼ 0:0820 þ 0:00638 ¼ 0:08838 However, the oxygen mole fraction in the air feed is yO2 ¼

0:0159 ¼ 0:153 ¼ 15:3% 0:104

and the nitrogen in the feed is yN2 ¼ 1  yO2 ¼ 0:847 ¼ 84:7% This does not compare favorably with the 21%/79% makeup normally assumed for air.

B

ILLUSTRATIVE EXAMPLE 15.16 Refer to the previous example. Calculate the percentage nitrogen recovery in the non-permeate. SOLUTION: The percentage recovery of nitrogen in the non-permeate stream can be calculated using Equation (15.42). Since the molar flow rates are specified in Figure 15.9, the N2 recovery is given by   F N2 ,NP 100; F ¼ qC % Recovery N2 ¼ F N2 ,F   0:0820 ¼ 100 0:0884 ¼ 92:8% ¼ 0:928

B

ILLUSTRATIVE EXAMPLE 15.17 Refer to Illustrative Example 15.16. Calculate the percent oxygen recovery in the permeate. SOLUTION: The percentage recovery of oxygen in the permeate stream is also calculated using Equation (15.42).   FO2 ,P 100 % Recovery O2 ¼ F O2 ,F   0:00472 ¼ 100 0:0159 ¼ 29:7% ¼ 0:297

B

References

437

ILLUSTRATIVE EXAMPLE 15.18 Refer to Illustrative Example 15.5. Calculate the separation factor based on the permeate and on the feed. SOLUTION: The separation factor, based on the permeate stream, is calculated using Equation (15.39). From Figure 15.9, 0:00472 ¼ 0:429 ¼ 0:43 0:011 0:0112 ¼ 0:012 ¼ 0:0932

yO2 ,P ¼ yO2 ,NP Substituting a0O2 ,N2 ¼

yO2 ,P =yN2 ,P 0:43=(1  0:43) ¼ 5:5322 ¼ yO2 ,NP =yN2 ,NP 0:12=(1  0:12)

The separation factor based on the feed stream is calculated using Equation (15.40). For this case yO2 ,F ¼

0:0159 ¼ 0:1528 ¼ 0:15 0:104

Therefore a00O2 ,N2 ¼

yO2 ,P =yN2 ,P 0:43=(1  0:43) ¼ 4:27 ¼ yO2 , f =yN2 , f 0:15=(1  0:15)

B

REFERENCES 1. S. SLATER, “Membrane Technology,” NSF Workshop Notes, Manhattan College, Bronx, NY, 1991 (adapted with permission). 2. P. SCHWEITZER, “Handbook of Separation Techniques for Chemical Engineers,” McGraw-Hill, New York City, NY, 1979. 3. L. E. APPLEGATE, “Membrane separation processes,” Chemical Engineering, pp. 64–89, New York City, NY, June 11, 1984. 4. G. PARKINSON, “Reverse osmosis: Trying for wider applications,” Chemical Engineering, New York City, NY, pp. 26–31, New York, May 30, 1983. 5. K. W. BROOKS, “Membranes’ push into separations,” Chemical Week, pp. 21– 24, Washington DC, January 16, 1985. 6. J. FLESCHE, Manhattan College Chemical Engineering Unit Operations Report, Manhattan College, Bronx, NY, 2000. 7. J. FAMULARO, et al., “Unit Operations Laboratory Manual,” Manhattan College, Bronx, NY, 1996. 8. F. V. KOSIKOWSKI, “Membrane separations in food processing,” in “Membrane Separations in Biotechnology” (W. C. MCGREGOR, ed.), Chapter 9, Marcel Dekker, Inc., New York City, NY, 1986. 9. A. GARCIA III, B. MEDINA, N. VERHOEK, and P. MOORE, “Ice cream components prepared with ultrafiltration and reverse osmosis membranes,” Biotechnology Progress, 5, pp. 46– 50, New York City, NY, 1989.

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10. J. MAUBOIS, “Recent developments of membrane ultrafiltration in the dairy industry,” in “Ultrafiltration Membranes and Applications” (A. R. COOPER, ed.), pp. 305–318, Plenum Press, New York City, NY, 1980. 11. P. C. WANKAT, “Rate-Controlled Separations,” Chapter 12, Chapman & Hall, Boston, MA, 1990. 12. M. C. PORTER, “Handbook of Industrial Membrane Technology,” Chapter 2, Noyes Publications, Park Ridge, NJ, 1990. 13. M. MULDER, “Basic Principles of Membrane Technology,” 2nd edition, Chapters VI – VII, Kluwer Academic Publishers, Boston, MA, 1996. 14. H. C. HOLLEIN, C. S. SLATER, R. L. D’AQUINO, and A. L. WITT, “Bioseparation via cross flow membrane filtration,” Chemical Engineering Education, 29, Washington D.C., pp. 86–93, 1995. 15. C. S. SLATER and H. C. HOLLEIN, “Educational initiatives in teaching membrane technology,” Desalination, 90, pp. 291– 302, 1993. 16. S. G. REDKAR and R. H. DAVIS, “Crossflow microfiltration of yeast suspensions in tubular filters,” Biotechnology Progress, 9, pp. 625 –634, 1993. 17. C. S. SLATER, H. C. HOLLEIN, P. P. ANTONNECHIA, L. S. MAZZELLA, and J. D. PACCIONE, “Laboratory experiences in membrane separation processes,” International Journal of Engineering Education, 5, pp. 369– 378, 1989. 18. C. S. SLATER, C. VEGA, and M. BOEGEL, “Experiments in gas permeation membrane processes,” International Journal of Engineering Education, 8, pp. 1– 7, 1992. 19. C. S. SLATER, M. BOEGEL, and C. VEGA, “Membrane gas separation experiments for a chemical engineering laboratory,” ASEE Conference Proceedings, Washington DC, pp. 648 –650, 1990. 20. “Prism Separators,” Bulletin No. PERM-6-008, Permea Inc., St. Louis, MO, 1986. 21. R. A. DAVIS and O. C. SANDALL, “A membrane gas separation experiment for the undergraduate laboratory,” Chemical Engineering Education, pp. 10– 21, Winter 1990. 22. L. D. CLEMENTS, M. M. OTTEN, and P. V. BHAT, “Laboratory Membrane Gas Separator—A New Teaching Tool,” Paper No. 53b presented at the AIChE Annual Meeting, Miami Beach, FL, 1986. 23. P. C. WANKAT, Rate-Controlled Separations, Chapter 13, Chapman & Hall, Boston, MA, 1990.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

16

Phase Separation Equipment INTRODUCTION Phase separation, as its name implies, simply involves the separation of one (or more) phase(s) from another phase. Most industrial equipment used for this class of processes involve the relative motion of the two phases under the action of various external forces (e.g., gravity, electrostatic, and so on). Consider the following. The separation of solids (particulates) from a gas (usually air) is based on the movement of solid particles through the gas. The objective is often their separation/removal in order for their recovery for economic reasons (e.g., coffee beans, gold dust, and so on) and/or to comply with applicable (environmental) standards and regulations. In order to accomplish this, the particle is subjected to one or more external forces, which are large enough to separate the solid phase from the gas phase during its residence time in the phase separation device. There are basically five phase separation processes: 1 Gas– Solid (G – S) 2 Gas– Liquid (G– L) 3 Liquid – Solid (L – S) 4 Liquid – Liquid (L – L); immiscible 5 Solid – Solid (S – S) As one might suppose, the major phase separation process encountered in industry is G – S. As such, the presentation to follow primarily addresses G –S subject matter, including equipment. However, it should be noted that the basic principle of G– S phase separation processes almost universally apply to G –L processes and equipment, and also often apply (but to a lesser degree) to L– S, L – L, and S– S phase separation processes. Traditional equipment for G– S separation processes include: 1 gravity settlers 2 centrifugal separators (cyclones)

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

439

440

Chapter 16 Phase Separation Equipment

3 electrostatic precipitators 4 wet scrubbers 5 baghouses Such equipment must be matched to process variables such as flow rate, temperature, nature and concentration of solids, and desired degree of separation. It is also well recognized that no universal separation method exists that will satisfy all problems and conditions. The choice of method depends on many technical and economic factors. Each separation problem is unique; therefore, some preliminary knowledge is required in order to design compatible equipment. To acquire this knowledge, particulate/phase information must be made available. Particulate properties most fundamental to the performance and choice of the equipment are: particle size, particle-size distribution, shape, structure, density, composition, electrical conductivity, abrasiveness, corrosiveness, flammability, hygroscopic properties, flowability, toxicity, and agglomeration tendencies. Also important is knowledge of the gas stream properties, including temperature, pressure, humidity, density, viscosity, dew point for condensable components, electrical conductivity, corrosiveness, toxicity, composition, and flammability. The process conditions include the allowable pressure drop, electrical power requirements, separation efficiency requirements, particle concentration, and gas volumetric flow rate. The plant factors include maintenance, space limitations, availability of utilities, applicable safety and health protection, potential disposal facilities, and materials of construction. Finally, knowledge of auxiliary equipment is required, including pumps, fans, compressors, motors, ducting, valves, control instrumentation, storage facilities, and conveying equipment. Although most of the properties and factors considered in the selection and design of separation equipment are self-explanatory, certain introductory G – S dynamics principles merit discussion and are now reviewed. This is followed by a review of the aforementioned G – S separation equipment; the chapter concludes with a discussion of G – L, L –S, L– L, and S – S separation equipment. There is some overlap with these phase separation processes and this is noted in the presentation. One of the authors(1,2) has employed several definitions for particulates: a small, discrete mass of solid or liquid matter; a fine liquid or a solid particle that is found in the air or emissions; any solid or liquid matter that is dispersed in a gas; or, insoluble solid matter dispersed in a liquid so as to produce a heterogeneous mixture. Further, particulate matter 10 (PM10) is defined as particulate matter with a diameter less than or equal to 10 micrometers (mm) while particulate matter 2.5 (PM2.5) is particulate matter with a diameter less than or equal to 2.5 micrometers (mm). Particle size is the single most important characteristic that affects the behavior of a particle. The range in size of particles observed in practice is remarkable. Some of the particles collected are as large as raindrops. However, some of the particles created in high-temperature incinerators and metallurgical processes can consist of a few molecules clustered together. These particles cannot be seen by sensitive light microscopes because they are extremely small in size. These sizes approach

Introduction

441

those of individual gas molecules (which range from 0.2– 1.0 nm). In fact, particles composed of a few molecules clustered together can exist in a stable form. Some of these industrial processes generate particles in the range of 10– 100 nm. However, particles in this size range can grow and agglomerate to yield particles in the þ100 nm range. The overall collection/removal process for particulates in a fluid essentially consists of four steps.(3) 1 An external force (or forces) must be applied that enables the particle to develop a velocity that will displace and/or direct it to a collection or retrieval section or area or surface. 2 The particle should be retained at this area with strong enough forces so that it is not re-entrained. 3 As collected/recovered particles accumulate, they are subsequently removed. 4 The ultimate disposition of the particles completes the process. Obviously, the first is the most important step. The particle collection mechanisms discussed below are generally applicable when the fluid is air; however, they may also apply if the fluid is water, as in the G – L, L– S, and L – L presentations. The forces listed below are basically the “tools” which may be used for particulate/recovery collection: 1 Gravity settling 2 Centrifugal action 3 Inertial impaction 4 Electrostatic attraction 5 Thermophoresis and diffusiophoresis 6 Brownian motion Note that all of these collection mechanism forces are strongly dependent on particle size. The above mechanisms 2 – 4 are briefly described below in Illustrative Examples 16.1– 16.3. ILLUSTRATIVE EXAMPLE 16.1 Describe how a “centrifugal force” is employed for particle capture/recovery. SOLUTION: So-called “centrifugal force” is one of the collection mechanisms used for particle capture. The shape or curvature of the collector causes the gas stream to rotate in a spiral motion. Larger particles move toward the outside wall by virtue of their momentum. The particles lose kinetic energy there and are separated from the gas stream. Particles are then acted upon by gravitational forces and are collected. Thus, both “centrifugal” and gravitational forces may be responsible for particle collection/recovery. B

442

Chapter 16 Phase Separation Equipment

Figure 16.1 Impaction.

ILLUSTRATIVE EXAMPLE 16.2 Describe inertial impaction. SOLUTION: Inertial impaction occurs when an object (e.g., a fiber or liquid droplet), placed in the path of a particulate-laden gas stream, causes the gas to diverge and flow around it. Larger particles, however, tend to continue in a straight path because of their inertia; they may impinge on the obstacle and be collected (as in Fig. 16.1). B

ILLUSTRATIVE EXAMPLE 16.3 Describe how electrostatic attraction can impact collection/recovery of particles. SOLUTION: Another primary particle collection mechanism involves electrostatic forces. The particles can be naturally charged, or, as in most cases involving electrostatic attraction, be charged by subjecting the particle to a strong electric field. The charged particles migrate to an oppositely charged collection surface. This is the collection mechanism responsible for particle capture in an electrostatic precipitator (see a later section). The electrostatic force FE experienced by a charged particle in an electric field is given by FE ¼ qEp (consistent units) where q ¼ particle charge (not to be confused with volume flowrate) and Ep ¼ the collection field intensity (electric field). B

Other factors affecting collection mechanisms include: 1 Nonspherical particles 2 Wall effects 3 Multiparticle effects 4 Multidimensional flow Details on (1) – (4) are available in the literature.(4)

FLUID– PARTICLE DYNAMICS Most industrial techniques used for the separation of particles from gases involve the relative motion of the two phases under the action of various external forces. The

Fluid –Particle Dynamics

443

recovery methods for particulates are based on the movement of solid particles (or liquid droplets) through a gas. Whenever a difference in velocity exists between a particle and its surrounding fluid, the fluid will exert a resistive force on the particle. Either the fluid (gas) may be at rest with the particle moving through it, or the particle may be at rest with the gas flowing past it. It is generally immaterial which phase (solid or gas) is assumed to be at rest; it is the relative velocity between the two that is important. The resistive force exerted on the particle by the gas is called the drag. In treating fluid flow through pipes, a friction factor term is used in many engineering calculations.(5) An analogous factor, called the drag coefficient, is employed in drag force calculations for flow past particles. Consider a fluid flowing past a stationary solid sphere. If FD is the drag force and r is the density of the gas, the drag coefficient CD is defined as CD ¼

FD =Ap rv2 =2gc

(16:1)

where Ap is given by pdp2 =4. In the following analysis, it is assumed that 1 the particle is a rigid sphere (with a diameter dp) surrounded by gas in an infinite medium (no wall or multiparticle effects), and 2 the particle or fluid is not accelerating. From dimensional analysis, one can show that the drag coefficient is solely a function of the particle Reynolds number, Re: CD ¼ CD (Re)

(16:2)

where Re ¼

dp vr m

(16:3)

and v is the relative velocity, m is the fluid (gas) viscosity and r is the fluid (gas) density. The quantitative use of the equation of particle motion (to be developed shortly) requires numerical and/or graphical values of the drag coefficient. Graphical values are presented in Figure 16.2. The drag force, FD, exerted on a particle by a gas at low Reynolds numbers is given in equation form by FD ¼ 6pmva=gc ¼ 3pmvdp =gc

(16:4)

Equation (16.4) is known as Stokes’ law and can be derived theoretically.(5,6) However, keep in mind that Stokes’ equation is valid only for very low Reynolds numbers—up to Re  0.5; at Re ¼ 1, it predicts a value for the drag force that is nearly 10% too low. In practical applications, Stokes’ law is generally assumed applicable up to a Reynolds number of 2.0. For this “creeping flow” around a particle, one obtains CD ¼ 24=Re

(16:5)

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Chapter 16 Phase Separation Equipment

Figure 16.2 Drag coefficient for spheres.

This is the straight-line portion of the log – log plot of CD vs Re (Fig. 16.2). For higher values of the Reynolds number, it is almost impossible to perform purely theoretical calculations. However, others(1) have managed to estimate, with a considerable amount of effort, the drag and/or drag coefficient at higher Reynolds numbers. If a particle is initially at rest in a stationary gas and is then set in motion by the application of a constant external force or forces, the resulting motion occurs in two stages. The first period involves acceleration, during which time the particle velocity increases from zero to some maximum velocity. The second stage occurs when the particle achieves this maximum velocity and remains constant. During the second stage, the particle is not accelerating. The final, constant, and maximum velocity attained is defined as the terminal settling velocity of the particle. Most particles can be shown to reach their terminal settling velocity almost instantaneously.(7) For this case, one can show that v¼

fdp2 rp 18m

(16:6)

for the Stokes’ law range. Equations are also available for higher Reynolds number flow.(1,5,6) Keep in mind that f denotes the external force per unit mass of particle. One consistent set of units for the equations above is ft/s2 for f, ft for dp, lb/ft3 for r, lb/ft . s for m, and ft/s for y . Another important consideration involves the Cunningham correction factor (CCF).(8) At very low values of the Reynolds number, when particles approach sizes comparable to the mean free path of the fluid molecules, the medium can no longer be regarded as continuous. For this condition, particles can fall between the molecules at a faster rate than predicted by the aerodynamic theories that led to the previous standard drag coefficients. To allow for this “slip,” Cunningham introduced a multiplying correction factor (C ) to the velocity in Stokes’ law.

Fluid –Particle Dynamics

445

ILLUSTRATIVE EXAMPLE 16.4 A fly ash particle settles through air. Calculate the particle’s terminal velocity and determine how far it will fall in 30 s. Assume that the particle is spherical. Use the data provided below: Fly ash particle diameter ¼ 40 mm Air temperature and pressure ¼ 2388F, 1 atm Specific gravity of fly ash ¼ 2.31 SOLUTION: For the problem at hand, the particle density is calculated using the specific gravity given: rp ¼ (2:31)(62:4) ¼ 144:14 lb=ft3 The density of air is

r ¼ P(MW)=RT ¼ (1)(29)=(0:7302)(238 þ 460) ¼ 0:0569 lb=ft3 The viscosity of air is

m ¼ 0:021 cP ¼ 1:41  105 lb=ft  s For a dp of 40 mm: v¼ ¼

gdp2 rp ; 18m

g ¼ gravity force

(16:6)

(32:2)[(40)=(25,400)(12)]2 (144) (18)(1:41  105 )

¼ 0:315 ft=s The distance, D, that the fly ash particle will fall in 30 s may now be calculated, D ¼ (30)(0:315); ¼ 9:45 ft

dp ¼ 40 mm B

There are many techniques available for measuring the particle size distribution of particulates. The wide size range covered, from nanometers to millimeters, cannot always be analyzed using a single measurement principle. Added to this are the usual constraints of capital costs vs operating costs, speed of operation, degree of skill required, and most important, the end-use requirement. A common method of specifying large particle sizes is to designate the screen mesh that has an aperture corresponding to the particle diameter. This received treatment in Chapter 14. Efficiency is the other characteristic quantity which warrants further discussion. The efficiency of a particulate recovery/control device is usually expressed as the percentage of mass collected by the unit compared with that entering the unit. It may be

446

Chapter 16 Phase Separation Equipment

calculated on a particle number basis: 

 particles collected 100 EN ¼ particles entering or on a total mass basis:   (inlet loading)  (outlet loading) E¼  100 inlet loading

(16:7)

(16:8)

It is extremely important to distinguish between the two. Larger particles, which possess greater mass and are more easily recovered in a device, will contribute more significantly to the efficiency calculated on a mass or weight basis.

ILLUSTRATIVE EXAMPLE 16.5 Consider an aerosol volume that contains 100 1-mm particles and 100 100-mm particles. If the efficiency of separation is 90% for 1-mm particles and 99% for 100-mm particles, calculate EN and E. SOLUTION: On a particle count basis, 90 1-mm and 99 100-mm particles will be removed out of a total of 200. This gives a particle count efficiency of 

 189 EN ¼ 100 ¼ 94:5% 200 On a mass basis, however, if a 1-mm particle has unit mass, a 100-mm particle has 106 mass unit. The mass efficiency is then given by   90(1) þ 99(106 ) E¼ 100 ¼ 99% 100(1) þ 100(106 ) Any expression of the efficiency of a particulate recovery device is therefore of little value without a careful description of the size spectrum of particles involved. B

GAS –SOLID (G –S) EQUIPMENT As noted above, this chapter will key on this section. The following G– S equipment receive treatment: 1 Gravity (settling) settlers 2 Cyclones 3 Electrostatic precipitators 4 Venturi scrubbers 5 Baghouses

Gas –Solid (G –S) Equipment

447

Equipment description and pertinent design/predictive equations are included below.(1) Illustrative examples complement the presentation.

Gravity Settlers The gravity settler was one of the first devices used to separate particulates from gases (primarily) and other fluids. It is an expansion chamber in which the gas velocity is reduced, thus allowing the particle to settle out under the action of gravity. One primary feature of this device is that the external force causing separation of the particles from the gas stream is provided free by nature. This chamber’s use in industry, however, is generally limited to the removal of larger-sized particles, for example, .50 mm in diameter. Inertial collectors, on the other hand, depend on another effect, in addition to gravity, to lead to a successful separation process. This other mechanism is an inertial or momentum effect. It arises by changing the direction of the velocity of the gas and imparting a downward motion to the particle. From a calculational point of view, this induced particle motion is superimposed on the motion arising as a result of gravity. An elutriator is a slight modification of the gravity settler. The unit consists of one or more vertical tubes or towers through which the dust-laden gas passes upward at a given velocity. The larger particles that settle at a velocity higher than that of the rising fluid air are collected at the bottom of the tube while the smaller particles are carried out the top. In order to vary the fluid velocity, several columns of different diameters may be used in series to bring about more refined separation. For capture to occur in a gravity settler, the particle must reach a collection surface a0 b0 c0 d0 in Figure 16.3 during its residence time in the unit. Theodore(1) has shown that dp ¼ (18mq=grp BL)0:5

(16:9)

The particle diameter above represents a limiting value since particles with diameters equal to or greater than this value will reach the collection surface and particles with diameters less than this value will escape from the unit. This limiting particle diameter may ideally be thought of as the minimum diameter of a particle that will automatically

Figure 16.3

Gravity settler nomenclature.

448

Chapter 16 Phase Separation Equipment

be captured for the above conditions. This diameter is normally denoted by dp or dp(min). Collection efficiencies of 100% were used to derive the equations for dp . The collection efficiency, E, for a monodispersed aerosol (particulates of one size) can be shown to be   grp BL 2 d (16:10) E¼ 18mq p The term in brackets in Equation (16.10) is often multiplied by a dimensionless empirical factor to correlate theoretical efficiencies with experimental data. If no information is available, it is suggested that 0.5 be used. Thus, Equation (16.10) can be written   grp BL 2 d (16:11) E ¼ 0:5 18mq p The process design variables for a settling chamber consist of length (L), width (B), and height (H ). These parameters are usually chosen by the chamber manufacturer in order to remove all particles above a specified size. The chamber’s design must provide conditions for sufficient particle residence time to capture the desired particle size range. This can be accomplished by keeping the velocity of the exhaust gas through the chamber as low as possible. If the velocity is too high, dust re-entrainment will occur. However, the design velocity must not be so low as to cause the design of the chamber volume to be exorbitant. Consequently, the units are designed for gas velocities in the range 1– 10 ft/s (0.30523.05 m/s). The development in this section only provides the theoretical collection efficiency of a settling chamber for a single-sized particle. Since the gas stream entering a unit consists of a distribution of particles of various sizes, a fractional efficiency curve must be used to determine the overall collection efficiency. This is simply a curve or equation describing the collection efficiency for particles of various sizes. As noted earlier, the overall efficiency can then be calculated using E ¼ S(Ei )(wi ) where

(16:12)

E ¼ overall collection efficiency Ei ¼ fractional efficiency of a specific size particle wi ¼ mass fraction of a specific size particle (in range)

ILLUSTRATIVE EXAMPLE 16.6 A hydrochloric acid mist in air at 258C is to be collected in a gravity settler. You are requested to calculate the smallest mist droplet (spherical in shape) that will definitely be collected by the settler. Assume the acid concentration to be uniform through the inlet cross section of the unit and Stokes’ law applies. Operating data and information on the gravity settler are given below: Dimensions of gravity settler ¼ 30 ft wide, 20 ft high, 50 ft long Actual volumetric flow rate of acidic gas ¼ 50 ft3/s

Gas –Solid (G –S) Equipment

449

Specific gravity of acid ¼ 1.6 Viscosity of air ¼ 0.0185 cP ¼ 1.243  1025 lb/ft . s Density of air ¼ 0.076 lb/ft3 SOLUTION:

For the problem at hand, first determine the density of the acid mist:

rp ¼ (62:4)(1:6) ¼ 99:84 lb=ft3 Calculate the minimum particle diameter in both feet and micrometers using Equation (16.6), assuming that Stokes’ law applies:  1=2 (18)(1:243  105 )(50) dp ¼ (32:2)(99:84)(30)(50) ¼ 4:82  105 ft There are 3.048  105 mm in 1 ft. Therefore d p ¼ 14:7 mm

B

Cyclones Cyclones provide a relatively low-cost method of removing particulate matter from exhaust gas streams. Cyclones are somewhat more complicated in design than simple gravity settling systems and their removal efficiency is accordingly much higher than that of settling chambers. However, cyclones are not as efficient as electrostatic precipitators, baghouses, and venturi scrubbers, but are often installed as precleaners before these more effective devices. The basic separation principle is simple. Particles enter the device with the flowing gas; the gas stream is forced to turn but the larger particles have more momentum and cannot turn with the gas. These larger particles impact and fall down the cyclone wall to be collected in a hopper. The gas stream actually turns a number of times in a helical pattern, much like the funnel of a tornado. The repeated turnings provide many opportunities for particles to pass through the streamlines, thus hitting the cyclone wall. There are other variations in the design of cyclones. They are usually characterized by where the gas enters and exits the cyclone body (tangentially, axially, or peripherally). There are four major parts to a cyclone: the inlet, the cyclone body, the dust discharge system, and the outlet. All affect the overall efficiency of the cyclone. A type of parallel arrangement uses the axial entry cyclone. Arrangements of high-efficiency, small-diameter axial cyclones can provide increases in collection efficiency with corresponding reductions in pressure drop, space, and cost. Such a unit is defined as a multiclone. Pressure drops commonly range from 4– 6 inches (10 – 15 cm) of water. Objects moving in circular paths tend to move away from the center of their motion. The object moves outward as if a force is pushing it out. This apparent force is known as centrifugal force. Note that, in actuality, centrifugal forces do not

450

Chapter 16 Phase Separation Equipment

exist—instead, physicists refer to the centripetal forces acting on an object. As noted above, the whirling motion of the gas in a cyclone causes particulate capture. Three important parameters can be used to characterize cyclone performance: dpc ¼ cut diameter E ¼ overall collection efficiency DP ¼ pressure drop Equations involving each of these parameters are provided below. The equations should be used with caution, however, since there are strict limitations on their applicability. The cut diameter is defined as the size (diameter) of particles collected with 50% efficiency. It is a convenient way of defining efficiency for a control device since it provides information on the effectiveness for a particle size range. A frequently used expression for cut diameter is d pc

where

9mBc ¼ 2pNvi (r p  r)

!0:5 (16:13)

m ¼ viscosity, lb/ft . s (Pa . s) N ¼ effective number of turns (5 – 10 for the common cyclone) vi ¼ inlet gas velocity, ft/s (m/s) rp ¼ particle density, lb/ft3 (kg/m3) r ¼ gas density, lb/ft3 (kg/m3) Bc ¼ inlet width, ft (m)

The cut diameter, dpc or [dp]cut, is a characteristic of the recovery/control device. Figure 16.4 shows a size efficiency curve and indicates the cut diameter and the critical diameter, [dp]crit, the particle size collected at 100% efficiency. As values of [dp]crit are difficult to obtain from such curves, the cut size is often determined instead.

Figure 16.4 Typical size efficiency curve.

Gas –Solid (G –S) Equipment

451

A number of formulas exist for the calculation of the cut diameter and critical diameter. A value of N, the number of turns, must be known in order to solve Equation (16.13) for [dp]cut. Given the volumetric flow rate, inlet velocity, and dimensions of the cyclone, N can be easily calculated. Values of N can vary from 1–10, with typical values in the 4 – 5 range. A number of equations have been developed for determining the fractional cyclone efficiency Ei for a given size particle. As noted earlier, fractional efficiency is defined as the fraction of particles of a given size collected in the cyclone, compared to those of that size entering the cyclone. The most popular method of calculating cyclone fractional efficiency and overall efficiency was developed by Lapple.(9,10) Lapple first computed the ratios dp/[dp]cut, the particle diameter vs the cut diameter ratio as determined from Equation 16.13. He found that cyclone efficiency correlates in a general way with this ratio. For a typical cyclone, efficiency will increase as the ratio increases as provided by Lapple in Figure 16.5. As a universal curve for common cyclones, the preceding correlation has been found to agree reasonably well with experimental data. To calculate fractional efficiencies, the procedure presented below should be completed. Lapple Calculation Procedure dp range wt fraction dp =½dp cut Ei for each dp from experiment in range or Lapple’s method; %

wt fraction  Ei

The sum of these products in the rightmost section of the box will yield the overall efficiency. Theodore(11) provides more detailed information. The pressure drop across a cyclone is an important parameter to the purchaser of such equipment. Increased pressure drop means greater costs for power to move an exhaust gas through the device. With cyclones, an increase in pressure drop usually means that there will be an improvement in collection efficiency. One exception to

Figure 16.5

Cyclone collection efficiency vs particle size ratio.

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Chapter 16 Phase Separation Equipment

this is the use of pressure recovery devices attached to the exit tube; these reduce the pressure drop but do not adversely affect collection efficiency. For these reasons, there have been many attempts to predict pressure drops from design variables. The concept is that having such an equation, one could work back and optimize the design of new cyclones. The most popular of the empirical pressure drop equation has the form DP ¼ Kc rv2i ;

consistent units

(16:14)

where Kc ¼ a proportionality factor. If DP is measured in inches of water, Kc can vary from 0.013 – 0.024, with 0.024 the norm. Velocities for cyclones range from 20– 70 ft/s (6 – 21 m/s), although common velocities range from 50– 60 ft/s (15 – 18 m/s). At velocities greater than 80 ft/s (24 m/s), turbulence increases in the cyclone and efficiency may actually decrease. Pressure drops for single cyclones vary depending on both size and design. Common ranges are listed below: Low-efficiency cyclones 2–4 in H2 O ð5–10 cm H2 OÞ Medium-efficiency cyclones 4–6 in H2 O ð10–15 cm H2 OÞ High-efficiency cyclones 8–10 in H2 O ð20–25 cm H2 OÞ ILLUSTRATIVE EXAMPLE 16.7 An engineer is requested to determine the cut size diameter and overall collection efficiency of a cyclone given the particle size distribution of a dust from a cement kiln. Particle size distribution and other pertinent data are provided in Table 16.1. Additional data: Gas viscosity ¼ 0.02 cP Specific gravity of the particles ¼ 2.9 Inlet gas velocity to cyclone ¼ 50 ft/s Table 16.1 Particle Size Distribution Data for Illustrative Example 16.7 Average particle size in range dp, mm 1 5 10 20 30 40 50 60 .60

Weight percent 3 20 15 20 16 10 6 3 7

Gas –Solid (G –S) Equipment

453

Effective number of turns within cyclone ¼ 5 Cyclone diameter ¼ 10 ft Cyclone inlet width ¼ 2.5 ft SOLUTION: Lapple’s method provides the collection efficiency as a function of the ratio of particle diameter to cut diameter, as presented in Figure 16.5. One may also use the Theodore– De Paola equation(1,11,12) 1:0 E¼ (16:15) 1:0 þ (dpc =dp )2 in place of Figure 16.5. For the problem at hand, determine the value of rp 2 r: rp  r ffi rp ¼ (2:9)(62:4) ¼ 181 lb=ft3 Calculate the cut diameter:

 dpc ¼

Substituting,

 dpc ¼

9mBc 2pNvi ðrp  rÞ

1=2 (16:13)

(9)(0:02)(6:72  104 )(2:5) (2p)(5)(50)(181)

1=2

¼ 3:26  105 ¼ 9:94 mm Table 16.2 Results for Illustrative Example 16.7 dp, mm 1 5 10 20 30 40 50 60 .60

wi

dp/dpc

Ei, %

wiEi, %

0.03 0.20 0.15 0.20 0.16 0.10 0.06 0.03 0.07

0.10 0.5 1.0 2.0 3.0 4.0 5.0 6.0 —

0 20 50 80 90 93 95 98 100

0.0 4.0 7.5 16.0 14.4 9.3 5.7 2.94 7.0

Table 16.2 is generated using Lapple’s method. Slightly more accurate results can be obtained by employing the Theodore –De Paola equation. The overall collection efficiency is therefore E ¼ Swi Ei ¼ 0 þ 4 þ 7:5 þ 16 þ 14:4 þ 9:3 þ 5:7 þ 2:94 þ 7 ¼ 66:84% ¼ 0:6684

B

454

Chapter 16 Phase Separation Equipment

Electrostatic Precipitators Electrostatic precipitators (ESPs) are unique among gas cleaning equipment in that the forces separating the particulates from the gas stream are applied directly to the particulates themselves, and hence the energy required to effect the separation may be considerably less than for other types of gas cleaning apparatus. Gas pressure drops through the precipitator may be of the order of 1 inch of water or less as compared with pressures of up to 10– 100 inches of water for scrubbers and baghouses. This fundamental advantage of electrostatic precipitation has resulted in its widespread use in applications where large gas volumes are to be handled and high efficiencies are required for collection of small particles. Burning low-sulfur coal produces a fly ash that has a high resistivity and is difficult to collect. The problem, similar to that for the smelter dusts, is that there is insufficient sulfur trioxide in the gas, which is often corrected by conditioning the gas with this chemical. This resistivity problem is discussed below and later in this chapter. The electrostatic precipitation process consists of three fundamental steps: 1 Particle charging 2 Particle collection 3 Recovery/removal of the collected dust Particle charging in precipitators is accomplished by means of a corona, which produces ions that become attached to the particles. Generation of a corona requires the development of a highly non-uniform electric field—a condition that occurs near the wire when a high voltage is applied between the wires and collection electrodes. The electric field near the wire accelerates electrons present in the gas to velocities sufficient to cause ionization of the gas in the region near the wire. The ions produced as a result of the corona migrate toward the collection electrode, and in the process, collide with and become attached to particles suspended in the gas stream. The attachment of ions results in the buildup of an electric charge, the magnitude of which is determined by the number of attached ions. The charge on the particles in the presence of an electric field results in a force in the direction of the collection electrode. The magnitude of the force is dependent on the charge and the field. This force causes particles to be deposited on the collection electrode where they are held by a combination of mechanical, electrical, and molecular forces. Once recovered, particles can be removed by coalescing and draining in the case of liquid aerosols, or by periodic impact or rapping in the case of solid material. In the latter case, a sufficiently thick layer of dust must be collected so that it falls into the hopper or bin in coherent masses (effectively like a sheet) to prevent excessive re-entrainment of the material into the gas stream. The physical arrangement of precipitators differ depending on the type of application. Wire and cylinder electrodes are used in some applications; however, for reasons of space economy, most commercial precipitators use plates as collection electrodes. The majority of precipitators are constructed so that the charging and collection steps take place within the same region. Precipitators of this type are termed

Gas –Solid (G –S) Equipment

455

single-stage. For some applications, charging takes place in one section which is followed by a section consisting of alternately charged plates. The collecting electric field is established independently of the corona field and such precipitators are termed two-stage. Once a particle is charged, it migrates toward the grounded collection electrode. An indicator of particle movement toward the collection electrode is denoted by the symbol w and is called the particle migration velocity or drift velocity. The particle migration velocity can be calculated, but most ESPs are designed using a particle migration velocity based on field experience rather than theory. Typical particle migration velocity rates such as those listed in Table 16.3 have been published by various ESP vendors. These values can be used to estimate the collection efficiency of the ESP. Probably the best way to gain insight into the process of electrostatic precipitation is to study the relationship known as the Deutsch – Anderson equation. This equation is used to determine the collection efficiency of the precipitator under ideal conditions. The simplest form of the equation is E ¼ 1  e(wA=q) where

(16:16)

E ¼ fractional collection efficiency of the precipitator A ¼ effective collecting plate area of the precipitator, ft2 (m2) q ¼ gas flow rate through the precipitator, acfs (acms) [actual ft3/s (actual m3/s)] w ¼ migration velocity, ft/s (m/s) Table 16.3 Typical Precipitation Rate Parameters for Various Applications Particle migration velocity Application Utility fly ash Pulverized coal fly ash Pulp and paper mills Sulfuric acid mist Cement (wet process) Cement (dry process) Gypsum Smelter Open-hearth furnace Blast furnace Hot phosphorus Flash roaster Multiple hearth roaster Catalyst dust Cupola

ft/s

cm/s

0.13 –0.67 0.33 –0.44 0.21 –0.31 0.19 –0.25 0.33 –0.37 0.19 –0.23 0.52 –0.64 0.06 0.16 –0.19 0.20 –0.46 0.09 0.25 0.26 0.25 0.10 –0.12

4.0–20.4 10.1–13.4 6.4–9.5 5.8–7.62 10.1–11.3 6.4–7.0 15.8–19.5 1.8 4.9–5.8 6.1–14.0 2.7 7.6 7.9 7.6 3.0–3.7

456

Chapter 16 Phase Separation Equipment

This equation has been used extensively for many years for theoretical collection efficiency calculations. Unfortunately the equation is not scientifically valid, and there are a number of operating parameters that can cause the results to be in error by a factor of 2. The Deutsch –Anderson equation neglects three significant process variables: 1 It completely ignores the fact that dust re-entrainment may occur during the rapping process. 2 It assumes that the particle size and, consequently, the migration velocity is uniform for all particles in the gas stream. 3 It assumes that the gas flow rate is uniform everywhere across the precipitator and that particle seepage through the hopper section does not occur. Therefore, this equation should only be used for making preliminary estimates of precipitation collection efficiency. When the desired collection efficiency and gas flow rate are specified, the required collecting area can be determined from the Deutsch – Anderson equation once an appropriate precipitation rate parameter has been chosen. More recently, better correlations with field data on high-efficiency ESPs have been obtained by raising the exponential term in Equation (16.16) to a power m using existing values of w, i.e., E ¼ 1  e(wA=q)

m

(16:17)

This provides a more accurate prediction of performance at high efficiency levels, but can become too pessimistic in certain situations. Typical values of m range between 0.4– 0.7, with 0.5 as the norm. Equation (16.17) is referred to as the Matts – Ohnfeldt equation. These and many other models have been proposed to predict particulate collection in an electrostatic precipitator, and while special advantages can be argued for each depending on which critical variables they account for, the fact remains that no model exists which accounts for all the variables that describe migration velocity in all situations. This is attributable mainly to the large number of inter-related variables that exist. In the final analysis, the goal is still to determine the correct amount of collecting surface area, which usually depends on the proper selection of w. ILLUSTRATIVE EXAMPLE 16.8 A small coal-fired power plant sends 2400 acfm through its electrostatic precipitation. The particle migration velocity is known to be 0.35 ft/s. What is the collection area if the overall ESP efficiency is 99.78%? SOLUTION:

Apply the Deutsch –Anderson equation: E ¼ 1  eðAw=qÞ

Substituting, one obtains 0:9978 ¼ e(A)(0:35)=(2400=60)

(16:16)

Gas –Solid (G –S) Equipment

457

Solving for A yields A ¼ 699 ft2

B

Venturi Scrubbers As the name “scrubber” implies, wet collectors or wet scrubbers are devices that use a liquid for separating particles (or polluted gases) from a gas stream. Water sprays can be injected into the gas stream; gas can be forced to pass through sheets or films of liquid; or, the gas can move through beds of plastic spheres covered with liquid. Each of these techniques can effectively collect/remove particulate matter from process gases. They can also effectively remove gases such as HCl or SO2, but removal/ recovery conditions must be right. In addition, gas –liquid contact can bring about gas conditioning, and to a lesser extent, liquid conditioning. In many cases, the best conditions for removing particulate matter are the poorest for removing gases. In this section, emphasis will be placed on the design and application of wet scrubbers, with particular emphasis on venturi scrubbers for the recovery of particulate matter. Optimum operating conditions for particulate matter removal will also be discussed. In a typical venturi, the velocity at the throat must increase in order to make up for the decrease in area at the throat. Velocities at such a constriction can range from 200 – 800 ft/s (from 61–244 m/s). If water is introduced into the throat, the gas is forced to move at a high velocity that will shear the water droplets. Particles in the gas stream then impact onto the droplets produced. Moving a large volume of gas through a small constriction gives a high-velocity flow, but also a large pressure drop across the system. The collection efficiency for most particles increases with increased velocities (and corresponding increased pressure drops) since the water is sheared into more smaller droplets than at lower velocities. The large number of small droplets, combined with the turbulence in the throat section, provides numerous impaction targets for particle collection. A number of performance and design equations have been developed from basic particle movement principles (theory) to explain the action of wet scrubbing systems. Many of these start from firm scientific concepts, but only give qualitative results when predicting collection efficiencies or pressure drops. The interaction of particulate matter having a given particle size distribution with water droplets having another size distribution is not easy to express in quantitative terms. As a result of this complexity, experimentally determined parameters are usually required in order to perform engineering calculations. One of the more popular and widely used collection efficiency equations is that originally suggested by Johnstone(13) E ¼ 1  ekR(c) where

0:5

E ¼ efficiency, fractional c ¼ inertial impaction parameter, dimensionless

(16:18)

458

Chapter 16 Phase Separation Equipment

R ¼ liquid-to-gas ratio, gal/1000 acf or gpm/1000 acfm k ¼ correlation coefficient, the value of which depends on the system geometry and operating conditions, typically 0.1– 0.2 acf/gal The term c is given by

c¼ where

Cdp2 rp vt 9mG d0

(16:19)

dp ¼ particle diameter, ft

rp ¼ particle density, lb/ft3 vt ¼ throat velocity, ft/s mG ¼ gas viscosity, lb/ft . s d0 ¼ mean droplet diameter, ft C ¼ Cunningham correction factor (CCF), dimensionless Values for the CCF are available in the literature.(1,5,6) However, this correction is usually neglected in scrubber calculations, but the effect becomes more pronounced as the particle size decreases, particularly below 1 mm. The mean droplet diameter, d0, for standard air and water in a venturi scrubber is given by the Nukiyama – Tanasawa relationship: d0 ¼

16,400 þ 1:45R1:5 vt

(16:20)

Available data indicate that venturis often operate with pressure drops in the 30– 100 in H2O range. Liquid-to-gas ratios for venturi scrubbers are usually in the range of 5 – 20 gal/1000 ft3 of gas. At many facilities, liquid-to-gas ratios ranging from 7 – 45 gal/1000 ft3 of gas have been reported. In many cases, a minimum ratio of 7.5 gal/1000 ft3 is needed to ensure that adequate liquid is supplied to provide good gas “sweeping”. Gas velocities for venturi scrubbers are in the 100 – 400 ft/s range. The low end of this range, 100 –150 ft/s, is typical of power plant applications, while the upper end of the range has been applied to lime kilns and blast furnaces.

ILLUSTRATIVE EXAMPLE 16.9 Calculate the overall efficiency of N scrubbers in parallel, assuming that the volumetric flow rates in each scrubber are q1, q2, . . . , qN and the corresponding efficiencies are E1, E2, . . . , EN, respectively. Assume that the gas is sufficiently well mixed so that the particle concentration (particles/volume) is the same at the inlet of each scrubber. Express the result in terms of the q values and the corresponding E values (see Fig. 16.6). SOLUTION:

If c1 is entering and c10 is leaving, Ei ¼ 1 

ci0 ci

Gas –Solid (G –S) Equipment

Figure 16.6

459

N scrubbers in parallel.

or ci0 ¼ ci (1  Ei ) Thus, E ¼1 ¼1

c10 q1 þ c20 q2 þ    þ cN0 qN c1 q1 þ c2 q2 þ    þ cN qN c1 (1  E1 )q1 þ c2 (1  E2 )q2 þ    þ cN (1  EN )qN c1 q1 þ c2 q2 þ    þ cN qN

P qi ci Ei ¼ P qi ci

(16:21)

If c1 ¼ c2 ¼    ¼ cN , the equation above reduces to P qi Ei E¼ P qi P qi Ei ¼ q

(16:22) B

ILLUSTRATIVE EXAMPLE 16.10 Calculate the overall efficiency of three venturi scrubbers operating in parallel and treating 10,000 acfm of gas. Data are provided in Table 16.4.

Table 16.4 Data for Three Scrubbers in Parallel Scrubber 1 2 3 Note: gr ¼ grain

q, acfm

c, gr/ft3

E

2500 5000 2500

2.0 4.0 2.0

0.996 0.985 0.996

460

Chapter 16 Phase Separation Equipment

SOLUTION:

Employ a modified form of the Equation (16.21)    E ¼ 1  S(qi ci Pi ) S(qi ci )

(16:23)

where Pi ¼ fractional penetration ¼ 12Ei For the flow rates, Inlet ¼ (2)(2500)(2:0) þ (5000)(4:0) ¼ 30,000 gr= min Outlet ¼ (2)(2500)(2:0)(0:004) þ (5000)(4:0)(0:015) ¼ 340 gr= min Thus, E¼

30,000  340 30,000

¼ 0:9887 ¼ 98:87%

B

ILLUSTRATIVE EXAMPLE 16.11 A consulting firm has been requested to calculate the throat area of a venturi scrubber to operate at a specified collection efficiency. Pertinent data are given below. Volumetric flow rate of process gas stream ¼ 11,040 acfm (at 688F) Density of dust ¼ 187 lb/ft3 Liquid-to-gas ratio ¼ 2 gal/1000 ft3 Average particle size ¼ 3.2 mm (1.05  1025 ft) Average water droplet size ¼ 48 mm (1.575  1024 ft) Johnstone scrubber coefficient k ¼ 0.14 Required collection efficiency ¼ 98% Viscosity of gas ¼ 1.23  1025 lb/(ft . s) CCF ¼ 1.0 SOLUTION:

Calculate the inertial impaction parameter c from Johnstone’s equation: 1=2

E ¼ 1  ekRc

1=2

0:98 ¼ 1  e(0:14)(2)c

(16:18)

Solving for c, one obtains

c ¼ 195:2 From the calculated value of c above, back calculate the gas velocity v at the venturi throat:

c¼

r p vt d 2p 9d0 mG

vt ¼

9cd0 m (9)(195:2)(1:575  104 )(1:23  105 ) ¼ r p d2p (187)(1:05  105 )2

¼ 165:1 ft=s

(16:19)

Gas –Solid (G –S) Equipment

461

Calculate the throat area S using the gas velocity at the venturi throat vt : S ¼ q=vt ¼ (11,040)=[(60)(165:1)] ¼ 1:114 ft2 Note that approximately 10 ft2 of throat area is generally required to treat 10,000–20,000 acfm. B

Baghouses(19) One of the oldest, simplest, and most efficient methods for recovering solid particulate from gas streams is by filtration through fabric media. The fabric filter is capable of providing high collection efficiencies for particles as small as 0.01 mm and will remove a substantial quantity of those particles as small as 0.01 mm. In its simplest form, the industrial fabric filter consists of a woven or felted fabric through which dust-laden gases are forced. A combination of factors results in the collection of particles on the fabric filters. When woven fabrics are used, a dust cake eventually forms; this, in turn, acts predominantly as a sieving mechanism. When felted fabrics are used, this dust cake is minimal or almost non-existent and the primary filtering mechanisms are a combination of inertial forces, impingement, and so on. These are essentially the same mechanisms that are applied to particle collection on wet scrubbers, where the collection media is in the form of liquid droplets rather than solid fibers. As particles are collected, the pressure drop across the fabric filtering media increases. Owing in part to fan limitations, the filter must be cleaned at predetermined intervals. Material is removed from the fabric by gravity and/or mechanical means. The fabric filters or bags are usually tubular or flat. The structure in which the bags are located (hang) is referred to as a baghouse and the number of bags in a baghouse may vary from less than a dozen to several thousand. Quite often, when great numbers of bags are involved, the baghouse is compartmentalized so that one compartment may be cleaned while others are still in service. The basic filtration process may be conducted in many different types of fabric filters in which the physical arrangement of hardware and the method of removing collected material from the filter media will vary. The essential differences may be related, in general, to the following: 1 Type of fabric 2 Cleaning mechanism(s) 3 Equipment geometry 4 Mode of operation The number of variables necessary to design a fabric filter is very large. Since fundamentals cannot treat all of these factors in the design and/or prediction of performance of a filter, this determination is basically left up to the experience and judgment of the design engineer. In addition, there is no one formula that can determine whether a fabric filter application is feasible. A qualitative description of the filtration process is possible, although quantitatively, the theories are far less successful. Theory, coupled with some experimental data, can help predict the performance and design of the unit.

462

Chapter 16 Phase Separation Equipment

The state-of-the-art of engineering process design is the selection of filter medium, superficial velocity, and cleaning method that will yield the best economic compromise. Industry relies on certain simple guidelines and calculations, which are usually considered proprietary information, to achieve this. Despite the progress in developing pure filtration theory, and in view of the complexity of the phenomena, the most common methods of correlation are based on predicting a form of a final equation that can be verified by experiment. The gas-to-cloth (G/C) ratio is a measure of the amount of gas passed through each square foot of fabric in the baghouse. It is given in terms of the number of cubic feet of gas per minute passing through one square foot of cloth. In other words, the G/C ratio ¼ gas volume rate/cloth area. Also note that this velocity is not the actual velocity through the openings in the fabric, but rather the apparent velocity of the gas approaching the cloth. As the G/C ratio increases, pressure drop (DP) also increases. In the United States, pressure drop in baghouse applications is generally measured in inches of water (in H2O). Despite several sophisticated formulas that have been developed, there is no satisfactory set of published equations that allows a designer to accurately calculate the efficiency of a prospective baghouse. However, there are three more heuristic formulas worth mentioning that can help baghouse designers: total inlet gas volume rate total filter cloth area in collector total inlet gas volume rate þ cleaning volume rate Net gas=cloth ratio ¼ on stream cloth Gross gas=cloth ratio ¼

Units gas=cloth ratio ¼

gas volume rate ft3=min ¼ ¼ ft=min cloth area ft2

(16:24) (16:25) (16:26)

As mentioned earlier, baghouse design is still very much an artform and nowhere is this more evident than in the selection of G/C ratios. Factors influencing the G/C ratio include the cleaning method, filter media, particle size and distribution, particle density, particle loading, and other factors unique to each situation. Because of their variability, it has never been possible to satisfactorily quantify all of these factors for every application. Once the G/C ratio, the cleaning method, and the filter medium have been selected, the essence of the flange-to-flange design selection process is complete. The only major consideration remaining is the baghouse material of construction. Typically, this is mild steel. In some specialty applications, stainless steel units are employed. An equation that can be used for determining the collection efficiency of a baghouse is(14) E ¼ 1  e(cLþft) where

c ¼ constant based on fabric, ft21 f ¼ constant based on cake, s21

(16:27)

Gas –Solid (G –S) Equipment

463

t ¼ time of operation to develop the cake thickness, s L ¼ fabric thickness, ft E ¼ collection efficiency, dimensionless The exit concentration (we) for the combined resistance system (the fiber and the cake) is we ¼ wi e(cLþft) where

(16:28)

we ¼ exit concentration, lb/ft3 wi ¼ inlet concentration, lb/ft3

A variation on Darcy’s formula for the flow of fluid through a porous bed has been developed for the flow of gases through a filter medium. The basic Darcy equation can be used to predict the pressure drop for an operating fabric filter with accumulated cake: DP ¼ SE v þ K2 c1 v2 t where

(16:29)

DP ¼ pressure drop, in H2O SE ¼ effective residual drag, in H2O v ¼ velocity, fpm K2 ¼ specific cake coefficient

The effect of bag failure on baghouse efficiency can be described by the following equation developed by Theodore and Reynolds: Pt ¼ Pt þ Ptc 0:582(DP)0:5 f q f¼ 2 LD (T þ 460)0:5

Ptc ¼

where

ð16:30Þ

Pt ¼ penetration after bag failure Pt ¼ penetration before bag failure Ptc ¼ penetration correction term; contribution of broken bags to Pt DP ¼ pressure drop, in. H2O f ¼ dimensional parameter (different from f in Eq. 16.28) q ¼ volumetric flow rate of contaminated gas, acfm L ¼ number of broken bags D ¼ bag diameter, inches T ¼ temperature, 8F

Refer to the literature for a detailed development of Equation (16.30).(15,16)

464

Chapter 16 Phase Separation Equipment

ILLUSTRATIVE EXAMPLE 16.12 A baghouse has been used to clean a particulate gas stream for nearly 30 years. There are 600 8-inch diameter bags in the unit and 50,000 acfm of dirty gas at 2508F enters the baghouse with a loading of 5.0 gr/ft3. The outlet loading is 0.03 gr/ft3. Local Environmental Protection Agency (EPA) regulations state that the outlet loading should not exceed 0.40 gr/ft3. If the system operates at a pressure drop of 6.0 in H2O, how many bags can fail before the unit is out of compliance? The Theodore–Reynolds equation applies and all contaminated gas emitted through the broken bags may be assumed the same as that passing through the tube sheet thimble. SOLUTION:

Calculate the efficiency E and penetration Pt before the bag failure(s): E ¼ [(inlet loading)  (outlet loading)]=(inlet loading) ¼ (5:0  0:03)=(5:0) ¼ 0:9940 ¼ 99:40% Pt ¼ 1  0:9940 ¼ 0:0060 ¼ 0:60%

The efficiency and penetration Pt based on regulatory conditions are E ¼ (5:0  0:4)=5:0 ¼ 0:920 ¼ 92:0%  Pt ¼ 1  0:920 ¼ 0:0800 ¼ 8:00% The penetration term Ptc associated with the failed bags is then Ptc ¼ 0:0800  0:0060 ¼ 0:0740 Since Ptc ¼

0:528(DP)0:5 f

(16:30)

and

f¼

LD2 (T

q þ 460)0:5

If neither the flow nor the concentration is uniformly distributed, the general equation developed in the previous section for the efficiency in a compartmentalized baghouse is used: P ci (1  Ei )qi P E ¼1 ci qi Substituting (see data), one obtains E ¼1

(2)(3:8)(2500)(1  0:995) þ (1)(4:25)(4000)(1  0:99) 36,000

¼ 0:9926 ¼ 99:26% Note that the penetration has increased by slightly over 10%.

B

Gas–Liquid (G – L) Equipment

465

GAS –LIQUID (G– L) EQUIPMENT The same equipment employed for G – S systems (not including venturi/wet scrubbers) may also be used for G –L systems. Gravity, centrifugal action, electrostatic effects, and filtering will perform the same separation function with G– L systems. Gas – liquid phase separation is typically required following some other separation process such as absorption, evaporation, gas – liquid reactions, and condensation. For example, mist eliminators are almost always employed with absorbers (see Chapter 10). Before separation techniques or equipment can be selected, the parameters of the separation must be defined. Information on flow rates and flow ratios of the phases, and dispersed-phase particle size (droplet or bubble size) should be known or estimated. The most common G – L separation processes usually involve the liquid dispersed in a gas such as the aforementioned absorbers as well as distillation columns, evaporators, or condensers. The usual objective is to prevent environmental emissions, downstream process contamination, or more importantly to recover a valuable material. Droplet size is the key to selecting the most effective separation technique, but size is difficult to measure. It must also be measured at the point where the separation is to be made owing to coalescence of collected particles. For new systems or where information of particle size and particle size distribution are not available, predictions or data from existing units can be used. Additional information is available in the literature.(17) ILLUSTRATIVE EXAMPLE 16.13 A sodium hydroxide spray in air at 308C is to be collected in a G– L separation unit that is essentially a gravity settler. The unit is 30 ft wide, 15 ft high, and 40 ft long. The volumetric flow rate of the gas is 42 ft3/s. Calculate the smallest mist droplet (spherical in shape) that will be entirely recovered by the settler. The specific gravity of the mist droplets may be assumed to be equal to 1.21. SOLUTION:

The important property data are tabulated below:

m ¼ 0:0185 cP ¼ 1:245  105 lb=ft  s r ¼ 0:0728 lb=ft3 Assume that Stokes’ law applies. The describing equation is 18mq d p (min) ¼ gr p BL Substituting the data, one obtains " d p (min) ¼

!0:5

18(1:245  105 lb=ft  s)(42 ft3 =s)

ð32:2 ft=s2 Þ 1:21  62:4 lb=ft3 (30 ft)(40 ft)

¼ 5:68  105 ft

(16:9)

#0:5

466

Chapter 16 Phase Separation Equipment

The reader should verify that the Stokes’ law assumption is valid. The reader may also choose to note the similarities between this example and Illustrative Example 16.6. B

ILLUSTRATIVE EXAMPLE 16.14 Calculate the smallest droplet in Illustrative Example 16.13 assuming the flow rate is halved. SOLUTION: If q is half of that in the previous example, then simply divide the answer by the basis for this is provided below. !0:5 18mq1 d p (min1 ) ¼ gr p BL d p (min2 ) ¼

18mq2 gr p BL

pffiffiffi 2;

!0:5

with q2 ¼ 12 q1 Substituting above yields d p (min2 ) ¼

!0:5 !0:5 18mq1 1 18mq1 ¼ pffiffiffi 2gr p BL 2 gr p BL

Therefore, d p (min2 ) ¼ ¼

d p (min1 ) pffiffiffi 2 5:68  105 ft pffiffiffi ¼ 4:016  105 ft 2

B

ILLUSTRATIVE EXAMPLE 16.15 Two mist eliminators operate in series with droplet recovery efficiencies of 90% and 99.5%, respectively. Calculate the overall efficiency of the two units. SOLUTION: This problem is best solved by noting the overall penetration given by the product of the fractional penetration for each device where For two units, P ¼ P1 P2 ;

P ¼ 1:0  E

According to the data given P1 ¼ 1  0:9 ¼ 0:1 and P2 ¼ 1  0:995 ¼ 0:005

(16:31)

Liquid –Solid (L –S) Equipment

467

Therefore, P ¼ (0:1)(0:005) ¼ 0:0005; fractional basis ¼ 0:05; percent basis and E ¼ 1  0:0005 ¼ 0:9995 ¼ 99:95%

B

LIQUID– SOLID (L –S) EQUIPMENT Liquid – solid separation that occurs under the action of gravity can be divided into two categories: Type I and Type II. Type I settling involves dilute mixtures where one can assume no solid particle-to-solid particle interaction. Type II settling involves mixtures that are sufficiently concentrated with the solids that they effectively settle as a mass, as with bulk filtration. Additional details follow. For Type I settling, all particles settle independently; consequently, if a particle size and size distribution is known, the settling rate of an individual particle can be calculated. The L– S unit can then be designed employing the same procedures provided for G – S equipment. Type II settling behavior generally produces a rather sharp line of demarcation between the clear liquid overflow and the settling solids. For these systems, the design is primarily dictated by the thickening capability of the unit, although the design must be adequate to provide sufficient overflow area to clarify the liquid overflow. Settlers for this category of mixtures are normally referred to as thickeners. They are sometimes constructed as a rectangular basin; however, they are most often of circular cross-section. This section examines three industrial separation techniques that exploit the density difference between a liquid and a solid. The driving force in these processes is usually the result of gravity, centrifugal action, and/or buoyant effects. Unlike the earlier section that treated gas-particle separation, this presentation will address topics keying on liquid– solid separation. Filtration is treated briefly at the end of this section.

Sedimentation Gravity sedimentation is a liquid – solid process that separates, under the effect of gravity, a feed slurry into an underflow slurry of higher solids concentration and an overflow of substantially clearer liquid. A difference in density between the solids and the suspended liquid is, as indicated, a necessary prerequisite. Nearly all commercial equipment for continuous sedimentation is built with relatively simple settling tanks. Distinction is commonly made depending on the purpose of the separation. If the clarity of the overflow is of primary importance, the process is called clarification and the feed slurry is usually dilute. If a thick underflow is the primary aim, the process is called thickening and the feed slurry is usually more concentrated.(18) The most commonly used thickener is the circular-basin type (shown in Fig. 16.7). The flocculant-treated feedstream enters the central feed well, which

468

Chapter 16 Phase Separation Equipment

Figure 16.7 The circular-basin continuous thickener.

dissipates the stream’s kinetic energy and disperses it gently into the thickener. The feed finds its height in the basin where its density matches the density of the suspension inside (the concentration increases downward in an operating thickener, giving stability to the process) and spreads out at that level. The settling solids move downward as does some liquid, with the liquid amount being determined by the underflow withdrawal rate. Most of the liquid goes upward and into the overflow. Typically, a thickener has three operating zones: clarification, zone settling, and compression (see Fig. 16.7).(18) Gravity clarifiers sometimes resemble circular thickeners but more often are rectangular basins with the feed at one end and overflow at the other. Settled solids are pushed to a discharge trench by paddles or blades on a chain mechanism. Flocculent may be added prior to the clarifier. Conventional thickeners are also used for clarification; however, the typically low feed concentrations hinder the benefits of zone settling and so the basin area is based on clarification-zone demands.(18) Thus, in a general sense, sedimentation is the process of removing solid particles heavier than water by gravity settling. It is the oldest and most widely used unit operation in water and wastewater treatment. Note that the terms sedimentation, settling, and clarification are often used interchangeably; the unit sedimentation basin may also be referred to as a sedimentation tank, clarifier, settling basin, or settling tank. In wastewater treatment, sedimentation is used to remove both inorganic and organic materials that are settleable in continuous flow conditions. It removes grit,

Liquid –Solid (L –S) Equipment

469

particulate matter in the primary settling tank, and chemical flocculants from a chemical precipitation unit. Sedimentation is also used for solids concentration in sludge thickeners.(19) Details on discrete, or individual, particle settling was presented earlier in this chapter. Here, Stokes’ law assumes that particles are present in relatively dilute concentrations and in a fluid medium of relatively large cross-section. When there are a large number of particles, the particles in close proximity will retard other particles. This is termed hindered settling. The effect is not significant at volumetric concentrations below 0.1%. However, when the particle diameter becomes appreciable with respect to the diameter of the container in which it is settling, the container walls will exert an additional retarding effect known as the wall effect. Farag(20) has reviewed and developed equations for the above situation. For hindered flow, the settling velocity is less than would be calculated from Stokes’ law. The density (employing SI units) of the fluid phase becomes the bulk density of the slurry, rm, which is defined as follows:

rm ¼ 1f rf þ (1  1f )rs

(16:32) 3

where rm is the density of slurry in kg (solid plus liquid)/m and 1f is the volume fraction of the liquid in the slurry mixture. The density difference is then:

rp  rm ¼ rp  [1f rf þ (1  1f )rp ] ¼ 1f (rp  rf )

(16:33)

The effective viscosity of the mixture, mm, is defined as:

mm ¼ mf b

(16:34)

where mf is the pure fluid viscosity and b is a dimensionless correction factor that is a function of 1f. The term b can be evaluated from Equation (16.35) b ¼ 101:82(11f ) log10 b ¼ 1:82(1  1f )

(16:35)

The settling velocity, v (with respect to the unit) is 1f times the velocity calculated by Stokes’ law. Substituting Equations (16.33) and (16.34) into Stokes’ law terminal velocity equation and multiplying by 1f for the relative velocity effect leads to: v¼

gdp2 (rp  rf ) 1f2 b 18mf

(16:36)

The Reynolds number, based on the settling velocity relative to the fluid is then: Re ¼

rm vdp gdp2 (rp  r)rm 1f ¼ mm 1f b2 18mf2

(16:37)

When the Reynolds number is less than 1, the settling is in the Stokes’ law regime. When the diameter, dp, of the particle becomes appreciable with respect to the diameter of the container, Dw, the terminal velocity is reduced. This is termed the aforementioned wall effect. In the case of settling in the Stokes’ law regime, the calculated terminal velocity is multiplied by a correction factor, kw.(20)

470

Chapter 16 Phase Separation Equipment

ILLUSTRATIVE EXAMPLE 16.16 Glass spheres are settling in water at 208C. The slurry contains 60 wt% solids and the particle diameter is 0.1554 mm. The glass density is 2467 kg/m3. Find the volume fraction, 1f, of the liquid and the bulk density of the slurry, rm. For water at 208C, rf ¼ 998 kg/m3 and mf ¼ 0.001 kg/m . s. SOLUTION: Start by assuming a basis of 100 kg of slurry. To determine the volume fraction of the liquid, divide the mass of liquid by its density. Since 100 kg of slurry is the basis and the slurry contains 60 wt% solid, mf ¼ 40 kg The volume of the fluid (water) is Vf ¼

mf 40 ¼ 0:040 m3 ¼ 998 rf

Similarly, mp ¼ 60 kg Vp ¼

mp 60 ¼ rp 2467 ¼ 0:0243 m3

Therefore, V ¼ Vf þ Vp ¼ 0:040 þ 0:0243 ¼ 0:0643 m3 and 1f ¼

Vf 0:040 ¼ 0:643 V ¼ 0:622

and 1p ¼ 1  1f ¼ 0:378 Calculate rm from Equation (16.32).

rm ¼ 1f rf þ 1p rp Substituting,

rm ¼ 0:622(998) þ 0:378(2467) ¼ 1553 kg=m3

ILLUSTRATIVE EXAMPLE 16.17 Refer to Illustrative Example (16.16). Calculate the Reynolds number.

B

Liquid –Solid (L –S) Equipment SOLUTION:

471

Employ Equation (16.37) to obtain: Re ¼

rm vdp mm 1f

Substituting, Re ¼

(1553)(0:00153)(1:554  104 ) (0:0049)(0:622)

¼ 0:121 B

Centrifugation Centrifugal force is widely used when a force greater than that of gravity is desired for the separation of solids and fluids of different densities, as in settling or for other forms of separation. Two terms need to be defined. A centrifugal force is created by moving a mass in a curved path and is “exerted” in the direction away from the center of curvature of the path. As noted earlier, centripetal force is the force applied to the moving mass in the direction toward the center of curvature that causes the mass to travel in a curved path. Centrifugation is therefore another process that uses density differences to separate solids from liquids (or an immiscible liquid from other liquids). The feed is subjected to forces that make the solids move radially through the liquid (outward if heavier, inward if lighter). In a sense, centrifugation is an extension of gravity sedimentation to particle sizes and to emulsions that are normally stable in a gravity field. The describing equations developed in an earlier section for particle motion again apply. The gravity force is replaced by a centripetal force, Fc (force/mass) Fc ¼

r v2 v2 ¼ t gc gc r

(16:38)

where r is the radius of curvature of the particle or heavier phase, v the angular velocity and vt the tangential velocity at the point in question. As indicated above, centrifugation attempts to increase the particle “settling” velocity many times higher than that of gravity by applying a centripetal force. The ratio of these two forces has been defined by some as the number of “Gs,” where G¼

r v2 g

(16:39)

Note that the g term in the denominator in Equation (16.39) is the acceleration due to gravity and not the term gc (the conversion constant) that appears in Equation (16.38). There are two main classes of centrifugation equipment: cyclones and centrifuges. Cyclones are generally classified as air particulate control/recovery equipment; this unit is primarily used for separating gas – solid systems. Centrifuges are primarily employed for liquid– solid separation; units in this category include basket, tubular,

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Chapter 16 Phase Separation Equipment

scroll-type, dish, and multiple chamber. The units may operate in the batch or continuous mode. Details on this equipment are available in the literature.(20–22) ILLUSTRATIVE EXAMPLE 16.18 A particle is spinning in a 3-inch ID (inside diameter) centrifuge with an angular velocity of 30 rad/s. Calculate the number of Gs for the particle. SOLUTION:

Employ Equation (16.39). G¼

r v2 g

Substituting yields: (3=12)(30)2 ¼ 7:0 32:2

B

Flotation Flotation processes are useful for the separation of a variety of species, ranging from molecular and ionic to microorganisms and mineral fines, from one another for the purpose of extraction of valuable products, as well as cleaning of waste waters. They are particularly attractive for separation problems involving very dilute solutions where most other processes usually fail. The success of flotation processes is primarily dependent on the tendency of surface-active species to concentrate at the water – fluid interface and on their capability to make selected non-surfaceactive materials hydrophobic by means of adsorption on them or association with them. Under practical conditions, the amount of interfacial area available for such concentration is increased by generating air bubbles or oil droplets in the aqueous solution.(22) Flotation is a gravity separation process based on the attachment of air or gas bubbles to solid (or liquid) particles that are then carried to the liquid surface where they accumulate as float and can be skimmed off. The process consists of two stages: the production of suitably small bubbles and their attachment to the particles. Depending on the method of bubble production, flotation is classified as dissolve-air, electrolytic, or dispersed-air, with dissolve-air primarily employed by industry.(18) The separation of solid particles into several fractions based upon their terminal velocities is called hydraulic classification. If there are particles of materials of different densities but of the same size, these may also be separated by the method of hydraulic separation or classification, i.e., by placing the particles of different densities in an upward-flowing stream of fluid (the fluid is often water). If the velocity of the water is adjusted so that it lies between the terminal falling velocities (or settling velocities) of the two particles, the slower particles will be carried upward and the particles of higher terminal velocity than the water velocity will move downward, and a separation is thereby attained.

Liquid –Solid (L –S) Equipment

473

ILLUSTRATIVE EXAMPLE 16.19 It is desired to separate quartz particles (specific gravity ¼ 2.65, with a size range of 40– 90 mm) from galena particles (specific gravity ¼ 7.5, with a similar size range of 40 –90 mm). The mixture is placed in a rising water flow (density, r ¼ 1000 kg/m3; viscosity, m ¼ 0.001 kg/m . s). See Figure 16.8. Calculate the water velocity to obtain pure galena. Will this pure galena be a top or bottom product? SOLUTION: Calculate the settling velocity of the largest quartz particle with a diameter dp ¼ 90 mm. Assume Stokes’ law flow regime applies. Therefore, from Equation (16.6), with the fluid density term rf retained,

vq ¼

gdp2 (rs  rf ) 9:807(9  105 )2 (2650  1000) ¼ 18(0:001) 18mf ¼ 0:0073 m=s

Calculating the settling velocity of the smallest galena particle with a diameter dp ¼ 4  1025 m,

vg ¼

Figure 16.8

gdp2 (rs  rf ) 9:807(4  105 )2 (7500  1000) ¼ 0:00567 m=s ¼ 18(0:001) 18mf

Particle separator.

474

Chapter 16 Phase Separation Equipment

To obtain pure galena, the upward velocity of the water must be equal to or greater than the settling velocity of the largest quartz particle. Therefore, vw ¼ 0:0073 m=s ¼ 7:3 mm=s Since the water velocity of 7.3 mm/s is greater than the settling velocity of the smallest galena particle, some galena will be washed up with the quartz. One may conclude that pure galena will be the bottom product; the top product will be the quartz plus some galena. B

ILLUSTRATIVE EXAMPLE 16.20 Refer to Illustrative Example 16.19. Determine the size range of the galena in the top product. SOLUTION: To determine the size range of the galena product, calculate the galena particle size that has a settling velocity of 7.3 mm/s. Assume Stokes’ law applies gdp2 (rs  rf ) 18m sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 18mf v 18(0:001)(0:0073) ¼ 4:54  105 m ¼ 45:4 mm ¼ dp ¼ 9:807(7500  1000) g(rs  rf ) v¼

(16:6)

The size ranges for the galena are therefore 40 –45.4 mm for the top washed product.

B

Units other than the traditional settler or L– S variety also find application. Dissolved air flotation is another phase separation process used to remove organic or inorganic solids suspended in feed streams such as waste-waters or slurries. The process effectively concentrates the feed by forming a sludge to remove valuable or toxic solids. Since chemical reagents are sometimes used to remove specific inorganic solids, the process can be chemical as well as physical. Flocculation and precipitation are physicochemical processes. In flocculation, fine suspended particles, which are difficult to settle out of the liquid, are brought together by flocculating chemicals to form larger, more easily collected particles. In precipitation, metals soluble in the liquid are made insoluble by use of precipitating chemicals. The solids removed by these two processes form a sludge that may be disposed of or treated to recover any valuable material. The operating costs for both processes are dependent on the amount and type of flocculating and precipitating chemicals used. Filtration is a popular liquid – solid separation process commonly used in industry. In wastewater treatment, it is used to purify the liquid by removing suspended solids. This is usually followed by flocculation or sedimentation for further solids removal. In sludge treatment, it is used to remove the liquid (sludge dewatering) and concentrate the solid, thereby reducing the sludge volume. This method is highly competitive with other sludge dewatering processes. Filtration may also be used in treating non-aqueous liquid streams.(5)

Liquid–Liquid (L – L) Equipment

475

In the filtration process, a liquid containing suspended solids is passed through a porous medium. The solids are trapped against the medium and the separation of solids from the liquid results. A filter press is the unit of choice in many applications, particularly if a Type II L– S system is to be treated. For large solid particles, a thick barrier such as sand may be used. This is known as granular media filtration. For smaller particles, a fine filter such as a filter cloth may be used. Fluid passage may be induced by gravity, positive pressure, or a vacuum. The filter is cleaned and the solids collected by passing a stream of liquid (often water) in the opposite direction of the stream flow; this is known as backwashing. Additional details are available in the literature.(5) Finally, the size and handling characteristics of the settled solids also have an impact on separator selection, design, and performance. The desired clarity of the liquid and the dryness of the solids are important additional selection criteria. For applications requiring the driest solids, a filtering centrifuge or filter press are the logical choices. Regarding equipment selection, it should be noted that for almost every application, more than one type of unit will accomplish the job. The selection procedure usually involves a study of the characteristics of the solids, process requirements, and consultation with several equipment manufacturers.

LIQUID– LIQUID (L –L) EQUIPMENT As noted earlier, the applicable theory, equipment, and operation of most phase separation processes are very similar regardless of whether the individual phases are gas, liquid, or solid. However, the separation of immiscible liquid phases (L – L) is different when compared to the others in that the important density difference between the phases is usually very small. Gas – solid (G – S), liquid –solid (L – S), or liquid –gas (L – G) separations typically have significant phase density differences. The separation of phases with a small density difference almost always requires large equipment, application of large forces, unique equipment geometries, or all of the above. As with G – S, L– S, and L– G separation processes, the common mechanisms are gravity, centripetal force, impaction, and electrostatic effects. The design for L– L separations is affected by temperature, pressure, presence of contaminants such as surfactants, and stream mixing effects. In order to design an immiscible liquid separation unit, the average drop size and droplet size range of the dispersed phase should be specified or approximated. The most common unit employed is a gravity settler (often referred to as a decanter) and may be operated in a batch or the more popular continuous mode. The unit may be designed vertically or horizontally. The vertical unit is rarely employed in practice. The horizontal unit is the usual choice for L– L separation with long vessels the most desirable geometry. In these units, the continuous phase (it may be either the lighter or heavier phase) flows horizontally and perpendicular to the settling phase. Turbulence and turbulent mixing is a problem, but it can be reduced or eliminated by reducing the superficial velocity of the continuous phase. The droplet settling

476

Chapter 16 Phase Separation Equipment

Figure 16.9 A horizontal decanter.

velocity is estimated from any of the equations provided in the Fluid – Particle Dynamics section at the beginning of this chapter but it is usually safe to assume Stokes’ law applies. A basic decanter layout is shown in Figure 16.9. The proper introduction of feed to a gravity separation is one of the keys to its performance. For optimum design, the feed should be: 1 introduced uniformly across the active cross-section of the decanter, and 2 accomplished in a manner so as to leave no residual jetting or turbulence. Forcing a direction change on the inlet flow can reduce the jet effect; a simple approach is to use a plate just inside the inlet at least two times the nozzle diameter and located one-half nozzle diameter in from the inlet. Four important decanter design factors include the following: 1 the velocities of both phases should be approximately equal 2 the feed mixture velocity should be below 2 ft/s 3 the flow for both phases should be laminar 4 the discharge velocity should not exceed ten times the superficial velocity in the separator The time for separation to occur (the “settling” time, either up or down) is obviously critical to the design, operation, and performance of the decanter. The aforementioned difficulty in accurately specifying droplet size(s) makes this calculation almost impossible. However, if this information is required, a conservative droplet size of 100 mm (microns) is recommended. As noted above, the droplet “settling” velocity may then be estimated from any of the equations provided in the Fluid – Particle Dynamics section at the beginning of this chapter, and one can usually safely assume Stokes’ law applies.

ILLUSTRATIVE EXAMPLE 16.21 Refer to Figure 16.9. If the volumetric feed rate is 26.7 ft3/h, calculate the length, L, of a decanter, 1.2 ft in diameter, to ensure an average residence time of 18 min.

Solid –Solid (S –S) Equipment SOLUTION:

477

The volume, V, of the decanter is  2 pD L V¼ 4 ¼ (0:785)(1:2)2 L ¼ 1:13L

The average residence time, t, is t¼ ¼

V q 1:13L 26:7

¼ 0:0423L; h Set t equal to the required residence time. (18=60) ¼ 0:0423L L ¼ 7:09 ft

B

SOLID –SOLID (S– S) EQUIPMENT This type of phase separation finds limited application in practice. There are two units that have received attention: the high-gradient magnetic separation (HGMS) device and solidification. Each are briefly discussed below.

High-Gradient Magnetic Separation High-gradient magnetic separation (HGMS) is a phase and component separation process still in the development stage. It could be used to remove magnetic or nonmagnetic materials from a variety of feeds such as slurries, sludges, and solids plus aqueous and non-aqueous liquids. Its potential use lies in the removal of ferromagnetic and paramagnetic particulates. It is being used on a small scale in wastewater treatment and coal desulfurization. The process works well when magnetic components make up only a small percentage of the total concentration, and solids make up 10 –15% of the total waste stream. The HGMS process uses a magnetized ferromagnetic filter to separate magnetic or paramagnetic particles from nonmagnetic particles. The filter captures the magnetic material, which is recovered when the filter is cleaned. The filter can also be used to remove nonmagnetic material by treating the feed with a magnetic seed. Capital investment may be low when only ferromagnetic material is removed but higher for the removal of nonmagnetic material. High-volume applications make the process more economically attractive.

Solidification Solidification is a process that transforms a feed into a solid product by fixation or encapsulation. In fixation, a chemical or physical process and a solidifying agent

478

Chapter 16 Phase Separation Equipment

are used to solidify the feed. Encapsulation is a process in which the feed is surrounded by a binder after it has been solidified by a chemical agent. Both produce a durable and impermeable product. Solidification has only recently become a popular form of feed treatment. Previously, it had been used solely for treating radioactive wastes because of its high cost compared to other disposal means. Stricter regulations have made solidification a more popular choice of feed treatment. Solidification processes may be separated into five general categories depending on the solidifying agent used: 1 silicate- and cement-based 2 lime-based 3 thermoplastic-based 4 organic polymer-based 5 encapsulation techniques All five of these processes are used for feed treatment. The cement- and lime-based methods are also used for sludges from stack-gas scrubbers after the sludges are chemically treated to precipitate out the metals. Solidification is unsuitable for treating organics, oxides, and toxic anions because these materials are difficult to solidify. Solidification has been used in treating feeds produced from steel mills, plating and lead-smelting plants, food production sludges, and sulfur residues. The silicate- and cement-based methods may be operated as batch or continuous processes. Portland cement and other additives are added to either a wet or sludge-like stream forming an impermeable, rock-like solid. The solidified product may then be used for land reclamation. The degree of solidification and strength of the product depends on the concentration of metals and organics present in the feed. This particular solidification process is the one most used in the United States. Some of its advantages are that the additives are not expensive, the process is well developed, equipment is available, and end product strength and permeability are easily controlled by the amount of cement used. The disadvantages are that feed weight and bulk are increased by the cement, and pretreatment or special additives may be needed to ensure proper solidification. The lime-based process is both chemical and physical. Lime, water, and siliceous material are added to the feed. A product known as pozzolanic cement is formed. Additives may be added to increase the strength of the end product. The degree of solidification depends on the reaction of the lime with the other components forming the cement. Once the product is formed, it is often used for landfill, mine reclamation fill, or capping material. The advantages and disadvantages of this process are similar to those of the cement-based process. Thermoplastic solidification is a two-step process. Initially, the feed, which is first dried, is combined with paraffins, bitumen, and polyethylene at a temperature above 2128F. Upon cooling, the mixture begins to solidify. In the second step, the solid waste is thermoplastically coated and then ready for disposal. The thermoplasticized product is fairly permeable to most aqueous solutions. In addition to toxic inorganics, the process is also used in treating nuclear wastes. The equipment used in

References

479

thermoplastic solidification is more specialized and therefore more costly than the equipment used in the cement- or lime-based processes. On the negative side, thermoplastics are flammable and the feed must be dried before processing takes place. In the organic polymer process, a monomer and a catalyst are combined with the feed and the polymer is allowed to form. The product is then containerized and ready for disposal. One popular technique is the urea – formaldehyde process that is used in treating nuclear wastes. Two advantages of the organic polymer process are that it may be used on wet or dry streams and that the end product weighs less than those produced from the cement-, lime-, or thermoplastic-based techniques. The disadvantages are that some organic polymers are biodegradable, the end product must be containerized, and the end product may not be as durable as the end products produced from the other methods. The encapsulation process is somewhat similar to the thermoplastic process. The dried feed is first chemically treated and then coated with a binder, usually polyethylene. The advantages of this process are that the encapsulated feed is very durable and resistant to water and deterioration, and the final product does not need to be stored in containers. The disadvantages are that the process is expensive, only small volumes of feed can be treated, and the sludge must be dried before processing. The various solidification treatments may be performed on-site, which requires a large investment, or at regional facilities, which requires transportation costs. As with any other option, the choice and final cost of the solidification process will depend upon the type of feed involved, the desired end product, and the investment capabilities of the user.(23) Finally, the separation of particle sizes using the methods described earlier can also be employed to separate particles of different (chemical) composition. This separation technique can only be employed if the solids in question have different densities but are of the same size.

REFERENCES 1. L. THEODORE, “Air Pollution Control Equipment Calculations,” John Wiley & Sons, Hoboken, NJ, 2008. 2. L. THEODORE: personal notes, 1982. 3. L. THEODORE: personal notes, 1976. 4. L. THEODORE, “Nanotechnology: Basic Calculations for Engineers and Scientists,” John Wiley & Sons, Hoboken, NJ, 2006. 5. J. ABULENCIA and L. THEODORE, “Fluid Flow for the Practicing Engineer,” John Wiley & Sons, Hoboken, NJ, 2009. 6. L. THEODORE: personal notes, 1960. 7. L. THEODORE: personal notes, 1966. 8. C. E. CUNNINGHAM, Proc. Roy. Soc. London, Ser. A, 83, 357, 1910. 9. C. LAPPLE, “Air Pollution Engineering Manual,” USEPA, AP-40, 94–99, 1951. 10. C. LAPPLE, Chem. Eng., New York City, NY, 58, 144, 1951. 11. L. THEODORE, “Engineering calculations: cyclone collection systems,” CEP, New York City, NY, Sept. 18–20, 2005. 12. L. THEODORE and V. DEPAOLA, “Predicting cyclone performance,” J. Air Poll. Control Assoc., Pittsburg, PA, 30, 1132–3, 1980.

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13. H. JOHNSTONE et al., Ist. Chem. Eng., New York City, NY, 46, 1601, 1954. 14. L. THEODORE: personal notes, 1981. 15. L. THEODORE and J. REYNOLDS, “Effect of bag failure on baghouse outlet loading,” J. Air Poll. Control Assoc., Pittsburg, PA, 870–872, August 1979. 16. L. THEODORE, “Engineering calculations: baghouse specification and operation simplified,” CEP, New York City, NY, 22, June, 2000. 17. D. GREEN and R. PERRY, “Perry’s Chemical Engineers’ Handbook,” 8th edition, McGraw-Hill, New York City, NY, 2008. 18. L. SVAROUSKY, “Sedimentation, centrifugation and flotation,” Chem. Eng., New York City, NY, July 16, 1979. 19. S. LIN, “Water and Wastewater Evaluations Manual,” McGraw-Hill, New York City, NY, 2001. 20. I. FARAG, “Fluid Flow,” A Theodore Tutorial, Theodore Tutorials, East Williston, NY, 1996. 21. J. REYNOLDS, J. JERIS, and L. THEODORE, “Handbook of Chemical and Environmental Engineering Calculations,” John Wiley & Sons, Hoboken, NJ, 2002. 22. R. ROUSSEAU (ed), “Handbook of Separation Process Technology,” John Wiley & Sons, Hoboken, NJ, 1987. 23. R. B. POJASEK, “Solid-waste disposal: solidification,” Chem. Eng., New York City, NY, 140 –145, August 13, 1979.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

96 Part Two Applications: Component and Phase Separation Processes the condensation of any vapor from a noncondensable gas. In crystallization (Chapter 14), a solution of dissolved solids is supersaturated, and allowed to crystallize out the excess solute, thereby forming a more pure crystalline solid. Membrane separation processes (Chapter 15) utilize a semi-permeable physical barrier to achieve a separation. Finally, phase separation processes receive treatment in Chapter 16. Problems for each chapter can be found online at www.wiley.com; follow links to this title.

Part Three

Other Topics This third and final part of the book provides an overview of six topics that deserve coverage in a book concerned with mass transfer. The motivation for their inclusion in this text should be readily apparent. Five of the six chapters deal with topics that the Accreditation Board for Engineering and Technology (ABET) have recently indicated should be included in any engineering curriculum. Topics covered include: 1 Other and Novel Separation Processes (Chapter 17) 2 Economics and Finance (Chapter 18) 3 Numerical Methods (Chapter 19) 4 Open-Ended Problems (Chapter 20) 5 Ethics (Chapter 21) 6 Environmental Management and Safety Issues (Chapter 22) Some introductory details on these topics are provided below. Chapter 17 reviews other mass-transfer processes not presented in Part II, including some novel ones. Topics include: freeze crystallization, ion exchange, liquid ion exchange, resin adsorption, evaporation, foam fractionation, dissociation extraction, electrophoresis, and vibrating screens. Chapter 18 is concerned with economics and finance. Economics and finance ultimately dictate many of the decisions made by practicing engineers and their companies. For example, a company may decide that because of the rising price of the feedstock to a distillation column, they will explore the possibility of producing the raw material from a cheaper raw material. A decision will then be based on whether it makes sense economically in the short- and long-term. Furthermore, economic evaluations are a major part of process and plant design. Chapter 19 is concerned with numerical methods. This subject was taught in the past as a means of providing engineers and scientists with ways to solve complicated mathematical expressions that they could not otherwise solve. However, with the advent of computers, these solutions are now readily obtained. A brief overview of numerical methods is given to provide the practicing engineer with some insight into what many of the currently used software packages (MathCad, Mathematica, MatLab, etc.) are actually doing. Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

481

482 Part Three Other Topics Chapter 20 introduces the reader to open-ended questions. Engineers, at their very essence, are problem solvers. Most problems in real life do not come fully defined with a prescribed methodology to obtain the solution. Thus, engineers need to unleash their creativity in order to obtain a solution. Open-ended questions are exercises in using brain power—and like any muscle, you use it or you lose it. Chapter 21 is concerned with ethics. For engineers, the question of ethics typically boils down to the struggle between the responsibility to ensure public health and safety and the responsibility to the employers, clients, and shareholders of the company. The case study approach is employed to make the reader think about ethical questions, to reflect on their past decisions, and to project forward to their future decisions a higher degree of thought and insight in determining when one is faced with an ethical dilemma. The final chapter is concerned with environmental management and safety issues. This chapter contains a broad discussion of environmental issues facing today’s engineers and presents some of the more recent technology to deal with the issues at hand. Since practicing engineers have an obligation to make safety a priority (accidents do happen), the chapter also deals with ways to ensure both employee and public safety, determination of the severity of accidents, and determining the causes and potential causes of accidents.

Chapter

17

Other and Novel Separation Processes As noted in Part Two, separation processes may be separated into two categories: phase separation and component separation processes; in the latter, a particular species is separated from a single-phase, multicomponent system. The treatments may fall into one or both of these categories. Component separation processes remove particular ionic or molecular species, generally without the use of chemicals. Phase separation processes are employed to separate phases, reduce feed volume, and to concentrate the feed into one phase before further treatment and material recovery are performed. Such feed streams, which can include slurries, sludges, and emulsions that contain more than one phase, are the usual candidates for this category. Many factors need to be considered when selecting a particular type of treatment. These include: the characteristics of the feed stream and the desired characteristics of the output stream, the technical feasibility of the different physical treatments when applied to a particular case, and economic, environmental, and energy considerations. In this chapter, the less established types of separation treatment methods used today are examined. Part Two of this book provided information, sometimes in significant detail, on the major mass transfer equipment and processes employed by industry. However, there are others involving either component or phase separation, or both, that did not receive any treatment but deserve some consideration. These topics include 1 Freeze crystallization 2 Ion exchange 3 Liquid ion exchange 4 Resin adsorption 5 Evaporation 6 Foam fractionation 7 Dissociation extraction Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

483

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Chapter 17 Other and Novel Separation Processes

8 Electrophoresis 9 Vibrating screens A brief review(1) of each of the above is provided in this chapter. Additional details are available in the literature where noted.

FREEZE CRYSTALLIZATION Freeze crystallization is a phase separation process that is in the process of becoming fully developed or achieving its full potential for significant commercial use. Most of the development has been for use in instant coffee and water desalination. On a laboratory scale, this process has been used in the treatment of feeds containing ammonium nitrate, paper mill bleach solutions, plating liquors, and arsenal redwater. The freeze crystallization process also has potential use in the recovery/removal of 1 –10% total dissolved solids (TDS) in aqueous streams. Contaminated wastewater can be subcooled during freeze crystallization in order to form purified ice crystals. The remaining liquid is more concentrated in soluble organic or inorganic materials. The crystals are subsequently removed, washed, and melted to recover the water. The liquid could be freeze crystallized again to further concentrate the feed and make treatment and/or disposal much easier.

ION EXCHANGE Ion exchange is generally used for the removal of dilute concentrations of heavy metals and anions from aqueous streams. Though the process is fully developed, it is not commonly used in industry. It has been used in recovering effluents from fertilizer manufacturing, the deionization of water, treating electroplating waste waters, and looks promising for the removal of cyanides and selected heavy metals from streams. Ion exchange is a two-step process. First, a solid material, the ion exchanger, collects specific ions after coming into contact with the aqueous stream. The exchanger is then exposed to another aqueous solution of a different composition that picks up the ions originally removed by the exchanger. The process is usually accomplished by sending the two aqueous streams through one or more fixed beds of exchangers. The ion-rich product stream may be recovered.

LIQUID ION EXCHANGE Liquid ion exchange (LIE) serves the same purpose as ion exchange, the removal of heavy metals and anions from aqueous streams. However, LIE may be used for treating a higher concentration of metals and ions. This process is also well-developed

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though not widely used in waste treatment; it has promising uses in the removal of cyanide from waste waters and in treating hydroxide slimes produced in electroplating. Liquid ion exchange uses an organic stream to carry out the transfer of ions from the aqueous stream to a second aqueous stream of different composition. The organic stream is immiscible in both aqueous streams. It removes the ionic or inorganic material from the feed stream and then passes it on to the second aqueous stream. The processing equipment is similar to that of liquid – liquid extraction (see Chapter 12). The ion-rich stream is usually treated further for recovery. The purified aqueous stream usually contains a dilute concentration of organic solvent and may therefore require further treatment. Capital investment is dependent upon the type of feed stream being treated. As one might suppose, high-volume applications improve the economic aspects of this process.

RESIN ADSORPTION Resin adsorption is used in the removal of organic solutes from aqueous streams. Solute concentrations may be as high as 8%. It is the preferred method when recovery of the adsorbate is desired since thermal regeneration of carbon destroys the organic material. It is also useful when there is a high concentration of dissolved inorganic salts in the stream. Synthetic resins may be used to remove hydrophobic or hydrophylic solutes, which may be recovered by chemical means. Resin adsorption is similar to carbon adsorption in that two filter beds are often used—one bed is used for adsorption while the other is being regenerated (see also Chapter 11). The stream flows downward in the system at a rate of 1 – 10 gal/min . ft2 of cross section, and adsorption stops when the bed becomes saturated or the effluent concentration reaches a certain level. Their applications include phenol recovery, fat removal, and color removal (or even change). Energy costs for this system depend on whether the resin is regenerated or not. If the resin is not regenerated, it must be disposed. Because the high cost of resins results in a high capital cost, regeneration is usually desirable to keep overall costs low.

EVAPORATION Evaporation is a common process used in both water desalination and the treatment of other feed streams. It may be used to treat a variety of these feeds: liquids, slurries, sludges, organic and inorganic streams, streams containing suspended or dissolved solids, and streams containing nonvolatile dissolved liquids. In other facilities, it has been used in processing radioactive wastes, discharges from paper mills and molasses distilleries, and side streams from trinitrotoluene (TNT) manufacturing. The process and equipment used in evaporation are similar to that of distillation (see Chapter 9) except that the vapor is not collected and condensed unless organic components are present. The stream usually flows through metal pipes that are heated by low-pressure steam outside the pipe walls. Other modes of operation that

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have been used are the solar evaporation from ponds or the heating of open vessels. The process concentrates the original feed stream and reduces its volume. Evaporation is an energy-intensive process and utility and equipment costs may be high.

FOAM FRACTIONATION By exploiting the difference between the surface and bulk concentrations of surfaceactive agents, it is possible to affect a separation by forming an air – water foam. If the operation is carried out in a column, it is possible to fractionate the foam phase from the liquid phase continuously. The foam is then collected and collapsed. Application of this operation to the effluents from industrial plants enables the separation of the detergents present in a discharge. Recycling of the foam fraction to the activated sludge tanks permits further degradation of the detergents. Other refractory contaminants that concentrate in the foam are also removed.(2) Extension of foam fractionation to the concentration of other ionic materials requires only the selection of a surfactant that will sequester the particular material. Recovery of the material from the foam fraction usually requires further treatment.

DISSOCIATION EXTRACTION In solvent extraction, separation of a liquid solution depends on the relative solubility of a component in another liquid that is immiscible with the solvent of the original solution. The separation of such liquid solutions may be accomplished in singlestage batch equipment or, as is more usual, in continuous multi-stage equipment. Separation of the extracted material usually requires removal of the solvent. The separation of mixtures of organic acids or bases can also be accomplished by distributing them between an aqueous solvent and an organic nonpolar solvent (see Chapter 14 for additional details). This technique has been used for the separation of mixtures of closely related organic acids. Occasionally, acids will not distribute themselves into the two phases; however, if their dissociation constants are different, it is possible to separate them. The extraction system necessary to accomplish this separation contains the organic acids, a nonpolar organic solvent, and aqueous caustic soda in an amount less than the equivalent of total acid. If a separation of organic bases is required, a similar system may be used, but an aqueous mineral acid replaces the caustic soda solution. Here, the base with the higher ionization constant is selectively concentrated in the aqueous phase.(2)

ELECTROPHORESIS Colloidal systems dispersed in buffered solutions have electrical charges surrounding the particles. These charges consist of a double layer that, in turn, consists of an inner

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487

layer due to the actual charge on the colloid, and an outer one due to the ionic charge from the solution. If an electrical potential is applied to such a system in a cell, the colloidal particles will migrate toward the electrodes according to their charge. This process is termed electrophoresis. In the electrophoresis of proteins, for example, movement is toward the anode in basic solutions and toward the cathode in acidic solutions. Different proteins show different mobilities and, hence, may be separated into pure components. Since the electrophoresis of colloid solutions produces density gradients in the electrode cell, spontaneous remixing of the components and the buffered solution may occur because of convection currents. To eliminate these currents, various porous media such as filter paper and starch are used to stabilize the separated components.(2)

VIBRATING SCREENS As the name implies, vibrating screens are screens that vibrate. The screen allows for separation of large-sized solid particles from gaseous (usually) or liquid (rarely) streams. The vibration increases the rate of the separation and helps reduce clogging and plugging. Obviously, the key design consideration for this separation process is the size opening of the screen. Equation (17.1) may be employed to calculate the percentage of open area, S, S ¼ (1  ND)(1  nd)

(17:1)

where N is number of wires per inch in one direction, n is number of wires per inch, D is the diameter of wires in the perpendicular direction (inches), and d is the diameter of those wires (inches). To calculate mesh, determine the space width and wire diameter (according to product requirements), then add them together and take the reciprocal. For example, if the space width is 0.048 inch and the wire diameter is 0.035 inch, the mesh is 1/0.083, or 12.

ILLUSTRATIVE EXAMPLE 17.1 Qualitatively describe design considerations for a ion-exchange system. SOLUTION: Although ion-exchange systems can be designed with the aid of only a few laboratory tests, a pilot plant operation of substantial duration is generally recommended. Ion exchangers are subject to fouling and loss of capacity in operation, which cannot be detected readily by laboratory tests. Therefore, it is desirable to operate a pilot plant throughout a number of cycles using the commercial plant regeneration procedures and, if at all possible, the actual stream to be treated in the commercial installation. B

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REFERENCES 1. S. DANATOS, “Unusual separations,” Chem. Eng., New York City, NY, December 7, 1964. 2. Author unknown, Chem. Eng., 38, New York City, NY, Sept. 30, 1963.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

18

Economics and Finance INTRODUCTION Chapter 18 is concerned with economics and finance. These two topics can ultimately dictate the decisions made by practicing engineers and their companies. As noted in the Introduction of Part Three, a company may decide that because of the rising price of the feedstock to a distillation column, they will explore the possibility of producing the raw material from a cheaper raw material. A decision will then be based on whether it makes sense economically in the short- and long-term. Furthermore, economic evaluations are a major part of process and plant design. This chapter provides introductory material to this vast field within engineering. The next two sections are devoted to a discussion on the need for economic analyses and definitions. This is followed with an overview of accounting principles. The chapter concludes with Illustrative Examples in the Applications section. Both the qualitative and quantitative viewpoint is emphasized in this chapter although it is realized that the broad subject of engineering economics cannot be fit into any rigid set of formulas. The material presented falls into roughly three parts: general principles, practical information, and applications. The presentation starts with the simplest situations and proceeds to more complicated formulations and techniques that may be employed if there are sufficient data available. Other texts referenced in the literature provide further details on the subject.

THE NEED FOR ECONOMIC ANALYSES A company or individual hoping to increase profitability must carefully assess a range of investment opportunities and select the most profitable options from those available. Increasing competitiveness also requires that efforts need to be made to reduce the costs of existing processes. In order to accomplish this, engineers should be fully aware of not only technical factors but also economic factors, particularly those that have the largest effect on profitability. In earlier years, engineers concentrated on the technical side of projects and left the financial studies to the economist. In effect, engineers involved in making Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

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estimates of the capital and operating costs have often left the overall economic analysis and investment decision-making to others. This approach is no longer acceptable. Some engineers are not equipped to perform a financial and/or economic analysis. Furthermore, many engineers already working for companies have never taken courses in this area. This short-sighted attitude is surprising in a group of people who normally go to great lengths to obtain all the available technical data before making an assessment of a project or study. The attitude is even more surprising when one notes that data are readily available to enable an engineer to assess the prospects of both his/her own company and those of his/her particular industry.(1) As noted above, the purpose of this chapter is to provide a working tool to assist the student or engineer in not only understanding economics and finance but also in applying technical information to the economic design and operation of processes and plants. The material to follow will often focus on industrial and/or plant applications. Hopefully, this approach will provide the reader with a better understanding of some of the fundamentals and principles. Bridging the gap between theory and practice is often a matter of experience acquired over a number of years. Even then, methods developed from experience must all too often be re-evaluated in the light of changing economic conditions if optimum designs are to result. The approach presented here therefore represents an attempt to provide a consistent and reasonably concise method for the solution of these problems involving economic alternatives.(2) The term economic analysis in engineering problems generally refers to calculations made to determine the conditions for realizing maximum financial return for a design or operation. The same general principles apply whether one is interested in the choice of alternatives for completing projects, in the design of plants so that the various components are economically proportioned, or in the economical operation of existing plants. General considerations that form the framework on which sound decisions must be made are often simple. Sometimes their application to the problems encountered in the development of a commercial enterprise involves too many intangibles to allow exact analysis; in that case, judgment must be intuitive. Often, however, such calculations may be made with a considerable degree of exactness. This chapter will attempt to develop a relatively concise method for applying these principles. Concern with maximum financial return implies that the criterion for judging projects involved is profit. While this is usually true, there are many important objectives which, though aimed at ultimate profit increase, cannot be immediately evaluated in quantitative terms. Perhaps the most significant of these is the recent increased concern with environmental degradation and sustainability. Thus, there has been some tendency in recent years to regard management of commercial organizations as a profession with social obligations and responsibilities; considerations other than the profit motive may govern business decisions. However, these additional social objectives are, for the most part, often not inconsistent with the economic goal of satisfying human wants with the minimum effort. In fact, even in the operation of primarily non-profit organizations, it is still important to determine the effect of various policies on profit.(2)

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The next section is devoted to definitions. This is followed with an overview of accounting principles and,finally, a section or applications.

DEFINITIONS Before proceeding to the applications, it would be wise to provide the reader with certain key definitions in the field. Fourteen (there are of course more) concepts that often come into play in an economic analysis are given below. The definitions have been drawn from the literature.(1–3) Note that some overlap of notations, e.g., P, exists in the material to follow.

Simple Interest The term interest can be defined as the money paid for the use of money. It is also referred to as the value or worth of money. Two terms of concern are simple interest and compound interest. Simple interest is always computed on the original principal. The basic formula to employ in simple interest calculations is: S ¼ P(1 þ ni)

(18:1)

where P ¼ original principal n ¼ time in years i ¼ annual interest rate S ¼ sum of interest and principal after n years Normally, the interest period is one year, in which case i is referred to as the effective interest rate.

Compound Interest Unlike simple interest, with compound interest, interest is added periodically to the original principal. The term conversion or compounding of interest simply refers to the addition of interest to the principal. The interest period or conversion period in compound interest calculations is the time interval between successive conversions of interest and the interest period is the ratio of the stated annual rate to this number of interest periods in one year. Thus, if the given interest rate is 10% compounded semi-annually, the interest period is 6 months and the interest rate per interest period is 5%. Alternately, if the given interest rate is 10% compounded quarterly, then the interest period is 3 months and the interest rate per interest period is 2.5%. One should always assume the interest is compounded annually unless otherwise stated. The basic formula to employ for compound interest is: S ¼ P(1 þ i)n

(18:2)

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If interest payments become due m times per year at compound interest, (m)(n) payments are required in n years. A nominal annual interest rate, i0 , may be defined by:   i0 mn (18:3) S¼P 1þ m In this case, the effective annual interest, i, is:   i0 m i¼ 1þ 1 m

(18:4)

In the limit (as m approaches infinity), such payments may be considered to be required at infinitesimally short intervals, in which case, the interest is said to be compounded continuously. Numerically, the difference between continuous and annual compounding is small. However, annual compounding may be significant when applied to very large sums of money.

Present Worth The present worth is the current value of a sum of money due at time n and at interest rate i. This equation is the compound interest equation solved for the present worth term, P P ¼ S(1 þ i)n

(18:5)

Evaluation of Sums of Money The value of a sum of money changes with time because of interest considerations. $1000 today, $1000 ten years from now, and $1000 ten years ago all have different meanings when interest is taken into account. $1000 today would be worth more ten years from now because of the interest that could be accumulated in the interim. On the other hand, $1000 today would have been worth less ten years ago because a smaller sum of money could have been invested then so as to yield $1000 today. Therefore, one must refer to the date as well as the sum of money. Summarizing, evaluating single sums of money requires multiplying by (1 þ i)n if the required date of evaluation is after the date associated with the obligation or multiplying by (1 þ i)2n if the required date of evaluation is before the date associated with the obligation. The term n is always the time in periods between the date associated with the obligation and the date of evaluation. The evaluation of sums of money may be applied to the evaluation of a uniform series of payments. A uniform series is a series of equal payments made at equal intervals. Suppose R is invested at the end of every interest period for n periods. The total value of all these payments, S, as of the date of the last payment, may be calculated from the equation S ¼ R[(1 þ i)n  1]=i The term S is then called the amount of the uniform series.

(18:6)

Definitions

493

Depreciation The term depreciation refers to the decrease in the value of an asset. Two approaches that can be employed are the straight line and sinking fund method. In the straight line method of depreciation, the value of the asset is decreased each year by a constant amount. The annual depreciation amount, D, is given by D ¼ (Original cost  Salvage value)=(Estimated life in years)

(18:7)

In the sinking fund method of depreciation, the value of the asset is determined by first assuming that a sinking fund consisting of uniform annual payments had been set up for the purpose of replacing the asset at the end of its estimated life. The uniform annual payment (UAP) may be calculated from the equation UAP ¼ (Original cost  Salvage value)(SFDF) where SFDF is the sinking fund deposit factor and is given by SFDF ¼ i=[(1 þ i)n  1]

(18:8)

The value of the asset at any time is estimated to be the difference between the original cost and the amount that would have accumulated in the sinking fund. The amount accumulated in the sinking fund is obtained by multiplying the SFDF by the compound amount factor (CAF) where CAF ¼ [(1 þ i)n  1]=i

(18:9)

Fabricated Equipment Cost Index A simple process is available to estimate the equipment cost from past cost data. The method consists of adjusting the earlier cost data to present values using factors that correct for inflation. A number of such indices are available; one of the most commonly used is the fabricated equipment cost index (FECI).   FECIyear B (18:10) Costyear B ¼ Costyear A FECIyear A Given the cost and FECI for year A, as well as the FECI for year B, the cost of the equipment in year B can be estimated.

Capital Recovery Factor In comparing alternative mass transfer processes or different options for a particular process from an economic point-of-view, one recommended procedure to follow is that the total capital cost can be converted to an annual basis by distributing it over the projected lifetime of the facility. The sum of both the annualized capital cost (ACC), including installation, and the annual operating cost (AOC), is called the total annualized cost (TAC) for the project or facility. The economic merit of the

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proposed facility, process, or scheme can be examined once the total annual cost is available. The conversion of the total capital cost (TCC) to an ACC requires the determination of an economic parameter known as the capital recovery factor (CRF). This parameter can be found in any standard economics textbook or calculated directly from the following equation: CRF ¼ i(1 þ i)n =[(1 þ i)n  1]

(18:11)

where n ¼ projected lifetime of the system i ¼ annual interest rate (as a fraction) The CRF is a positive, fractional number. Once this factor has been determined, the ACC can be calculated from the following equation: ACC ¼ (TCC)(CRF)

(18:12)

The annualized capital cost reflects the cost associated with recovering the initial capital expenditure over the depreciable life of the system.

Present Net Worth There are various approaches that may be employed in the economic selection of the best of several alternatives. A single sum is calculated that would provide for all expenditures over a common time period for each alternative in the present net worth (PNW) method of economic selection. The alternative having the least PNW of expenditures is selected as the most economical. The equation to employ is PNW ¼ CC þ PN þ PWD  PWS

where

(18:13)

CC ¼ capital cost PN ¼ future renewals PWD ¼ other disbursements PWS ¼ salvage value

If the estimated lifetimes differ for the various alternatives, employ a period of time equal to the least common multiple of the different lifetimes for renewal purposes.

Perpetual Life Capitalized cost can be viewed as present worth under the assumption of perpetual life. Computing capitalized cost involves, in a very real sense, finding the present worth of an infinite series of payments. To obtain the present worth of an infinite series of payments of $R at the end of each interest period forever, one needs simply to

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495

divide R by i, where i is the interest rate per interest period. Thus, to determine what sum of money, P, would have to be invested at 8.0% to provide payments of $100,000 at the end of each year forever, P would have to be such that the interest on it each period would be $100,000. Withdrawal of the interest at the end of each period would leave the original sum intact to again draw $100,000 interest at the end of the next period. For this example, P ¼ 100,000=0:08 ¼ $1,250,000 The $1,250,000 would be the present worth of an infinite series of payments of $100,000 at the end of each year forever, assuming money is worth 8%. To determine the present worth of an infinite series of payments of $R at the end of each n periods forever, first multiply by the SFDF to convert to an equivalent single period payment and then divide by i to obtain the present worth.

Break-Even Point From an economic point-of-view, the break-even point of a process operation is defined as that condition when the costs (C) exactly balance the income (I ). The profit (PR) is therefore, PR ¼ I  C

(18:14)

At break-even, the profit is zero.

Approximate Rate of Return Rate of return can be viewed as the interest that will make the present worth of net receipts equal to the investment. The approximate rate of return (ARR) (defined by some as p), may be estimated from the equation below: p ¼ ARR ¼ Average annual profit or earnings=Initial total investment

(18:15)

To determine the average annual profit, simply divide the difference between the total money receipts (income) and the total money disbursements (expenses) by the number of years in the period of the investment.

Exact Rate of Return Using the approximate rate of return as a guide, one can generate the exact rate of return (ERR). This is usually obtained by trial-and-error and an interpolation calculation of the rate of interest that makes the present worth of net receipts equal to the investment. The approximate rate of return will tend to overestimate the exact rate of return when all or a large part of the receipts occur at the end of a period of investment.

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The approximate rate will tend to underestimate the exact rate when the salvage value is zero and also when the salvage value is a high percentage of the investment.

Bonds A bond is a written promise to pay both a certain sum of money (redemption price) at a future date (redemption date) and equal interest payments at equal intervals in the interim. The holder of a $1000, 5% bond, redeemable at 105 (bond prices are listed without the last zero) in 10 years, with interest payable semi-annually would be entitled to semi-annual payments of $1000  0.025 or $25 for 10 years and 105% of $1000, that is $1050, at the end of 10 years when the bond is redeemed. The interest payment on a bond is found by multiplying the face value of the bond by the bond interest rate per period. From above, the face value is $1000 and the bond interest rate per period is 0.025. Therefore, the periodic interest payment is $25. Redeemable at 105 means that the redemption price is 105% (1.05) of the face value of the bond. The purchase price of a bond depends on the yield rate, i.e., the actual rate of return on the investment represented by the bond purchase. Therefore, the purchase price of a bond is the present worth of the redemption price plus the present worth of future interest payments, all computed at the yield rate. The bond purchase price formula is: V ¼ C(1 þ i)n þ R[1  (1 þ i)n ]=i

(18:16)

where V ¼ purchase price C ¼ redemption price R ¼ periodic interest payment n ¼ time in periods to maturity i ¼ yield rate

Incremental Cost By definition, the average unit increment cost is the increase in cost divided by the increase in production. Only those cost factors that vary with production can affect the average unit increment cost. In problems involving decisions as to whether to stay in production or temporarily shut down, the average unit increment cost may be compared with the unit increment cost or the unit selling price.

PRINCIPLES OF ACCOUNTING(4) Accounting is the science of recording business transactions in a systematic manner. Financial statements are both the basis for and the result of management decisions.

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497

Such statements can tell a manager or an engineer a great deal about a company, provided that one can interpret the information correctly. Since a fair allocation of costs requires considerable technical knowledge of operations in the chemical process industries, a close liaison between the senior process engineers and the accountants in a company is desirable. Indeed, the success of a company depends on a combination of financial, technical, and managerial skills. Accounting is also the language of business, and the different departments of management use it to communicate within a broad context of financial and cost terms. The engineer who does not take the trouble to learn the language of accountancy denies oneself the most important means available for communicating with top management. He/she may be thought by them to lack business acumen. Some engineers have only themselves to blame for their lowly status within the company hierarchy since they seem determined to displace themselves from business realities behind the screen of their specialized technical expertise. However, more and more engineers are becoming involved in decisions that are business related. Engineers involved in feasibility studies and detailed process evaluations are dependent for financial information on the company accountants, especially for information on the way the company intends to allocate its overhead costs. It is vital that the engineer should correctly interpret such information and that he/she can, if necessary, make the accountant understand the effect of the chosen method of allocation. The method of allocating overheads can seriously affect the assigned costs of a project and, hence, the apparent cash flows for that project. Since these cash flows are often used to assess profitability by such methods as the NPV (net present value), unfair allocation of overhead costs can result in a wrong choice between alternative projects. In addition to understanding the principles of accountancy and obtaining a working knowledge of its practical techniques, the engineer should be aware of possible inaccuracies of accounting information in the same way that he/she allows for errors in any technical data. At first acquaintance, the language of accountancy appears illogical to most engineers. Although the accountant normally expresses information in tabular form, the basis of all practice can be simply expressed by: Capital ¼ Assets  Liabilities

(18:17)

Assets ¼ Capital þ Liabilities

(18:18)

or

Capital, often referred to as net worth, is the money value of the business, since assets are the money values of things the business owns while liabilities are the money value of the things the business owes. Most engineers have great difficulty in thinking of capital (also known as ownership) as a liability. This is easily overcome once it is realized that a business is a legal entity in its own right, owing money to the individuals who own it. This realization is absolutely essential when considering large companies with stockholders, and is used

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for consistency even for sole ownerships and partnerships. If a person, say FR, puts up $10,000 capital to start a business, then that business has a liability to repay $10,000 to that person. It is even more difficult to think of profit as being a liability. Profit is the increase in money available for distribution to the owners and effectively represents the interest obtained on the capital. If the profit is not distributed, it represents an increase in capital by the normal concept of compound interest. Thus, if the business makes a profit of $5000, the liability is increased to $15,000. With this concept in mind, Equation (18.18) can be expanded to: Assets ¼ Capital þ Liabilities þ Profit

(18:19)

where the capital is considered as the cash investment in the business and is distinguished from the resultant profit in the same way that principal and interest are separated. Profit (as referred to above) is the difference between the total cash revenue from sales and the total of all costs and other expenses incurred in making those sales. With this definition, Equation (18.19) can be further expanded to: Assets þ Expenses ¼ Capital þ Liabilities þ Profit þ Revenue from sales

(18:20)

Some engineers have the greatest difficulty in regarding an expense as being equivalent to an asset, as is implied by Equation (18.20). However, consider FR’s earnings. During the period in which he made a profit of $5000, his total expenses excluding his earnings were $8000. If he assessed the worth of his labor to the business at $12,000, then the revenue required from sales would be $25,000. Effectively, FR has a personal income of $17,000 in the year, but he has apportioned it to the business as $12,000 expense for his labor and $5000 return on his capital. In larger businesses, there will also be those who receive salaries but do not hold stock and therefore, receive no profits, and stockholders who receive profits but no salaries. Thus, the difference between expenses and profits is very practical. The period covered by the published accounts of a company is usually one year, but the details from which these accounts are compiled are often entered daily in a journal. The journal is a chronological listing of every transaction of the business, with details of the corresponding income or expenditure. For the smallest businesses, this may provide sufficient documentation but, in most cases, the unsystematic nature of the journal can lead to computational errors. Therefore, the usual practice is to keep accounts that are listings of transactions related to a specific topic such as “Purchase of Distillation Crude.” This account would list the cost of each purchase of crude oil, together with the date of purchase, as extracted from the journal. The traditional work of accountants has been to prepare balance sheets and income statements. Nowadays, accountants are becoming increasingly concerned with forward planning. Modern accountancy can roughly be divided into two branches: financial accountancy and management or cost accountancy.

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Financial accountancy is concerned with stewardship. This involves the preparation of balance sheets and income statements that represent the interest of stockholders and are consistent with the existing legal requirements. Taxation is an important element of financial accounting. Management accounting is concerned with decision-making and control. This is the branch of accountancy closest to the interest of most process engineers. Management accounting is concerned with standard costing, budgetary control, and investment decisions. Accounting statements only present facts that can be expressed in financial terms. They do not indicate whether a company is developing new products that will ensure a sound business future. A company may have impressive current financial statements and yet may be heading for bankruptcy in a few years’ time if provision is not being made for the introduction of sufficient new products or services.

APPLICATIONS The remainder of the chapter is devoted to Illustrative Examples, many of which contain technical developmental material. A good number of these mass transfer related applications have been drawn from the National Science Foundation (NSF) literature(4–8) and two other key sources.(9,10)

ILLUSTRATIVE EXAMPLE 18.1 A plant manager spends $10,000 on new packing for a tower that recovers a valuable product in a liquid stream from a distillation column. The manager decides to depreciate the equipment at $1430/yr (7-yr straight-line method depreciation) and estimates that the equipment will generate $1500/yr in annual profit. Define the commonly accepted formula for payout time and calculate the payout time for this system. SOLUTION: The payout time is calculated as the fixed capital investment divided by the sum of the annual profit plus the annual depreciation Payout time ¼

Fixed capital investment Annual profit þ Annual depreciation

When the formula is applied to the data, the following result is obtained: Payout time ¼

$10,000 $1500 þ $1430

¼ 3:44 yr

B

ILLUSTRATIVE EXAMPLE 18.2 A mass transfer process emits 50,000 acfm of gas containing metal particulate at a loading of 2.0 gr/ft3. A particulate recovery device is employed for particle capture and the metal

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captured from the unit is worth $0.03/lb. Experimental data have shown that the collection efficiency, E, is related to the system pressure drop, DP, by the formula: E¼ where

DP DP þ 15:0

E ¼ fractional collection efficiency DP ¼ pressure drop lbf/ft2

If the overall fan is 55% efficient (overall) and electric power costs $0.18/kWh, at what collection efficiency is the cost of power equal to the value of the recovered material? What is the pressure drop in inches of water (in. H2O) at this condition? SOLUTION: The value of the recovered material (RV) may be expressed in terms of the collection efficiency E, the volumetric flow rate q, the inlet metal loading w, and the value of the metal (MV): RV ¼ (q)(w)(MV)(E) Substituting yields      50,000 ft3 2:0 gr 1 lb 0:03$ RV ¼ (E) ¼ 0:429E $=min 7000 gr lb min ft3 The recovered value can be expressed in terms of pressure drop by replacing E by DP: RV ¼

(0:429)(DP) $=min DP þ 15:0

The cost of power (CP) in terms of DP, q, the cost of electricity (CE) and the fan efficiency, Ef, is CP ¼ (q)(DP)(CE)=(Ef ) Substitution yields        50,000 ft3 DP lbf 0:18$ 1 min  kW 1 1h CP ¼ kWh 44,200 ft  lbr 0:55 60 min min ft2 ¼ 0:006DP $=min The pressure drop at which the cost of power is equal to the value of the recovered material is found by equating RV with CP: RV ¼ CP Solving gives DP ¼ 66:5 lbf =ft2 ¼ 12:8 in: H2 O Figure 18.1 shows the variation of RV, CP, and profit with pressure drop. The collection efficiency corresponding to the above calculated DP is DP DP þ 15:0 66:5 ¼ 66:5 þ 15:0

E¼

¼ 0:82 ¼ 82:0%

Applications

Figure 18.1

501

Profit as a function of pressure drop.

The reader should note that operating below this efficiency (or the corresponding pressure drop) will produce a profit; operating above this value leads to a loss. The operating condition for maximum profit can be estimated from Figure 18.1. Calculating this value is left as an exercise for the reader. [Hint: Set the first derivative of the profit (i.e., RV 2 CP) with respect to DP equal to zero. The answer is 13.9 lbf/ft2.] B

ILLUSTRATIVE EXAMPLE 18.3 Three different mass transfer devices are available for the recovery of a valuable chemical from a stream. The service life is 10 years for each device. Their capital and annual operating costs are shown in Table 18.1. Which is the most economical unit? Employ a straight-line depreciation method of analysis. SOLUTION: To select the most economical recovery device, a comparison can be performed among the three units based on the total annualized cost (TAC). Table 18.2 can be used to simplify these calculations. A comparison among units A, B, and C indicates that unit C has the lowest TAC and should be selected as the most economical unit of the three being evaluated. A similar result would be obtained if a return on investment (ROI) method of analysis were employed. B

Table 18.1 Device A B C

Initial and Operating Cost Initial cost

Annual operating cost

Salvage value in year 10

$300,000 $400,000 $450,000

$50,000 $35,000 $25,000

0 0 0

502

Chapter 18 Economics and Finance Table 18.2 Total Annual Cost Unit Capital investment Depreciation Operating costs Total annual cost

A

B

C

$300,000 $30,000 $50,000 $80,000

$400,000 $40,000 $35,000 $75,000

$450,000 $45,000 $25,000 $70,000

ILLUSTRATIVE EXAMPLE 18.4 As noted earlier, the break-even point of a process operation is defined as that condition when the costs (C ) exactly balance the income (I ). The profit (P) is therefore P ¼ I 2 C. At breakeven, the profit is zero. The cost and income (in dollars) for a mass transfer operation are given by the following equations: I ¼ $60,000 þ 0:021N C ¼ $78,000 þ 0:008N where N is the yearly production units of mass of the chemical being manufactured. Calculate the break-even point for this operation. SOLUTION:

Write the equation relating C to I. Note that at break-even operation, P ¼ 0. I¼C

Substitute for C and I in terms of N: $60,000 þ 0:021N ¼ $78,000 þ 0:008N Solving for N at the break-even point: N ¼ 1,384,600 Calculating the cost at the break-even point: C ¼ $78,000 þ 0:008N ¼ $78,000 þ 0:008(1,384,600) ¼ $89,077 The reader should note that as N decreases below 1,384,600 units of mass, P is negative (there is a loss). Higher values of N lead to a profit. B

ILLUSTRATIVE EXAMPLE 18.5 A small packed absorption tower is being designed to remove ammonia from air by scrubbing with water at 688F and one atmosphere. It was decided that either 1 inch Raschig rings or 1 inch

Applications

503

Tellerettes will be used as the packing in the tower. The value of NOG is 6.0 for either packing, but HOG is 2.5 ft for Raschig rings and 1.0 ft for the Tellerettes. The cost of 1-inch Raschig rings is $7.56/ft3 and for Tellerettes is $26.40/ft3. The installed cost of the tower shell is $83 per foot of tower height. Determine which of the packings will be the more economical for the system. Assume a tower diameter of 20 inches for both cases. SOLUTION:

Calculate the required packing height, Z (ft), for a Raschig ring (RR) tower: Z(RR) ¼ N OG H OG ¼ (6:0)(2:5) ¼ 15:0 ft

Calculate the RR packed volume in ft3 and the packing cost PC: V(RR) ¼ (pD2=4)(Z) ¼ (p=4)(20=12)2 (15) ¼ 32:7 ft3 PC(RR) ¼ (32:7)(7:56) ¼ $247 Calculate the RR tower shell cost TC in $: TC(RR) ¼ (15)(83) ¼ $1245 Calculate the total TCC cost of the RR tower: TCC(RR) ¼ 247 þ 1245 ¼ $1492 Repeat the calculation for the required packing height, Z (ft), for a Tellerette (T) tower: Z(T) ¼ N OG H OG ¼ (6:0)(1:0) ¼ 6:0 ft V(T) ¼ (p=4)(20=12)2 (6:0) ¼ 13:1 ft3 PC(T) ¼ (26:40)(13:1) ¼ $346 TC(T) ¼ (83)(6:0) ¼ $498 TCC(T) ¼ 346 þ 498 ¼ $844 The cost of a Raschig ring tower is 1.77 times the cost of a Tellerette packed tower in this service. Select the Tellerette tower. B

504

Chapter 18 Economics and Finance

ILLUSTRATIVE EXAMPLE 18.6 An adsorber cost $852,644 in 1982. A company intends to install a similar type of unit in its facility in 2010. Estimate the cost of the new adsorber. If the total (direct plus indirect) installation cost is 60% of the adsorber cost, what is the annualized capital cost of this unit? The expected life of the unit is 10 years. Assume an interest rate of 10%. Data: 1982 FECI ¼ 306.2 2010 FECI ¼ 418.2 (estimated) SOLUTION:

Estimate the capital cost, CC, of the adsorber in 2010: Cost2010 ¼ Cost1982 (2010 FECI=1982 FECI) CC ¼ (852,644)(418:2=306:2) ¼ $1,165,000

Calculate the installation cost, IC: IC ¼ (0:60)(1,165,000) ¼ $698,700 Calculate the total capital cost, TCC: TCC ¼ 1,165,000 þ 698,700 ¼ $1,863,700 Obtain the capital recovery factor, CRF: CRF ¼ i(1 þ i)n =[(1 þ i)n  1]; i ¼ 0:1 and n ¼ 10 ¼ (0:1)(1:1)10 =[(1:1)10  1] ¼ 0:16275 Finally, calculate the annualized capital cost, ACC, of the adsorber in $/yr: ACC ¼ (TCC)(CRF) ¼ (1,863,700)(0:16275) ¼ 303,300 $=yr

B

ILLUSTRATIVE EXAMPLE 18.7 Prior to being processed in an adsorber, a 200,000 acfm stream of particulate contaminated air is to be treated using one of three devices: an electrostatic precipitator, a venturi scrubber, or a baghouse. The data contained in Table 18.3 were obtained from a reliable vendor. Which air pollution control device should be selected for this operation?

Applications Table 18.3

Economic Data for Illustrative Example 18.7 ESP

Total capital cost Operating cost Maintenance cost Equipment lifetime Interest

SOLUTION: the acfm:

505

Venturi scrubber

$17.5/acfm $0.30/acfm-yr $120,000/yr 10 yr 10%

$14.0/acfm $0.35/acfm-yr $150,000/yr 10 yr 10%

Baghouse $16.0/acfm $0.40/acfm-yr $130,000/yr 12 yr 10%

Obtain the total capital costs, TCC, for each control device by multiplying by TCC(ESP) ¼ (200,000)(17:5) ¼ $3,500,000 TCC(VEN) ¼ (200,000)(14:0) ¼ $2,800,000 TCC(BAG) ¼ (200,000)(16:0) ¼ $3,200,000

Calculate the capital recovery factor for each control device: CRF ¼ (i)(1 þ i)n =[(1 þ i)n  1] CRF(ESP) ¼ (0:1)(1 þ 0:1)10 =[(1 þ 0:1)n  1] ¼ 0:16275 CRF(VEN) ¼ 0:16275 CRF(BAG) ¼ 0:14676 Calculate the annual capital cost, ACC, for each control device: ACC(ESP) ¼ (3,500,000)(0:16275) ¼ 569,625 $=yr ACC(VEN) ¼ (2,800,000)(0:16275) ¼ 455,700 $=yr ACC(BAG) ¼ (3,200,000)(0:14676) ¼ 469,632 $=yr Calculate the annual operating cost, AOC, for each device: AOC(ESP) ¼ (200,000)(0:30) ¼ 60,000 $=yr AOC(VEN) ¼ (200,000)(0:35) ¼ 70,000 $=yr AOC(BAG) ¼ (200,000)(0:40) ¼ 80,000 $=yr Calculate the total annual cost, TAC, for each device. See Table 18.4.

506

Chapter 18 Economics and Finance Table 18.4 Total Annual Cost

ACC ($/yr) AOC ($/yr) AMC ($/yr) TAC ($/yr)

ESP

VEN

BAG

569,600 60,000 120,000 749,600

455,700 70,000 150,000 675,700

469,600 80,000 130,000 679,600

Finally, select the most economically attractive control device. According to this analysis, either the venturi scrubber or the baghouse is the most attractive economically; the difference in the TAC is marginal. B

ILLUSTRATIVE EXAMPLE 18.8 Plans are underway to construct and operate some type of mass transfer unit that will serve to purify a product stream from a chemical reactor. The company is still undecided as to whether to install a distillation column or an extraction unit. The extraction unit is less expensive to purchase and operate than a comparable distillation system, primarily because of energy costs. However, projected income from the distillation unit is higher since it will handle a larger quantity of process liquid and provide a purer product. Based on the economic and financial data given in Table 18.5, select the mass transfer unit that will yield the higher annual profit. Calculations should be based on an interest rate of 12% and a process lifetime of 12 years for both units. SOLUTION:

Calculate the capital recovery factor, CRF:  CRF ¼ (0:12)(1 þ 0:12)12 [(1 þ 0:12)12  1] ¼ 0:1614

Determine the annual capital and installation costs for the extraction (EXT) unit: COST(EXT) ¼ (2,625,000 þ 1,575,000)(0:1614) ¼ $677,880=yr Table 18.5 Economic Data for Illustrative Example 18.8 Costs/credits Mass transfer unit Peripherals Total capital Installation Operation Maintenance Income

Extraction

Distillation

$750,000 $1,875,000 $2,625,000 $1,575,000 $400,000/yr $650,000/yr $2,000,000/yr

$800,000 $2,175,000 $2,975,000 $1,700,000 $550,000/yr $775,000/yr $2,500,000/yr

Applications Table 18.6

507

Cost Analysis

Total installed ($/yr) Operation ($/yr) Maintenance ($/yr) Total annual cost ($/yr) Income credit ($/yr)

Extraction

Distillation

678,000 400,000 650,000 1,728,000 2,000,000

755,000 550,000 775,000 2,080,000 2,500,000

Determine the annual capital and installation costs for the distillation (DST) unit: COST(DST) ¼ (2,975,000 þ 1,700,000)(0:1614) ¼ $754,545=yr See Table 18.6 for a comparison of costs and credits for both devices. Finally, calculate the profit for each unit on an annualized basis: PROFIT(EXT) ¼ 2,000,000  1,728,000 ¼ þ272,000 $=yr PROFIT(DST) ¼ 2,500,000  2,080,000 ¼ þ420,000 $=yr A distillation unit should be selected based on the above economic analysis. Detailed cost estimates are beyond the scope of this text. Such procedures are capable of producing accuracies in the neighborhood of +5%. However, such estimates generally require many months of engineering work. This type of analysis is designed to give the reader a basis for a preliminary cost analysis only. B

ILLUSTRATIVE EXAMPLE 18.9(10) Karen Tschinkel, an undergraduate student from Manhattan College’s prestigious chemical engineering program was given the assignment to design and operate a liquid extraction unit in the most cost-effective manner. The design is to be based on an outlet solute concentration in the solvent that will result in the maximum annual profit for the process. A line diagram of the proposed countercurrent unit is provided in Figure 18.2. Having completed a mass transfer course, Karen realized that there are two costs which need to be considered: 1 the extraction unit employed for solvent recovery 2 the value of solvent recovered She also notes that the higher the concentration C (mg/L) of the solvent recovery stream, the smaller will be the concentration difference driving force, and the higher the size requirement of the extractor (resulting in higher equipment cost). Alternatively, with a higher C, the concentration of recovered solute is higher, thus leading to an increase in profits. Based on a similar system design, Ricci and Theodore Consultants (RAT) have provided the following annual economic models: Recovered solute profit: A(C2100); A ¼ $/yr(mg/L) Liquid extraction unit cost: B/(5002C ); B ¼ $(mg/L)/yr

508

Chapter 18 Economics and Finance

Figure 18.2 Proposed liquid extraction unit.

For the above system, RAT suggest values for the coefficients in the model be set at: A ¼ 10 B ¼ 100; 000 Employing the above information, Karen has been asked to calculate an outlet concentration, C, that will 1 provide breakeven (BE) operation, and 2 maximize profits (MP). SOLUTION: Since there are two contributing factors to the cost model, one may write the following equation for the profit as a function of exit solvent concentration: P ¼ A(C  100) 

B 500  C

For breakeven operation, set P ¼ 0, so that (500  C)(C  100) ¼ B=A This may be expanded and rewritten as C2  (500 þ 100)C þ [(B=A) þ (500)(100)] ¼ 0 C2  600C þ (10,000 þ 50,000) ¼ 0 The solution to this quadratic equation for A ¼ 10 and B ¼ 100,000 is pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (600) + (600)2  (4)(1)(60,000) C¼ 2(1) ¼

600 + 342 2

Applications

Figure 18.3

509

First derivative test.

The two solutions for BE operation are: C ¼ f473 mg=L, 127 mg=Lg To maximize (or minimize) the profit, take the first derivative of P with respect to C and set it equal to zero, i.e., dP B ¼A ¼0 dC (500  C)2 Solving: (500  C)2 ¼ B=A pffiffiffiffiffiffiffiffiffi 500  C ¼ + B=A C ¼ f400 mg=L, 600 mg=Lg However, based on the physical interpretation of these roots, it is readily apparent that the exit solvent concentration, C, cannot be greater than the inlet feed concentration. Hence, the root C ¼ 600 mg/L has no physical meaning and may be neglected. In order to determine if this root is the relative maxima, the first derivative test must be employed. By qualitatively examining the value of the derivative (þ/2) on both sides of each root, an inference can be made as to whether the original function contained a relative maxima or relative minima at the root. For instance, if the value of the derivative changes in sign from positive to negative as the concentration is increasing at the root, the slope of the line tangent to the original function changes from positive to negative at this point, and hence the root is a relative maxima in the original function (this assumes continuity about the point of interest in the original function). Inversely, should the sign change from negative to positive in the derivative, then the original function realizes a relative minima at the root. In this particular example, the values of the derivative arbitrarily close to each root have the following sign shown in Figure 18.3. A relative maxima in the profit equation, P(C), is realized at C ¼ 400 mg/L. Thus, for MP, the exit concentration is equal to 400 mg/L. Alternatively, the second derivative test may be employed, which examines a function’s change about its point(s) of inflection. This test would yield the same result. B

ILLUSTRATIVE EXAMPLE 18.10 Refer to Illustrative Example 18.9. Karen has also been asked to perform the calculations if A ¼ 10, B ¼ 400 and A ¼ 10, B ¼ 400,000. Finally, an analysis of all the results is requested.

510

Chapter 18 Economics and Finance

Figure 18.4 Profit-discharge concentration plot. SOLUTION: that:

Using a similar procedure as in Illustrative Example 18.9, Karen determined

for A ¼ 10, B ¼ 400 C ¼ f499 mg=L, 101 mg=Lg for BE C ¼ 480 mg=L for MP for A ¼ 10, B ¼ 400,000 C ¼ 300 mg=L for BE C ¼ 300 mg=L for MP In terms of analysis, a qualitative sketch of the three scenarios are provided in Figure 18.4.

B

ILLUSTRATIVE EXAMPLE 18.11 A distillation column processes two types of crude oil streams: North Texas and Venezuelan. Because of consumer demand, the production of diesel fuel, home heating oil, and gasoline must be limited. Based on the information in Table 18.7, estimate the optimum daily usage rate of each crude oil to maximize the profit, P. The profit generated on processing North Texas (A) and Venezuelan Crude (B) is $2.00/gal and $1.60/gal, respectively.

References

511

Table 18.7 Crude Oil Usage and Production Information

Diesel fuel Home heating oil Gasoline

North Texas, %

Venezuelan, %

Maximum production rate, gal/day

8 29 63

11 54 35

1500 5500 11,000

SOLUTION: Set NA and NB equal to the daily usage rate of North Texas and Venezuelan crude. One may then write the following based on the problem statement: P ¼ 2:00NA þ 1:60NB ;

N ¼ gal=day

Constraints: 0:08NA þ 0:11NB  1500 0:29NA þ 0:54NB  5500 0:63NA þ 0:35NB  11,000 By trial-and-error, or from an optimization program, one obtains (approximately): NA ¼ 16,800 gal=day NB ¼ 1150 gal=day P ¼ $35,500=day

B

REFERENCES 1. L. THEODORE and K. NEUSER, “Engineering Economics and Finance,” A Theodore Tutorial, Theodore Tutorials, East Williston, NY, 1996. 2. J. REYNOLDS, J. JERIS, and L. THEODORE, “Handbook of Chemical and Environmental Engineering Calculations,” John Wiley & Sons, Hoboken, NJ, 2002. 3. L. THEODORE and J. BARDEN, “Mass Transfer Operations,” A Theodore Tutorial, Theodore Tutorials, East Williston, NY, 1995. 4. F. HOLLAND, F. WATSON, and J. WILKINSON, “Financing principles of accounting,” Chem. Eng., New York City, NY, July 8, 1974. 5. R. DUPONT, L. THEODORE, and J. REYNOLDS, “Accident and Emergency Management: Problems and Solutions,” VCH Publishers, New York City, NY, 1991. 6. L. THEODORE, R. DUPONT, and J. REYNOLDS, “Pollution Prevention: Problems and Solutions,” Gordon and Breach Publishers, Amsterdam, Holland, 1994. 7. K. GANESON, L. THEODORE, and J. REYNOLDS, “Air Toxics: Problems and Solutions,” Gordon and Breach Publishers, Amsterdam, Holland, 1996. 8. R. DUPONT, T. BAXTER, and L. THEODORE, “Environmental Management: Problems and Solutions,” CRC Press, Boca Raton, FL, 1998. 9. J. REYNOLDS, R. DUPONT, and L. THEODORE, “Hazardous Waste Incineration Calculations: Problems and Software,” John Wiley & Sons, Hoboken, NJ, 1991. 10. L. THEODORE, F. RICCI, and T. VAN VLIET, “Thermodynamics for the Practicing Engineer,” John Wiley & Sons, Hoboken, NJ, 2009.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

19

Numerical Methods INTRODUCTION Early in one’s career, the engineer/scientist learns how to use equations and mathematical methods to obtain exact answers to a large range of relatively simple problems. Unfortunately, these techniques are often not adequate for solving realworld problems, although the reader should note that one rarely needs exact answers in technical practice. Most real-world applications are usually inexact because they have been generated from data or parameters that are measured, and hence represent only approximations. What one is likely to require in a realistic situation is not an exact answer but rather one having reasonable accuracy from an engineering point of view. The solution to an engineering or scientific problem usually requires an answer to an equation or equations, and the answer(s) may be approximate or exact. Obviously an exact answer is preferred, but because of the complexity of some equations, exact solutions may not be attainable. Furthermore, to engineers, an answer that is precise may not be necessary and wastes time. For this condition, one may resort to another method that has come to be defined as a numerical method. Unlike the exact solution, which is continuous and in closed form, numerical methods provide an inexact (but often reasonably accurate) solution. The numerical method leads to discrete answers that are almost always acceptable. The numerical methods referred to above provide a step-by-step procedure that ultimately leads to an answer and a solution to a particular problem. The method usually requires a large number of calculations and is therefore ideally suited for digital computation. High speed computing equipment has had a tremendous impact on engineering design, scientific computation, and data processing. The ability of computers to handle large quantities of data and to perform mathematical operations described above at tremendous speeds permits the examination of many more cases and more engineering variables than could possibly be handled on the slide rule—the trademark of engineers of yesteryear. Scientific calculations previously estimated in lifetimes of computation time are currently generated in seconds and, in many instances, microseconds.(1) Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

513

514

Chapter 19 Numerical Methods

This chapter is concerned with numerical methods. This subject was taught in the past as a means of providing engineers with ways to solve complicated mathematical expressions that they could not solve otherwise. A brief overview of numerical methods is given to provide the practicing engineer with some insight into what many of the currently used software packages (MathCad, Mathematica, MatLab, etc.) are actually doing. The authors have not attempted to cover all the topics of numerical methods. Topics that traditionally fall in the domain of this subject include: Regression analyses (Illustrative Example 19.1) Differentiation (Illustrative Example 19.2) Integration (Illustrative Example 19.3) Simultaneous linear algebraic equations (Illustrative Example 19.4) Nonlinear algebraic equations (Illustrative Example 19.5) Ordinary differential equation(s) (Illustrative Example 19.5) Partial differential equation(s) (Illustrative Example 19.6) Optimization (Illustrative Example 19.7) Since detailed treatment of each of the above topics is beyond the scope of this mass transfer text, the reader is referred to the literature(2–4) for a more extensive analysis and additional information. The remainder of this chapter consists of an illustrative example section on applications that examine all the topics listed above. Also note that the illustrative examples primarily address mass transfer topics.

APPLICATIONS ILLUSTRATIVE EXAMPLE 19.1 Two component diffusivity data (D) for air –water vapor has been extracted from the literature, as presented in Table 19.1. The literature suggests that diffusivity varies with the temperature to the 1.5 power, i.e., D / T 1:5 Regress(5) the above data to a model of the form D ¼ A þ B(T)1:5 and indicate how well the model fits the data. Table 19.1 Diffusivity –Temperature Data T, 8C

D, ft2/s

0 25.9 42.0 59.0

0.22 0.258 0.288 0.305

Applications SOLUTION:

515

A regression analysis (employing EXCEL) shows that A ¼ 0:2208 B ¼ 0:0015

so that D ¼ 0:2208 þ 0:0015x; x ¼ T 1:5 with D in ft2/s and T in 8C. The correlation coefficient is 0.99554, indicating an excellent fit. B

ILLUSTRATIVE EXAMPLE 19.2 Refer to Table 19.2. Generate expressions for the benzene concentration gradient, dC/dt, at t ¼ 4.0 s. Use several differentiation methods.(6) SOLUTION:

Method 1 The first method consists of choosing any three data points and calculating the slope of the two extreme points. This slope is approximately equal to the slope at the point lying in the middle. The value obtained will be the equivalent of the derivative at that point 4. Using data points from 3.0 to 5.0, Slope ¼ ¼

C5  C3 t5  t3 1:63  2:70 ¼ 0:535 5:0  3:0

Method 2 The second method involves taking the average of two slopes. Using the same points chosen above, two slopes are calculated, one for points 3 and 4 and the other for points 4 and 5. Adding the two results and dividing them by two will provide an approximation of the derivative Table 19.2 Illustrative Example 19.2 Data Time (s) 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0

Concentration of benzene (mg/L) 7.46 5.41 3.80 2.70 2.01 1.63 1.34 1.17

516

Chapter 19 Numerical Methods

at point 4. For the points used in this method, the results are: Slope1 ¼

C4  C3 t4  t3

2:01  2:70 ¼ 0:69 4:0  3:0 C5  C4 Slope2 ¼ t5  t4 ¼

1:63  2:01 ¼ 0:38 5:0  4:0 0:69 þ (0:38) ¼ 0:535 ¼ 2 ¼

Slopeavg

Method 3 Method three consists of using any three data points (in this case the same points chosen before) and fitting a curve to it. The equation for the curve is obtained by employing a second-order equation and solving it with the three data points. The derivative of the equation is then calculated and evaluated at any point (here, point 4 is used): C ¼ 0:155t 2  1:775t þ 6:63 dC ¼ 0:31t  1:775 dt Evaluated at t ¼ 4.0 s dC ¼ 0:31(4)  1:775 ¼ 0:535 dt

Method 4 The following method uses the method of least squares. In this case, all data points are used to generate a second-order polynomial equation. This equation is then differentiated and evaluated at the point where the value of the derivative is required. For example, Microsoft Excel can be employed to generate the regression equation. Once all the coefficients are known, the equation has only to be analytically differentiated: C ¼ 0:1626t2  1:9905t þ 7:3108 dC ¼ 0:3252t  1:9905 dt Evaluated at t ¼ 4.0 s: dC ¼ 0:3252(4:0)  1:9905 ¼ 0:6897 dt

Method 5(7) The last two methods are very similar to each other. They are based on five data points used to generate coefficients. For this development, represent C and t by f and x (as it appeared in the literature(7)), respectively.

Applications

517

The fifth method uses five data points to generate a five coefficient (fourth order) model using an equation of the form f ¼ A þ Bx þ Cx2 þ Dx3 þ Ex4 equation. This method is known as interpolating. A set of equations is used to evaluate numerical derivatives from the interpolating polynomial. The equations are listed below: (25f0 þ 48f1  36f2 þ 16f3  3f4 ) 12h (3f  10f þ 18f2  6f3 þ f4 ) 0 1 f 0 (x1 ) ¼ 12h ( fi2  8fi1 þ 8fiþ1  fiþ2 ) f 0 (xi ) ¼ 12h (fn4 þ 6fn3  18fn2 þ 10fn1 þ 3fn ) f 0 (xn1 ) ¼ 12h (3fn4  16fn3 þ 36fn2  48fn1 þ 25fn ) 0 f (xn ) ¼ 12h f 0 (x0 ) ¼

(19:1) (19:2) (19:3) (19:4) (19:5)

where h ¼ xiþ1 2 xi fi ¼ function evaluated at i For example, the equation obtained for “the five data set” from 1.0 to 5.0 s, i.e., t ¼ 1.0, 2.0, 3.0, 4.0 and 5.0 s, using the equations given above is f (x) ¼ 0:0012x4 þ 0:002x3 þ 0:2616x2  2:34x þ 7:467 All these equations are evaluated for each value of x and f (x). The value obtained for point 4.0 is 20.5448.

Method 6(7) The last method also uses five data points but only three coefficients are generated for a secondorder polynomial equation of the form f ¼ A þ Bx þ Cx2 . Another set of equations are used to evaluate the derivative at each point using this method. The equations are provided below: (54f0 þ 13f1 þ 40f2 þ 27f3  26f4 ) 70h (34f0 þ 3f1 þ 20f2 þ 17f3  6f4 ) 0 f (x1 ) ¼ 70h (2f  f þ fiþ1 þ 2fiþ2 ) i2 i1 f 0 (xi ) ¼ 10h (6fn4  17fn3  20fn2  3fn1 þ 34fn ) f 0 (xn1 ) ¼ 70h (26f  27f  40fn2  13fn1 þ 54fn ) n4 n3 f 0 (xn ) ¼ 70h f 0 (x0 ) ¼

(19:6) (19:7) (19:8) (19:9) (19:10)

At point 4.0, the solution for the derivative using this method is 20.6897. Comparing all the values obtained for the derivative at t ¼ 4.0 s, it can be observed that the answers are very close to each other. It is important to remember that these are approximate values and that they vary depending on the approach and the number of data points used to generate the equations. B

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Chapter 19 Numerical Methods

ILLUSTRATIVE EXAMPLE 19.3 The volume of a reactor undergoing conversion X for the reactant (the principal component from the bottom of a distillation column) is described by the following integral: V ¼ 6:0  10

3

0:45 ð

(1  0:125X)3 dX ; 104 (1  X)(1  0:5X)2

liters

0

Calculate the volume using the trapezoidal rule method of integration. Discuss the effect of varying the increment in DX (e.g., if DX ¼ 0.45, 0.09, 0.05, 0.01, 0.005, 0.001).(8) SOLUTION: An algorithm for the trapezoid rule is given in Figure 19.1. For an increment size of DX ¼ 0.45, the step size for the reactor volume is: h ¼ X1  X0 ¼ 0:45  0:00 ¼ 0:45 Evaluate the function at X0 and X1: V0 ¼ f (X0 ) ¼ 6:0  103



(1  0:125X)3 dX 4 10 (1  X)(1  0:5X)2

 X¼0:0

(1)3 ¼ 6:0  103 4 10 (1)(1)2 ¼ 60 V1 ¼ f (X1 ) ¼ 6:0  103



(1  0:125X)3 dX 4 10 (1  X)(1  0:5X)2

 X¼0:45

3

¼ 6:0  103

[1  0:125(0:45)] 104 (1  0:45)[1  0:5(0:45)]2

¼ 152:67 The two-point trapezoid rule is given by Xð1

h f (X) dX ¼ [ f (X0 ) þ f (X1 )] 2

X0

Therefore, h V ¼ [ f (X0 ) þ f (X1 )] 2 h ¼ [V0 þ V1 ] 2 0:45 [60 þ 152:67] ¼ 2 ¼ 47:85 L The trapezoid rule is often the quickest but least accurate way to perform a numerical integration by hand. However, if the step size is decreased, the answer should converge to the analytical solution. Note that for smaller step sizes, the results of each numerical integration must be

Applications

∫

≠

∫

Figure 19.1

Algorithm for trapezoid rule.

519

520

Chapter 19 Numerical Methods Table 19.3 Trapezoid Rule for Various Step Sizes Step size (DX )

Volume (L)

0.45 0.09 0.05 0.01 0.005 0.001

47.85094 43.13532 42.98842 42.92521 42.92324 42.92260

added together to obtain the final answer. The results (to seven significant figures) for various step sizes are listed in Table 19.3. B

ILLUSTRATIVE EXAMPLE 19.4 The concentration, C, variation with length, z, in a 3 ft continuous contact mass transfer device is described by the equation: dC k ¼  dz C v where k is 50 h21, v is 50 ft/h, C is in lbmol/ft3, z is in feet and C ¼ 1.0 lbmol/ft3 at z ¼ 0. It has been proposed to represent the above with 10 staged units 0.3 ft in length. Develop solutions to this problem using a finite difference method of solving an ordinary differential equation and a lumped parameter model employing a method of solution of simultaneous linear algebraic equations. SOLUTION:

Finite Difference Method A finite difference procedure is now applied to the equation. The first derivative of C with respect to z is equivalent to the finite difference in the z-direction, i.e., dC DC Cnþ1  Cn ¼ ¼ dz Dz Dz Inserting this into the describing equation and noting that the average concentration over the increment should be used leads to   Cnþ1  Cn k Cav ¼ vz Dz Knowing the initial condition, each successive concentration may be found by rearranging the above equation:   kDz Cav þ Cn Cnþ1 ¼  vz The application of this equation requires a trial-and-error procedure where the average concentration over the increment is approximated and then checked when Cnþ1 is

Applications

Figure 19.2

521

Lumped-parameter method.

calculated. One should note that the approximation part of this problem can be removed by replacing Cav by Cnþ1 þ Cn 2 and solving for Cnþ1 directly in terms of Cn.

Lumped-parameter method In the lumped-parameter method, the device is divided into equal stages and each segment is considered to be a perfectly mixed vessel. This is shown schematically in Figure 19.2. The equation is now written as Cout  Cin k ¼  Cout vz Dz Rearranging the above equation gives Cout ¼

Cin kDz 1þ vz

If the initial concentration is C0 and the term 1þ

kDz ¼R vz

then the following set of equations are generated for each stage: C1  RC0 ¼ 0; C2  RC1 ¼ 0;

first stage second stage

.. . Cfinal  RCfinal1 ¼ 0;

final stage

The above represents a series of (simultaneous) linear algebraic equations. These equations can be solved by a direct step-by-step hand calculation. Alternatively, a Gauss –Jordan or Gauss– Seidel method can also be used. The results of the analyses by the two methods are combined and presented in tabular form for comparison along with the analytical solution (see Table 19.4). B

522

Chapter 19 Numerical Methods Table 19.4 Concentration Profile via Three Methods z, ft

Analytical

Finitedifference

Lumpedparameter

0.0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7

1.0000 0.8354 0.6980 0.5830 0.4870 0.4070 0.3400 0.2840 0.2370 0.1980

1.0000 0.8358 0.7001 0.5863 0.4910 0.4112 0.3444 0.2884 0.2415 0.2022

1.0000 0.8474 0.7181 0.6086 0.5157 0.4371 0.3704 0.3139 0.2660 0.2254

ILLUSTRATIVE EXAMPLE 19.5 The equation describing the concentration of a pollutant in a flowing river is given by dC ¼ 0:580C þ 6  105 dt with C0 ¼ 5.0  1025 gmol/cm3 at t ¼ 0 min. Estimate C in gmol/cm3 at 1 and 5 minutes. Use the Runge–Kutta method of analysis.(8) SOLUTION: The Runge– Kutta (R –K) method is one of the most widely used techniques for solving first-order differential equations. For the equation dy ¼ f (x, y) dx

(19:11)

h ynþ1 ¼ yn þ (D1 þ 2D2 þ 2D3 þ D4 ) 6

(19:12)

the solution takes the form

where

D1 ¼ hf (x, y)   h D1 D2 ¼ hf xn þ , yn þ 2 2   h D2 D3 ¼ hf xn þ , yn þ 2 2 D4 ¼ hf (xn þ h, yn þ D3 )

The term h represents the increment in x. The term yn is the solution to the equation at xn, and ynþ1 is the solution to the equation at xnþ1 where xnþ1 ¼ xn þ h. Thus, the R –K method provides a straightforward means for developing expressions for Dy in terms of the function f(x, y) at various “locations” along the interval in question.

Applications

523

For a simple equation of the form dC ¼ a þ bC dt

(19:13)

where at t ¼ 0, C ¼ C0, the R– K algorithm given above becomes (for t ¼ h) h C1 ¼ C0 þ (D1 þ 2D2 þ 2D3 þ D4 ) 6 where

(19:14)

D1 ¼ hf (x, y) ¼ h(a þ bC0 )   h D1 D2 ¼ hf xn þ , yn þ ¼ h[a þ b(C0 þ D1 =2)] 2 2   h D2 D3 ¼ hf xn þ , yn þ ¼ h[a þ b(C0 þ D2 =2)] 2 2 D4 ¼ hf (xn þ h, yn þ D3 ) ¼ h[a þ b(C0 þ D3 )]

The same procedure is repeated to obtain values for C2 at t ¼ 2h, C3 at t ¼ 3h, and so on. Based on the data provided, evaluate the R –K coefficients for t ¼ h: D1 D2 D3 D4

¼ 1:0(6:0  2:9) ¼ 3:1 ¼ 1:0[6:0  0:58(5:0 þ 3:1=2)] ¼ 2:2 ¼ 1:0[6:0  0:58(5:0 þ 2:2=2)] ¼ 2:46 ¼ 1:0[6:0  0:58(5:0 þ 2:46)] ¼ 1:67

Calculate C1: h C1 ¼ C0 þ (D1 þ 2D2 þ 2D3 þ D4 ) 6 1 ¼ 5:0 þ [3:1 þ 2(2:2) þ 2(2:46) þ 1:67] 6 ¼ 7:35 gmol=cm3 To calculate C2: D1 ¼ 1:0[6:0  0:58(7:35)] ¼ 1:74 D2 ¼ 1:24 D3 ¼ 1:38 D4 ¼ 0:94 and h C1 ¼ C0 þ (D1 þ 2D2 þ 2D3 þ D4 ) 6 1 ¼ 7:34 þ [1:74 þ 2(1:24) þ 2(1:38) þ 0:94] 6 ¼ 8:66 gmol=cm3

524

Chapter 19 Numerical Methods

Also calculate C3, C4, and C5: at t ¼ 3 min C3 ¼ 9:40 C4 ¼ 9:81 at t ¼ 4 min C5 ¼ 10:04 at t ¼ 5 min The reader is left the exercise of comparing the numerical solution above (including that at t ¼ 1) with that provided by the analytical solution. The R–K method can also be used if the function also contains the independent variable. Consider the following equation: dC ¼ f1 (C, t) dt

(19:15)

1 C1 ¼ C0 þ (D1 þ 2D2 þ 2D3 þ D4 ) 6

(19:16)

For this situation,

with D1 ¼ hf (C, t)   D1 h D2 ¼ hf C0 þ , t0 þ 2 2   D2 h D3 ¼ hf C0 þ , t0 þ 2 2 D4 ¼ hf (C0 þ D3 , t0 þ h) If, for example, dC ¼ 10C  ect dt then     D1 D2 ¼ h 10 C0 þ  e[C0 þ (D1 =2)][t0 þ (h=2)] 2 Situations may arise when there is a need to simultaneously solve more than one ordinary differential equation (ODE). In a more general case, one could have n dependent variables y1, y2, . . . , yn, with each related to a single independent variable x by the following system of n simultaneous first-order ODEs: dy1 ¼ f1 (x, y1 , y2 , . . . , yn ), dx dy2 ¼ f2 (x, y1 , y2 , . . . , yn ), dx .. .

(19:17)

dyn ¼ fn (x, y1 , y2 , . . . , yn ), dx Note that the equations in Equation (19.17) are interrelated, i.e., they are dependent on each other. This is illustrated in the following development.

Applications

525

Consider the following two equations: dC ¼ AeET=R C ¼ f (C, t) dt dT DH ¼ kC ¼ g(C, t) dt rCP

(19:18)

dy ¼ f (x, y, z); (e:g:, xyz) dx dz ¼ g(x, y, z); (e:g:, x2 y2 ez ) dt

(19:19)

or, in a more general sense,

The R –K algorithm for Equation (19.19) is 1 y1 ¼ y0 þ (RY1 þ 2RY2 þ 2RY3 þ RY4 ) 6 1 z1 ¼ z0 þ (RZ1 þ 2RZ2 þ 2RZ3 þ RZ4 ) 6

(19:20)

where y1 2 y0 ¼ Dy, z1 2 z0 ¼ Dz, h ¼ Dx and RY1 ¼ h  f (x0 , y0 , z0 ) RZ1 ¼ h  g(x0 , y0 , z0 ) RY2 RZ2 RY3 RZ3 RY4 RZ4

¼ h  f (x0 þ h=2, y0 þ RY1 =2, z0 þ RZ1 =2) ¼ h  g(x0 þ h=2, y0 þ RY1 =2, z0 þ RZ1 =2) ¼ h  f (x0 þ h=2, y0 þ RY2 =2, z0 þ RZ2 =2) ¼ h  g(x0 þ h=2, y0 þ RY2 =2, z0 þ RZ2 =2) ¼ h  f (x0 þ h, y0 þ RY3 , z0 þ RZ3 ) ¼ h  g(x0 þ h, y0 þ RY3 , z0 þ RZ3 )

Although the R– K approach (and other companion methods) have traditionally been employed to solve first-order ODEs, it can also treat higher ODEs. The procedure requires reducing an nth order ODE to a first-order ODE. For example, if the equation is of the form(9) d2 y ¼ f ( y, x) dx2

(19:21)

set z¼

dy dx

(19:22)

so that dz d2 y ¼ dx dx2

(19:23)

526

Chapter 19 Numerical Methods

The second-order equation in Equation (19.21) has now been reduced to the two first-order ODEs in Equation (19.24): d2 y dz ¼ ¼ f ( y, x) dx2 dx dy ¼z dx

(19:24)

The procedure set forth in Equations (19.19) and (19.20) can be applied to generate a solution to Equation (19.21). Note, however, that the first derivative (i.e., dy/dx or its estimate) is required at the start of the integration. Extending the procedure to higher-order equations is left as an exercise for the reader. B

ILLUSTRATIVE EXAMPLE 19.6 The vapor pressure, p0 , for a new synthetic chemical at a given temperature has been determined to take the form: p0 ¼ T 3  2T 2 þ 2T If p0 ¼ 1, one may then write f (T) ¼ T 3  2T 2 þ 2T  1 ¼ 0 where the term T is in K. The actual temperature, t, is given by t ¼ 103 T Solve the above equation for the actual temperature in K for p0 ¼ 1.0. Earlier studies indicate that t is in the 1000–1200 K range. SOLUTION: The subject of the solution to a nonlinear algebraic equation is considered in this example. Although several algorithms are available, the presentation will key on the Newton– Raphson method of evaluating the root(s) of a nonlinear algebraic equation. The solution to the equation f (x) ¼ 0

(19:25)

is obtained by guessing a value for x (xold) that will satisfy the above equation. This value is continuously updated (xnew) using the recursion equation xnew ¼ xold 

f (xold ) f 0 (xnew )

(19:26)

until either little or no change in (xnew 2 xold)/xold is obtained. One can express this operation graphically (see Fig. 19.3).

Applications

527

′

Figure 19.3

Newton– Raphson method.

Noting that f 0 (xold ) ¼

df (x) Df (x) f (xold )  0  ¼ dx Dx xold  xnew

(19:27)

one may rearrange Equation (19.27) to yield Equation (19.26). The xnew then becomes the xold in the next calculation. This method is also referred to as Newton’s method of tangents and is a widely used method for improving a first approximation to a root to the aforementioned equation of the form f (x) ¼ 0. The above development can be rewritten in subscripted form to accommodate a computer calculation. Thus, f 0 (xn ) ¼

f (xn ) xn  xnþ1

(19:28)

f (xn ) f 0 (xn )

(19:29)

from which xnþ1 ¼ xn 

where xnþ1 is again the improved estimate of xn and the solution to the equation f (x) ¼ 0. For this procedure, the value of the function and the value of the derivative of the function are determined at x ¼ xn, and the new approximation to the root, xnþ1, is obtained. The same procedure is repeated, with the new approximation, to obtain a still better approximation of the root. This continues until successive values for the approximate root differ by less than a prescribed small value, 1, which controls the allowable error (or tolerance) in the root. Relative to the previous

528

Chapter 19 Numerical Methods

Figure 19.4 Failure of the Newton–Raphson method. estimate, 1¼

xnþ1  xn xn

(19:30)

Despite its popularity, the method suffers for two reasons. First, an analytical expression for the derivative [i.e., f 0 (xn)] is required. In addition to the problem of having to compute an analytical derivative value at each iteration, one would expect Newton’s method to converge fairly rapidly to a root in the majority of cases. However, as is common with most numerical methods, it may fail occasionally. A possible initial oscillation followed by a displacement away from a root is illustrated in Figure 19.4. Note, however, that the method would have converged (in this case) if the initial guess had been somewhat closer to the exact root. Thus, it can be seen that the initial guess may be critical to the success of the calculation. With reference to the problem, assume an initial temperature t1. Set t1 ¼ 1100 so that T1 ¼ (1100)(103 ) ¼ 1:1 Obtain the analytical derivative, f 0 (T ): f 0 (T) ¼ 3T 2  4T þ 2 Calculate f (T1) and f 0 (T1): f (1:1) ¼ T 3  2T 2 þ 2T  1 ¼ (1:1)3  2(1:1)2 þ 2(1:1)  1 ¼ 0:111 f 0 (1:1) ¼ 3T 2  4T þ 2 ¼ 3(1:1)2  4(1:1) þ 2 ¼ 1:23 Use the Newton–Raphson method to estimate T2. Employ Equation (19.26): T2 ¼ T1 

f (T1 ) 0:111 ¼ 1:0098 ¼ 1:1  f 0 (T1 ) 1:23

Calculate T3: f (T2 ) ¼ 0:0099 f 0 (T2 ) ¼ 1:0198 T3 ¼ 1:0001 Finally, calculate the best estimate (based on two iterations) of t: t ¼ 1000:1 K

Applications

529

Other methods that may be employed include: 1 2 3 4

Wegstein’s method False-position Half-interval Second-order Newton –Raphson

Details are available in the literature.(2,3)

B

ILLUSTRATIVE EXAMPLE 19.7 Qualitatively discuss methods of solving partial differential equations. SOLUTION: Many practical problems in engineering involve at least two independent variables, i.e., the dependent variable is defined in terms of (or is a function of) more than one independent variable. The derivatives describing these independent variables are defined as partial derivatives. Differential equations containing partial derivatives are referred to as partial different equations (PDEs). Contrary to a widely accepted myth, an engineer’s mathematical obligations do not end after formulating an equation for a problem, where it may be given to a mathematician to solve. Even if such an ideal situation should exist, it is still necessary for engineers to have a reasonable understanding of the mathematical methods and their limitations employed in the solution in order to interpret results. It has been said that “the solution of a partial differential equation is essentially a guessing game.” In other words, one cannot expect to be given a formal method that will yield exact solutions for all partial differential equations.(8) Fortunately, numerical methods for solving these equations were developed during the middle and latter part of the 20th century. The three main PDEs encountered in engineering practice are briefly introduced below employing C (e.g., the concentration as the independent variable). The parabolic equation: @C @2 C ¼ 2 @t @x

(19:31)

@2 C @2 C þ 2 ¼0 @x2 @y

(19:32)

@2 C @2 C ¼ 2 @t 2 @x

(19:33)

The elliptical equation:

The hyperbolic equation:

The preferred numerical method of solution involves finite differencing. Only the parabolic and elliptical equations are considered below. Examples of parabolic PDEs include @C @2 C ¼D 2 @t @x

(19:34)

530

Chapter 19 Numerical Methods

and (the two-dimensional)  2  @C @ C @2 C ¼D þ @t @x2 @y2

(19:35)

Ketter and Prawel,(3) as well as many others, have reviewed the finite difference approach to solving Equation (19.34). B

ILLUSTRATIVE EXAMPLE 19.8 Qualitatively discuss optimization. SOLUTION: Optimization is viewed by many as a tool in decision-making. It often aids in the selection of values that allow the practicing engineering to better solve a problem. This brief answer provides a qualitative look at optimization. In its most elementary and basic form, one may say that optimization is concerned with the determination of the “best” solution to a given problem. This process is required in the solution of many general problems in engineering and applied science—in the maximization (or minimization) of a given function(s), in the selection of a control variable to facilitate the realization of a desired condition, in the scheduling of a series of operations or events to control completion dates of a given project, in the development of optimal layouts of organizational units within a given design space, etc.(4) The optimization problem has been described succinctly by Aris(10) as “getting the best you can out of a given situation.” Problems amenable to solution by mathematical optimization techniques have the general characteristics that 1 there are one or more independent variables whose values must be chosen to yield a viable solution, and 2 there is some measure of “goodness” available to distinguish between the many viable solutions generated by different choices of these variables. Mathematical optimization techniques are used for guiding the problem solver to that choice of variables that maximizes the goodness measure (profit, for example) or that minimizes some badness measure (cost, for example). One of the most important areas for the application of mathematical optimization techniques is in engineering design. Applications include: 1 2 3 4 5 6 7

the generation of “best” functional representations (curve fitting, for example), the design of optimal control systems, determining the optimal height (or length) of a mass transfer unit, determining the optimal diameter of a unit, finding the best equipment materials of construction, generating operating schedules, and selecting operating conditions.

Once a particular subject or process scheme has been selected for study, it is common practice to optimize the process from a capital cost and O&M (operation and maintenance) standpoint.

References

531

There are many optimization procedures available, most of them too detailed for meaningful application in a text of this nature. These sophisticated optimization techniques, some of which are routinely used in the design of conventional chemical and petrochemical plants, invariably involve computer calculations. However, use of these techniques in the majority of industrial applications is not warranted.(11) B

REFERENCES 1. M. MOYLE, “Introduction to Computers for Engineers,” John Wiley & Sons, Hoboken, NJ, 1967. 2. B. CARNAHAN and J. WILKES, “Digital Computing and Numerical Methods,” John Wiley & Sons, Hoboken, NJ, 1973. 3. R. KETTER and S. PRAWEL, “Modern Methods of Engineering Computations,” McGraw-Hill, New York City, NY, 1969. 4. J. REYNOLDS, J. JERIS, and L. THEODORE, “Handbook of Chemical and Environments Engineering Calculations,” John Wiley & Sons, Hoboken, NJ, 2004. 5. S. SHAEFER and L. THEODORE, “Probability and Statistics Applications for Environmental Science,” CRC Press/Taylor & Francis Group, Boca Raton, FL, 2007. 6. L. PEREZ, Homework Assignment, Manhattan College, Bronx, NY, 2003. 7. F. LAVERY, “The perils of differentiating engineering data numerically,” Chem. Eng., New York City, NY, Jan. 15, 1979. 8. L. THEODORE, class notes, Manhattan College, Bronx, NY, 1991. 9. L. THEODORE, class notes, Manhattan College, Bronx, NY, 1971. 10. R. ARIS, “Discrete Dynamic Programming,” Blaisdell, New York City, NY, 1964. 11. J. SANTOLERI, J. REYNOLDS, and L. THEODORE, “Introduction to Hazardous Waste Incineration,” 2nd edition, John Wiley & Sons, Hoboken, NJ, 2000.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

20

Open-Ended Problems INTRODUCTION The educational literature provides frequent references to individuals, particularly engineers and scientists, that have different learning styles; and, in order to successfully draw on these different styles, a variety of approaches can be employed. One such approach involves the use of open-ended problems. The term “open-ended problem” has come to mean different things to different people. It basically describes an approach to the solution of a problem and/or situation for which there is usually not a unique solution. Three literature sources(1–3) provide sample problems that can be used when this educational tool is employed. One of the authors of this text has applied this somewhat unique approach and has included numerous open-ended problems in several course offerings at Manhattan College. Student comments for the graduate course “Accident and Emergency Management” were recently tabulated. Student responses (unedited) to the question, “What aspects of this course were most beneficial to you?” are listed below. 1 “The open-ended questions gave engineers a creative license. We don’t come across many of these opportunities.” 2 “Open-ended questions allowed for candid discussions and viewpoints that the class may not have been otherwise exposed to.” 3 “The open-ended questions gave us an opportunity to apply what we were learning in class with subjects we have already learned which gave us a better understanding of the course.” 4 “Much of the knowledge that was learned in this course can be applied to everyday situations and in our professional lives.” 5 “Open-ended problems made me sit down and research the problem to come up with ways to solve them.” 6 “I thought the open-ended problems were inventive and got me to think about problems in a better way.” 7 “I felt that the open-ended problems were challenging. I, like most engineers, am more comfortable with quantitative problems vs qualitative.” Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

533

534

Chapter 20 Open-Ended Problems

The remainder of this chapter addresses a host of topics involved with open-ended problems. The remaining sections are entitled: Developing Students’ Power of Critical Thinking Creativity Brainstorming Inquiring Minds Failure, Uncertainty, Success: Are They Related? Angels on a Pin The chapter concludes with an Applications section containing open-ended Illustrative Examples, primarily in the mass transfer field.

DEVELOPING STUDENTS’ POWER OF CRITICAL THINKING(4) It has often been noted that we are living in the middle of an information revolution. For more than a decade, that revolution has had an effect on teaching and learning. Teachers are hard-pressed to keep up with the advances in their fields. Often their attempts to keep the students informed are limited by the difficulty of making new material available. The basic need of both teacher and student is to have useful information readily accessible. Then comes the problem of how to use this information properly. The objectives of both teaching and studying such information are: to assure comprehension of the material and to integrate it with the basic tenets of the field it represents; and, to use comprehension of the material as a vehicle for critical thinking, reasoning, and effective argument. Information is valueless unless it is put to use; otherwise, it becomes mere data. To use information most effectively, it should be taken as an instrument for understanding. The process of this utilization works on a number of incremental levels. Information can be: absorbed, comprehended, discussed, argued in reasoned fashion, written about, and integrated with similar and contrasting information. The development of critical and analytical thinking is the key to the understanding and use of information. It is what allows the practicing engineer to discuss and argue points of opinion and points of fact. It is the basis for the formation of independent ideas. Once formed, these ideas can be written about and integrated with both similar and contrasting information.

CREATIVITY Engineers bring mathematics and science to bear on practical problems, molding materials and harnessing technology for human benefit. Creativity is often a key component in this synthesis; it is the spark in motivating efforts to devise solutions to novel problems, design new products, and improve existing practices. In the competitive

Creativity

535

marketplace, it is a crucial asset in the bid to win the race to build better equipment and machines, decrease product delivery times, and anticipate the needs of future generations.(5) Although one of the keys to the success of an engineer or a scientist is to generate fresh approaches, processes and products, they also need to be creative. Gibney(5) has detailed how some schools and institutions are attempting to use certain methods that essentially share the same objective: open students’ minds to their own creative potential. Gibney(5) provides information on “The Art of Problem Definition” developed by the Rensselaer Polytechnic Institute. To stress critical thinking, they teach a seven step methodology for creative problem development. These steps are provided below: 1 Define the problem 2 State objective 3 Establish functions 4 Develop specifications 5 Generate multiple alternatives 6 Evaluate alternatives 7 Build In addition, Gibney(5) identified the phases of the creative process set forth by psychologists. They essentially break the process down into five basic stages: 1 Immersion 2 Incubation 3 Insight 4 Evaluation 5 Elaboration Psychologists have ultimately described the creative process as recursive. At any one of these stages, a person can double back, revise ideas, or gain new knowledge that reshapes his or her understanding. For this reason, being creative requires patience, discipline, and hard work. Finally, Dellafemina(6) recently outlined five secrets regarding the creative process: 1 Creativity is ageless 2 You don’t have to be Einstein 3 Creativity is not an eight hour job 4 Failure is the mother of all creativity 5 Dead men don’t create The reader is left with a thought from Theodore:(7) Creativity usually experiences a quick and quiet death in rooms that house large conference tables.

536

Chapter 20 Open-Ended Problems

BRAINSTORMING Panitz(8) has demonstrated how brainstorming strategies can help engineers generate an outpouring of ideas. Brainstorming guidelines include: 1 Carefully define the problem upfront 2 Allow individuals to consider the problem before the group tackles it 3 Create a comfortable environment 4 Record all suggestions 5 Appoint a group member to serve as a facilitator 6 Keep brainstorming groups small A checklist for change was also provided, as detailed below: 1 Adapt 2 Modify 3 Magnify 4 Minify 5 Put to other uses 6 Substitute 7 Rearrange 8 Reverse 9 Combine

INQUIRING MINDS In an exceptional and well-written article by Lih(9) entitled Inquiring Minds, Lih commented on inquiring minds saying “You can’t transfer knowledge without them.” His thoughts (which have been edited) on the inquiring or questioning process follow: 1 Inquiry is an attitude—a very important one when it comes to learning. It has a great deal to do with curiosity, dissatisfaction with the status quo, a desire to dig deeper, and having doubts about what one has been told. 2 Questioning often leads to believing—there is a saying that has been attributed to Confucius: “Tell me, I forget. Show me, I remember. Involve me, I understand.” It might also be fair to add: “Answer me, I believe.” 3 Effective inquiry requires determination to get to the bottom of things. 4 Effective inquiry requires wisdom and judgment. This is especially true for a long-range intellectual pursuit that is at the forefront of knowledge. 5 Inquiry is the key to successful life-long learning. If one masters the art of questioning, independent learning is a breeze.

Failure, Uncertainty, Success: Are They Related?

537

6 Questioning is good for the questionee as well. It can help clarify issues, uncover holes in an argument, correct factual and/or conceptual errors, and eventually lead to a more thoughtful outcome. 7 Teachers and leaders should model the importance of inquiry. The teacher/ leader must allow and encourage questions, and demonstrate a personal thirst for knowledge.

FAILURE, UNCERTAINTY, SUCCESS: ARE THEY RELATED? After many years of uncertainty, failures, and some successes, Theodore(7) ultimately came to the conclusion that failure is a unifying theme of success. This experience can allow one to proceed into uncharted waters, confident in his/her decision-making and problem-solving ability. Obviously, there are positive aspects associated with failures. Theodore(10) discusses this subject in his “As I See It” column entitled “On Failure.” Here is an excerpt from that article. I am amazed at how often we as a society employ the word failure (or fail) in a negative way. Here are some recent headlines or opening sentences/statements from the media: ‘local students fail to pass . . . a failure in Olympic gold effort . . . Charismatic fails to win Triple Crown . . . math scores a failure’ . . . , etc. Failure is used routinely to describe an event or activity. Unfortunately, some people have now come to be frightened of not only failure and the possibility of failing but also by the word itself. This kind of mentality can obviously have a negative impact on an individual. There is a thin line separating failure from success. For me, failure is an integral and positive part of life; success is often achieved following what others would describe as failure(s). As indicated above, failures are often encountered along the path of success. I often remind my students that my undergraduate GPA was 2.4/4.0 and that I was viewed as a failure by at least one of my professors. Later as a professor myself, my first 19 proposals (approximately 100 pages each) to the National Science Foundation, U.S. Environmental Protection Agency, etc., were rejected before my first award. Those were tough times, particularly for my fragile ego, but it proved to be a real learning experience. Today, I rarely prepare and submit a proposal unless I am wired for the award. I really believe that providing numerous examples of failures to my students has significantly helped them to achieve success. In any event, when you or someone you care about fails, remember the lines from DeCervantes’ Don Quixote: ‘Fortune may have yet a better success in reserve for you, and they who lose today may win tomorrow.’ Yep, it is okay to fail.

Ultimately, the degree to which one succeeds (or fails) is based in part on one’s state of mind or attitude. As President Lincoln once said: “Most people are about as happy as they make their minds to be.” William James once wrote: “The greatest discovery of my generation is that human beings can alter their lives by altering

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their attitude of mind.” So, no matter what one does, it is in the hands of that individual to make it a meaningful, pleasurable, and positive experience. This experience can ultimately bring success.

ANGELS ON A PIN(11) There is a tale that appeared in print many years ago that dissected the value of an open-ended approach to a particular problem. That story is presented below. Some time ago I received a call from a colleague who asked if I would be the referee on the grading of an examination question. He was about to give a student a zero for his answer to a physics question while the student claimed he should receive a perfect score and would if the system were not set up against the student. The instructor and the student agreed to submit this to an impartial arbitrater and I was selected. I went to my colleague’s office and read the examination question: “Show how it is possible to determine the height of a tall building with the aid of a barometer.” The student had answered: “Take a barometer to the top of the building, attach a long rope to it, lower the barometer to the street and then bring it up, measuring the length of the rope. The length of the rope is the height of the building.” I pointed out that the student really had a strong case for full credit since he had answered the question completely and correctly. On the other hand, if full credit was given, it could well contribute to a high grade for the student in his physics course. A high grade is supposed to certify competence in physics but the answer did not confirm this. I suggested that the student have another try at answering the question. I was not surprised that my colleague agreed but I was surprised that the student did. I gave the student six minutes to answer the question with the warning that the answer should show some knowledge of physics. At the end of five minutes, he had not written anything. I asked if he wished to give up but he said no. He had many answers to this problem; he was just thinking of the best one. I excused myself for interrupting him and asked him to please go on. In the next minute, he dashed off his answer which read: “Take the barometer to the top of the building and lean over the edge of the roof. Drop that barometer, timing its fall with a stopwatch. Then using the formula S ¼ 1/2at 2, calculate the height of the building.” At this point, I asked my colleague if he would give up. He conceded and I gave the student almost full credit. In leaving my colleague’s office, I recalled that the student had said he had many other answers to the problem and so I asked him what they were. “Oh yes,” said the student. “There are a great many ways of getting the height of a tall building with a barometer. For example, you could take the barometer out on a sunny day and measure the height of the barometer and the length of its shadow, and the length of the shadow of the building and by the use of a simple proportion, determine the height of the building.”

Applications

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“Fine,” I asked. “And the others?” “Yes,” said the student. “There is a very basic measurement method that you will like. In this method, you take the barometer and begin to walk up the stairs. As you climb the stairs, you mark off the length of the barometer along the wall. You then count the number of marks and this will give you the height of the building in barometer units. A very direct method.” “Of course, if you want a more sophisticated method, you can tie the barometer to the end of a string, swing it as a pendulum, and determine the value of ‘g’ at the street level and at the top of the building. From the difference of the two values of ‘g’ the height of the building can be calculated.” Finally, he concluded, there are many other ways of solving the problem. “Probably the best,” he said, “is to take the barometer to the basement and knock on the superintendent’s door.” When the superintendent answers, you speak to him as follows: “Mr. Superintendent, here I have a fine barometer. If you tell me the height of this building, I will give you this barometer.” At this point I asked the student if he really did know the conventional answer to this question. He admitted that he did, said that he was fed up with high school and college instructors trying to teach him how to think using the “scientific method,” and to explore the deep inner logic of the subject in a pedantic way, as is often done in the new mathematics, rather than teaching him the structure of the subject. With this in mind, he decided to revive scholasticism as an academic lark to challenge the Sputnik-panicked classrooms of America.

APPLICATIONS Several of the open-ended Illustrative Examples in this chapter have been drawn from the literature(1–3) and from the class notes of Theodore,(12) keying primarily on mass transfer issues.

ILLUSTRATIVE EXAMPLE 20.1 Your boss at work wishes to speed up the process used to determine the column diameter in the design of an absorber. You have been asked to convert the flooding line from U.S. Stoneware’s pressure drop correlations to equation form. You have also be asked to integrate this equation into a computer spreadsheet. The spreadsheet should calculate the diameter for the user once supplied with the appropriate system values (such as L, G, F, fraction of flooding velocity, and so on). SOLUTION: Below is one equation that relates the liquid and gaseous flow rates in the tower with the column cross-sectional area at the flooding point (13) Y ¼ 3:84  1:06X  0:119X 2

(20:1)

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Chapter 20 Open-Ended Problems "

Where

# wG2 F cm0:2 L Y ¼ ln rG rL gc Af2 "   # L rG 0:5 X ¼ ln G rL wG ¼ gas flow rate, lb/h L ¼ liquid superficial mass velocity at the bottom of packing, lb/h . ft2 G ¼ gas superficial mass velocity at the bottom of packing, lb/h . ft2

rL ¼ liquid density, lb/ft3 rG ¼ gas density, lb/ft3 F ¼ packing factor, dimensionless Af ¼ cross-sectional area of the column, ft2

mL ¼ liquid viscosity, cP c ¼ ratio, density of water/liquid density, dimensionless gc ¼ constant ¼ 4.173  108 lb . ft/lbf . h This equation is applicable in the range 0.01 , e X , 10 or 24.6 , X , 2.3.

B

ILLUSTRATIVE EXAMPLE 20.2 A twenty-four-year-old perforated plate distillation column is no longer delivering the degree of separation required for the process. Rather than replace the unit, you have been asked to recommend what other possible steps can be taken to the existing unit to get it back “on line.” Unfortunately, it is not currently possible to completely replace the column. Some other solutions need to be found. SOLUTION: The issue with the current operation is its inability to perform the separation at its designed level. The first thing that needs to be investigated is what is the current state of the column? How was it originally designed? Were safety margins included in the design that allow for some “tweaking” of the actual structure of the column? How much change in pressure (drop) is allowed? What will the materials stand up to? A possibility for the column is changing the flow rate of gas through the column. If the gas is made to pass through the column at a higher velocity, it should result in more foaming of the liquid on the trays. This will increase the mass transfer area. This will also help to separate the products. Unfortunately, this could create new problems with flooding and entrainment. This needs to be examined very closely before it is implemented. It also might be possible to change the reboiler and condenser conditions. It is possible after 24 years of service that these two devices are not working at their optimum settings. An examination should be made of their contribution to the failure to separate the products. Is the column failing because these devices are no longer functioning? If so, then they need to be replaced with newer models. The composition and conditions of the feed stream also need to be examined. Is it the same as what the column was originally designed to separate? Has the process been altered upstream in

Applications

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such a way that the separation is no longer feasible? Perhaps it has changed through the years in such a manner that the column is no longer designed to deal with the stream. This stream could be altered to adhere to the original conditions to make the column operate correctly again. Also, the feed tray (location) could be changed; the inlet stream quality (defined as q in Chapter 9) might also be changed. Adding trays is the most difficult thing to implement but would probably have the largest impact. The trays could be added near the top of the column if a safety margin of headspace had been designed into the column. A possibility would be the complete removal of all the trays and reinstalling them with smaller tray spacing between them. If that is done, the company should consider replacing the trays with a better tray design. Significant work has gone into tray design in the past 24 years. Improvements have been made to their efficiency and capabilities. Finally, it is possible that the column needs cleaning and that could increase the column’s efficiency. B

ILLUSTRATIVE EXAMPLE 20.3(14) Consider a chemical plant that uses two liquid feeds of different densities as provided in Table 20.1. It produces four different liquid chemical products of varying density following chemical reaction and separation, see Table 20.2. The plant storage requirements call for maintaining 4–5 weeks supply of each feed, 4–6 weeks supply of products A, B, and C, and 1– 2 weeks supply of product D. The plant operates year round but each tank must be emptied once a year for a week for maintenance. Tanks are normally dedicated to one feed or product and one or two could be used as “swing” tanks; however, one day of cleaning is required between uses with different liquids. Specify a set of tanks from the “standard” sizes given in Table 20.3 to minimize this plant’s needs. Table 20.1 Feed Data Feed 1 Feed 2

110,000 lb/day 50,000 lb/day

r ¼ 49 lb/ft3 r ¼ 68 lb/ft3

Table 20.2 Product Data Product A Product B Product C Product D

40,000 lb/day 25,000 lb/day 10,000 lb/day 95,000 lb/day

r ¼ 52 lb/ft3 r ¼ 62 lb/ft3 r ¼ 52 lb/ft3 r ¼ 47 lb/ft3

Table 20.3 Tank Data Standard Tank Sizes (gallons): 2800 11,200 56,100 5600 16,800 140,000 8400 28,100 281,000

561,000 1,123,000

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Chapter 20 Open-Ended Problems

Table 20.4 Storage Requirements Material

lb/day

Gal/day

Days

Gals storage required

Feed 1 Feed 2 Product Product Product Product

110,000 50,000 40,000 25,000 10,000 95,000

16,800 5500 5750 3020 1440 15,120

28 –35 28 –35 28 –42 28 –42 28 –42 7–14

470,200– 587,800 154,000– 192,500 161,100– 241,700 84,500– 126,700 40,300– 60,400 105,800– 211,700

A B C D

SOLUTION: There is no single, simple method for determining the optimum mix of stage tanks for a chemical plant. Most often, estimates are made of the minimum and maximum amounts of feeds, intermediates, and products that must be kept on hand. Then some additional allowance is made to permit periodic cleaning and maintenance of the tanks. The minimum number of tanks may not always be optimum if the tanks are extremely large. Several smaller tanks may cost somewhat more initially but they offer more flexibility in use. One solution and some calculational details follow. Determine the maximum and minimum amounts of each material to be stored (see Table 20.4). The conversion to gallons requires dividing the rate in lb/day by the density in lb/ft3, then multiplying by 7.481 gal/ft3. For each material, select a set of tanks: Feed 1: Feed 2: Product Product Product Product

A: B: C: D:

Use 2 Use 4 Use 2 Use 2 Use 1 Use 2

281,000 gallon 56,100 gallon 140,000 gallon 56,100 gallon 56,100 gallon 140,000 gallon

tanks tanks tanks tanks tanks tanks

This set is 5% short on the maximum of Feed 1, 11% short on Product B, and 7% short on Product C. For most situations, this would be acceptable. Select spare and/or “swing” tanks to provide for maintenance: Feed 1: Feed 2 and all products:

Use an additional 281,000 gallon tank Use one 56,100 gallon “swing” tank

This combination provides adequate auxiliary storage for maintenance periods: Total tanks required: 3 4 8

281,000 gallons 140,000 gallons 56,100 gallons

The number of tanks required will be quite large in this application. If market forces such as fluctuating demand require this much storage, they may all be necessary. More modern commercial operations such as “just-in-time” manufacturing, call for reducing in-plant inventory to the absolute minimum possible. B

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543

ILLUSTRATIVE EXAMPLE 20.4 Consider the absorber system shown in Figure 20.1. It is designed to operate with a maximum discharge concentration of 50 ppm. Once the unit is installed and running, the unit operates with a discharge of 60 ppm. Rather than purchase a new unit, what options are available to bring the unit into compliance with the specified design concentration? SOLUTION: 1 Use a smaller packing size. A different size may produce a better packing factor. 2 Use a different type of packing. A different packing type may produce a better packing factor. 3 Make sure there is no channeling inside the column. The packing should be randomly distributed and there should be no open spots where water will accumulate. 4 Increase the water flow entering the column. If the water flow is increased, there will be a greater mass transfer between the liquid and gas phase and the efficiency of the column will increase. In effect, the slope of the operating line will increase and provide a greater driving force for mass transfer. 5 Use a liquid other than water to scrub the gas. Another liquid may cause better mass transfer between the gas and liquid phase.

Figure 20.1

Faulty design?

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Chapter 20 Open-Ended Problems

6 Check to see if flooding is occurring. If the column is near the flooding point, there will not be much mass transfer. The water will be pushed up the column by the gas rather than the water coming down the column. 7 Lower the initial concentration in the carrier gas (if possible). 8 Increase the height of the bed. This will provide more surface area for mixing and more time in the absorber. 9 Increase the pressure. This will reduce the slope of the operating line and lead to a greater driving force. 10 Decrease the temperature. This will reduce the slope of the operating line and lead to a greater driving force. 11 Increase the pressure and decrease the temperature. 12 Additional packing height could be added at the top of the column. This would increase the amount of the solute removed from the gaseous phase. 13 Add sprays at the bottom of the column. 14 Modify the process producing the problem. Since the details of this process are not included in the problem statement, no specific recommendations can be mentioned in this discussion. 15 Finally, before considering changes to the system, one should undertake a thorough inspection of the unit and surrounding components. The emission monitoring system should be recalibrated. All valves, fittings, and pipes should be checked for plugging or leaks. The liquid distributor should be checked to make sure it is functioning properly. The distribution of the packing should be inspected to make sure it is as uniform as possible. Any problems encountered during this inspection should be corrected immediately. Following the maintenance check, the performance of the unit should be re-evaluated. The reader should note that some of the recommendations above could lead to higher pressure drops and potential problems with the flow. An example of this problem would be the implementation of suggestion 1. The reader is left the exercise of determining what other steps could lead to flow/pressure drop problems. Hint: There are at least six suggestions that fall into this category, including firing the engineer who designed the absorber. B

ILLUSTRATIVE EXAMPLE 20.5 Refer to Illustrative Example 20.4. A similar situation exists with a gaseous adsorption unit. Rather than purchase a new unit, what options are available to bring the unit into compliance with the specified design concentration? Note: The packing is activated carbon and there is now no liquid flow. SOLUTION: 1 Check/increase the depth of the adsorption bed. Make up any adsorbent lost to carryover (as necessary). Increasing the depth of the bed will increase the removal capacity of the column. 2 Change the adsorbent type or size. Use an adsorbent with a higher saturation capacity. 3 Reducing the flow will increase the residence time and may allow for more adsorption. 4 Decrease the inlet temperature. This will favor the adsorption equilibrium process. 5 Increase the system pressure. This will favor the adsorption equilibrium process.

Applications

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6 Regenerate the adsorbent for a longer period. 7 Replace the adsorbent. 8 Consider changing the flow direction. Again, pressure drop considerations should be included in the analysis.

B

ILLUSTRATIVE EXAMPLE 20.6 Refer to Illustrative Example 20.5. A similar situation exists with a liquid adsorption unit. Rather than purchase a new unit, what options are available to bring the unit into compliance with the specified design concentration? SOLUTION: The solution to this problem, in many respects is similar to that presented in Illustrative Examples 20.4 and 20.5. B

ILLUSTRATIVE EXAMPLE 20.7 Comment on the various techniques available for the treatment of solid wastes. SOLUTION: There are many ways to manage solid waste. A few methods are discussed below. One option in solid waste treatment is incineration. There are two major types of incinerators: stoker types and fluidized bed types. Stokers have the capability to stabilize the combustion for many types of materials, including infectious materials. The advantage of the fluidized bed incinerator is its ability to achieve nearly isothermal conditions within the bed. The high mass and heat transfer rates lower the temperature necessary to achieve the required removal/ destruction efficiency. Fluidized bed incineration is used mainly for hazardous waste. There may be direct energy recovery. Plasma membranes are an efficient way of disposing of wastes, including infectious materials. This process decomposes complex organic molecules into hydrogen, carbon dioxide, coke and hydrogen chloride. The waste is broken into elemental atoms in plasma. The gases are cooled and sprayed with caustic to remove particulates and acid gases. This method is used for the complete destruction of PCBs and dioxins. Instead of complete disposal, there are alternatives such as recycling. There are several applications of recycling. The first is direct onsite reuse where the site/plant reuses the waste it generated. Offsite recovery can be utilized when the refuse generated by the site is too little to warrant an onsite recovery unit. Another option in recycling is the sale of refuse offsite. In this option, the plant sells the waste from their plant for use as raw materials in another plant. Yet another way to dispose of solid waste is landfilling. This solution partially appeases the aesthetic problem waste brings to a society. In this solution, the waste is buried and out of sight. Burying the waste also aids in the ultimate disposal of it. The organic parts of the waste can decompose and hopefully, once again become part of the Earth. B

ILLUSTRATIVE EXAMPLE 20.8 A certain petrochemical company plans on installing an adsorption unit to remove a specific volatile organic compound (VOC). Rather than rely on existing design methods and procedures entirely, the company wishes to conduct a pilot scale test to aid in the design. You have been

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Chapter 20 Open-Ended Problems

asked to plan this test. Discuss how you will do this. How will you simulate actual conditions at the pilot scale? What measurements need to be taken? SOLUTION: This study would obviously require a detailed plan of action. A summary of the key steps are provided below. Prepare a bench scale unit for testing. Prepare a small pilot scale unit for testing. Use the results of steps (2) and (3) above for scale up purposes. During testing, measure inlet and outlet concentrations, and pressure drop for a variety of operating conditions. 5 Consider another mass transfer option. B 1 2 3 4

ILLUSTRATIVE EXAMPLE 20.9 You are an engineer at an Aquafina plant. As stated on a bottle of Aquafina water, it is purified using reverse osmosis (RO). One of the RO units at your plant is no longer operating to the required “separation” efficiency. What steps should be taken to avoid purchasing a new purification unit? SOLUTION: 1 Check the membrane to see if it needs to be replaced. 2 If the membrane is satisfactory, check the concentration of the feed stream to the unit. The feed stream may be more concentrated due to a problem before the RO unit. 3 Check that the appropriate pressure is being applied in order to overcome the osmotic pressure (i.e., check the pumps). 4 Check that none of the other components of the system have been damaged by the membrane. B

ILLUSTRATIVE EXAMPLE 20.10 List some advantages and disadvantages of employing distillation vs liquid–liquid extraction to separate a two-component liquid stream. SOLUTION: Distillation is normally employed for large-scale processing and where there is a need for a high degree of separation. Maintenance costs are usually lower. It almost always is the preferred choice. Liquid–liquid extraction is less expensive to operate and there are no heat transfer requirements. It is usually employed for small-scale operations. Some specific comments are provided below.

Distillation: 1 A separation technique that is based on difference in boiling points/volatilities of the individual components. 2 The feed flows down the column to the reboiler where it is heated; the vapor is allowed to flow up the column and the liquid product is removed in the reboiler. The vapor product is condensed and removed from the top of the column as the distillate.

References

547

Liquid – liquid extraction: 1 A separation process that is based on the relative solubilities of solutes in immiscible solvents. 2 The solution containing the desired solute is contacted with another solvent that is immiscible with the solution. 3 The solvent is chosen so that the solute in the solution has more affinity for the added solvent. Mass transfer of the solute from the solution to the solvent occurs. 4 The solvent containing the extracted solute leaves the top of the column and is referred to as the extract stream. The solution exits the bottom of the column containing only small amounts of solute and is called the raffinate. Refer to Chapters 9 and 12 for additional details.

B

REFERENCES 1. A. M. FLYNN and L. THEODORE, “An Air Pollution Control Equipment Design Course for Chemical and Environmental Engineering Students Using an Open-Ended Problem Approach,” ASEE Meeting, Rowan University, NJ, 2001. 2. A. M. FLYNN, J. REYNOLDS, and L. THEODORE, “Courses for Chemical and Environmental Engineering Students Using an Open-Ended Problem Approach,” AWMA Meeting, San Diego, CA, 2003. 3. L. THEODORE, personal notes, 2002–2004. 4. Manhattan College Center for Teaching, “Developing Students’ Power of Critical Thinking,” Bronx, NY, January, 1989. 5. K. GIBNEY, “Awakening creativity,” ASEE Promo, Washington, DC, March, 1988. 6. J. DELLAFEMINA, “Jerry’s Rules,” Modern Maturity, March–April, 2000. 7. L. THEODORE, personal notes, 1998. 8. B. PANITZ, “Brain storms,” ASEE Promo, Washington, DC, March, 1998. 9. M. LIH, “Inquiring minds,” ASEE Promo, Washington, DC, December, 1998. 10. L. THEODORE, “On Failure,” AS I SEE IT Column, The Williston Times, East Williston, NY, July 2, 1999. 11. “Angels on a Pin,” www.lhup.edu/~dsimanek/angelpin.htm 12. L. THEODORE, class notes, Manhattan College, Bronx, NY, 1999–2003. 13. J. REYNOLDS, personal notes, Manhattan College, Bronx, NY, 1987. 14. D. KAUFFMAN, “Process Synthesis and Design,” A Theodore Tutorial, Theodore Tutorials, East Williston, NY, 1992.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Chapter

21

Ethics INTRODUCTION The primary responsibility of an engineering professional is to protect public health and safety. However, engineering professionals also have a responsibility to their employers or clients, to their families, to themselves, and to the environment. Meeting these responsibilities will challenge the practicing engineer to draw upon a system of ethical values. Well, what about ethics? Ethics means “doing the right thing” as opposed to “what you have the right to do.” But doing the right thing is not always obvious or easy. In fact, ethical decisions are often difficult and may involve a certain amount of self-sacrifice. Doing the right thing for a practicing engineer can be especially challenging. Furthermore, the corporate and government world has confused this concept by developing ethics programs that emphasize only what you have the right to do. An organization, for example, may have a list—often called a Code of Ethics or Code of Conduct—of what an employee can and cannot get away with. Employees are required to sign an acknowledgment that they have read and understood the list. The company unfortunately calls this “ethics training.”(1) One difficulty in some situations is recognizing when a question of ethics is involved. Frequently, in the area of environmental management, a breach of ethics involves a practice that endangers public health and safety or covers up a violation of a rule or regulation. Occasionally, however, a breach may involve a case of the exact opposite. This might seem an unlikely scenario. How can someone be too honest, too caring, or too professional? Regarding the above, one example is lying to save a life. Suppose you are standing on a street and a woman runs past you chased by two men. She screams, “They are trying to attack me!” as she dashes into the entry of a building around a corner. The men ask you, “Which way did she go?” What do you tell them? Clearly, the right thing is to lie. In this case, the value of caring overrides the value of honesty. This situation is exaggerated to illustrate that sometimes it is appropriate to violate certain values to protect public health and safety. In doing the right thing, ideally one should not have to make snap decisions and should take the time to

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

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Chapter 21 Ethics

investigate all of the facts, e.g., whether or not the woman was a thief and the men were police. Sometimes one must decide how much to sacrifice to ensure public health and safety. In establishing environmental regulations, the regulating agency must decide how safe and how stringent to make the regulations. For example, in the case of air toxic regulations, one standard may result in 10 cancer cases per one million people. But why isn’t it for one or none? Who should decide?(2) This chapter, in line with its title, addresses ethics and ethical issues relevant to the practicing engineer. Topics covered include: Teaching Ethics Case Study Approach Integrity Moral Issues Guardianship Engineering and Environmental Ethics Future Trends The chapter concludes with three Illustrative Examples in the mass transfer arena. These case studies have been primarily drawn from the work of Wilcox and Theodore.(3)

TEACHING ETHICS Professionals are often skeptical about the value or practicality of discussing ethics in the workplace. When students hear that they are required to take an ethics course or if they opt for one as an elective in their schedules, they frequently wonder whether ethics can be taught. They share the skepticism of the practitioners about such discussion. Of course, both groups are usually thinking of ethics as instruction in goodness, and they are rightly skeptical, given their own wealth of experience with or knowledge of moral problems. They have seen enough already to know that you cannot change a person’s way of doing things simply by teaching about correct behavior. The teaching of ethics is not a challenge if ethics is understood only as a philosophical system. Parks(4) notes that teaching ethics is important but “if we are concerned with the teaching of ethics that is understood as the practice of accountability to a profession vital to the common good, the underlying and more profound challenge before all professional schools [and other organizations] is located in the question—How do we foster the formation of leadership characterized, in part, by practice of moral courage?”(4) Moral courage requires knowing and acting. College and university educators, as well as those charged with ethics training in the private sector, develop a sense of uneasiness when topics such as “fostering leadership formation,” “moral courage,” or “knowing and acting on that knowledge” are mentioned. Such terms resurrect images of theological indoctrination, Sunday school recitations, or pulpit sermonizing.

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These images contrast sharply with what the present-day professor envisions as the groves of academic freedom and dispassionate analysis. Perhaps out of fear of disrespecting the dignity of students and devaluing their critical reasoning powers or their ability to understand where the truth lies, faculty will take a dim view of academic goals that go beyond those strictly cognitive. The consequence of such values among the professoriate is the further erosion of a moral commons where an agreed-upon set of values and beliefs allows for discourse on ethics. Of course, the erosion has continued steadily from the inception of the Enlightenment Project in the seventeenth century until the present day wherever industrialized and postindustrialized societies have been subject to rapid cultural, economic, political, and technological change. It is not simply an erosion in the realm of higher education. Practitioners in the engineering and scientific communities experience the same erosion of the moral commons taking place in society as a whole. The authors are certainly in agreement with their colleagues in higher education and those who do ethics training in the private sector, that individuals are not to be manipulated or indoctrinated. However, they are also convinced not only that students and other participants in ethics analysis must have a body of knowledge but also that they have a responsibility for the civic life of American society. Such responsibility requires leadership, moral courage, and action. Of course, none of these characteristics can be demanded or forced, only elicited. That is the great, yet delicate challenge facing the professoriate and all those charged with ethics training in other sectors. Eliciting a sense of civic responsibility as a goal of ethics analysis can be realized only as a derivative of cognitive processes and not as a direct goal. In sum, the formation of personal character and the practice of virtue are not to be subject to external control and the diminution of individual freedom through manipulation or indoctrination.

CASE STUDY APPROACH The authors believe that the case study method is a valuable way to take seriously Parks’s response to the question “Can ethics be taught?” They also consider the method an important tool in investigating the relationship among assumptions, values, and the moral life, as well as ethical reflection on those three aspects of life. The authors are convinced that the case study method is one of the most useful ways of teaching ethics and of achieving the goals of ethics education outlined by the Hastings Center.(5) 1 Stimulating the moral imagination. The concreteness of the case study appeals to the learning style of most people. While a certain amount of ambiguity is essential to evoke interest and discussion, it is also a stimulus to enlivening the knowledge. Hopefully, the participant will begin to appreciate the moral complexity of a situation that in the past might have been thought of only as a technical or managerial problem. Practice in the art of case discussion has the larger intent of leading the individual to bring an ethical frame of reference to bear on the variety of problems faced in the discipline studied. Stimulating

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the moral imagination is similar to putting on a pair of glasses that are tinted. The result is that the world is seen through that tint. As a consequence of the case study method, the editors and authors of the cases hope that each individual will see his or her field of study through the interpretive glasses of engineering and environmental ethics. He or she would then routinely ask: “What is the moral issue here?” 2 Recognizing ethical issues. The case analyst should not be content with a good “imagination.” The further challenge is the recognition of specific moral problems and how they differ from one another in terms of immediacy or urgency. Concreteness is an important asset of the case study and clearly assists in achieving this second goal. Comparing and contrasting a variety of cases through discussion is essential to recognition and leads to achievement of the next goal. 3 Developing analytical skills. Differentiation, comparison, contrasts—all of these must be related to an enhanced ability to solve the problem. To achieve this goal, the student of ethics is taught to bring the skills developed in his or her major field of study to bear on the ambiguous situation, the moral dilemma, or the competing values that must be addressed. Analytic skills are best honed through the use of examples or cases. The technical ability to analyze all dimensions of an environmental spill will have an impact on how the moral aspect of the problem is understood in terms of resolving the problem. Of course, ethical systems that emphasize the importance of consequences, the obligations inherent in a duty-based ethic, as well as theories of justice or virtue will enhance the ability to use technical or discipline-based analytic skills in resolving the problem. Knowing, however, is related to acting. This leads to the fourth goal. 4 Eliciting a sense of moral obligation and personal responsibility. Much has already been said about the importance of this goal. However, it should be clear that a sense of moral obligation does not mean there is one set of absolute answers. Dictating a solution is quite different from an internalization process whereby the individual commits himself or herself to be a “seeker,” one who takes personal responsibility for addressing and resolving the moral problems facing engineers or scientists. Both professions constitute the “guardians of the system” in the technical community. They are the first line of response to the problems and dilemmas facing the professions as such. To point to the Environmental Protection Agency, the Occupational Safety and Health Administration, the Federal Bureau of Investigation, congressional formulators of public policy, or other sovereign countries as the parties responsible for resolving acute problems is to abnegate one’s moral responsibility as a professional person. To say this is not to dictate solutions, but to alert individuals to their personal responsibility for the integrity of the respective field. Eliciting a sense of responsibility depends on an assessment of the assumptions or “images at the core of one’s heart.” Assessment of ethical systems or normative frames of reference must be connected to the actual assumptions or images that constitute a

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person’s world view. Challenging the individual to examine that world view in relation to a case and ethical systems is the first step in joining doing to knowing. Closely related to the achievement of this goal is the following one. 5 Tolerating—and resisting—disagreement and ambiguity. An essential component of case discussion is the willingness to listen carefully to the pointsof-view held by others. Cases, by their nature, are ambiguous. They are bare-boned affairs meant more to be provocative than to lead to a clear-cut jury decision. The purpose of the case is to stimulate discussion and learning among individuals. As a result, there will be much disagreement surrounding the ethical issues in the case and the best option for resolving it. Toleration does not mean “putting up with people with whom I disagree.” Respect for the inherent dignity of the person and a willingness to understand not only another position, but also a person’s reasons for or interest in that point-ofview, should be part of the case discussion. Toleration does not mean all opinions must be of equal value and worth. It is true that respect for and listening to another person’s argument may lead one to change a position. However, a careful description and discussion of the other person’s position may also lead to a greater conviction that one’s own position is correct. What is clearly of central concern is the belief that the free flow of ideas and carefully wrought arguments, presented from all sides without fear of control manipulation, threat, or disdain, is at the core of human understanding and development. This hallowed concept of academic freedom is the catalyst that allows human communities to be committed to the search for truth without at the same time declaring absolute possession of the truth.

INTEGRITY Scenarios are, for the most part, designed to reflect ambiguity in work situations. The ethicist hopes to get his or her hands dirty, dealing with the bottom-line motives of survival, competitiveness, and profitability as well as the mixed motives of selfinterest, respect for the rights of others, and altruism. Obtaining an ethical solution to a difficult moral problem or dilemma is based on much more than choosing the correct ethical framework with its normative frame of reference. One must also be ready to examine fundamental assumptions and the values to which the assumptions give rise. Carter has made this point recently in a discussion of “integrity.”(6) 1 Honesty in relation to integrity. Carter explores integrity in relation to the value society places on honesty. On this subject, one of the best-known and most popular ethics books of the last few decades is Bok’s Lying: Moral Choice in Public and Private Life, 1978.(7) Without taking away from the merits of Lying, Carter notes: “Plainly, one cannot have integrity without being honest (although, as we shall see, the matter gets complicated), but one can certainly be honest and yet have little integrity.”(6) Honesty is far easier to practice than the tough work of figuring out what it takes to have

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integrity in a situation. Integrity requires a high degree of moral reflectiveness. Honesty may result in harm to another person. Furthermore, “if forthrightness is not preceded by discernment, it may result in the expression of an incorrect moral judgment.”(6) The racist may be transparently honest, Carter declares, but he certainly lacks integrity because his beliefs, deeply held as they might be, are wrong. He has not engaged in the hard work of examining his fundamental assumptions, values, beliefs. 2 Personal integrity without public responsibility? It would appear that one cannot have integrity without responsibility since any consideration of integrity addresses the effects of our conduct on other people. In our work life and our community life, we have public responsibilities for our clients and fellow citizens.

MORAL ISSUES(8) The conflict of interest between Chief Seattle (and Native Americans in general) and President Pierce (and the European –American expansion) provides a perfect example of how ethics and the resulting codes of behavior they engender can differ drastically from culture to culture, from religion to religion, and even from person to person. This enigma, too, is noted again and again by Seattle:(9) I do not know. Our ways are different from your ways. . . . But perhaps it is because the red man is a savage and does not understand. . . . The air is precious to the red man, for all things share the same breath . . . the white man does not seem to notice the air he breathes. . . . I am a savage and do not understand any other way. I have seen a thousand rotting buffaloes on the prairie, left by the white man who shot them from a passing train. I am a savage and I do not understand how the smoking iron horse can be more important than the buffalo we kill only to stay alive.

Chief Seattle sarcastically uses the European word “savage” and all its connotations throughout his address. When one finishes reading the work, it becomes obvious which viewpoint (President Pierce’s or his own) Chief Seattle feels is the savage one. What his culture holds dearest (the wilderness), the whites see as untamed, dangerous, and savage. What the whites hold in highest regard (utilization of the Earth and technological advancement), the Native Americans see as irreverent of all other living things. Each culture maintains a distinct and conflicting standard for the welfare of the world. Opposing viewpoints and moralities such as these are prevalent throughout the world and have never ceased to present a challenge to international, national, state, community, and interpersonal peace. It is generally accepted, however, that any historical ethic can be found to focus on one of four different underlying moral concepts: 1 Utilitarianism focuses on good consequences for all. 2 Duties Ethics focus on one’s duties. 3 Rights Ethics focus on human rights. 4 Virtue Ethics focus on virtuous behavior.

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Utilitarians hold that the most basic reason why actions are morally right is that they lead to the greatest good for the greatest number. “Good and bad consequences are the only relevant considerations, and, hence all moral principles reduce to one: ‘We ought to maximize utility.’”(9) Duties Ethicists concentrate on an action itself rather than the consequences of that action. To these ethicists there are certain principles of duty such as “Do not deceive” and “Protect innocent life” that should be fulfilled even if the most good does not result. The list and hierarchy of duties differs from culture to culture, religion to religion. For Judeo-Christians, the Ten Commandments provide an ordered list of duties imposed by their religion.(9) Often considered to be linked with Duties Ethics, Rights Ethics also assesses the act itself rather than its consequences. Rights Ethicists emphasize the rights of the people affected by an act rather than the duty of the person(s) performing the act. For example, because a person has a right to life, murder is morally wrong. Rights Ethicists propose that duties actually stem from a corresponding right. Since each person has a right to life, it is everyone’s duty to not kill. It is because of this link and their common emphasis on the actions themselves that Rights Ethics and Duty Ethics are often grouped under the common heading Deontological Ethics.(10) The display of virtuous behavior is the central principle governing Virtue Ethics. An action would be wrong if it expressed or developed vices—for example, bad character traits. Virtue Ethicists, therefore, focus upon becoming a morally good person. To display the different ways that these moral theories view the same situation, one can explore their approach to the following scenario that Martin and Schinzinger present:(9) On a midnight shift, a botched solution of sodium cyanide, a reactant in organic synthesis, is temporarily stored in drums for reprocessing. Two weeks later, the day shift foreperson cannot find the drums. Roy, the plant manager, finds out that the batch has been illegally dumped into the sanitary sewer. He severely disciplines the night shift foreperson. Upon making discreet inquiries, he finds out that no apparent harm has resulted from the dumping. Should Roy inform government authorities, as is required by law in this kind of situation?

If a representative of each of the four different theories on ethics just mentioned were presented with this dilemma, their decision-making process would focus on different principles. The Utilitarian Roy would assess the consequences of his options. If he told the government, his company might suffer immediately under any fines administered and later (perhaps more seriously) due to exposure of the incident by the media. If he chose not to inform authorities, he risks heavier fines (and perhaps even worse press) in the event that someone discovers the cover-up. Consequences are the utilitarian Roy’s only consideration in his decision-making process. The Duties Ethicist Roy would weigh his duties and his decision would probably be more clear-cut than his utilitarian counterpart. He is obliged foremost by his duty to obey the law and must inform the government. The Rights Ethicist mind-frame would lead Roy to the same course of action as the duties ethicist—not necessarily because he has a duty to obey the law but because the people in the community have the right to informed consent. Even though Roy’s

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inquiries informed him that no harm resulted from the spill, he knows that the public around the plant has the right to be informed of how the plant is operating. Vices and virtues would be weighed by the Virtue Ethicist Roy. The course of his thought process would be determined by his own subjective definition of what things are virtuous, what things would make him a morally good person. Most likely, he would consider both honesty and obeying the law virtuous, and withholding information from the government and the public as virtueless and would, therefore, tell the authorities.

GUARDIANSHIP Despite the great teaching advantage that comes with case use, there are two important questions that case discussants must keep in mind when they assess the ethical problem: 1 Who are the guardians of the system? This question addresses the issue of who, among engineering or science professionals, is responsible for the ethical standards in the organization. If professionals point the finger at senior management, the legal department, the Environmental Protection Agency, or the Department of Justice, they have indeed misunderstood the nature of a professional calling. The first line of defense is the willingness of professionals themselves to maintain and enhance the integrity of the engineering or scientific profession through their own personal adherence to the highest standards of conduct and to assume responsibility for commitment to these standards within the companies where they work. Moreover, ethics is a positive task, not a list of dos and don’ts. To achieve excellence in one’s work presumes a commitment to the client’s contract, public safety, and environmental integrity, among several factors that are all too often thought of as “management” issues. They are, in reality, the ethical standards of the work itself. Thus, the ethical engineer or scientist is the one who identifies with the profession and all that is involved in the work assigned or contracted. 2 Who gives support to the guardians of the system? This second issue goes to the heart of the assessment problem, but also has an impact on the first issue. Unless the organization backs those who assume positive responsibility for the ethical tenor of the group, very little will change. Why would someone risk ostracism or retaliation by confronting a person engaging in unethical behavior or illegal behavior if there is no institutional support for the one assuming responsibility? Effective guardianship is facilitated if: 1 There are clear-cut standards of behavior and high expectations of the membership. 2 The standards are brought to the attention of the members through a welldeveloped training program.

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3 The standards are taken seriously by the senior leadership team of the firm. They must demonstrate that seriousness by taking an active role in the training, without, at the same time, creating a chilly climate stifling discussion and participation in the training. The ethics program must be seen not as frosting on the cake or as a value added on to forestall legal problems through better compliance. The CEO needs to demonstrate a commitment to the values and principles that drive the business. Ethics training is no add-on. Ethics is what drives the organization: trust, integrity, fidelity to the client. 4 It is evident that the leadership “walks the talk” in all aspects of its decision making and actions. 5 There are mechanisms in place to address the concerns of the members, mechanisms such as an ombudsperson, a hotline, etc. 6 Those who adversely affect the integrity of the business are effectively and fairly disciplined. Another way of addressing the question of who supports the guardians is to emphasize the importance of organizational or corporate culture. A positive response to the six points just raised has a great impact on the culture of the organization. Unless there is what is sometimes called a “thick” culture, wherein respect for and adherence to guardianship and the tenets of integrity and trust are palpable, individually ethical persons can do very little to raise the moral climate. An organization is more than the sum total of the individuals who constitute the membership. The attitudes conveyed, values expressed, and ways of doing business in an organization profoundly affect the perceptions of the members therein and set the tone of the company. Having a positive impact on culture is a great challenge which is not easily achieved. Culture is so subtle that one often does not even realize or understand its dimensions until a significantly different culture is experienced. An organization will not have effective guardianship of the system unless there is a concerted attempt to create, enhance, or reinforce a culture where values and ethics are clear and fully supported. There is little doubt, however, that the twin issues of guardianship and culture are much more difficult to address than the institutionalization of the ethics program itself.

ENGINEERING AND ENVIRONMENTAL ETHICS(11) In the ethical theories presented here, established hierarchies of duties, rights, virtues, and desired consequences exist so that situations where no single course of action satisfies all of the maxims can still be resolved. The entry of environmentalism into the realm of ethics raises questions concerning where it falls in this hierarchy. Much debate continues over these questions of how much weight the natural environment should be given in ethical dilemmas, particularly in those where ecological responsibility seems to oppose economic profitability and technological advances.

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Those wrapped up in this technology/economy/ecology debate can generally be divided into three groups: 1 Environmental extremists. 2 Technologists to whom ecology is acceptable provided it does not inhibit technological or economic growth. 3 Those who feel technology should be checked with ecological responsibility. Each is briefly discussed below. After his year-and-a-half of simple living on the shores of Walden Pond, Henry David Thoreau rejected the pursuit of technology and industrialization. While most would agree with his vision of nature as being inspirational, few would choose his way of life. Even so, the movement rejecting technological advances in favor of simple, sustainable, and self-sufficient living is being embraced by more and more people who see technology as nothing but a threat to the purity and balance of nature. Often called environmental extremists by other groups, they even disregard “environmental” technologies that attempt to correct pollution and irresponsibilities, past and present. They see all technology as manipulative and uncontrollable and choose to separate themselves from it. To them, the environment is at the top of the heirarchy. On the other extreme are the pure technologists. They view the natural world as a thing to be subdued and manipulated in the interest of progress—technological and economic. This is not to say one won’t find technologists wandering in a national park admiring the scenery. They do not necessarily deny the beauty of the natural environment but they see themselves as separate from it. They believe that technology is the key to freedom, liberation, and a higher standard of living. It is viewed, therefore, as inherently good. They see the environmental extremists as unreasonable and hold that even the undeniably negative side effects of certain technologies are best handled by more technological advance. The technologists place environmental responsibility at the bottom of their ethical heirarchy. Somewhere in the middle of the road travels the third group. While they reap the benefits of technology, they are concerned much more deeply than the technologists with the environmental costs associated with industrialization. It is in this group that most environmental engineers find themselves. They are unlike the environmental extremists since, as engineers, they inherently study and design technological devices and have faith in the ability of such devices to have a positive effect on the condition of the environment. They also differ from the technologists. They scrutinize the effects of technologies much more closely and critically. While they may see a brief, dilute leak of a barely toxic chemical as an unacceptable side effect of the production of a consumer product, the technologists may have to observe destruction—the magnitude of that caused by Chernobyl—before they consider rethinking a technology they view as economically and socially beneficial. In general, this group sees the good in technology but stresses that it cannot be reaped if technological growth goes on unchecked. The ethical behavior of engineers is more important today than at any time in the history of the profession. The engineers’ ability to direct and control the technologies

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they master has never been stronger. In the wrong hands, the scientific advances and technologies of today’s engineer could become the worst form of corruption, manipulation, and exploitation. Engineers, however, are bound by a code of ethics that carry certain obligations associated with the profession. Some of these obligations include: 1 Support ones professional society 2 Guard privileged information 3 Accept responsibility for one’s actions 4 Employ proper use of authority 5 Maintain one’s expertise in a state-of-the-art world 6 Build and maintain public confidence 7 Avoid improper gift exchange 8 Practice conservation of resources and pollution prevention 9 Avoid conflict of interest 10 Apply equal opportunity employment 11 Practice health, safety, and accident prevention 12 Maintain honesty in dealing with employers and clients There are many codes of ethics that have appeared in the literature. The preamble for one of these codes is provided below:(9) Engineers, in general, in the pursuit of their profession, affect the quality of life for all people in our society. Therefore, an Engineer, in humility and with the need for Divine guidance, shall participate in none but honest enterprises. When needed, skill and knowledge shall be given without reservation for the public good. In the performance of duty and in fidelity to the profession, Engineers shall give utmost.

FUTURE TRENDS(11) Although the environmental movement has grown and matured in recent years, its development is far from stagnant. To the contrary, change in individual behavior, corporate policy, and governmental regulations are occurring at a dizzying pace. Because of the Federal Sentencing Guidelines, the Defense Industry Initiative, as well as a move from compliance to a values-based approach in the marketplace, corporations have inaugurated company-wide ethics programs, hotlines, and senior line positions responsible for ethics training and development. The Sentencing Guidelines allow for mitigation of penalties if a company has taken the initiative in developing ethics training programs and codes of conduct. In the near future, these same Guidelines will apply to infractions of environmental law. As a result, the corporate community will undoubtedly welcome ethics integration in engineering and science programs generally, but more so in those that emphasize environmental issues. Newly hired employees, particularly those in the

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environmental arena who have a strong background in ethics education, will allay fears concerning integrity and responsibility. Particular attention will be given to the role of public policy in the environmental arena as well as in the formation of an environmental ethic. Regulations instituted by federal, state, and local agencies continue to become more and more stringent. The deadlines and fines associated with these regulations encourage corporate and industrial compliance of companies (the letter of the law) but it is the personal conviction of the corporate individuals that lies in the spirit of the law, and the heart of a true ecological ethic. To bolster this conviction of the heart, there must be the emergence of a new dominant social paradigm.(10) This is defined as “the collection of norms, beliefs, values, habits, and survival rules that provide a framework of reference for members of a society. It is a mental image of social reality that guides behavior and expectations.”(10) The general trend in personal ethics is steadily “greener” and is being achieved at a sustainable pace with realistic goals. A modern day author suggests the following. “The flap of one butterfly’s wings can drastically affect the weather.”(12) While this statement sounds much like one concepturalized by a romantic ecologist, it is actually part of a mathematical theory explored by the contemporary mathematician Gleick(13) in his book Chaos, Making a New Science. The “butterfly” theory illustrates that the concept of interdependence, as Chief Seattle professed it, is emerging as more than just a purely environmental one. This embracing of the connectedness of all things joins the new respect for simplified living and the emphasis on global justice, renewable resources, and sustainable development (as opposed to unchecked technological advancement) as the new, emerging social paradigm. The concept of environmentalism is now widely held; its future is becoming deeply held. Finally, one should note that an ethical analysis generally produces no absolute answers. This can be quite disconcerting because engineers expect valid, correct, and useful answers to problems. When one studies statics, for example, one knows what the rules are and all agree on the right answer. In ethics, however, the best an individual can do is argue that some answers are better than others, and, of course, these answers are always open to disagreement. Engineers, as professionals, have a special responsibility to the public, and this responsibility is often expressed in terms of professional ethics. Engineers invariably face situations where values become variables in the decision-making process. Indeed, the ethical aspects of a decision often prove more difficult than the technical. Recognizing this, the engineering profession has strongly encouraged engineering schools to introduce more professional ethics into engineering curricula. Professional ethics has rightly become an integral part of engineering education. Engineering ethics is tricky enough when it concerns only how engineers relate to each other and the public; it becomes trickier still when one also considers how engineers ought to react to the non-human environment. That is, when one asks “what is the engineers’ environmental ethic?” Environmental ethics, to an even greater degree than ordinary ethics, is a subject without definition and without consensus. And yet, every person on this planet makes

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everyday decisions that relate to environmental ethics. Questions as simple as “What should I eat?” or “How should I move from place to place?” all raise environmental and ethical issues. Environmental ethics is especially important for engineers because so much of their work affects the environment. How should the engineer balance the human gains of development against environmental damage? When should the engineer maintain client confidentiality in the face of potential environmental problems?

APPLICATIONS The three Illustrative Examples below have been drawn from the work of Wilcox and Theodore,(3) keying primarily on mass transfer issues. Each application is presented in case-study format, containing both a Fact Pattern and Questions for Discussion. ILLUSTRATIVE EXAMPLE 21.1(3) Fact Pattern Laura is an engineer working in a chemical plant. She has recently received a job offer from another company, which she accepts because she knows that the new job could be a big step in her career. Laura is responsible for one of the production lines in the plant she will soon be leaving. She has always been a reliable worker and an effective manager. However, having handed in her letter of resignation, she has been less attentive to her work over the past couple of weeks. She figures that there is no need to worry about this job anymore; she has to concentrate on her future. On Laura’s next-to-last day of work at the plant, Harry, a coworker on the same production line, finds out that there is a problem with the purity of the product: The level of impurities is a little higher than acceptable. Harry decides to consult Laura. He says, “The product coming out is below the required purity. I think you should investigate it so we can solve this problem.” Laura replies, “I would love to help you, Harry, but tomorrow is my last day here. I don’t want to start dealing with this problem because it could take a while to solve. Let my replacement worry about it.” Harry answers, “Laura, if we let this problem go, we’ll continue to have a product that doesn’t meet regulation. The problem could also get worse. You are the expert here, so you could easily fix this mess.” “Harry, you’re a friend of mine. Please don’t ask me to get involved in this problem. It’s not my concern anymore. I just want to relax during my last two days at work,” pleads Laura. “It’s not like the plant will blow up. Wait for two days. You can pretend that you didn’t notice anything until then.” Reluctantly, Harry agrees. “I know you’re really looking forward to your new job. It’s just that I’ll feel guilty knowing that something is wrong, and I’m not doing anything about it. But I guess I can wait for two days.”

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“Harry, don’t worry. Take it easy for a couple of days. Just think of it as a minor delay,” replies Laura.

Questions for Discussion: 1 2 3 4 5

What are the facts in this case? Do you think Laura should stay focused on her current job? Should Laura handle the problem? Do you think it’s okay for Harry to ignore the problem for the next two days? Should Harry consult someone else now that Laura has refused to deal with the problem? B

ILLUSTRATIVE EXAMPLE 21.2(3) Fact Pattern Joe Murray, a chemical engineer at the HairMagic Company, is in charge of the production of a new hair coloring product, which is expected to be ready for mass production in a month. Joe has been reviewing the results of the hair coloring tests conducted on volunteers, which show that when the product was used on dark-haired men and women, it changed the color of the hair with an accuracy of 99.2 percent. However, on fair- and light-colored hair types, it had an accuracy of only 89.3 percent. Joe realizes that the product needs to be refined further and that some substitutions should be made in its chemical makeup. Joe decides to go to his manager and explain that production will have to be delayed. His manager, on the other hand, feels that production should be initiated and that further refining can be done during production so as not waste any more money or time. Ethically, Joe feels strongly that this is wrong: the refining needs special attention and production should not even be considered when the product is not yet perfected. Joe’s only alternative is to go above his manager and try to convince upper management to delay production. However, doing this could anger his manager and could also lead to Joe’s dismissal. But it might also make him appear to be confident and conscientious, and could also serve to publicize his abilities as an engineer.

Questions for Discussion: 1 2 3 4

What are the facts in this case? What issues are involved? What are the choices Joe must make? What could be the consequences of each decision he could make?

B

ILLUSTRATIVE EXAMPLE 21.3(3) Fact Pattern Tom is preparing for his final exam in mass transfer operations, a course that he has been struggling with all semester long. He desperately needs to pass this exam because as it stands he has a D in the course and not passing this final exam means he will have to repeat the class.

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One night Tom sees a janitor that he is acquainted with and they begin to talk about his course difficulties. The janitor, whose name is Mike, likes Tom and thinks he’s a good kid. Mike offers to help Tom with his mass transfer exam. The student, bewildered as to how the janitor can help, asks, “Did you take mass transfer while you were in school, Mike?” “No, Tom, I went only as far as high school but I know for a fact that your professor keeps all of his exam papers on his desk. I saw them when I was cleaning his office one night.” Mike says he can let Tom into his professor’s office to look around for the final exam. Tom is excited about the idea of getting it ahead of time and feels a sense of relief. On the other hand, Tom realizes that if he gets caught, he can get thrown out of school and that would not be good. He has two options: to study hard for the exam (and possibly still not do well) or obtain the exam (at the risk of getting caught). Which is the best alternative?

Questions for Discussion: 1 What are the facts in this case? 2 What is the ethical problem with what Tom is doing? 3 Do you think Tom should try to get the exam ahead of time?

B

REFERENCES 1. H. TABACK, “Ethics corner: doing the right thing,” Environmental Management, Pittsburgh, PA, January, 2002. 2. H. TABACK, “Ethics corner: when ego gets in the way,” Environmental Management, Pittsburgh, PA, January, 2002. 3. J. WILCOX and L. THEODORE, “Engineering and Environmental Ethics: A Case Study Approach,” John Wiley & Sons, Hoboken, NJ, 1998. 4. S. PARKS, “Professional ethics, moral courage, and the limits of personal virtue,” in “Can Virtue be Taught,” (B. Darling-Smith, ed.), University of Notre Dame Press, 1993. 5. “The Teaching of Ethics in Higher Education,” The Hastings Center, Hastings on the Hudson, NY, 1980. 6. S. L. CARTER, “The insufficiency of honesty,” Altoulie Monthly, 1996. 7. S. BOK, “Lying: Moral Choice in Public and Private Life,” Pantheon, New York City, NY, 1978. 8. G. BURKE, B. SINGH, and L. THEODORE, “Handbook of Environmental Management and Technology,” 2nd edition, John Wiley & Sons, Hoboken, NJ, 2000. 9. M. MARTIN and R. SCHINZINGER, “Ethics in Engineering,” McGraw-Hill, New York City, NY, 1989. 10. I. BARBOUR, “Ethics in an Age of Technology,” Harper, San Francisco, CA, 1993. 11. M. K. THEODORE and L. THEODORE, “Introduction to Environmental Management,” CRC Press/Taylor & Francis Group, Boca Raton, FL, 2009. 12. J. GLEICK, “Chaos, Making a New Science,” Viking Press, New York City, NY, 1987.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

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Environmental Management and Safety Issues INTRODUCTION In the last four decades, people have become aware of a wide range of environmental issues. All sources of air, land, and water pollution are under constant public scrutiny. Increasing numbers of professionals are being confronted with problems related to environmental management. Because many of these issues are of concern, practicing engineers must develop a proficiency and an improved understanding of technical and scientific issues regarding environmental management and safety issues in order to cope with these challenges. The problem of what to include and what to omit has been particularly difficult for this chapter. However, every attempt has been made to offer material to the reader at a level that should enable them to better cope with some of the complex problems encountered in environmental management today. The chapter is divided into the following five sections: Introduction Environmental Issues of Concern Health Risk Assessment Hazard Risk Assessment Illustrative Examples The next section provides a broad overview of all the key environmental issues. This is followed by two sections on risk assessment—one concerned with health and the other concerned with hazards. The chapter concludes with a section that contains 19 Illustrative Examples.

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

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ENVIRONMENTAL ISSUES OF CONCERN(1) The degradation of the environment is not a problem that is restricted to the United States or even to developed countries. On the contrary, under-developed countries are also struggling with several environmental issues that have already been resolved in many developed countries. In the United States, the Environmental Protection Agency (EPA) as well as individual states are working hard to implement regulations addressing areas of environmental concern. Generators and sources of pollutants are being identified so that solutions may be targeted to specific areas. There are several different areas of concern related to air pollutants and their control. Atmospheric dispersion of pollutants can be mathematically modeled to predict where pollutants emitted from a particular source, such as a combustion facility stack, will settle to the ground and at what concentration. Pollution control equipment can be added to various sources to reduce the amount of pollutants before they are emitted into the air. Acid rain, the greenhouse effect, and global warming (climate change) are all indicators of adverse effects to the air, land, and sea, which result from the excessive amount of pollutants being released into the air. Two topics that few people are aware of are the issue of indoor air quality and vapor intrusion. Inadequate ventilation systems in homes and businesses directly affect the quality of health of the people within the buildings. For example, the episode of Legionnaires’ disease that occurred in Philadelphia in the 1970s was related to microorganisms in the cooling water of the air-conditioning system. Noise pollution, although not traditionally an air pollution topic, is included in this topic area. The effects of noise pollution are not generally noticed until hearing is impaired. And, although impairment of hearing is a commonly known result of noise pollution, few people realize that stress is also a significant result of excessive noise exposure. The human body enacts its innate physiologic defensive mechanisms under conditions of loud noise and the fight to control these physical instincts causes tremendous stress on the individual. Pollutants entering rivers, lakes, and oceans come from a wide variety of sources, including stormwater runoff, industrial discharges, and accidental spills. It is important to understand how these substances disperse in order to determine how to control them. Municipal and industrial wastewater treatment systems are designed to reduce or eliminate problem substances before they are introduced into natural water systems, industrial use systems, drinking water supply, and other water systems. Often, wastewater from industrial plants must be pretreated before it can be discharged into a municipal treatment system. Programs to reduce and dispose of municipal waste include reuse, reduction, recycling, and composting, in addition to incineration and landfilling. Potentially infectious waste generated in medical facilities must be specially packaged, handled, stored, transported, treated, and disposed of to ensure the safety of both the waste handlers and the general public. Radioactive waste may have far more serious impacts on human health and the environment, and treatment and disposal requirements for radioactive substances must be strictly adhered to. Incineration has been a typical treatment method for hazardous waste for many years. The Comprehensive Environmental Response, Compensation, and Liability Act (CERCLA), also known as Superfund, was enacted to identify and remedy

Environmental Issues of Concern

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uncontrolled hazardous waste sites. It also attempts to place the burden of cleanup on the generator rather than on the federal government. Asbestos, household hazardous wastes, used oil, metals, and underground storage tanks either contain, or inherently are, hazardous materials that require special handling and disposal. Furthermore, it is important to realize that small quantity generators of hazardous wastes are regulated as well as large generators. Pollution prevention, both domestic and industrial, can be accomplished through 1 proper residential and commercial building design, 2 proper heating, cooling, and ventilation systems, 3 energy conservation, 4 reduction of water consumption, and 5 attempts to reuse or reduce materials before they become wastes. Domestic and industrial solutions to environmental problems result from considering ways to make homes and workplaces more energy-efficient as well as ways to reduce the amount of wastes generated within them. Additional environmental concerns include electromagnetic fields that emanate from power distribution systems. Items related to both worker and community health and safety and training have been brought to the forefront by the increasingly stringent regulations developed by the Occupational Safety and Health Administration (OSHA) and other federal and state regulatory agencies. The best way to prevent a dangerous situation is to be informed of the possible outcomes ahead of time and to be prepared to respond to an emergency situation. Guidelines on how to monitor the results of an environmental action are needed to determine how well an existing cleanup effort is proceeding or how present background levels will affect discharges from new facilities. Economic considerations also play a large role in the implementation of an environmental strategy. Three of the newer waste remediation technologies include: 1 Bioremediation is a process that utilizes microorganisms to transform harmful substances into nontoxic compounds. It may be used to treat contaminated soil or groundwater and it is one of the most promising new technologies for treating chemical spills and hazardous wastes. 2 Soil vapor extraction is used to remove volatile organic compounds from soil. A vacuum is applied to the soil, causing the movement of vapors toward extraction wells. Volatiles are then readily removed from the subsurface of the soil through the extraction wells. 3 Biofiltration is a process that exploits the ability of microorganisms to remove and treat biodegradable substances in air (gas) streams. In the past, it has been used successfully in Europe to remove odors from wastewater treatment plants and compost factories, and it is now being used to remove volatile organic compounds. Practicing engineers need to be informed on how to make decisions about associated risks and how to communicate these risks and their effects on the environment to

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the public. Risk related topics include short-term and long-term threats to human health and the environment. Risk assessment is the most important consideration for remediation of harmful effects stemming from the presence of a hazardous substance and risk-based decision-making is a tool that is now routinely being used to select a clean-up alternative. There are four topics that are relatively new in the area of environmental management: 1 ISO 14000 is an international certification standard for an organization’s environmental management system. It ensures that the objectives, targets, procedures, and systems of the environmental management system are part of the organization’s routine operations. 2 Environmental audits provide a means of assessing the environmental condition of the organization to prevent health risks. 3 Environmental justice is a new term for describing the disproportionate distribution of environmental risks in minority and low-income communities. Federal attention is now being focused on environmental and human health conditions in these areas, with the goal of achieving equality of environmental protection for all communities. 4 Environmental ethics relates to rules of proper environmental conduct (see previous chapter for details.) This chapter is not intended to be all-encompassing. Rather, it is to be used as a starting point. References are provided throughout that provide more detailed information on each topic.

HEALTH RISK ASSESSMENT(2 – 4) There are many definitions for the word risk. It can be described as a triplet combination of event, probability, and consequences. It can also be described as a measure of economic loss or human injury in terms of both the incident likelihood and the magnitude of the loss or injury. People face all kinds of risks everyday, some voluntarily and others involuntarily. Therefore, risk plays a very important role in today’s world. Studies on cancer caused a turning point in the world of risk because it opened the eyes of risk scientists and health professionals to the world of risk assessments. Since 1970, the field of risk assessment has received widespread attention within both the scientific and regulatory committees. It has also attracted the attention of the public. Properly conducted risk assessments have received fairly broad acceptance, in part because they put into perspective the terms toxic, hazard, and risk. Toxicity is an inherent property of all substances. It states that all chemical and physical agents can produce adverse health effects at some dose or under specific exposure conditions. In contrast, exposure to a chemical has the capacity to produce a particular type of adverse effect. Risk, however, is the probability or likelihood that an adverse outcome will occur in a person or a group that is exposed to a particular concentration or dose of

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569

the hazardous agent. Therefore, risk is generally a function of exposure or dose. Consequently, health risk assessment is defined as the process or procedure used to estimate the likelihood that humans or ecological systems will be adversely affected by a chemical or physical agent under a specific set of conditions.(5) The term risk assessment is not only used to describe the likelihood of an adverse response to a chemical or physical agent, but it has also been used to describe the likelihood of any unwanted event. This subject is treated in more detail in the next section. These include risks such as: explosions or injuries in the workplace; natural catastrophes; injury or death due to various voluntary activities such as skiing, skydiving, flying, and bungee jumping; diseases; death due to natural causes; and many others.(6) Risk assessment and risk management are two different processes but they are intertwined. Risk assessment and risk management give a framework not only for setting regulatory priorities but also for making decisions that cut across different environmental areas. Risk management refers to a decision-making process that involves such considerations as risk assessment, technology feasibility, economic information about costs and benefits, statutory requirements, public concerns, and other factors. Therefore, risk assessment supports risk management in that the choices on whether and how much to control future exposure to the suspected hazards may be determined. Regarding both risk assessment and risk management, this section will primarily address this subject from a health perspective. The next section will primarily address this subject from a safety and accident perspective. The reader should note that two general types of potential health risk exist. These are classified as: 1 Acute. Exposures that occur for relatively short periods of time, generally from minutes to one or two days. Concentrations of (toxic) air contaminants are usually high relative to their protection criteria. In addition to inhalation, airborne substances might directly contact the skin, or liquids and sludges may be splashed on the skin or into the eyes, leading to adverse health effects. This subject area falls, in a general sense, in the domain of hazard risk assessment (HZRA). 2 Chronic. Continuous exposure occurring over long periods of time, generally several months to years. Concentrations of inhaled (toxic) contaminants are usually relatively low. This subject area falls in the general domain of health risk assessment (HRA) and it is this subject that is addressed in this section. Thus, in contrast to the acute (short-term) exposures that predominate in hazard risk assessment, chronic (long-term) exposures are the major concern in health risk assessments. Health risk assessments provide an orderly, explicit, and consistent way to deal with scientific issues in evaluating whether a problem exists and what the magnitude of the problem may be. This evaluation typically involves large uncertainties because the available scientific data are limited and the mechanisms for adverse health impacts or environmental damage are only imperfectly understood. When one examines risk,

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how does one decide how safe is safe or how clean is clean? To begin with, one has to look at both sides of the risk equation; that is, both the toxicity of a pollutant and the extent of public exposure. Information is required at both the current and potential exposure, considering all possible exposure pathways. In addition to human health risks, one needs to look at potential ecological or other environmental effects. In conducting a comprehensive risk assessment, one should remember that there are always uncertainties and these assumptions must be included in the analysis.

Risk Evaluation Process for Health In recent years, several guidelines and handbooks have been produced to help explain approaches for doing health risk assessments. As discussed by a special National Academy of Sciences committee convened in 1983, most human or environmental health hazards can be evaluated by dissecting the analysis into four parts: health (problem) identification, dose-response assessment or toxicity assessment, exposure assessment, and risk characterization (see Figure 22.1). For some perceived problems, the risk assessment might stop with the first step, identification, if no adverse effect is identified or if an agency elects to take regulatory action without further analysis. Regarding identification, a problem is defined as a toxic agent or a set of conditions that has the potential to cause adverse effects to human health or the environment. Identification involves an evaluation of various forms of information in order to identify the different problems. Dose-response or toxicity assessment is required in an overall assessment: responses/effects can vary widely since all chemicals and contaminants vary in their capacity to cause adverse effects. This step frequently requires

Figure 22.1 The health risk evaluation process.

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571

that assumptions be made to relate experimental data for animals and humans. Exposure assessment is the determination of the magnitude, frequency, duration, and routes of exposure of human populations and ecosystems. Finally, in risk characterization, toxicology and exposure data/information are combined to obtain a qualitative or quantitative expression of risk. Risk assessment involves the integration of the information and analysis associated with the above four steps to provide a complete characterization of the nature and magnitude of risk and the degree of confidence associated with this characterization. A critical component of the assessment is a full elucidation of the uncertainties associated with each of the major steps. Under this broad concept of risk assessment are encompassed all of the essential problems of toxicology. Risk assessment takes into account all of the available dose-response data. It should treat uncertainty not by the application of arbitrary safety factors, but by stating them in quantitatively and qualitatively explicit terms, so that they are not hidden from decision-makers. Risk assessment, defined in this broad way, forces an assessor to confront all the scientific uncertainties and to set forth in explicit terms the means used in specific cases to deal with these uncertainties.(7)

HAZARD RISK ASSESSMENT(2 – 4,8) Risk evaluation of accidents serves a dual purpose. It estimates the probability that an accident will occur and also assesses the severity of the consequences of an accident. Consequences may include damage to the surrounding environment, financial loss, or injury to life. This section is primarily concerned with the methods used to identify hazards and the causes and consequences of accidents. Issues dealing with health risks have been explored in the previous section. Risk assessment of accidents provides an effective way to help ensure either that a mishap does not occur or that the likelihood of an accident is reduced. The result of the risk assessment allows concerned parties to take precautions to prevent an accident before it happens. Regarding definitions, the first thing an individual needs to know is what exactly is an accident. An accident is an unexpected event that has undesirable consequences. The causes of accidents have to be identified in order to help prevent accidents from occurring. Any situation or characteristic of a system, plant, or process that has the potential to cause damage to life, property, or the environment is considered a hazard. A hazard can also be defined as any characteristic that has the potential to cause an accident. The severity of a hazard plays a large part in the potential amount of damage a hazard can cause if it occurs as noted earlier. Risk is the probability that human injury, damage to property, damage to the environment, or financial loss will occur. An acceptable risk is one whose probability is unlikely to occur during the lifetime of the plant or process. An acceptable risk can also be defined as an accident that has a high probability of occurring but with negligible consequences. Risks can be ranked qualitatively in categories of high, medium, and low. Risk can also be ranked quantitatively as an annual number of fatalities per million affected individuals. This is normally denoted as a number per one million, for example, 3  1026. This number indicates that on average, three workers will die every year

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out of one million individuals. Another quantitative approach that has become popular in industry is the Fatal Accident Rate (FAR) concept. This determines or estimates the number of fatalities over the lifetime of 1000 workers. The lifetime of a worker is defined as 105 hours, which is based on a 40-hour work week for 50 years. A reasonable FAR for a chemical plant is 3.0 with 4.0 usually taken as a maximum. A FAR of 3.0 means that there are 3 deaths for every 1000 workers over a 50-year period. Interestingly, the FAR for an individual at home is approximately 3.0. Some of the Illustrative Examples in this chapter compliment many of the hazard concepts described below with technical calculations and elaborations.

Risk Evaluation Process for Accidents As with Health Risk Assessment (HRA), there are four key steps involved in a Hazardous Risk Assessment (HZRA). These are presented in Figure 22.2. A more detailed flowchart is presented in Figure 22.3 if the system in question is a chemical plant. These steps are detailed below: 1 A brief description of the equipment and chemicals used in the plant is needed. 2 Any hazard in the system has to be identified. Hazards that may occur in a chemical plant include: a Fire b Toxic vapor release c Slippage d Corrosion e Explosions f Rupture of a pressurized vessel g Runaway reactions 3 The event or series of events that will initiate an accident has to be identified. An event could be a failure to follow correct safety procedures, improperly repaired equipment, or a safety mechanism. 4 The probability that the accident will occur has to be determined. For example, if a chemical plant has a given life, what is the probability that the temperature in a reactor will exceed the specified temperature range? The probability can be ranked from low to high. A low probability means that it is unlikely for the event to occur in the life of the plant. A medium probability suggests that there is a possibility that the event will occur. A high probability means that the event will probably occur during the life of the plant. 5 The severity of the consequences of the accident must be determined. 6 The information from steps 4 and 5 are combined. If the probability of the accident and the severity of its consequences are low, then the risk is usually deemed acceptable and the plant should be allowed to operate. If the probability of occurrence is too high or the damage to the surrounding area is too great, then the risk is usually unacceptable and the system needs to be modified to minimize these effects.

Hazard Risk Assessment

Figure 22.2

Hazard risk assessment flowchart.

Figure 22.3

Chemical plant hazard risk assessment flowchart.

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The heart of the hazard risk assessment algorithm provided is enclosed in the dashed box of Figure 22.3. The algorithm allows for re-evaluation of the process if the risk is deemed unacceptable (the process is repeated starting with either step 1 or 2). As is evident from the lessons of past accidents, it is essential for industry to abide by stringent safety procedures. The more knowledgeable the personnel, from the management to the operators of a plant, and the more information that is available to them, the less likely a serious incident will occur. The new regulations, and especially Title III of CERCLA, will help to ensure that safety practices are up to standard.(1,8) However, these regulations should only provide a minimum standard. It should be up to the companies, and specifically the plants, to see that every possible measure is taken to ensure the safety and well-being of the community and the environment in the surrounding area. It is also up to the community itself, under Title III, to be aware of what goes on inside local industry, and to prepare for any problems that might arise.

APPLICATIONS The remainder of this chapter is devoted to Illustrative Examples, many of which contain technical development material. A good number have been drawn from National Science Foundation (NSF) literature(9–14) and two other sources.(15,16) The last example provides a detailed analysis of both health and hazard risk assessment.

ILLUSTRATIVE EXAMPLE 22.1 Explain why a large open bottle of a liquid waste with a finite vapor pressure ultimately fills the room with the odor of that waste. SOLUTION: Through the process of mass transfer, the waste evaporates from the open bottle because of its vapor pressure. Then it diffuses through the air in the room from locations of high concentrations (e.g., at the mouth of the open bottle), to locations of lower concentrations (e.g., at the far ends of the room). Diffusion will continue throughout the room. Given the sensitivity of the human nose plus the nature of the waste evaporated, the odor of waste would then be detected throughout the room. B

ILLUSTRATIVE EXAMPLE 22.2 A 106 gal/day (1.0 MGD) wastewater from a treatment plant contains 0.2 mg suspended solids (SS) per cubic meter of wastewater. The separated sludge from the plant consists of the SS. If 10% by weight of lime is required to stabilize the sludge treatment and 80% of the solids are captured, calculate the daily and annual lime requirements. SOLUTION:

The sludge flow rate is _ S ¼ (0:2)(106 )(8:34)=(1000) m ¼ 1668 lb=day

Applications

575

The treated sludge is _ TS ¼ (0:8)(1668) m ¼ 1334 lb=day The lime requirement is therefore _ L ¼ (0:1)(1334) m ¼ 133:4 lb=day The annual requirement is _ L ¼ (133:4)(365) m ¼ 48,691 lb=yr

B

ILLUSTRATIVE EXAMPLE 22.3 Estimate the required landfill area for a community with a population of 260,000. Assume that the following conditions apply: 1 Solid waste generation ¼ 7.6 lb/capita . day 2 Compacted specific gravity (density) of solid wastes in landfall ¼ 830 lb/yd3 3 Average site depth of compacted solid wastes ¼ 20 ft SOLUTION:

Determine the daily solid wastes generation rate in tons per day: Generation rate ¼

(260,000 people)(7:6 lb=capita  day) 2000 lb=ton

¼ 988 ton=day The required area is determined as follows: Volume required=day ¼

(988 ton=day)(2000 lb=ton) 830 lb=yd3

¼ 2381 yd3 =day Area required=yr ¼

(2381 yd3 =day)(365 day=yr)(27 ft3 =yd3 ) (20 ft)(43,650 ft2 =acre)

¼ 26:88 acre=yr Area required=day ¼

(2381 yd3 =day)(27 ft3 =yd3 ) ¼ 26:88=365 (20 ft)(43,650 ft2 =acre)

¼ 0:074 acre=day The actual site requirements will be greater than the value computed because additional land is required for a buffer zone, office and service building, access roads, utility access, etc. Typically, this allowance varies from 20 –40%. Thus, if an allowance of 30% over the lifetime of the facility is employed, the daily area requirement becomes Area required=day ¼ (0:074)(1:3) ¼ 0:096 acre=day

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Chapter 22 Environmental Management and Safety Issues B

B

B

Figure 22.4 Flow diagram for Illustrative Example 22.4. A more rigorous approach to the determination of the required landfill area involves consideration of the contours of the completed landfill and the effects of gas production and overburden compaction. B

ILLUSTRATIVE EXAMPLE 22.4 A liquid stream contaminated with a pollutant is being cleansed with a stripping control device. If the liquid has 600 ppm of pollutant and it is permissible to have 50 ppm of this pollutant in the discharge stream, what fraction of the liquid can bypass the control device? SOLUTION: Using a basis of 1 lb of liquid fed to the control device, the flow diagram in Figure 22.4 applies. Note that: B ¼ fraction of liquid bypassed 1  B ¼ fraction of liquid treated Performing a pollutant balance around point 2 in Figure 22.4 yields (1  B)(0) þ 600B ¼ (50)(1:0) Solving gives B ¼ 0:0833 Note that in some operations, a process does a more complete job than is required. For example, if moist air is passed through a fresh silica gel dryer, the air will leave the system almost bone dry. If it were desirable to have air containing some moisture, one would have to reintroduce water vapor into the air. This would be a wasteful process compared to bypassing the proper amount of original moist air. In general, a finished product is made only as good as it has to be to meet competition and/or to satisfy the user. “Product quality giveaway” is costly and is often minimized by bypassing. B

ILLUSTRATIVE EXAMPLE 22.5 A municipality in the Midwest has a population of 50,000 and generates 100,000 yd3 of municipal waste annually. The waste is made up of 30% compacted waste and 70% uncompacted waste. Assume that the waste has a density of 1000 lb/yd3 compacted and 400 lb/yd3 uncompacted. How many pounds of waste are generated by this city each year and by each person each year?

Applications

577

SOLUTION: Based on the waste densities given in the problem statement, the following generation rates are determined: Waste generated=yr ¼ (0:3)(100,000 yd3 )(1000 lb=yd3 ) þ (0:7)(100,000 yd3 )(400 lb=yd3 ) ¼ 30,000,000 lb þ 28,000,000 lb ¼ 58,000,000 lb=yr Per capita generation rate ¼

58,000,000 lb=yr 50,000 people

¼ 1160 lb=person  yr ¼ 3:2 lb=person  day

B

ILLUSTRATIVE EXAMPLE 22.6 An incinerator burns mercury-contaminated waste. The waste material has an ash content of 1%. The solid waste feed rate is 1000 lb/h and the gas flow rate is 20,000 dscfm. It is reported that the average mercury content in the particulates was 2.42 mg/g when the vapor concentration was 0.3 mg/dscm. For the case where incinerator emissions meet the particulate standard of 0.08 gr/dscf (0.1832 g/dscm) with a 99.5% efficient electrostatic precipitator (ESP), calculate the amount of mercury bound to the fly ash that is captured in the ESP in lb/day. SOLUTION: The amount of ash leaving the stack is       0:08 gr 1 lb 20,000 dscf 60 min 24 h ¼ 329 lb=day dscf 7000 gr 1 min h day The amount of ash collected in the ESP is (329 lb=day)=(1  0:995 collected) ¼ 65,800 lb=day

B

ILLUSTRATIVE EXAMPLE 22.7 Refer to the previous example. Calculate the amount of mercury leaving the stack as a vapor and with the fly ash in grams/day. SOLUTION:

The amount of mercury leaving the stack with the fly ash is

(329 lb ash=day)(2:42  106 g Hg=g ash) ¼ 7:96  104 lb Hg=day ¼ 0:361 g Hg=day The amount of mercury leaving the stack as vapor is       0:3  103 g Hg 20,000 dscf 1 m3 60 min 24 h ¼ 244:8 g=day dscm 1 min h day 35:3 ft3 Total mercury leaving the stack ¼ 244:8 þ 0:361 ¼ 245:2 g=day

B

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Chapter 22 Environmental Management and Safety Issues

ILLUSTRATIVE EXAMPLE 22.8 Some wastewater and water standards and regulations are based on a term defined as parts per million (ppm) or parts per billion (ppb). Define the two major classes of these terms and describe the inter-relationship from a calculational point-of-view. Also convert 10 calcium parts per million parts of water on a mass basis to parts per million on a mole basis. SOLUTION: A water stream seldom consists of a single component. It may also contain two or more phases (a dissolved gas and/or suspended solids), or a mixture of one or more solutes. For mixtures of substances, it is convenient to express compositions in mole fractions or mass fractions. The following definitions are often used to represent the composition of component A in a mixture of components: mass of A ¼ mass fraction of A total mass of water stream moles of A ¼ mole fraction of A yA ¼ total moles of water stream

wA ¼

Trace quantities of substances in water streams are often expressed in parts per million (ppmw) or as parts per billion (ppbw) on a mass basis. These concentrations can also be provided on a mass per volume basis for liquids and on a mass per mass basis for solids. Gas concentrations are usually represented on a mole or volume basis (e.g., ppmm or ppmv). The following equations apply: ppmw ¼ 106 wA ¼ 103 ppbw ppmv ¼ 106 yA ¼ 103 ppbv The two terms ppmw and ppmm are related through the molecular weight. To convert 10 ppmw Ca to ppmm, select a basis of 106 g of solution. The mass fraction of Ca is first obtained by the following equation: Mass of Ca ¼ 10 g Moles Ca ¼ Moles H2 O ¼

10 g ¼ 0:25 mol 40 g=mol 106 g  10 g ¼ 55,555 mol 18 g=mol

Mole fraction Ca ¼ yCa ¼

0:25 mol ¼ 4:5  106 0:25 mol þ 55,555 mol

ppmm of Ca ¼ 106 yCa ¼ (106 )(4:5  106 ) ¼ 4:5

B

Applications

579

ILLUSTRATIVE EXAMPLE 22.9 Calculate the Theoretical Chemical Oxygen Demand (ThCOD) of a 100 mg/L solution of glucose. Also calculate the total organic content.(15,16) SOLUTION:

First, balance the chemical equation for the reaction of glucose and oxygen: C6 H12 O6 þ 6O2 ! 6CO2 þ 6H2 O

Second, determine the oxygen/substrate stoichiometric ratio for the oxygen reaction: Ratio ¼ (6)(32)=[(1)(180)] ¼ 0:067 mg O2 =mg glucose Third, calculate the ThCOD for a 100-mg/L solution of glucose. This is equal to the product of the mass concentration of glucose and the stoichiometric ratio: ThCOD ¼ (100 mg glucose=L)(1:067 mg O2 =mg glucose) ¼ 106:7 mg O2 =L or 106:7 mg COD=L

B

ILLUSTRATIVE EXAMPLE 22.10 The following 5-day biochemical oxygen demand (BOD5) and total suspended solids (TSS) data were collected from a clarifier at a local municipal wastewater treatment plant over a 7-day period.(15,16) The National Pollutant Discharge Elimination System (NPDES)(15,16) permit limitations for BOD5 and TSS effluent concentrations from this wastewater treatment plant are 45 mg/L on a 7-day average. Based on this information (see Table 22.1), is the treatment plant within its NPDES permit limits? SOLUTION: The BOD5 7-day average concentration based on the data tabulated in the problem statement is (BOD5 )7 ¼ (45 þ 79 þ 64 þ 50 þ 30 þ 25 þ 21)=7 ¼ 44:9 mg=L Table 22.1 Daily BOD5 and TSS Effluent Concentration Data Collected Over a 7-Day Period at a Municipal Wastewater Treatment Plant Day 1 2 3 4 5 6 7

BOD (mg/L) 45 79 64 50 30 25 21

TSS (mg/L) 20 100 50 42 33 25 15

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Chapter 22 Environmental Management and Safety Issues

The 7-day average concentration for TSS is (TSS)7 ¼ (20 þ 100 þ 50 þ 42 þ 33 þ 25 þ 15)=7 ¼ 40:7 mg=L The wastewater treatment plant is still within its NPDES permit limit (but only marginally) of an average 7-day maximum concentration of 45 mg/L for both BOD5 and TSS. B

ILLUSTRATIVE EXAMPLE 22.11 The average gasoline tank in an automobile has a 14-gal capacity. Every time the gas tank is filled, the vapor space in the tank is displaced to the environment. Since all forms of hydrocarbons in the atmosphere contribute to the formation of ozone and need to be controlled, this problem attempts to quantify some of these emissions. Assume the automobile tank vapor space, the air, and the gasoline supply is all at 208C. The vapor space is saturated with gasoline. The vapor-phase mole fraction of gasoline under these conditions is approximately 0.4. The lost vapor has a molecular weight of about 70 g/gmol and a liquid specific gravity of 0.62. 1 Calculate the amount of gasoline (in gallons of liquid) that is lost to the air during a 10-gal fill. 2 How much is lost annually from 50 million cars filled once each week with 10 gal of gasoline. SOLUTION:

The vapor specific volume in m3/kgmol is V RT ¼ n P (8:314)(293) ¼ 101:3 ¼ 24:05 m3 =kgmol

The amount of gasoline vapor in the tank in kgmol is n¼

(0:4)(10 gal) (264:1 gal=m3 )(24:05 m3 =kgmol)

¼ 6:298  104 kgmol The liquid volume of the gasoline vapor in the tank in gallons is V1 ¼

(6:298  104 kgmol)(70 kg=kgmol)(264:1 gal=m3 ) (0:62  1000 kg=m3 )

¼ 0:01878 gal The gasoline loss in gallons per car per year can now be calculated: Lost ¼ (0:01878 gal=fill)(52 fills=yr) ¼ 0:976 gal=car  yr

Applications

581

The estimated annual loss (AL) arising because of the vapor displaced during filling is AL ¼ [0:976 gal=(car  yr)](50,000,000 cars) ¼ 4:88  107 gal=yr

B

ILLUSTRATIVE EXAMPLE 22.12 Describe the two methods utilized to perform an exposure assessment. SOLUTION: The two methods by which an exposure assessment is performed are by direct measurement and computer modeling. Direct measurement involves using receptors or analyzers placed at various locations around a specific area to measure the time-averaged concentration of an agent. Computer models are utilized to predict possible pathways of exposure, generally from a point source release. B

ILLUSTRATIVE EXAMPLE 22.13 Discuss the significance of Figure 22.5. SOLUTION: The figure below allows one to compare the relative cost of the detriment (associated with an accident) with the cost of improved protection. As can be seen on the graph, for low levels of cost protection, the costs are unreasonably high. However, for high levels of cost protection, the cost of detriment is significantly low. Therefore, a cost-benefit analysis should be performed in order to determine a reasonable cost for an acceptable level of protection while keeping the detrimental costs to a minimum. From a plant’s perspective, the level of protection should be set in or around A. From a purely economic point-of-view, this point roughly represents the minimum cost. Other factors such as regulatory requirements, good will, and so on, can change this. See Chapter 18 for additional details on economic considerations. B

Figure 22.5 Cost/protection analysis.

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ILLUSTRATIVE EXAMPLE 22.14 Two factory workers at a nail polish manufacturering facility are exposed to acetone at the following concentrations and durations: Employee A

Employee B

1000 ppm for 180 minutes 500 ppm for 120 minutes 200 ppm for 180 minutes

2000 ppm for 120 minutes 700 ppm for 180 minutes

The 8-hour time-weighted average (TWA) acetone exposure limits for the American Conference of Government Industrial Hygienists (ACGIH), Occupational Health and Safety Administration (OSHA), and the National Institute of Occupational Safety and Health (NIOSH) are 250 ppm, 750 ppm, and 250 ppm, respectively. Calculate each worker’s respective 8-hour TWA exposure. What do these results indicate about the worker’s exposure at the company? SOLUTION:

Determine the 8-hour TWA exposure, E, for each employee: E (Employee A) ¼ (C 1 T 1 þ C2 T 2 þ    C n T n )=8 ¼ [(1000)(3) þ (500)(2) þ (200)(3)]=8 ¼ 575 ppm E (Employee B) ¼ (C 1 T 1 þ C2 T 2 ) þ    Cn T n )=8 ¼ [(2000)(2) þ (700)(3) þ (0)(3)]=8 ¼ 763 ppm

Compare each employee’s calculated 8-hour TWA exposure to the limits established by each reference source. Employee A’s 8-hour TWA exposure level is below the OSHA permissible exposure level (PEL) and above both the ACGIH recommended exposure limit (REL) and NIOSH threshold limit value (TLV). Employee B’s 8-hour TWA exposure level is above all three reference sources. Since the OSHA PELs are the only legally enforceable standards, administrative actions/engineering controls to reduce the air quality concentrations of acetone in the work area or to reduce the time of exposure are required by law. If the employee’s time of exposure is reduced, consideration should also be given to exceedance of OSHA short-term exposure limit (STEL) concentrations. B

ILLUSTRATIVE EXAMPLE 22.15 Discuss exposure guidelines. SOLUTION: The exposure guidelines discussed above are primarily based on industrial custom, toxicological studies, human exposure data, or a combination of these. The guidelines were developed for workers in an industrial environment and, in certain states, for municipal employees. Thus, they are not meant to be used for general air quality levels for exposure to the public. Furthermore, there is a limitation on the use of the exposure guidelines as a relative index of toxicity because the exposure limits are based on different effects for different chemicals. For example, the TLV-TWA for acetone is chosen to prevent irritation to the eyes and respiratory system, while the TLV-TWA for acrylonitrile is chosen to reduce the risk of cancer.

Applications

583

Exposures to these chemicals at other concentration levels could lead to other deleterious effects. Thus, when evaluating the risk of chemical exposure, all toxicological data should be thoroughly reviewed and evaluated. B

ILLUSTRATIVE EXAMPLE 22.16 Two large bottles of flammable solvent were ignited by an undetermined ignition source after being knocked over and broken by a janitor while cleaning a 10 ft  10 ft  10 ft research laboratory. The laboratory ventilator was shut off and the fire was fought with a 10 lb CO2 fire extinguisher. As the burning solvent had covered much of the floor area, the fire extinguisher was completely emptied in putting the fire out. The Immediate Danger to Life or Health (IDLH) level for CO2 set by NIOSH is 50,000 ppm. At that level, vomiting, dizziness, disorientation, and breathing difficulties occur after a 30-minute exposure. At a 10% level (100,000 ppm), death can occur after a few minutes even if the oxygen in the atmosphere would otherwise support life. Calculate the concentration of CO2 in the room after the fire extinguisher is emptied. Does it exceed the IDLH value? Assume that the gas mixture in the room is uniformly mixed, that the temperature in the room is 308C (warmed by the fire above the normal room temperature of 208C) and that the ambient pressure is 1 atm. SOLUTION:

First, calculate the number of moles of CO2 discharged by the fire extinguisher: moles of CO2 ¼ (10 lb CO2 )(454 g=lb)=(44 g=gmol CO2 ) ¼ 103 gmol CO2

Calculate the volume of the room: Room volume ¼ (10 ft)(10 ft)(10 ft)(0:0283 m3 =ft3 ) ¼ 28:3 m3 ¼ 28,300 L Next, calculate the total number of moles of gas in the room: moles of gas ¼ PV=RT ¼ (1 atm)(28,300 L)=(0:08206 atm  L=gmol  K)(303 K) ¼ 1138 gmol gas Calculate the concentration, or mole fraction, of CO2 in the room: mole fraction ¼ gmol CO2 =gmol gas ¼ 103 gmol CO2 =1138 gmol gas ¼ 0:0905 Convert this fraction to a percent and compare to the IDLH and lethal levels: %CO2 ¼ (mole fraction)(100) ¼ (0:0905)(100) ¼ 9:05% The IDLH level is 5.0% and the lethal level is 10.0%. Therefore, the level in the room of 9.05% does exceed the IDLH level for CO2. It is also dangerously close to the lethal level. The person extinguishing the fire is in great danger and should take appropriate safety measures. B

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Chapter 22 Environmental Management and Safety Issues

ILLUSTRATIVE EXAMPLE 22.17 Briefly describe HAZOP and HAZAN. SOLUTION: A Hazard and Operability (HAZOP) Study is a systematic approach to recognizing and identifying possible hazards that may cause the failure of a piece of equipment or part of a process system in a new or existing facility. This qualitative enterprise is primarily conducted by a team of technical experts in plant design and operation. A HAZOP study may be applied to operating process plants or it may be performed at various stages throughout the design. An early start will lead to a safer, more efficient design, and ultimately, higher profits. Before any action is taken, the goals of the study should be defined. There are five objectives to most HAZOP studies: 1 To identify areas of the design that may possess a significant hazard potential. 2 To identify and study features of the design that influence the probability of a hazardous incident occuring. 3 To familiarize the study team with the design information available. 4 To ensure that a systematic study is made of the areas of significant hazard potential. 5 To identify pertinent design information not currently available to the team. The recommended procedure for implementing a HAZOP study is as follows. The section of the process to be studied is first identified; generally, the focus is on a major piece of equipment although a pump or a valve may be chosen depending on the hazardous nature of the materials being handled and the operating conditions. Once the intended operation has been defined, a list of possible deviations from the intended operations is developed. The degrees of deviation from normal operation are conveyed by the use of guide words, some of which are listed below. The purpose of these guide words is to develop the thought process and encourage discussion that is related to any potential deviations in the system. When a possible deviation is recognized, the possible causes and consequences are determined. Alterations and appropriate action to be taken are then recommended. Final steps in the methodology include issuing formal reports and following up on recommendations.

Guide words No or not More or less As well as Part of Reverse Other than

Meaning No part of the intention is achieved, but nothing else happens. Quantitative increases or decreases to the intended activity. All of the intention is achieved, but some additional activity occurs. Only part of the intended is achieved; part is not. The opposite of the intention occurs. No part of the intention is achieved. Something different happens.

Examples No flow, no agitation, no reaction. More flow, higher pressure, lower temperature, less time. There is an additional component, contaminant, extra phase. Component omitted, part of multiple destinations omitted. Reverse flow, reverse order of addition. Wrong component, startup, shutdown, utility failure.

Applications

585

After the serious hazards have been identified via a HAZOP study or some other qualitative approach, a quantitative examination should be performed. Hazard quantification or hazard analysis (HAZAN) involves the estimation of the expected frequencies or probabilities of events with adverse or potentially adverse consequences. It logically ties together historical occurrences, experience, and imagination. Once a hazard has been identified, an event tree and fault tree analysis may be used to evaluate the causes and consequences, and mitigate the possible effects. Event tree analyses (ETA) are used to represent possible failure sequences and to analyze the sequence of events that lead to an accident or failure. When the potential hazard has been identified, an analysis of the consequences can be initiated by selecting an appropriate model. The event tree model is started from the initial occurrence and built upon by sequencing the possible events and safety systems that come into play. The model displays at a glance branches of events that relate the proper functioning or failure of a safety device or system and the ultimate consequence. The model also allows quick identification of the various hazards that can result from the single initial event. Fault tree analysis (FTA) begins with the ultimate consequences and works backward to the possible cause and failures. It is based on the most likely or most credible events that lead to the accident. FTA demonstrates the mitigating or reducing effects and can include causes stemming from human error as well as equipment failure. The task of constructing a fault tree is tedious and requires a probability background to handle common failures. B

ILLUSTRATIVE EXAMPLE 22.18 A storage tank at a refinery contains a large volume of contaminated oil waiting to be processed via distillation. The tank is protected against emissions of vapors by a nitrogen blanket. In the event that the nitrogen blanket fails, there is a vent system that includes a canister of activated carbon to adsorb toxic fumes from the vented gas. The tank itself is situated on a high-traffic plant area and is in danger of tank failure resulting from collisions with vehicles improperly operated by plant personnel. Annual failure rate estimates are given below for the (hazardous) oil storage tank subjected to a variety of scenarios: From vent Nitrogen blanket, and Adsorbent canister Tank rupture, or Truck collision, or Driver on drugs out, or Driver drunk, or Brakes fail, or Metal fatigue, or Earthquake Tank overfill, or Miscellaneous tank failure

0.05/y 0.05/y

0.1/y 0.8/y 0.2/y 0.05/y 0.01/y 0.25/y 0.3/y

Prepare a fault tree(14,16) and calculate the overall failure rate for emissions from the storage tank. Which failure mode(s) is (are) most important in terms of potential hazard mitigation and should be addressed first? Use the principle that the failure rate for the independent events (across branches) is the sum (OR gate) of the probabilities, while for dependent events (down branches), the failure rate is the product (AND gate) of the probabilities.

586

Chapter 22 Environmental Management and Safety Issues

≡ ≡

Figure 22.6 Fault tree, Illustrative Example 22.18. SOLUTION: Prepare a fault tree. See Figure 22.6. Calculate the overall failure rate (OFR) for the emissions from the storage tank: OFR ¼ (0:1 þ 0:8 þ 0:2) þ (0:01 þ 0:05) þ (0:25 þ 0:3) ¼ 1:71 Determine which failure mode is most important in terms of potential hazard mitigation: emission < tank rupture < truck collision < driver drunk The most significant contribution to failure is seen to be the hazard associated with drunken truck drivers. Mitigation of this hazard is a good place to start in the process of overall hazard reduction. Alcohol and substance abuse programs can provide a positive step toward mitigation of driver generated accidents. In addition, construction of a physical barrier between

Applications

587

the tank and the roadway would lower the frequency of toxic emissions drastically by preventing the collisions. B

ILLUSTRATIVE EXAMPLE 22.19(18) A risk assessment being conducted at a chemical plant is concerned with the consequences of two incidents that are defined as follows: I. A distillation column explosion resulting from detonation of an unstable chemical (ethylene oxide). II. A continuous 240 g/s release of the toxic chemical (ethylene oxide) at an elevation of 125 m. Both incidents occur at approximately the same location in the plant. Two weather conditions are envisioned, namely a northeast wind and a southwest wind (6.0 miles/h) with stability category “B”. Associated with these two wind directions are events IIA, and IIB, respectively, defined as follows: IIA: Toxic cloud to the southwest IIB: Toxic cloud to the northeast Based on an extensive literature search, the probabilities and conditional probabilities of the occurrence of the defined events in any given year are estimated as follows: P(I) ¼ 106 P(II) ¼ 1=33,333 P(IIAjII) ¼ 0:33 P(IIBjII) ¼ 0:67 The consequences of events I, IIA, and IIB, in terms of number of people killed, are estimated as follows: I: All persons within 200 meters of the explosion center are killed; all persons beyond this distance are unaffected. IIA: All persons in a pie-shaped segment 22.5 degrees width (downwind of the source) are killed if the concentration of the toxic gas is above 0.33 mg/L; all persons outside of this area are unaffected. IIB: Same as IIA. Three people are located within 200 m of the explosion center but not in the pie-shaped segment described above. Five people are located within the pie-shaped segment southwest of the discharge center; three are 350 m downwind, two are 600 m away at the plant fence (boundary). Another four people are located 500 m away outside the pie-shaped segment but within the plant boundary. All individuals are at ground level. Calculate the average annual individual risk based on the number of individuals potentially affected. Also calculate the average risk based on all the individuals within the plant boundary. Hint: Perform atmospheric dispersion calculations at a distance of 300, 500, 800, 1000, 1500, and 2000 m from the emission source.

588

Chapter 22 Environmental Management and Safety Issues

≤

Figure 22.7 Plant personnel location. SOLUTION: Draw a line diagram (see Figure 22.7) of the plant layout and insert all pertinent data and information. An event tree for the process [including a possible (though negligible) vapor cloud explosion] is provided in Figure 22.8. The probability of event IIA and event IIB occurring is: P(IIA) ¼ P(II)P(IIAjII) ¼ (1=33,333)(0:33) ¼ 1=100,000 ¼ 105 P(IIB) ¼ P(II)P(IIBjII) ¼ (1=33,333)(0:67) ¼ 2=100,000 ¼ 2  105 Perform a dispersion calculation to determine the zones where the concentration of the nanochemical exceeds 0.33 mg/L. Assume a continuous emission for a point source.

Applications

Figure 22.8

589

Event tree.

To maintain consistent units, convert wind speed from mi/h to m/s and concentration from mg/ L to g/m3: 0:33 mg

1g

103 L

L

106 mg

m3

6:0 mi

5280 ft

h

0:3048 m

h

mi

3600 s

ft

¼ 3:3  104 g=m3

¼ 2:68 m=s

Set up the Pasquill –Gifford model using the data and calculations provided above.(17) C(x, 0, 0, H) ¼ m exp[0:5(H=sz )2 ]=psy sz u

590

Chapter 22 Environmental Management and Safety Issues Table 22.2 Dispersion Calculations

where

x (m)

sy (m)

sz (m)

C (g/m3)

300 500 800 1000 1500 2000

52 83 128 156 225 295

30 51 86 110 170 235

3.101026 3.341024 9.011024 8.711024 5.691024 3.571024

C ¼ 3.3  1024 g/m3 m ¼ 240 g/s u ¼ 2.68 m/s H ¼ 125 m

sy, sz ¼ f (x) Employing the sy and sz values recommended in the literature for stability category “B”,(17) calculate the downwind concentrations that satisfy the Pasquill– Gifford equation. The results for the recommended downwind distances are shown in Table 22.2. Note that the concentration goes through a maximum that is in excess of 0.33 mg/L; thus, there are two solutions. A linear interpolation (or plotting the results on a graph) indicates that the maximum GLC is approximately 9.9  1024 g/m3 and is located at a downwind distance of about 850 m. The “critical” zone is located between 500 m and 2175 m. One may determine which individuals within the pie-shaped segment downwind from the source will be killed if either Accident I or II occurs. Refer to the problem statement or Figure 22.7. One can conclude that three individuals within the 200 m radius will die from Accident I. Two individuals located in the pie-shaped segment and 600 m southwest of the emission source will die from Accident II. The total annual risk (TAR) for the process may now be determined. The total risk, measured in terms of the average annual total number of people killed, is obtained by multiplying the number of people in each impact zone by the sum of the probabilities of the events affecting that zone, and summing the results: TAR ¼ (3)P(I) þ (2)P(IIA) ¼ (3)(106 ) þ (2)(105 ) ¼ 2:3  105 The average annual risk (AAR) based only on the “potentially affected” people can be calculated. The average annual individual risk for the eight people in the impact zone is obtained by dividing the total annual risk by 8. Since only eight people are “affected,” AAR ¼ 2:3  105 =8 ¼ 2:9  106

References

591

Finally, calculate the average annual individual risk for all the individuals within the plant (fence) boundary. The average is now based on 12 rather than eight individuals. AAR ¼ 2:3  105 =12 ¼ 1:9  106

B

REFERENCES 1. G. BURKE, B. SINGH, and L. THEODORE, “Handbook of Environmental Management and Technology,” 2nd edition, John Wiley & Sons, Hoboken, NJ, 2001. 2. L. THEODORE, M. HYLAND, Y. MCGUINN, L. SCHOEN, and F. TAYLOR, “Principles of Accident and Emergency Management,” USEPA Manual, Air Pollution Training Institute, RTP, NC, 1988. 3. A. M. FLYNN and L. THEODORE, Personal notes, 1999. 4. A. M. FLYNN and L. THEODORE, “Health, Safety and Accident Management in the Chemical Process Industries,” Marcel Dekker, New York City, NY (acquired by Taylor & Francis), Boca Raton, FL, 2002. 5. D. PAUSTENBACH, “The Risk Assessment of Environmental and Human Health Hazards: A Textbook of Case Studies,” John Wiley & Sons, Hoboken, NJ, 1989. 6. Manual of Industrial Hazard Assessment Techniques, Office of Environmental and Scientific Affairs, The World Bank, October, 1985, London. 7. J. RODRICKS and R. TARDIFF, “Assessment and Management of Chemical Risks,” ACS, Washington DC, 1984. 8. M. K. THEODORE and L. THEODORE, “Introduction to Major Environmental Issues,” CRC Press, Boca Raton, FL, 2010. 9. J. REYNOLDS, R. DUPONT, and L. THEODORE, “Hazardous Waste Incineration Calculations: Problems and Software,” John Wiley & Sons, Hoboken, NJ, 1991. 10. R. DUPONT, L. THEODORE, and J. REYNOLDS, “Accident and Emergency Management: Problems and Solutions,” VCH Publishers, New York City, NY, 1991. 11. L. THEODORE, R. DUPONT, and J. REYNOLDS, “Pollution Prevention: Problems and Solutions,” Gordon and Breach Publishers, Amsterdam, Holland, 1994. 12. K. GANESON, L. THEODORE, and J. REYNOLDS, “Air Toxics—Problems and Solutions,” Gordon and Breach Publishers, Amsterdam, Holland, 1996. 13. R. DUPONT, T. BAXTER, and L. THEODORE, “Environmental Management: Problems and Solutions,” CRC Press, Boca Raton, FL, 1998. 14. S. SHAEFER and L. THEODORE, “Probability and Statistics for Environmental Scientists,” CRC Press, Boca Raton, FL, 2007. 15. L. STANDER and L. THEODORE, “Environmental Regulatory Calculations Handbook,” John Wiley & Sons, Hoboken, NJ, 2008. 16. J. REYNOLDS, J. JERIS, and L. THEODORE, “Handbook of Chemical and Environmental Engineering Calculations,” John Wiley & Sons, Hoboken, NJ, 2002. 17. L. THEODORE, “Air Pollution Center Equipment Calculations,” John Wiley & Sons, Hoboken, NJ, 2008. 18. B. BARKWILL, C. RODESCHIN, and L. THEODORE, “Is it a Health or Hazard Risk Problem . . . or Both?” AWMA Proceedings, Calgary, 2010.

NOTE: Additional problems are available for all readers at www.wiley.com. Follow links for this title. These problems may be used for additional review, homework, and/or exam purposes.

Appendix A UNITS B MISCELLANEOUS TABLES C STEAM TABLES

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

593

Appendix

A

Units A.1

THE METRIC SYSTEM

A.2

THE SI SYSTEM

A.3

SEVEN BASE UNITS

A.4

TWO SUPPLEMENTARY UNITS

A.5

SI MULTIPLES AND PREFIXES

A.6

CONVERSION CONSTANTS

A.7

SELECTED COMMON ABBREVIATIONS

A.1 THE METRIC SYSTEM The need for a single worldwide coordinated measurement system was recognized over 300 years ago. In 1670, Gabriel Mouton, Vicar of St. Paul’s church in Lyon, proposed a comprehensive decimal measurement system based on the length of one minute of arc of a great circle of the Earth. In 1671, Jean Picard, a French astronomer, proposed the length of a pendulum beating seconds as the unit of length. (Such a pendulum would have been fairly easy to reproduce, thus facilitating the widespread distribution of uniform standards.) Other proposals were made, but over a century elapsed before any action was taken. In 1790, in the midst of the French Revolution, the National Assembly of France requested the French Academy of Sciences to “deduce an invariable standard for all the measures and weights.” The Commission appointed by the Academy created a system that was, at once, simple and scientific. The unit of length was to be a portion of the Earth’s circumference. Measures for capacity (volume) and mass (weight) were to be derived from the unit of length, thus relating the basic units of the system to each other and to nature. Furthermore, the larger and smaller versions of each unit were to be created by multiplying or dividing the basic units by 10 and its multiples. This feature provided a great convenience to users of the system by Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

595

596

Appendix A

Units

eliminating the need for such calculating and dividing by 16 (to convert ounces to pounds) or by 12 (to convert inches to feet). Similar calculations in the metric system could be performed simply by shifting the decimal point. Thus the metric system is a base-10 or decimal system. The Commission assigned the name metre (which we now spell meter) to the unit of length. This name was derived from the Greek word metron meaning “a measure.” The physical standard representing the meter was to be constructed so that it would equal one ten-millionth of the distance from the north pole to the equator along the meridian of the Earth running near Dunkirk in France and Barcelona in Spain. The metric unit of mass, called the gram, was defined as the mass of one cubic centimeter (a cube that is 1/100 of a meter on each side) of water at its temperature of maximum density. The cubic decimeter (a cube 1/10 of a meter on each side) was chosen as the unit of fluid capacity. This measure was given the name liter. Although the metric system was not accepted with enthusiasm at first, adoption by other nations occurred steadily after France made its use compulsory in 1840. The standardized character and decimal features of the metric system made it well suited to scientific and engineering work. Consequently, it is not surprising that the rapid spread of the system coincided with an age of rapid technological development. In the United States, by Act of Congress in 1866, it was made “lawful throughout the United States of America to employ the weights and measures of the metric system in all contracts, dealings, or court proceedings.” By the late 1860s, even better metric standards were needed to keep pace with scientific advances. In 1875, an international treaty, the “Treaty of the Meter,” set up well-defined metric standards for length and mass, and established permanent machinery to recommend and adopt further refinements in the metric system. This treaty, known as the Metric Convention, was signed by 17 countries, including the United States. As a result of the Treaty, metric standards were constructed and distributed to each nation that ratified the Convention. Since 1893, the internationally agreed to metric standards have served as the fundamental weights and measures standards of the United States. By 1900, a total of 35 nations—including the major nations of continental Europe and most of South America—had officially accepted the metric system. Today, with the exception of the United States and a few small countries, the entire world is using predominantly the metric system or is committed to such use. In 1971, the Secretary of Commerce, in transmitting to Congress the results of a 3-year study authorized by the Metric Study Act of 1968, recommended that the U.S. change to predominant use of the metric system through a coordinated national program. The International Bureau of Weights and Measures located at Sevres, France, serves as a permanent secretariat for the Metric Convention, coordinating the exchange of information about the use and refinement of the metric system. As measurement science develops more precise and easily reproducible ways of defining the measurement units, the General Conference of Weights and Measures (the diplomatic organization made up of adherents to the Convention) meets periodically to ratify improvements in the system and the standards.

A.3 Seven Base Units

597

A.2 THE SI SYSTEM In 1960, the General Conference adopted an extensive revision and simplification of the system. The name Le Systeme International d’Unites (International System of Units), with the international abbreviation SI, was adopted for this modernized metric system. Further improvements in and additions to SI were made by the General Conference in 1964, 1968, and 1971. The basic units in the SI system are the kilogram (mass), meter (length), second (time), Kelvin (temperature), ampere (electric current), candela (the unit of luminous intensity), and radian (angular measure). All are commonly used by the engineer. The Celsius scale of temperature (08C –273.15 K) is commonly used with the absolute Kelvin scale. The important derived units are the newton (SI unit of force), the joule (SI unit of energy), the watt (SI unit of power), the pascal (SI unit of pressure), and the hertz (unit of frequency). There are a number of electrical units: coulomb (charge), farad (capacitance), henry (inductance), volt (potential), and weber (magnetic flux). One of the major advantages of the metric system is that larger and smaller units are given in powers of ten. In the SI system, a further simplification is introduced by recommending only those units with multipliers of 103. Thus for lengths in engineering, the micrometer (previously micron), millimeter, and kilometer are recommended, and the centimeter is generally avoided. A further simplification is that the decimal point may be substituted by a comma (as in France, Germany, and South Africa), while the other number, before and after the comma, will be separated by spaces between groups of three, that is, one million dollars will be $1 000 000,00. More details are provided below.

A.3 SEVEN BASE UNITS a. Length—meter (m) The meter (common international spelling, metre) is defined as 1 650 763.00 wavelengths in vacuum of the orange-red line of the spectrum of krypton-86. The SI unit of area is the square meter (m2). The SI unit of volume is the cubic meter (m3). The liter (0.001 cubic meter), although not an SI unit, is commonly used to measure fluid volume. b. Mass—kilogram (kg) The standard for the unit of mass, the kilogram, is a cylinder of platinum – iridium alloy kept by the International Bureau of Weights and Measures at Paris. A duplicate in the custody of the National Bureau of Standards serves as the mass standard for the United States. This is the only base unit still defined by an artifact. The SI unit of force is the newton (N). One newton is the force which, when applied to a 1 kilogram mass, will give the kilogram mass an acceleration of 1 (meter per second) per second. 1 N ¼ 1 kg . m/s2. The SI unit for pressure is the pascal (Pa). 1 Pa ¼ 1 N/m2. The SI unit for work and energy of any kind is the joule (J). 1 J ¼ 1 N . m. The SI unit for power of any kind is the watt (W). 1 W ¼ 1 J/s.

598

Appendix A

Units

c. Time—second (s) The second is defined as the duration of 9 192 632 770 cycles of the radiation associated with a specified transition of the cesium-133 atom. It is realized by tuning an oscillator to the resonance frequency of cesium-133 atoms as they pass through a system of magnets and a resonant cavity into a detector. The number of periods or cycles per second is called frequency. The SI unit for frequency is the hertz (Hz). One hertz equals one cycle per second. The SI unit for speed is the meter per second (m/s). The SI unit for acceleration is the (meter per second) per second (m/s2). d. Electric current—ampere (A) The ampere is defined as that current which, if maintained in each of two long parallel wires separated by one meter in free space, would produce a force between the two wires (due to their magnetic fields) of 2  1027 newton for each meter of length. The SI unit of voltage is the volt (V), 1 V ¼ 1 W/A. The SI unit of electrical resistance is the ohm (V). 1 V ¼ 1 V/A. e. Temperature—Kelvin (K) The Kelvin is defined as the fraction 1/273.15 of the thermodynamic temperature of the triple point of water. The temperature 0 K is called absolute zero. On the commonly used Celsius temperature scale, water freezes at about 08C and boils at about 1008C. The 8C is defined as an interval of 1 K and the Celsius temperature 08C is defined as 273.15 K. In Fahrenheit scale, 1.8 degrees are equal to 1.08C or 1.0 K; the Fahrenheit scale uses 328F as a temperature corresponding to 08C. f. Amount of substance—mole (mol) The mole is the amount of substance of a system that contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles. The SI unit of concentration (of amount of substance) is the mole per cubic meter (mol/m3). g. Luminous intensity—candela (cd) The candela is defined as the luminous intensity of 1/600 000 of a square meter of a blackbody at the temperature of freezing platinum (2045 K). The SI unit of light flux is the lumen (lm). A source having an intensity of 1 candela in all directions radiates a light flux of 4p lumens.

A.4 TWO SUPPLEMENTARY UNITS a. Phase angle—radian (rad) The radian is the plane angle with its vertex at the center of a circle that is subtended by an arc equal in length to the radius.

A.6

Conversion Constants (SI)

599

b. Solid angle—steradian (sr) The steradian is the solid angle with its vertex at the center of a sphere that is subtended by an area of the spherical surface equal to that of a square with sides equal in length to the radius.

A.5 SI MULTIPLES AND PREFIXES Multiples and submultiples 1 000 000 000 000 1 000 000 000 1 000 000 1 000 100 10 Base unit 1 0.1 0.01 0.001 0.000 001 0.000 000 001 0.000 000 000 001 0.000 000 000 000 001 0.000 000 000 000 000 001

1012 109 106 103 102 101 100 1021 1022 1023 1026 1029 10212 10215 10218

Prefixes

Symbols

tera (ter’a) giga ( ji’ga) mega (meg’a) kilo (kil’o) hecto (hek’to) deka (dek’a)

T G M k h da

deci (des’i) centi (sen’ti) milli (mil’i) micro (mi’kro) nano (nan’o) pico (pe’ko) femto (fem’to) atto (at’to)

d c m m n p f a

A.6 CONVERSION CONSTANTS (SI) To convert from Length m m m m m m m ft ft

To cm mm microns (mm) ˚) angstroms (A in. ft mi in. m

Multiply by 100 1000 106 1010 39.37 3.281 6.214  1024 12 0.3048 (Continued)

600

Appendix A

Units

To convert from

To

Multiply by

ft ft

cm mi

30.48 1.894  1024

Mass kg kg kg kg kg lb lb lb lb lb

g lb oz ton grains oz ton g kg grains

1000 2.205 35.24 2.268  1024 1.543  104 16 5  1024 453.6 0.4536 7000

Time s s s s s

min h day week yr

0.01667 2.78  1024 1.157  1027 1.653  1026 3.171  1028

Force N N N N N lbf lbf lbf lbf

kg . m/s2 dynes g . cm/s2 lbf lb . ft/s2 N dynes g . cm/s2 lb . ft/s2

1 105 105 0.2248 7.233 4.448 4.448  105 4.448  105 32.17

Pressure atm atm atm atm atm atm atm atm atm

N/m2 (Pa) kPa bars dynes/cm2 lbf/in2 (psi) mm Hg at 08C (torr) in Hg at 08C ft H2O at 48C in H2O at 48C

1.013  105 101.3 1.013 1.013  106 14.696 760 29.92 33.9 406.8 (Continued)

A.6

To convert from

To

Conversion Constants (SI)

Multiply by

psi psi psi in H2O at 48C in H2O at 48C in H2O at 48C

atm mm Hg at 08C (torr) in H2O at 48C atm psi mm Hg at 08C (torr)

6.80  1022 51.71 27.70 2.458  1023 0.0361 1.868

Volume m3 m3 m3 m3 m3 ft3 ft3 ft3 ft3

L cm3 (cc, mL) ft3 gal (U.S.) qt in3 gal (U.S.) m3 L

1000 106 35.31 264.2 1057 1728 7.48 0.02832 28.32

Energy J J J J J J J cal cal cal Btu Btu Btu Btu ft . lbf ft . lbf ft . lbf

N.m erg dyne . cm kW . h cal ft . lbf Btu J Btu ft . lbf ft . lbf hp . h cal kW . h cal J Btu

1 107 107 2.778  1027 0.2390 0.7376 9.486  1024 4.186 3.974  1023 3.088 778 3.929  1024 252 2.93  1024 0.3239 1.356 1.285  1023

Power W W W W

J/s cal/s ft . lbf/s kW

1 0.2390 0.7376 1023 (Continued)

601

602

Appendix A

Units

To convert from

To

Multiply by

kW kW hp hp hp hp

Btu/s hp ft . lbf/s kW cal/s Btu/s

0.949 1.341 550 0.7457 178.2 0.707

Concentration mg/m3 mg/m3 mg/m3 gr/ft3 gr/ft3 lb/ft3 lb/ft3 lb/ft3

lb/ft3 lb/gal gr/ft3 mg/m3 g/m3 mg/m3 mg/L lb/gal

6.243  10211 8.346  10212 4.370  1027 2.288  106 2.288 1.602  1010 1.602  107 7.48

Viscosity P (poise) P P P P P lb/ft . s lb/ft . s lb/ft . s lb/ft . s

g/cm . s cP (centipoise) kg/m . h lb/ft . s lb/ft . h lb/m . s P g/cm . s kg/m . h lb/ft . h

1 100 360 6.72  1022 241.9 5.6  1023 14.88 14.88 5.357  103 3600

Heat Capacity cal/g . 8C cal/g . 8C cal/g . 8C cal/gmol . 8C J/g . 8C Btu/lb . 8F Btu/lb . 8F Btu/lb . 8F

Btu/lb . 8F kcal/kg . 8C cal/gmol . 8C Btu/lbmol . 8F Btu/lb . 8F cal/g . 8C J/g . 8C Btu/lbmol . 8F

1 1 Molecular weight 1 0.2389 1 4.186 Molecular weight

A.7

Selected Common Abbreviations

A.7 SELECTED COMMON ABBREVIATIONS ˚, A A abs amb app. MW, M atm, ATM at. wt. b.p. bbl Btu cal cg cm cgs system conc. cc, cm3 cu ft, ft3 cfh cfm cfs m3, M3 8 8C 8F 8R ft ft . lb fpm fps fps system f.p. gr g, gm h in kcal kg km liq L log ln m.p.

angstrom unit of length absolute ambient apparent molecular weight atmospheric atomic weight boiling point barrel British thermal unit calorie centigram centimeter centimeter-gram-second system concentrated, concentration cubic centimeter cubic feet cubic feet per hour cubic feet per minute cubic feet per second cubic meter degree degree Celsius, degree Centigrade degree Fahrenheit degree Reamur, degree Rankine foot foot pound feet per minute feet per second foot-pound-second system freezing point grain gram hour inch kilocalorie kilogram kilometer liquid liter logarithm (common) logarithm (natural) melting point

603

604

Appendix A

Units

m, M mm mks system mph mg ml mm mm min mol wt, MW, M oz ppb pphm ppm lb psi psia psig rpm s sp gr, SG sp ht sp wt sq scf STP temp wt

meter micrometer (micron) meter-kilogram-second system miles per hour milligram milliliter millimeter millimicron minute molecular weight ounce parts per billion parts per hundred million parts per million pound pounds per square inch pounds per square inch absolute pounds per square inch gage revolutions per minute second specific gravity specific heat specific weight square standard cubic foot standard temperature and pressure temperature weight

Appendix

B

Miscellaneous Tables B.1

COMMON ENGINEERING CONVERSION FACTORS

B.2

PROPERTIES OF SELECTED GASES AT 1 ATM AND 208C (688F)

B.3

PROPERTIES OF SELECTED LIQUIDS AT 1 ATM AND 208C (688F)

B.4

PROPERTIES OF WATER AT 1 ATM (CRITICAL POINT 3748C, 22.09 MPa)

B.5

PROPERTIES OF AIR AT 1 ATM

B.6

DIMENSIONS, CAPACITIES, AND WEIGHTS OF STANDARD STEEL PIPES

B.7

DIMENSIONS OF HEAT EXCHANGER TUBES

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

605

606

Appendix B

Miscellaneous Tables

Table B.1 Common Engineering Conversion Factors Length 1 ft ¼ 12 in ¼ 0.3048 m, 1 yard ¼ 3 ft 1 mi ¼ 5280 ft ¼ 1609.344 m 1 nautical mile (nmi) ¼ 6076 ft Mass 1 slug ¼ 32.174 lb ¼ 14.594 kg 1 lb ¼ 0.4536 kg ¼ 7000 grains Acceleration and Area 1 ft/s2 ¼ 0.3048 m/s2 1 ft2 ¼ 0.092903 m2 Mass flow and Mass flux 1 slug/s ¼ 14.594 kg/s, 1 lb/s ¼ 0.4536 kg/s 1 kg/m2 . s ¼ 0.2046 lb/ft2 . s ¼ 0.00636 slug/ft2 . s Pressure 1 lbf/ft2 ¼ 47.88 Pa, 1 torr ¼ 1 mm Hg 1 psi ¼ 144 psf, 1 bar ¼ 105 Pa 1 atm ¼ 2116.2 psf ¼ 14.696 psi ¼ 101,325 Pa ¼ 29.9 in Hg ¼ 33.9 ft H2O Power 1 hp ¼ 550 (ft . lbf )/s ¼ 745.7 W 1 (ft . lbf )/s ¼ 1.3558 W 1 Watt ¼ 3.4123 Btu/h ¼ 0.00134 hp Specific weight 1 lbf/ft3 ¼ 157.09 N/m3 Viscosity 1 slug/(ft . s) ¼ 47.88 kg/(m . s) ¼ 478.8 poise (P) 1 P ¼ 1 g/(cm . s) ¼ 0.1 kg/(m . s)¼ 0.002088 slug/(ft . s) Temperature scale readings 8F ¼ (9/5)8C þ 32 8C ¼ (5/9)(8F 2 32) 8R ¼ 8F þ 459.69 K ¼ 8C þ 273.16 8R ¼ (1.8) K Specific heat or Gas constant 1 (ft . lbf )/(slug . 8R) ¼ 0.16723 (N . m)/ (kg . K) 1 Btu/(lb . 8R) ¼ 4186.8 J/(kg . K) 

Volume 1 ft3 ¼ 0.028317 m3 ¼ 7.481 gal, 1 bbl ¼ 42 U.S. gal 1 U.S. gal ¼ 231 in3 ¼ 3.7853 L ¼ 4 qt ¼ 0.833 Imp. gal 1 L ¼ 0.001 m3 ¼ 0.035315 ft3 ¼ 0.2642 U.S. gal Density 1 slug/ft3 ¼ 515.38 kg/m3, 1 g/cm3 ¼ 1000 kg/m3 1 lb/ft3 ¼ 16.0185 kg/m3, 1 lb/in3 ¼ 27.68 g/cm3 Velocity 1 ft/s ¼ 0.3048 m/s, 1 knot ¼ 1 nmi/h ¼ 1.6878 ft/s 1 mi/h ¼ 1.4666666 ft/s ¼ 0.44704 m/s Volume flow 1 gal/min ¼ 0.002228 ft3/s ¼ 0.06309 L/s 1 million gal/day ¼ 1.5472 ft3/s ¼ 0.04381 m3/s Force and Surface tension 1 lbf ¼ 4.448222 N ¼ 16 oz, 1 dyne ¼ 1 g . cm/s2 ¼ 1025 N 1 kgf ¼ 2.2046 lbf ¼ 9.80665 N 1 U.S. (short) ton ¼ 2000 lbf, 1 N ¼ 0.2248 lbf 1 N/m ¼ 0.0685 lbf/ft Energy and Specific energy 1 ft . lbf ¼ 1.35582 J, 1 hp . h ¼ 2544.5 Btu 1 Btu ¼ 252 cal ¼ 1055.056 J ¼ 778.17 ft . lbf 1 cal ¼ 4.1855 J, 1 ft . lbf/lb ¼ 2.9890 J/kg Heat flux 1 W/m2 ¼ 0.3171 Btu/(h . ft2) Kinematic viscosity 1 ft2/h ¼ 2.5061025 m2/s, 1 ft2/s ¼ 0.092903 m2/s 1 stoke (st) ¼ 1 cm2/s ¼ 0.0001 m2/s ¼ 0.001076 ft2/s Thermal conductivity 1 cal/(s . cm . 8C) ¼ 242 Btu/(h . ft . 8R) 1 Btu/(h . ft . 8R) ¼ 1.7307 W/(m . K)

Note that the intervals in absolute (Kelvin) and 8C are equal. Also, 18R ¼ 18F. Latent heat: 1 J/kg ¼ 4.2995  1024 Btu/lb ¼ 10.76 lbf . ft/slug ¼ 0.3345 lbf . ft/lb, 1 Btu/lb ¼ 2325.9 J/kg. Heat transfer coefficient: 1 Btu/(h . ft2 . 8F) ¼ 5.6782 W/(m2 . 8C). Heat generation rate: 1 W/m3 ¼ 0.09665 Btu/(h . ft3). Heat transfer per unit length: 1 W/m ¼ 1.0403 Btu/(h . ft). Mass transfer coefficient: 1 m/s ¼ 11.811 ft/h, 1 lbmol/(h . ft2) ¼ 0.0013562 kgmol/(s . m2).

607

26 28.96 17.03 39.944 58.1 44.01 28.01 70.91 30.07 28 4.003 2.016 36.5 34.1 16.04 50.5 19.5 28.02 30.01 44.02 32.00 44.1 64 18.02

1.09 1.20 0.74 1.66 2.49 1.83 1.16 2.95 1.25 1.17 0.166 0.0838 1.53 1.43 0.667 2.15 0.804 1.16 1.23 1.83 1.36 1.88 2.66 0.749

Density, r, kg/m3 8.3 15.0 13.6 13.5 8.09 15.7 3.49 6.8 8.3 118.7 108.0 8.76 8.67 20.1

15.2 15.4 7.92 14.7 5.2 13.6

1.48 1.82 1.03 0.85 0.97 1.97 0.905 1.34 1.24 1.34

1.76 1.90 1.45 2.00 1.38 1.02

Kinematic, n, m2/s (106 )

0.97 1.80 1.01 2.24

Dynamic, m, kg/m . s (105 )

Viscosity

1.30 1.40 1.31 1.67 1.11 1.30 1.40 1.34 1.19 1.22 1.66 1.41 1.41 1.30 1.32 1.20 1.27 1.40 1.40 1.31 1.40 1.15 1.29 1.33

Ratio of specific heats, k

33.5 65.0 71.7 49.7 42.0 77.8 218.3

37.5 72.9 34.5 76.1 48.2 50.5 2.26 12.8 81.5 88.9 45.8 65.8

425.2 304 133 417 305 283.1 5.26 33 324.6 373.6 190 416.1 126 179 309 154 369.9 430 647

61.6 37 111.3

Pcrit, atm

309.5 133 405

Tcrit, K

Example: At 208C, the properties of argon gas are: molecular weight ¼ 39.944, density ¼ 1.66 kg/m3 (0.00322 slug/ft3 ¼ 0.104 lb/ft3), dynamic viscosity ¼ 0.0000224 kg/m . s (0.0224 cP ¼ 4.68  1027 slug/ft . s ¼ 1.51  1025 lb/ft . s), kinematic viscosity ¼ 13.5  1026 m2/s (13.5 cSt ¼ 1.45  1024 ft2/s ¼ 0.523 ft2/h), specific heat ratio ¼ 1.67.

Acetylene Air (dry) Ammonia Argon Butane Carbon dioxide Carbon monoxide Chlorine Ethane Ethylene Helium Hydrogen Hydrogen chloride Hydrogen sulfide Methane Methyl chloride Natural gas Nitrogen Nitrogen oxide (NO) Nitrous oxide (N2O) Oxygen Propane Sulfur dioxide Water vapor

Gas

Molecular weight

Table B.2 Properties of Selected Gases at 1 atm and 208C (688F)

608

Acetic acid Acetone Ammonia Benzene Carbon disulfide Carbon tetrachloride Castor oil Crude oil Engine oil (unused) Ethanol (ethyl alcohol) Ethylene glycol Freon-12 Fuel oil, heavy Fuel oil, medium Gasoline

Liquid

785 608 881 1272 1590 970 856 888 789 1117 1330 908 854 680

Density, kg/m3 0.403 0.362 0.739 0.608 927.8 8.4 900.2 1.4 19.16 0.198 145.9 3.82 0.429

9.67 9000 72 7994 11 214 2.63 1324 32.7 2.92

Kinematic viscosity, n, m2/s 106

3.16 2.20 6.51

Absolute viscosity, m, kg/m . s 10,000

5.7

55.1

2.16

27.6 910 10.1 47.9 1.20

Vapor pressure, kPa

2.28 3.27 1.58

3.0

2.70

2.31 2.13 2.88

Surface tension, N/m 100

Table B.3 Properties of Selected Liquids at 1 atm and 208C (688F)

1144 1644

924 1474

1298

1174

Sound velocity, m/s

1880 2433 2382

2051.6 2210 4798 1700

Heat capacity, J/kg . K

0.145 0.182 0.250

0.18 0.18

0.521 0.159

Thermal conductivity, k, W/m . K

(Continued)

0.65

0.7

2.45

Coefficient of thermal expansion, b, 1000  K21

609

1260 804 13,500 791 1041 1030 919 624 919 917 917 1025 862 998

1183 0.239 0.115 0.756 1.34 2.06 91.4 43.5 113.4 316.2 1.04 1.73 1.06

14,900 1.92 15.6 5.98 14 21.2 840 400 1040 2900 10.7 14.9 10.0 2.34 2.34

7.28

58.9

0.14 3.11 11  1027 13.4

3.6 3.5 7.28

6.33 2.8 48.4 2.25

Properties of Selected Liquids at 1 atm and 208C (688F) (Continued )

1498

1535

1909 1320 1450 1103

4186

13.9 2512

2386

0.215

0.286 0.149

0.5

Example: At 208C, the properties of liquid methanol are: density ¼ 791 kg/m3 (or SG ¼ 0.791), dynamic viscosity ¼ 0.000598 kg/m . s (or 0.598 cP), kinematic viscosity ¼ 0.756  1026 m2/s (0.756 cP ¼ 8.14  1026 ft2/s), surface tension ¼ 0.0225 N/m (0.00154 lbf/ft), vapor (or saturation) pressure ¼ 13,400 Pa (1.943 psi), sound velocity ¼ 1104 m/s, heat capacity ¼ 2512 J/kg . K, thermal conductivity ¼ 0.215 W/m . K.

Glycerin Kerosene Mercury Methanol Milk (skim) Milk (whole) Olive oil Pentane Soybean oil SAE 10 oil SAE 30 oil Seawater Turpentine Water

TABLE B.3

610

32 41 50 59 68 77 86 104 122 140 158 176 194 212

0 5 10 15 20 25 30 40 50 60 70 80 90 100

1000 1000 1000 999 998 997 996 992 988 983 978 972 965 958

kg/m

3

1.940 1.940 1.940 1.938 1.937 1.934 1.932 1.925 1.917 1.908 1.897 1.886 1.873 1.859

slug/ft

Density, r

3

1.788 1.518 1.307 1.139 1.003 0.890 0.799 0.657 0.548 0.467 0.405 0.355 0.316 0.283

0.373 0.317 0.273 0.238 0.209 0.186 0.167 0.137 0.114 0.975 0.846 0.741 0.660 0.591

5

kg/m . s (10 ) slug/ft . s (10 )

3

Absolute (dynamic) viscosity, m

1.788 1.519 1.307 1.139 1.005 0.893 0.802 0.662 0.555 0.475 0.414 0.365 0.327 0.295

1.925 1.635 1.407 1.226 1.082 0.961 0.864 0.713 0.597 0.511 0.446 0.393 0.352 0.318

ft2/s (105)

Kinematic viscosity, n m /s (106)

2

0.0756 0.0749 0.0742 0.0735 0.0728 0.0720 0.0712 0.0696 0.0679 0.0662 0.0644 0.0626 0.0608 0.0589

0.611 0.87 1.227 1.70 2.337 3.17 4.242 7.375 12.34 19.92 31.16 47.35 70.11 101.33

9.40 7.87 6.79 5.86 5.15

13.25

Surface tension, Vapor Prandtl g, N/m pressure, kPa number, Pr

Example: At 508C (1228F) r ¼ 988 kg/m3 (1.917 slug/ft3), m ¼ 0.548  1023 kg/m . s (0.114  1025 slug/ft . s), n ¼ 0.555  1026 m2/s (0.597  1025 ft2/s), g ¼ 0.0679 N/m (0.00465 lbf/ft), vapor pressure ¼ 12,340 Pa (1.79 psi).

8F

8C

Temperature

Table B.4 Properties of Water at 1 atm (Critical Point 3748C, 22.09 MPa)

611

1.52 1.40 1.29 1.248 1.205 1.165 1.128 1.09 1.060 1.000 0.946 0.835 0.746 0.675 0.616 0.525 0.457

1.51 1.61 1.71 1.76 1.81 1.86 1.90 1.95 2.00 2.09 2.17 2.38 2.57 2.75 2.93 3.25 3.55

Viscosity, m, kg/m . s (105) 0.98 1.15 1.32 1.41 1.50 1.60 1.68 1.79 1.87 2.09 2.30 2.85 3.45 4.07 4.76 6.19 7.77

Kinematic viscosity, n, m2/s (105) 1004.8 1004.8 1004.8 1004.8 1004.8 1004.8 1007.0 1009.0 1009.0 1009.0 1017.0 1025.8 1034.1

Heat capacity, cp, J/kg . K

0.709

5.73

0.715 0.713

Prandtl number, Pr

0.686 0.680

3.65 3.53

Thermal expansion coefficient, b, K (103)

3.12 3.53 3.88 4.24

2.7 2.8

2.0 2.21 2.42 2.49

Thermal conductivity, k, W/m . K (102)

Example: At 508C, the air properties are: density ¼ 1.09 kg/m3 (0.00211 slug/ft3 ¼ 0.679 lb/ft3), dynamic viscosity ¼ 0.0000195 kg/m . s (4.073  1027 slug/ft . s ¼ 1.31  1025 lb/ft . s), thermal conductivity, k ¼ 0.028 W/m . K, coefficient of thermal expansion, b ¼ 1/T ¼ 1/(273 þ 50) ¼ 0.0031 K21. The Prandtl number, Pr ¼ cpm/k ’ 0.7.

240 220 0 10 20 30 40 50 60 80 100 150 200 250 300 400 500

Temperature, 8C

Density, r, kg/m3

Table B.5 Properties of Air at 1 atm

612

0.405

0.540

0.675

0.840

1.050

1.315

1.660

1.900

2.375

1 4

3 8

1 2

3 4

1

1 14

1 12

2

Outside diameter, in.

1 8

Nominal pipe size, in. 0.068 0.095 0.088 0.119 0.091 0.126 0.109 0.147 0.113 0.154 0.133 0.179 0.140 0.191 0.145 0.200 0.154 0.218

80 40 80 40 80 40 80 40 80 40 80 40 80 40 80 40 80

Wall thickness, in

40

Schedule no.

1.500 2.067 1.939

1.278 1.610

0.742 1.049 0.957 1.380

0.423 0.622 0.546 0.824

0.215 0.364 0.302 0.493

0.269

ID, in

Table B.6 Dimensions, Capacities, and Weights of Standard Steel Pipes

1.069 1.075 1.477

0.881 0.800

0.433 0.494 0.639 0.668

0.217 0.250 0.320 0.333

0.093 0.125 0.157 0.167

0.072

Cross-sectional area of metal, in2

0.01225 0.02330 0.02050

0.00891 0.01414

0.00300 0.00600 0.00499 0.01040

0.00098 0.00211 0.00163 0.00371

0.00025 0.00072 0.00050 0.00133

0.00040

Inside sectional area, ft2

(Continued)

3.63 3.65 5.02

3.00 2.72

1.47 1.68 2.17 2.27

0.74 0.85 1.09 1.13

0.31 0.42 0.54 0.57

0.24

Pipe weight, lb/ft

613

5.563

6.625

8.625

5

6

8

12.75

4.500

4

12

4.000

3 12

10.75

3.500

3

10

2.875

0.203 0.276 0.216 0.300 0.226 0.318 0.237 0.337 0.258 0.375 0.280 0.432 0.322 0.500 0.365 0.594 0.406 0.688

40 80 40 80 40 80 40 80 40 80 40 80 40 80 40 80 40 80

3.364 4.026 3.826 5.047 4.813 6.065 5.761 7.981 7.625 10.020 9.562 11.938 11.374

2.323 3.068 2.900 3.548

2.469

Dimensions, Capacities, and Weights of Standard Steel Pipes (Continued )

2 12

TABLE B.6

3.678 3.17 4.41 4.30 6.11 5.58 8.40 8.396 12.76 11.91 18.95 15.74 26.07

2.254 2.228 3.016 2.680

1.704

0.06170 0.08840 0.07986 0.1390 0.1263 0.2006 0.1810 0.3474 0.3171 0.5475 0.4987 0.7773 0.7056

0.02942 0.05130 0.04587 0.06870

0.03322

12.51 10.79 14.98 14.62 20.78 18.97 28.57 28.55 43.39 40.48 64.40 53.36 88.57

7.66 7.58 10.25 9.11

5.79

614

Appendix B

Miscellaneous Tables

Table B.7 Dimensions of Heat Exchanger Tubes

Tube OD, in

Surface area, per foot of length, ft

B.W.G. gauge

Thickness, in

Tube ID, in

Flow area, in2

External

Internal

1 4 1 4

22

0.028

0.194

0.0295

0.0655

0.0508

24

0.022

0.206

0.0333

0.0655

0.0539

1 2 1 2 1 2

18

0.049

0.402

0.1269

0.1309

0.1052

20

0.035

0.430

0.1452

0.1309

0.1126

22

0.028

0.444

0.1548

0.1309

0.1162

3 4 3 4 3 4 3 4

10

0.134

0.482

0.1825

0.1963

0.1262

14

0.083

0.584

0.2679

0.1963

0.1529

16 18

0.065 0.049

0.620 0.652

0.3019 0.3339

0.1963 0.1963

0.1623 0.1707

1 1 1 1

8 14 16 18

0.165 0.083 0.065 0.049

0.670 0.834 0.870 0.902

0.3526 0.5463 0.5945 0.6390

0.2618 0.2618 0.2618 0.2618

0.1754 0.2183 0.2278 0.2361

1 14 1 14

8

0.165

0.920

0.6648

0.3272

0.2409

1 14

14 16

0.083 0.065

1.084 1.120

0.9229 0.9852

0.3272 0.3272

0.2838 0.2932

1 14

18

0.049

1.152

1.042

0.3272

0.3016

2 2 2 2

11 12 13 14

0.120 0.109 0.095 0.083

1.760 1.782 1.810 1.834

2.433 2.494 2.573 2.642

0.5236 0.5236 0.5236 0.5236

0.4608 0.4665 0.4739 0.4801

(1 in ¼ 25.4 mm; 1 in2 ¼ 645.16 mm2; 1 ft ¼ 0.3048 m; 1 ft2 ¼ 0.0929 m2).

Appendix

C

Steam Tables C.1

SATURATED STEAM

C.2

SUPERHEATED STEAM

C.3

SATURATED STEAM – ICE

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

615

616

Saturated Steam

Absolute pressure, lbf/in2, P

0.08854 0.09995 0.12170 0.14752 0.17811 0.2563 0.3631 0.5069 0.6982 0.9492 1.2748 1.6924 2.2225 2.8886 3.718 4.741 5.992 7.510 9.339 11.526

Table C.1

Temperature, 8F, T

32 35 40 45 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200

0.01602 0.01602 0.01602 0.01602 0.01603 0.01604 0.01606 0.01608 0.01610 0.01613 0.01617 0.01620 0.01625 0.01629 0.01634 0.01639 0.01645 0.01651 0.01657 0.01663

Saturated liquid, vl 3306 2947 2444 2036.4 1703.2 1206.6 867.8 633.1 468.0 350.3 265.3 203.25 157.32 122.99 97.06 77.27 62.04 50.21 40.94 33.62

Evaporation difference 3306 2947 2444 2036.4 1703.2 1206.7 867.9 633.1 468.0 350.4 265.4 203.27 157.34 123.01 97.07 77.29 62.06 50.23 40.96 33.64

Saturated vapor, vg

Specific volume, ft3/lb

0.00 3.02 8.05 13.06 18.07 28.06 38.04 48.02 57.99 67.97 77.94 87.92 97.90 107.89 117.89 127.89 137.90 147.92 157.95 167.99

Saturated liquid, hl 1075.8 1074.1 1071.3 1068.4 1065.6 1059.9 1054.3 1048.6 1042.9 1037.2 1031.6 1025.8 1020.0 1014.1 1008.2 1002.3 996.3 990.2 984.1 977.9

Evaporation difference

Enthalpy, Btu/lb

1075.8 1077.1 1079.3 1081.5 1083.7 1088.0 1092.3 1096.6 1100.9 1105.2 1109.5 1113.7 1117.9 1122.0 1126.1 1130.2 1134.2 1138.1 1142.0 1145.9

Saturated vapor, hg 0.0000 0.0061 0.0162 0.0262 0.0361 0.0555 0.0745 0.0932 0.1115 0.1295 0.1471 0.1645 0.1816 0.1984 0.2149 0.2311 0.2472 0.2630 0.2785 0.2938

Saturated liquid, sl 2.1877 2.1709 2.1435 2.1167 2.0903 2.0393 1.9902 1.9428 1.8972 1.8531 1.8106 1.7694 1.7296 1.6910 1.6537 1.6174 1.5822 1.5480 1.5147 1.4824

2.1877 2.1770 2.1597 2.1429 2.1264 2.0948 2.0647 2.0360 2.0087 1.9826 1.9577 1.9339 1.9112 1.8894 1.8685 1.8485 1.8293 1.8109 1.7932 1.7762

Saturated vapor, sg

(Continued)

Evaporation difference

Entropy, Btu/lb . 8R

617

210 212 220 230 240 250 260 270 280 290 300 310 320 330 340 350 360 370 380 390 400 410 420 430 440 450

TABLE C.1

14.123 14.696 17.186 20.780 24.969 29.825 35.429 41.858 49.203 57.556 67.013 77.68 89.66 103.06 118.01 134.63 153.04 173.37 195.77 220.37 247.31 276.75 308.83 343.72 381.59 422.6

Saturated Steam

0.01670 0.01672 0.01677 0.01684 0.01692 0.01700 0.01709 0.01717 0.01726 0.01735 0.01745 0.01755 0.01765 0.01776 0.01787 0.01799 0.01811 0.01823 0.01836 0.01850 0.01864 0.01878 0.01894 0.01910 0.01926 0.0194

27.80 26.78 23.13 19.365 16.306 13.804 11.746 10.044 8.628 7.444 6.449 5.609 4.896 4.289 3.770 3.324 2.939 2.606 2.317 2.0651 1.8447 1.6512 1.4811 1.3308 1.1979 1.0799

27.82 26.80 23.15 19.382 16.323 13.821 11.763 10.061 8.645 7.461 6.466 5.626 4.914 4.307 3.788 3.342 2.957 2.625 2.335 2.0836 1.8633 1.6700 1.5000 1.3499 1.2171 1.0993

178.05 180.07 188.13 198.23 208.34 218.48 228.64 238.84 249.06 259.31 269.59 279.92 290.28 300.68 311.13 321.63 332.18 342.79 353.45 364.17 374.97 385.83 396.77 407.79 418.90 430.1

971.6 970.3 965.2 958.8 952.2 945.5 938.7 931.8 924.7 917.5 910.1 902.6 894.9 887.0 879.0 870.7 862.2 853.5 844.6 835.4 826.0 816.3 806.3 796.0 785.4 774.5

1149.7 1150.4 1153.4 1157.0 1160.5 1164.0 1167.3 1170.6 1173.8 1176.8 1179.7 1182.5 1185.2 1187.7 1190.1 1192.3 1194.4 1196.3 1198.1 1199.6 1201.0 1202.1 1203.1 1203.8 1204.3 1204.6

0.3090 0.3120 0.3239 0.3387 0.3531 0.3675 0.3817 0.3958 0.4096 0.4234 0.4369 0.4504 0.4637 0.4769 0.4900 0.5029 0.5158 0.5286 0.5413 0.5539 0.5664 0.5788 0.5912 0.6035 0.6158 0.6280

1.4508 1.4446 1.4201 1.3901 1.3609 1.3323 1.3043 1.2769 1.2501 1.2238 1.1980 1.1727 1.1478 1.1233 1.0992 1.0754 1.0519 1.0287 1.0059 0.9832 0.9608 0.9386 0.9166 0.8947 0.8730 0.8513

(Continued)

1.7598 1.7566 1.7440 1.7288 1.7140 1.6998 1.6860 1.6727 1.6597 1.6472 1.6350 1.6231 1.6115 1.6002 1.5891 1.5783 1.5677 1.5573 1.5471 1.5371 1.5272 1.5174 1.5078 1.4982 1.4887 1.4793

618

Temperature, 8F, T

0.0196 0.0198 0.0200 0.0202 0.0204 0.0209 0.0215 0.0221 0.0228 0.0236 0.0247 0.0260 0.0278 0.0305 0.0369

Saturated liquid, vl 0.9748 0.8811 0.7972 0.7221 0.6545 0.5385 0.4434 0.3647 0.2989 0.2432 0.1955 0.1538 0.1165 0.0810 0.0392

Evaporation difference 0.9944 0.9009 0.8172 0.7423 0.6749 0.5594 0.4649 0.3868 0.3217 0.2668 0.2201 0.1798 0.1442 0.1115 0.0761

Saturated vapor, vg

Specific volume, ft3/lb

441.4 452.8 464.4 476.0 487.8 511.9 536.6 562.2 588.9 617.0 646.7 678.6 714.2 757.3 823.3

Saturated liquid, hl 763.2 751.5 739.4 726.8 713.9 686.4 656.6 624.2 588.4 548.5 503.6 452.0 390.2 309.9 172.1

Evaporation difference

Enthalpy, Btu/lb

1204.6 1204.3 1203.7 1202.8 1201.7 1198.2 1193.2 1186.4 1177.3 1165.5 1150.3 1130.5 1104.4 1067.2 995.4

Saturated vapor, hg 0.6402 0.6523 0.6645 0.6766 0.6887 0.7130 0.7374 0.7621 0.7872 0.8131 0.8398 0.8679 0.8987 0.9351 0.9905

Saturated liquid, sl

0.8298 0.8083 0.7868 0.7653 0.7438 0.7006 0.6568 0.6121 0.5659 0.5176 0.4664 0.4110 0.3485 0.2719 0.1484

Evaporation difference

1.4700 1.4606 1.4513 1.4419 1.4325 1.4136 1.3942 1.3742 1.3532 1.3307 1.3062 1.2789 1.2472 1.2071 1.1389

Saturated vapor, sg

Entropy, Btu/lb . 8R

A bridged from Thermodynamic Properties of Steam, by Joseph H. Keenan and Frederick G. Keyes. Copyright 1936, by Joseph H. Keenan and Frederick G. Keyes, Published by John Wiley & Sons, Inc., Hoboken, NJ.



466.9 514.7 566.1 621.4 680.8 812.4 962.5 1133.1 1325.8 1542.9 1786.6 2059.7 2365.4 2708.1 3093.7

Absolute pressure, lbf/in2, P

460 470 480 490 500 520 540 560 580 600 620 640 660 680 700

Saturated Steam

TABLE C.1

619

40 (267.25)

20 (227.96)

14.696 (212.00)

10 (193.21)

5 (162.24)

1 (101.74)

v h s v h s v h s v h s v h s v h s v

Absolute pressure, lbf/ in2 (Saturated temperature)

392.6 1150.4 2.0512 78.16 1148.8 1.8718 38.85 1146.6 1.7927

200

404.5 1159.5 2.0647 80.59 1158.1 1.8857 40.09 1156.2 1.8071 27.15 1154.4 1.7624

220

Table C.2 Superheated Steam

452.3 1195.8 2.1153 90.25 1195.0 1.9370 45.00 1193.9 1.8595 30.53 1192.8 1.8160 22.36 1191.6 1.7808 11.040 1186.8 1.6994 7.259

300 482.2 1218.7 2.1444 96.26 1218.1 1.9664 48.03 1217.2 1.8892 32.62 1216.4 1.8460 23.91 1215.6 1.8112 11.843 1211.9 1.7314 7.818

350 512.0 1241.7 2.1720 102.26 1241.2 1.9942 51.04 1240.6 1.9172 34.68 1239.9 1.8743 25.43 1239.2 1.8396 12.628 1236.5 1.7608 8.357

400 541.8 1264.9 2.1983 108.24 1264.5 2.0205 54.05 1264.0 1.9436 36.73 1263.5 1.9008 26.95 1262.9 1.8664 13.401 1260.7 1.7881 8.884

450 571.6 1288.3 2.2233 114.22 1288.0 2.0456 57.05 1287.5 1.9689 38.78 1287.1 1.9261 28.46 1286.6 1.8918 14.168 1284.8 1.8140 9.403

500

550 601.4 1312.0 2.2468 120.19 1311.7 2.0692 60.04 1311.3 1.9924 40.82 1310.9 1.9498 29.97 1310.5 1.9160 14.93 1308.9 1.8384 9.916

Temperature, 8F

631.2 1335.7 2.2702 126.16 1335.4 2.0927 63.03 1335.1 2.0160 42.86 1334.8 1.9734 31.47 1334.4 1.9392 15.688 1333.1 1.8619 10.427

600 690.8 1383.8 2.3137 138.10 1383.6 2.1361 69.01 1383.4 2.0596 46.94 1383.2 2.0170 34.47 1382.9 1.9829 17.198 1381.9 1.9058 11.441

700 750.4 1432.8 2.3542 150.03 1432.7 2.1767 74.98 1432.5 2.1002 51.00 1432.3 2.0576 37.46 1432.1 2.0235 18.702 1431.3 1.9467 12.449

800

869.5 1533.5 2.4283 173.87 1533.4 2.2509 86.92 1533.2 2.1744 59.13 1533.1 2.1319 43.44 1533.0 2.0978 21.70 1532.4 2.0214 14.454

1000

(Continued)

809.9 1482.7 2.3923 161.95 1482.6 2.2148 80.95 1482.4 2.1383 55.07 1482.3 2.0958 40.45 1482.1 2.0618 20.20 1481.4 1.9850 13.452

900

620

180 (373.06)

160 (363.53)

140 (353.02)

120 (341.25)

100 (327.81)

80 (312.03)

60 (292.71)

h s v h s v h s v h s v h s v h s v h s v

Absolute pressure, lbf/ in2 (Saturated temperature)

TABLE C.2

200

220

Superheated Steam

1181.6 1.6492

300 1208.2 1.6830 5.803 1204.3 1.6475 4.592 1200.1 1.6188 3.783 1195.7 1.5944

350 1233.6 1.7135 6.220 1230.7 1.6791 4.937 1227.6 1.6518 4.081 1224.4 1.6287 3.468 1221.1 1.6087 3.008 1217.6 1.5908 2.649 1214.0 1.5745 2.361

400 1258.5 1.7416 6.624 1256.1 1.7078 5.268 1253.7 1.6813 4.363 1251.3 1.6591 3.715 1248.7 1.6399 3.230 1246.1 1.6230 2.852 1243.5 1.6077 2.549

450 1283.0 1.7678 7.020 1281.1 1.7346 5.589 1279.1 1.7085 4.636 1277.2 1.6869 3.954 1275.2 1.6683 3.443 1273.1 1.6519 3.044 1271.0 1.6373 2.726

500

550 1307.4 1.7926 7.410 1305.8 1.7598 5.905 1304.2 1.7339 4.902 1302.5 1.7127 4.186 1300.9 1.6945 3.648 1299.3 1.6785 3.229 1297.6 1.6642 2.895

Temperature, 8F

1331.8 1.8162 7.797 1330.5 1.7836 6.218 1329.1 1.7581 5.165 1327.7 1.7370 4.413 1326.4 1.7190 3.849 1325.0 1.7033 3.411 1323.5 1.6894 3.060

600 1380.9 1.8605 8.562 1379.9 1.8281 6.835 1378.9 1.8029 5.683 1377.8 1.7822 4.861 1376.8 1.7645 4.244 1375.7 1.7491 3.764 1374.7 1.7355 3.380

700 1430.5 1.9015 9.322 1429.7 1.8694 7.446 1428.9 1.8443 6.195 1428.1 1.8237 5.301 1427.3 1.8063 4.631 1426.4 1.7911 4.110 1425.6 1.7776 3.693

800

1531.9 1.9762 10.830 1531.3 1.9442 8.656 1530.8 1.9193 7.207 1530.2 1.8990 6.172 1529.7 1.8817 5.396 1529.1 1.8667 4.792 1528.6 1.8534 4.309

1000

(Continued)

1480.8 1.9400 10.077 1480.1 1.9079 8.052 1479.5 1.8829 6.702 1478.8 1.8625 5.738 1478.2 1.8451 5.015 1477.5 1.8301 4.452 1476.8 1.8167 4.002

900

621

400 (444.59)

350 (431.72)

300 (417.33)

280 (411.05)

260 (404.42)

240 (397.37)

220 (389.86)

200 (381.79)

h s v h s v h s v h s v h s v h s v h s v h s

TABLE C.2

Superheated Steam 1210.3 1.5594 2.125 1206.5 1.5453 1.9276 1202.5 1.5319

1240.7 1.5937 2.301 1237.9 1.5808 2.094 1234.9 1.5686 1.9183 1232.0 1.5573 1.7674 1228.9 1.5464 1.6364 1225.8 1.5360 1.3734 1217.7 1.5119 1.1744 1208.8 1.4892

1268.9 1.6240 2.465 1266.7 1.6117 2.247 1264.5 1.6003 2.063 1262.3 1.5897 1.9047 1260.0 1.5796 1.7675 1257.6 1.5701 1.4923 1251.5 1.5481 1.2851 1245.1 1.5281

1295.8 1.6513 2.621 1294.1 1.6395 2.393 1292.4 1.6286 2.199 1290.5 1.6184 2.033 1288.7 1.6087 1.8891 1286.8 1.5998 1.6010 1282.1 1.5792 1.3843 1277.2 1.5607

1322.1 1.6767 2.772 1320.7 1.6652 2.533 1319.2 1.6546 2.330 1317.7 1.6447 2.156 1316.2 1.6354 2.005 1314.7 1.6268 1.7036 1310.9 1.6070 1.4770 1306.9 1.5894

1373.6 1.7232 3.066 1372.6 1.7120 2.804 1371.5 1.7017 2.582 1370.4 1.6922 2.392 1369.4 1.6834 2.227 1368.3 1.6751 1.8980 1365.5 1.6563 1.6508 1362.7 1.6398

1424.8 1.7655 3.352 1424.0 1.7545 3.068 1423.2 1.7444 2.827 1422.3 1.7352 2.621 1421.5 1.7265 2.442 1420.6 1.7184 2.084 1418.5 1.7002 1.8161 1416.4 1.6842

1476.2 1.8048 3.634 1475.5 1.7939 3.327 1474.8 1.7839 3.067 1474.2 1.7748 2.845 1473.5 1.7662 2.652 1472.8 1.7582 2.266 1471.1 1.7403 1.9767 1469.4 1.7247

1528.0 1.8415 3.913 1527.5 1.8308 3.584 1526.9 1.8209 3.305 1526.3 1.8118 3.066 1525.8 1.8033 2.859 1525.2 1.7954 2.445 1523.8 1.7777 2.134 1522.4 1.7623

622

Absolute pressure, lbf/in2, P

0.0885 0.0808 0.0505 0.0309 0.0185 0.0108 0.0062 0.0035

Temperature, 8F, T

32 30 20 10 0 210 220 230

Table C.3 Saturated Steam –Ice

0.01747 0.01747 0.01745 0.01744 0.01742 0.01741 0.01739 0.01738

Saturated ice, vi 3.306 3.609 5.658 9.05 14.77 24.67 42.2 74.1

Saturated steam, vg  1023

Specific volume, ft3/lb Sublimation difference 1219.1 1219.3 1219.9 1220.4 1220.7 1221.0 1221.2 1221.2

Saturated ice, hi 2143.35 2144.35 2149.31 2154.17 2158.93 2163.59 2168.16 2172.63

Enthalpy, Btu/lb

1075.8 1074.9 1070.6 1066.2 1061.8 1057.4 1053.0 1048.6

Saturated steam, hg 20.2916 20.2936 20.3038 20.3141 20.3241 20.3346 20.3448 20.3551

Saturated ice, si

2.4793 2.4897 2.5425 2.5977 2.6546 2.7143 2.7764 2.8411

Sublimation difference

Entropy, Btu/lb . 8R

2.1877 2.1961 2.2387 2.2836 2.3305 2.3797 2.4316 2.4860

Saturated steam, sg

Index

Absorption, 187 absorption factor, A, 211, 212 Chen equation (column diameter), 208 Chen equation (number of theoretical trays), 229 Coburn’s equation, 211 column diameter, 207 –210 column height, 210– 212 flooding (flooding velocity), 207 height of a single transfer unit, HOG, 210, 212 key equations for absorption calculation, 242, 243 Kremser– Brown– Sounders equation (number of theoretical trays), 229 loading, 207 minimum liquid –to–gas ratio, 202 number of overall transfer units, NOG, 210 –212 NOG for column with constant absorption factor, 211 overall efficiency of bubble-cap tray absorbers, 230, 231 packing height as function of efficiency & packing size (ceramic), 223 packing height as function of efficiency & packing size (plastic), 222 pressure drop through packed column, 224 theoretical stage, 228 weeping, 228 Absorption equipment, 189 absorption column, 191 packed vs. trayed column comparison, 241, 242

packed column absorbers, 189– 192, 200–227 typical packings, 193, 194 “rules of thumb” for design of packed columns, 224 tray column absorbers, 196, 227–235 bubble-cap tray absorbers, 197, 198 sieve tray absorbers, 199, 200 Activity coefficient, 61 NRTL equation, 62 parameters, 64 Wilson equation, 62 parameters, 63 Adsorption, 245 breakthrough point, 266 activated alumina, properties of, 248 activated carbon pressure drop curves (EPA chart), 270 activated carbon, properties of, 248 adsorbate/adsorbent, 245 adsorbent activation, 246 adsorbent capacity, 246 adsorption equilibria, 250 adsorption isotherms of carbon dioxide on activated carbon, 255 breakthrough capacity (BC), 266 breakthrough curve, 265 chemisorption, 246 critical diameter, 246 equilibrium capacity (CAP), 266 Freundlich equation, 253 heat capacities of common adsorbents (ambient conditions), 269 heat of adsorption (chemisorption vs. physisorption), 252 HEEL, 266 Langmuir isotherms, 253, 254

Mass Transfer Operations for the Practicing Engineer. By Louis Theodore and Francesco Ricci Copyright # 2010 John Wiley & Sons, Inc.

623

624 Index Adsorption (Continued ) mass transfer zone (MTZ), 265 molecular sieve pressure drop chart, 270 molecular sieves, properties of, 249 physisorption, 245, 246 regeneration of adsorbent, 251, 266 saturation capacity (SAT), 266 silica gel, properties of, 249 “sorption” 245 vapor/solid equilibrium (Adsorption) isotherms carbon dioxide on molecular sieves, 252 carbon tetrachloride on activated carbon, 65 selected hydrocarbons on activated carbon, 251 working capacity/charge (WC), 266, 267 Adsorption equipment, 257 adsorption cycle vs. desorption cycle, 259 classification of adsorption/desorption cycles, 283 contact condensers, 263 continuous rotary bed, 261 overall design procedure, 271 surface condensers, 263 Adsorption isotherm, See also “Adsorption” adsorbent capacity, 65 adsorption equilibrium, 64 BET isotherm, 66 Freundlich isotherm, 65, 66 Langmuir isotherm, 66 Polanyi potential theory, 66, 67 Antoine equation, 47– 49, 166 coefficients, 48 Apparent rejection, 423 Atomic mass units, 23 Atomic weight, 23 Avogadro’s number, 23 Baghouse, 461 collection efficiency of a baghouse, 462 Darcy equation (pressure drop), 463 gas-to-cloth (G/C) ratio, 462 Barometer, 22 Batch operation, 108 Boiling

Bubble point calculations bubble point pressure (BPP), 50, 166, 175 bubble point temperature (BPT), 167 Buffer region, 81 Cascade, 116 Celsius scale, 20, 21 Centrifugation, 471 Chemical engineering, 3 history of, 4 Chemical reactions, 39, 40 Conservation law, 8 Continuous-contact operation, 116, 117 Controlling resistance, 89 –90 controlling films for various systems, 90 Convection convective transfer, 8 Conversion factors, See “Units” Cooling towers, 340–343 Critical properties critical temperature, 42 Crystallization, 371 agglomeration, 381, 389 crystal growth, 379, 380 crystal size distribution (CSD), 381, 386, 387 crystallization processes, types of, 372 design considerations, 397, 398 efficiency of crystallization separation, 390, 391 magma, 393 mixed magma underflow, 381 mother liquor, 371, 381 nucleation, 372 occlusion, 393 schematic diagram of generalized crystallization process, 380 seeds, 380 surface area ration (SAR), 389 volume ratio (VR), 389 Crystallization equipment crystallizers, types of, 392, 393 forced circulation crystallizer, 399 Swenson –Walker crystallizer, 392 evaporative crystallizer line diagram, 394 Cunningham correction factor (CCF), 444, 458 Cyclones, 449 critical diameter, 450

Index cut diameter, 450 Lapple calculation procedure, 451 multiclone, 449 Theodore–De Paola Equation, 453 Decanter, 264, 475 horizontal decanter schematic, 476 settling time, 476 Degrees of freedom, 44, 45, 375 Density of air, 25 of water, 25 Dew point, 41 Differential element, 9 Diffusion Fick’s first law, 73, 74 molecular diffusion, 43 Diffusion, steady state molecular in gases, 75 diffusion in multicomponent mixtures, 76 diffusion of A through non-diffusing B, 75, 84, 85 equimolar counterdiffusion, 76, 83, 84 equimolar counterdiffusion and/or diffusion in dilute solutions, 88–90 in liquids, 79 diffusion of A through non-diffusing B, 80 equimolar counterdiffusion, 80 Diffusion, thermal diffusion, 102 Diffusion, turbulent diffusion, 80 Diffusivity, 74, 82 estimation methods for gases, 78 –79 estimation methods for liquids: Wilke–Chang equation, 80 of common gases at, 08C, 78 of water vapor in air at, 258C and, 1 atm, 366 Distillate, 119 Distillation, 119 approach to flooding velocity, 159 batch distillation, 127 –133 boil-up ratio, RB, 135 Chang correlation (substitute for Gilliland correlation), 170, 171, 178 column diameter, 159 –161 column height, 161

625

constant molal overflow, 142 distributed vs. undistributed components, 163 entrainment (entrainment flooding point), 140, 159 equilibrium stage, 122 ethanol/water equilibrium diagram (via Wilson equation, 1atm), 124 Fair flooding correlation, 159, 180, 181 feed condition factor, q, 146, 147 feed tray, graphical location of, 149 Fenske equation (minimum theoretical stages), 168, 177 Fenske –Underwood–Gilliland (FUG) shortcut method, 161–173 flash distillation, 120– 127 graphical solution of binary flash, 122 multicomponent flash, 125–127 fractional recovery, 164 Gilliland correlation (number of theoretical stages), 169–170, 178 height equivalent to a theoretical plate (HETP), 184, 185 internal reflux ratio, 143 key components, 162 Kirkbride equation (theoretical feed tray location), 171, 179 McCabe –Thiele graphical method, 142–158 step-by-step procedure, 152–154 overall efficiency, 154, 171–173 O’Connell correlation, 171–173, 179, 180 partial condensation, 121 phase equilibrium constant, K, 121 pinch point, 152 q-line, 148 Rachford –Rice equation, 125–127 Rayleigh equation, 127–133 reboiler pressure, 167 rectification section, 119, 135 rectification section operating line (ROL), 144 reflux ratio, R, 134 minimum reflux ratio, Rmin 150 optimum reflux ratio, 152, 153 reflux ratio optimization multipliers, m, 153

626 Index Distillation (Continued ) relative volatility, 121, 141, 167 geometric mean relative volatility, 167, 168 sharp separation, 163 side streams, 136 stripping section, 119, 135 stripping section operating line (SOL), 145 surge volume, 161, 183 theoretical stage, 141 total reflux, 149, 150 transfer unit, 184 tray efficiency, 154 Underwood equations (minimum reflux ratio), 169, 177 weeping, 140, 159 Distillation equipment overflow weir, 137 packed column distillation, 184, 185 partial condenser, 140 partial reboiler, 140 setting the column pressure, 164 –167 shortcut design methods (binary or multicomponent), 161 sieve tray column (single crossflow), 138, 139 tray, types of bubble-cap tray, 137 sieve tray, 138 valve tray, 138 tray, single crossflow configuration, 137 active area, 139 bubbling area, 160 net area, 160 tray spacing, 182 Drag coefficient, 443 for spheres, 444 Drag force, 443 Drying, 347 constant rate period, 349 drop diameters, 364 falling rate period, 349 Friedman and Marshall heat transfer coefficient equation, 356 moisture content critical moisture content, 348, 349 equilibrium moisture content, 347

final moisture content, 349 free moisture content, 350 rotary dryer flight arrangements, 354 rotary dryers, 352–356 rotary dryer unit (Manhattan College), 353 spray dryers, 361 spray dryer unit (Manhattan College), 362 Economic analysis, 490 Eddies Eddy transfer, 80 Efficiency fractional stage efficiency, 116 overall efficiency of bubble-cap tray absorbers, 230, 231 overall efficiency of trayed distillation columns, 154, 171– 173 Electrodialysis, 101 Electrophoresis, 486 Electrostatic attraction, 442 Deutsch –Anderson equation, 455, 456 Matts –Ohnfeldt equation, 456 particle migration velocity, 455 typical precipitation rate parameters for various applications, 455 Electrostatic precipitators (ESPs), 454 Elutriator, 447 Encapsulation, 478 Engineering economics, 489 bonds, 496 break-even point, 495 compound interest, 491 depreciation, 493 fabricated equipment cost index (FECI), 493 incremental cost, 496 perpetual life, 494 present net worth (PNW), 494 present worth, 492 profit, 498 rate of return, 495, 496 simple interest, 491 Environmental issues of concern, 566–568 Equilibrium, 37 –40 equilibrium considerations, 37, 38

Index Ergun equation, 269 Ethics, 549 code of ethics, 549 engineering and environmental ethics, 557 –559 Evaporation (as a novel mass transfer operation), 485 Extraction, 293 analytical calculation procedures, 304 –309 Baker equation, 322 Chen equation (number of theoretical stages), 324 countercurrent extraction, 296, 309 crosscurrent extraction, 296, 305 dissociation extraction, 486 equilibrium data for n-butanol/acetic acid/water system at, 308C, 302 extract, 295 fractional extraction, 294 Kremser equation, 309 leaching/lixiviation, 293 liquid –liquid extraction, 294 overflow, 316 variable vs. constant overflow, 317, 318 phase ratio, 297 plait point, 300, 301 raffinate, 295 solid –liquid extraction, 312 types of, 313 solvent selectivity, 298 ternary equilibrium diagram, 300, 301 theoretical stage, 295 underflow, 316 Extraction equipment liquid –liquid extraction column (Manhattan College), 297 multistage devices, 296 single stage units, 295, 296, 306 solid –liquid extraction equipment, 315, 316 Fahrenheit scale, 20, 21 Filtration, 474 Flocculation, 474 Flotation processes, 472 Flow patterns cocurrent flow, 109 –111

627

countercurrent flow, 111, 112 crossflow, 112, 113 Flux, 74 net flux, 75 Foam fractionation, 486 Freeze crystallization, 484 Freezing point, 41 Gas laws Boyle’s law, 31, 32 Charle’s law, 31, 32, 34 Dalton’s law, 35 ideal gas law, 31–35 Gas permeation, 432 describing equations, 433–435 “stage cut” 434 Gibb’s phase rule, 375 Gravitational constant, gc, 17 Gravity sedimentation, 467 circular-basin continuous thickener, 468 clarification vs. thickening, 467 hindered settling, 469 thickener operating zones, 468 wall effect, 469 Gravity settlers, 447, See “decanter” Heat duty of a condenser, 150, 151 Heat exchangers dimensions of heat exchanger tubes, 614 Heat transfer classic equation for, 393 Henry’s law, 187, 200 High-gradient magnetic separation (HGMS), 477 Hindered settling, 469 Humidification, 327 adiabatic saturation temperature, 329, 333 cooling ponds, 343 cooling towers, 340–343 dew point temperature, Tdp, 329 humid enthalpy, 330 humid heat capacity, 328 humid volume, 328, 330 humidity absolute humidity, 327 molal humidity, 327 relative humidity, 328

628 Index Humidification (Continued ) psychrometric chart, 329 high temperature, 332 low temperature, 331 saturation curve, 330 spray columns, 343 wet-bulb temperature, Twb, 329 Ideal gas law, 31 –35 Ideal stage, See “Theoretical stage” Inertial collectors, 447 Inertial impaction, 442 Ion exchange, 484 Kelvin scale, 20, 21 Kinetic analysis, 39 Laminar film, 81 Laminar flow, 28 Liquid ion exchange (LIE), 484 Liquid–solid equilibrium (LSE), 68, 69. See also “Adsorption” Logarithmic mean logarithmic mean humidity difference, 361 logarithmic mean temperature difference, 360 Lognormal probability distribution, 385, 386 Manometer, 22 Mass fraction, 24 Mass transfer macroscopic approach, 8, 10 microscopic approach, 9 molecular approach, 9 Mass transfer, theories of boundary layer theory, 81 empirical approaches, 81 surface renewal theory, 81 two film theory, 81, 82 Mass transfer coefficients, 80 experimental mass transfer coefficients, 90 individual mass transfer coefficients, 81– 87 overall mass transfer coefficients, 87 Mass transfer equipment absorption, 104

adsorption, 104 crystallization, 392–393 distillation, 103, 104 extraction, 104– 105 humidification and drying, 105 Mass transfer operations classification of, 97– 102 contact of immiscible phases, 98– 101 direct contact of miscible phases, 102 miscible phases separated by a membrane, 101 Maximization/minimization first derivative test, 509 Melting point, 41 Membrane separation, 407 desalination (via reverse osmosis), 410 electrodialysis, 408 gas permeation, 408, 432 describing equations, 433– 435 microfiltration, 407, 427 describing equations, 428– 430 nanofiltration, 407 permeate, 407, 409 retentate, 407, 409 reverse osmosis (RO), 407–414 describing equations, 414– 418 ultrafiltration (UF), 420 describing equations, 421– 425 Membrane separation equipment dialysis membrane, 408 reverse osmosis, hollow fine fiber (HFF) for, 409, 410 Microfiltration, 427 average transmembrane pressure (ATP), 428 describing equations, 428–430 Mixing macroscopic, 28 molecular, 28 Mole fraction, 24 Molecular diffusion, 8 Molecular weight, 23, 24 Moles, 23, 24 Momentum, 17 Nonideal solutions, 61 –64 Normal boiling point, 41 Normal probability distribution, 385 Novel separation processes, 483

Index Numerical methods, 513 differentiation methods, 515– 517 finite difference method, 520 lumped-parameter method, 521 method of least squares, 516 Newton –Raphson method, 526–528 regression analysis, 515 Runge–Kutta method, 522 –526 trapezoidal rule, 518 –520 Optimization, 530 Osmosis osmotic pressure, 412 Partial differential equations, 529 Particulate matter, 10 (PM10), 440 Particulate matter, 2.5 (PM2.5), 440 pH, 29, 30 Phase diagram, 41 eutectic point, 373 isothermal invariant point, 376 lever rule, 376 plait point, 300, 301 ternary equilibrium diagram, 300, 301 two component solid –liquid system, 373 Phase equilibrium, 41 Gibbs phase rule, 44, 45 Phase separation vs. component separation processes, 483 Piping dimensions, capacities and weights of standard steel pipes, 612, 613 Poise, 25 Pressure, 22, 23 interfacial partial pressure, 82 vapor pressure, 31 Process variables, 19 Properties intensive vs. extensive, 20 of air at, 1 atm, 611 of selected gases at, 1 atm and, 208C (688F), 607 of selected liquids at, 1 atm and, 208C (688F), 608, 609 of water at, 1atm, 610 physical vs. chemical, 19 saturated steam tables, 616 –618 saturated steam-ice tables, 622 superheated steam tables, 619 –621

629

Psychrometric chart, 329 definitions of psychrometric terms, 330 high temperature, 332 low temperature, 331 Radiation radiative transfer, 8 Rankine scale, 20, 21 Rate considerations, 38, 39, 71 Relative volatility, 121 Resin adsorption, 485 Reverse osmosis (RO), 407– 414 describing equations, 414–418 Van’t Hoff equation, 414 Reynolds number, 28, 29 Risk assessment cost/protection analysis, 581 hazard analysis (HAZAN), 585 hazard and operability study (HAZOP), 584, 585 hazard risk assessment, 571–574 health risk assessment, 568–571 Scientific notation, 17, 18 Screen mesh, 382, 383 Screen scales, Tyler and US Standard, 383 Semi-batch operation, 108 Significant figures, 17, 18 Simpson’s, 3-point rule, 128 Solidification, 477 Stagewise operation, 116 Standard conditions standard vs. actual conditions, 33 –35 STP, 32, 33 Standard deviation (sample), 384 State function, 151 Steady state, 108 Steam tables, 616–622 Steel pickling, 212 Stoke’s law, 443 Stripping, 235 key equations for stripping calculation, 242, 243 number of transfer units (NTU), 240 removal efficiency, 240 stripping of an EO-water system, 238 Stripping column, 103, 137 Sublimation, 99

630 Index Temperature, 20 Temperature scales, 20, 21 Theoretical stage, 110, 111, 141 Transient operation, See “Unsteady state” Transport phenomena, 7 transport equations, 9 transport phenomena approach, 9 Triple point, 42 Turbulent flow, 28 Ultrafiltration, 420 concentration polarization, 423 describing equations, 421 –425 gel formation, 423 Unit conversion, 15, 16 Unit operations definition of, 5 unit operations approach, 7 unit operations concept, 4, 5 Units, 11 –14 abbreviations, selected common, 603, 604 conversion factors, 15, 16 conversion factors, common engineering, 606 English engineering units, 11, 13 prefixes, 14 SI units, 11, 13 SI conversion constants, 599 –602 SI multiples and prefixes, 599 Universal gas constant, R, 32, 33 Unsteady state, 108 Vacuum, 22 Vapor, 42 Vapor pressure, 31 Antoine equation, 47–49 coefficients of selected substances, 48 Clapeyron equation, 47, 48 coefficients of selected substances, 47 Vapor-liquid equilibrium (VLE) Henry’s law, 45, 53– 59 Henry’s law constants for gases in water, 55

P-x,y diagram methanol-water system at, 408C (Raoult’s law), 52 Raoult’s law, 45–53 modified Raoult’s law, 61 Raoult’s law vs. Henry’s Law, 59– 61 T-x,y diagram ethanol-water system at, 1 atm, 51 x,y diagram methanol –water system at, 408C (Raoult’s law), 53 Vapor– solid equilibrium (VSE) adsorbent capacity, 65 adsorption equilibrium, 64 BET isotherm, 66 Freundlich isotherm, 65, 66 Langmuir isotherm, 66 Polanyi potential theory, 66, 67 vapor/solid equilibrium isotherms carbon dioxide on molecular sieves, 252 carbon tetrachloride on activated carbon, 65 selected hydrocarbons on activated carbon, 251 Venturi scrubber, 457 gas velocity, typical values for, 458 Johnstone collection efficiency equation, 457 Nukiyama –Tanasawa relationship, 458 overall efficiency of N scrubbers in parallel, 458, 459 Vibrating screens (as a novel mass transfer operation), 487 Viscosity absolute viscosity, 25, 26 of air, 26 of water, 26 kinematic viscosity, 25, 26 Saybolt seconds, 26 Saybolt universal viscometer, 26 Volume fraction, 24