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Introduction to Optimization for Environmental and Chemical Engineers

Introduction to Optimization for Environmental and Chemical Engineers

By

Louis Theodore and Kelly Behan

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2018 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper International Standard Book Number-13: 978-1-138-48912-7 (Hardback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http:// www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice:  Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at  http://www.taylorandfrancis.com and the CRC Press Web site at  http://www.crcpress.com

TO Donald J. Trump who may yet save our great nation from enemies within and abroad (L.T.) My mother and father who inspire me to be extraordinary (K.B.)

Contents Preface............................................................................................................................................ xiii About the Book............................................................................................................................ xvii Authors.......................................................................................................................................... xix

Part I  Optimization Fundamentals and Principles 1 Optimization Overview.........................................................................................................3 1.1 History of Optimization...............................................................................................4 1.2 The Computer Age.........................................................................................................6 1.3 The Scope of Optimization...........................................................................................7 1.4 Conventional/Established Optimization Procedures..............................................7 1.5 Contemporary Optimization: Linear Programming................................................ 8 References................................................................................................................................ 11 2 Mathematical Operations.................................................................................................... 13 2.1 The Quadratic Equation.............................................................................................. 15 2.2 Interpolation and Extrapolation................................................................................ 17 2.3 Significant Figures and Approximate Numbers..................................................... 20 2.4 Errors.............................................................................................................................22 2.5 Differentiation.............................................................................................................. 23 2.6 Numerical Integration................................................................................................. 26 2.6.1 Trapezoidal Rule............................................................................................. 27 2.6.2 Simpson’s Rule................................................................................................ 28 2.7 Simultaneous Linear Algebraic Equations............................................................... 29 2.7.1 Gauss–Jordan Reduction...............................................................................30 2.7.2 Gauss Elimination..........................................................................................30 2.7.3 Gauss–Seidel Approach................................................................................. 31 2.8 Nonlinear Algebraic Equations................................................................................. 31 2.9 Ordinary Differential Equations............................................................................... 32 2.10 Partial Differential Equations.................................................................................... 37 2.10.1 Parabolic PDE.................................................................................................. 37 2.10.2 Elliptical PDE................................................................................................... 39 References................................................................................................................................ 40 3 Perturbation Techniques...................................................................................................... 41 3.1 One Independent Variable..........................................................................................42 3.2 Two Independent Variables........................................................................................ 45 3.3 Three Independent Variables..................................................................................... 48 References................................................................................................................................ 49 4 Search Methods..................................................................................................................... 51 4.1 Interval Halving Method............................................................................................ 53 4.2 The Bisection Method.................................................................................................. 56 vii

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4.3 The Golden Section Method....................................................................................... 57 4.4 Method of Steepest Ascent/Descent......................................................................... 60 References................................................................................................................................63 5 Graphical Approaches..........................................................................................................65 5.1 Rectangular Coordinates............................................................................................ 66 5.1.1 Quadrants........................................................................................................ 67 5.1.2 Linear versus Non-Linear.............................................................................. 67 5.1.3 Intercepts.......................................................................................................... 68 5.2 Logarithmic-Logarithmic (Log-Log) Coordinates.................................................. 68 5.3 Semi-Logarithmic (Semi-Log) Coordinates............................................................. 73 5.3.1 Plotting Straight Lines on Semi-Logarithmic Paper.................................. 73 5.3.2 Other Graphical Coordinates........................................................................ 76 5.4 Methods of Plotting Data............................................................................................ 76 5.5 Applications..................................................................................................................77 References................................................................................................................................ 81 6 Analytical Methods...............................................................................................................83 6.1 Breakeven Considerations..........................................................................................83 6.2 One Independent Variable..........................................................................................85 6.3 General Analytical Formulation of the Optimum.................................................. 88 6.4 Two Independent Variables........................................................................................ 89 6.5 Three Independent Variables..................................................................................... 93 References................................................................................................................................ 94 7 Linear Programming............................................................................................................. 95 7.1 Definitions..................................................................................................................... 96 7.2 Basic Concepts of Optimization................................................................................. 97 7.3 Applied Mathematical Concepts in Linear Programming.................................... 99 7.4 Applied Engineering Concepts in Linear Programming.................................... 102 7.5 Other Real-World Applied Concepts in Linear Programming........................... 105 References.............................................................................................................................. 107

Part II  Environmental Engineering Applications 8 Air Pollution Management................................................................................................ 111 8.1 Outdated Air Pollution Control Device.................................................................. 111 8.2 Coal-Fired Power Plant Options.............................................................................. 113 8.3 Particulate Control Equipment Options................................................................. 114 8.4 Recovering Dust......................................................................................................... 117 8.5 Incinerating Mercury-Contaminated Waste.......................................................... 119 8.6 Optimizing Incinerator Performance...................................................................... 121 References.............................................................................................................................. 123 9 Water Pollution Management........................................................................................... 125 9.1 Break-Even Point Operation..................................................................................... 125 9.2 Wastewater Treatment Equipment Options........................................................... 126

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9.3 Water Disinfection Options...................................................................................... 127 9.4 Water Plant Operation............................................................................................... 131 9.5 Generating Waste Revenues from Waste Streams................................................ 132 9.6 Membrane Construction Costs................................................................................ 133 References.............................................................................................................................. 135 10 Solid Waste Management.................................................................................................. 137 10.1 Solid Waste Treatment............................................................................................... 137 10.2 Cement Kiln versus Rotary Kiln Incinerator......................................................... 139 10.3 Sludge Waste Receiving Tank Size.......................................................................... 142 10.4 Discounted Cash Flow Application......................................................................... 144 10.5 Maximum Mercury Waste Emission Constraints................................................. 147 10.6 Structural Considerations of Concrete Mix........................................................... 149 References.............................................................................................................................. 151 11 Health Risk Assessment..................................................................................................... 153 11.1 Minimum Purging Time for a Vessel to Permit Entry......................................... 153 11.2 Minimizing the Cost of Cancer-Fighting Drugs................................................... 155 11.3 Reducing Insurance Costs........................................................................................ 157 11.4 Reducing Health Concerns Associated with Flue Gas Emissions...................... 158 11.5 Vitamin Caloric and Protein Requirements........................................................... 160 11.6 Pharmaceutical Cancer Drug................................................................................... 161 References.............................................................................................................................. 163 12 Hazard Risk Assessment................................................................................................... 165 12.1 Safety Expenditures versus Return of Safety Investments................................. 165 12.2 Plant Safety Proposal Comparisons........................................................................ 167 12.3 Effect of Failure Distribution Rate........................................................................... 169 12.4 Estimating the Effect of Worst-Case Explosions...................................................172 12.5 Economics of Fire Protection Equipment............................................................... 174 12.6 Optimizing the Operation of a Cement Incinerator............................................. 177 References.............................................................................................................................. 180

Part III  Chemical Engineering Applications 13 Fluid Flow Applications..................................................................................................... 183 13.1 Fan Selection............................................................................................................... 183 13.2 Pump Selection........................................................................................................... 185 13.3 Pipe Diameter Selection............................................................................................ 188 13.4 Two-Stage Compressor.............................................................................................. 189 13.5 Ventilation Models..................................................................................................... 191 13.6 Three-Stage Compressor........................................................................................... 195 References.............................................................................................................................. 197 14 Chemical Reactor Applications........................................................................................ 199 14.1 Comparing a Continuous Stirred Tank Reactor to a Tubular Flow Reactor Performance��������������������������������������������������������������������������������������������������������������� 199

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14.2 Optimizing Fluidized-Bed Reactor Performance.................................................. 204 14.3 Two Reactors in Series: Achieving Maximum Conversion.................................. 208 14.4 Maximizing Selectivity............................................................................................. 210 14.5 Optimizing Batch Reactor Performance................................................................. 212 14.6 Optimizing Operating Schedules............................................................................ 214 References.............................................................................................................................. 215 15 Mass Transfer Applications............................................................................................... 217 15.1 Optimum Reflux Ratio.............................................................................................. 217 15.2 Optimizing a Filter Press Operation....................................................................... 219 15.3 Optimizing Production Rates.................................................................................. 221 15.4 Process and Equipment Modifications................................................................... 223 15.5 Economic Analysis for an Absorber versus an Extractor..................................... 227 15.6 Processing Crude Oil................................................................................................. 232 References.............................................................................................................................. 233 16 Heat Transfer Applications................................................................................................ 235 16.1 Steam Options............................................................................................................ 235 16.2 Optimum Insulation Thickness............................................................................... 238 16.3 Selecting the Most Profitable Exchanger................................................................ 239 16.4 Critical Insulation Thickness................................................................................... 242 16.5 Recovering Quality Energy...................................................................................... 246 16.6 Maximizing Profit through Energy Recovery....................................................... 247 References.............................................................................................................................. 248 17 Plant Design Applications................................................................................................. 249 17.1 Shipping Facilities...................................................................................................... 249 17.2 Tank Farms................................................................................................................. 252 17.3 Cyclone Selection and Design..................................................................................254 17.4 Minimizing the Cost of a Batch Plant Operation..................................................258 17.5 Ventilation Models with System Variables............................................................. 259 17.6 Plant Structure Design..............................................................................................261 References.............................................................................................................................. 262

Part IV  Select Optimization Applications 18 Select Environmental Engineering Applications......................................................... 265 18.1 Air Management........................................................................................................ 265 18.2 Water Management.................................................................................................... 268 18.3 Solid Waste Management......................................................................................... 271 18.4 Health Risk Assessment............................................................................................ 273 18.5 Hazard Risk Assessment.......................................................................................... 279 18.5.1 DRaT II Model............................................................................................... 281 18.5.2 DRaT IV Model.............................................................................................. 282 18.5.3 DRaT V Model............................................................................................... 283 References.............................................................................................................................. 283

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19 Select Chemical Engineering Applications................................................................... 285 19.1 Fluid Flow................................................................................................................... 285 19.2 Chemical Reactors...................................................................................................... 287 19.3 Mass Transfer Operations......................................................................................... 293 19.4 Heat Exchangers......................................................................................................... 295 19.5 Plant Design................................................................................................................ 297 References..............................................................................................................................300 20 Advanced Optimization Term Projects.......................................................................... 303 20.1 Sulfuric Acid Process................................................................................................. 303 20.2 Optimum Indoor Ventilation Flow Rates...............................................................304 20.3 Fluid Transportation System....................................................................................306 20.4 Mass Transfer Application........................................................................................ 311 References.............................................................................................................................. 315 Index.............................................................................................................................................. 317

Preface It is no secret that in recent years, the number of people entering the engineering field has increased. Most are beginning college students and some had earlier chosen a non-technical major/career path. A large number of these individuals are today seeking degrees in environmental and/or chemical engineering. These prospective students will require an understanding and appreciation of the numerous mathematical and optimization methods that are routinely employed in practice. This technical stepping stone to a successful career is rarely provided at institutions that award engineering degrees. This introductory text on optimization attempts to supplement existing curricula with a sorely needed tool to eliminate this void. More on the need for this book. The question often arises as to the educational background required for engineering students to possess meaningful analysis capabilities since technology has changed the emphasis that is placed on certain technical subjects. Before computer usage became popular, instruction in engineering analysis was (and still is in many places) restricted to simple systems, and most of the effort was devoted to solving a few elementary equations that were derived. These cases were mostly of academic interest, and because of their simplicity, were of little practical value. To this end, a considerable amount of time is now required to acquire skills in mathematics, especially in numerical methods, statistics, and optimization. In fact, most engineers are given courses in classical mathematics, but experience shows that very little of this knowledge is retained after graduation for the simple reason that these mathematical methods are not adequate for solving most systems of equations encountered in industry. In addition, advanced mathematic skills are either not provided in courses or are forgotten through sheer disuse since they are not related to engineering practice. The material in this book was prepared primarily for environmental and chemical engineering students and, to a lesser extent, for engineering professionals who wish to obtain a better understanding of the various optimization methods that can be employed in solving technical problems. In presenting the text material, the authors have stressed the pragmatic approach in the application of mathematical tools to later assist the reader in grasping the role of optimization in engineering problem-solving situations. As noted previously, this book serves two purposes. It may be used as a textbook for beginning environmental and chemical students or as a “reference” book for practicing engineers involved with optimization applications. For both audiences, it is assumed that the reader has already taken basic courses in physics and chemistry, and should have a minimum background in mathematics through elementary calculus. The authors’ ultimate aim is to offer the reader the fundamentals of several optimization methods with accompanying practical engineering applications. The reader is encouraged through the references to continue his or her own development beyond the scope of the presented material. As is usually the case in preparing any text, questions of what to include and what to omit have been particularly difficult. As noted above, the material in this book attempts to address optimization calculations common to both the environmental and chemical engineering disciplines. The book provides the reader with numerous solved illustrative examples so that the interrelationship between both theory and application is emphasized in nearly all of the chapters. One key feature of this book is that the xiii

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solutions to the problems are presented in a stand-alone manner. Throughout the book, the illustrative examples are laid out in such a way as to develop the reader’s technical understanding of optimization, with more difficult examples located at or near the end of each chapter. The book is divided into four (IV) parts (see also Table of Contents):

Part I: Part II: Part III: Part IV:

Optimization Fundamentals and Principles Environmental Engineering Applications Chemical Engineering Applications Select Industrial Applications

It should be noted that Part IV was an afterthought. After completing the manuscript, the authors felt that it lacked some real-world applications drawn from both the environmental and chemical engineering industries. As such, three chapters were added to this work—one concerned with environmental engineering, one concerned with chemical engineering, and one concerned with term projects. Chapter 18, titled Select Environmental Engineering Applications, is an extension of Part II and contains five sections—air management, water management, solid waste management, health risk assessment, and hazard risk assessment. Chapter  19, titled Select Chemical Engineering Applications, is on the other hand an extension of Part III and also contains five sections—fluid flow, chemical reactors, mass transfer applications, heat transfer applications, and plant design. Chapter 20 contains four sections that address more advanced optimization material. Most chapters contain a short introduction to the topic in question, which is followed by developmental material, which, in turn, is followed by several illustrative examples. Since this book offers material not only to individuals with limited technical background but also to those with extensive engineering industrial experience, this book can be used as a text in either a general introductory optimization course or (perhaps) as a training tool in industry for challenged engineering professionals. Hopefully, the text is simple, clear, to the point, and imparts a basic understanding of the theory and application of some of the important optimization methods employed in practice. It should also assist the reader in helping master the difficult task of explaining what was once a very complicated subject matter in a way that is easily understood. The authors feel that this distinguishes this text from the numerous others in this field. Is should also be noted that the authors have long advocated that basic science courses—particularly those concerned with mathematics—should be taught to engineers by an engineer; and the books adopted for use in these courses should also be written by an engineer. For example, a mathematician may lecture on differentiation— say dy /dx —not realizing that in a real-world application involving an estuary, y  could refer to concentration while x  could refer to time. The readers of this book will not encounter this problem. The reader should also note that parts of the material in the book were drawn from one of the author’s notes of yesteryear. In a few instances, the original source was not available for referencing purposes. Any oversight will be corrected in a later printing/edition. The authors wish to express appreciation to those who have contributed suggestions for material covered in this book. Their comments have been very helpful in the selection and presentation of the subject matter. Special appreciation is extended to Drs. Walter Matystik and Francesco Ricci.

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Finally, the authors are especially interested in learning the opinions of those who read this book concerning its utility and serviceability in meeting the needs for which it was written. Corrections, improvements, and suggestions will be considered for inclusion in any later editions. Lou Theodore  East Williston, NY  Kelly Behan  Mineola, NY 

About the Book As noted in both the Preface and Table of Contents, this optimization book is divided into four parts:

Part I: Part II: Part III: Part IV:

Optimization Fundamentals and Principles Environmental Engineering Applications Chemical Engineering Applications Select Industrial Applications

Following an optimization overview (Chapter 1) and a review of mathematical operations (Chapter  2), Part  I then examines the five optimization approaches covered in the book: perturbation techniques (Chapter  3), search methods (Chapter  4), graphical techniques (Chapter 5), analytical methods (Chapter 6), and the general subject of linear programming (Chapter  7). Chapters 3–6 primarily examine traditional optimization from an unconstrained  perspective, while Chapter 7 approaches optimization from a constrained  perspective. The above approach provides optimization topics ranging from the optimization of a single variable function to more detailed nonlinear constrained problems. Note also that although the optimization methods are provided with computers in mind, programming and coding are not addressed. The applications that follow in the next three Parts (II, III, IV) discuss how to improve existing systems and/or designing improved and more efficient new systems. Part II keys on optimization applications in environmental engineering. The five topics addressed are air management (Chapter 8), water management (Chapter 9), solid waste management (Chapter 10), health risk assessment (Chapter 11), and hazard risk assessment (Chapter 12). Each chapter contains a brief introduction to the subject in question, which is then followed by a host of examples dealing with industrial applications that employ some—if not all—of the optimization methods introduced in Part I. Also note that the performance criteria employed can vary from classical economic choices to technological factors, such as energy conservation, degree of purity, etc. Part III keys on optimization applications in chemical engineering. The five types addressed are fluid flow (Chapter 13), chemical reactors (Chapter 14), mass transfer (Chapter 15), heat transfer (Chapter 16), and plant design (Chapter 17). Each chapter contains a brief introduction to the subject in question, which is then followed by a host of examples that key on industrial applications that employ some—if not all—of the optimization methods introduced in Part I. Part IV, contains three chapters that are concerned with what the authors have termed, “Select Industrial Applications”. The first chapter covers environmental engineering while the second chapter keys on chemical engineering. Each of these two chapters contains five applications—one for each of the subject areas addressed earlier in Parts II and III. The third chapter provides more advanced optimization material that some might refer to as term projects, and it contains four illustrative examples. Regarding the subject of references, the authors recommend the following six texts for what is described as “suggested reading” for topics associated with environmental and chemical engineering: xvii

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1. R. Perry and D. Green, Perry’s Chemical Engineers’ Handbook , 8th edition, McGrawHill, New York City, NY, 2008. 2. L. Theodore, Chemical Engineering: The Essential Reference , McGraw-Hill, New York City, NY, 2014. 3. J. Reynolds, J. Jeris, and L. Theodore, Handbook of Chemical and Environmental Engineering Calculations , John Wiley & Sons, Hoboken, NJ, 2004. 4. L. Theodore, R. Dupont, and K. Ganesan, Unit Operations in Environmental Engineering , Wiley‑Scrivener, Salem, MA, 2018. 5. M. K. Theodore and L. Theodore, Introduction to Environmental Management , CRC Press/Taylor & Francis, Boca Raton, FL, 2000. 6. G. Burke, B. Singh, and L. Theodore, Handbook of Environmental Management and Technology , 2nd edition, John Wiley & Sons, Hoboken, NJ, 2000. These six books provide extensive background information, developmental material, illustrative examples and problems, and other references for many of the chapters in each Part of this book. The four recommended texts concerned with mathematical methods are

1. R. Ketter and S. Prawel, Modern Methods in Engineering Computation , McGraw-Hill, New York City, NY, 1969. 2. H. Mickley, T. Sherwood, and C. Reed, Applied Mathematics in Chemical Engineering , McGraw-Hill, New York City, NY, 1957. 3. M. Moyle, Introduction to Computers for Engineers , John Wiley & Sons, Hoboken, NJ, 1967. 4. J. Happel, Chemical Process Economics , John Wiley & Sons, Hoboken, NJ, 1958.

The beginning of each chapter (not including Part IV) contains additional references that are topic specific. The reader, particularly the beginning student, should consider reviewing (and perhaps reading) these references before attempting to grasp the fundamentals, principals, concepts, and so on, associated with each chapter. Finally, an important point needs to be made. There have been numerous occasions during one of the author’s 50-year tenure as an educator when students solved a problem using a packaged program such as Excel, MathCAD, and so on. For example, the problem could have involved the solution to a differential equation or the regression of some data. On being questioned how the packaged program performed the calculation, the student almost always responded with something to the effect of, “I don’t know, and I don’t care.” For this reason, the reader should note that no detailed attempt was made to introduce and explain the packaged computer programs presently available that provide easy access to trouble-free solutions of (even) complex mathematical problems, including optimization. The emphasis in this introductory text on optimization was to provide the reader with an understanding of fundamental principles in order to learn how these methods can be used to obtain answers to questions within one’s own subject matter specialty; becoming “computer literate” in this field was not the objective. Thus, anyone privy to applicable software can obtain detailed solutions to large and complex optimization problems by following relatively simple instructions associated with the software/computer. It should also be noted that significant material of an example/problem nature was drawn from lecture notes prepared for several courses offered by one of the authors during his career as an educator; the source of some of this material is no longer known.

Authors Over the past 50  years, Dr. Louis Theodore was a successful educator at Manhattan College (holding the rank of full professor of chemical engineering), graduate program director, researcher, professional innovator, and communicator in the engineering field. During this period, he was primarily responsible for his program achieving a number two ranking by US News & World Report. He has authored nearly 100 text/reference books and over 100 technical papers. He currently serves as a part-time consultant to the US Environmental Protection Agency (EPA) and runs Theodore Tutorials. He is a member of Phi Lambda Upsilon, Sigma Xi, Tau Beta Pi, American Chemical Society, American Society of Engineering Education, Royal Hellenic Society, and a fellow of the International Air & Waste Management Association (AWMA). Dr. Theodore is the recipient of the AWMA’s prestigious Ripperton award that is “presented to an outstanding educator, who, through example, dedication, and innovation has so inspired students to achieve excellence in their professional endeavors.” He was also the recipient of the American Society for Engineering Education’s (ASEE) AT&T Foundation award for “excellence in the instruction of engineering students.” Ms. Behan was employed in the utility industry and worked in the area of developing growth jobs and analyzing site plans. She also worked for the New York City Fire Department and was primarily involved in physiological research on World Trade Center victims. Ms. Behan previously served as Editorial Manager for a book on basketball coaching. She is currently involved with structural and civil engineering activities. In addition, her recent activities have centered on structural engineering with Turner Construction.

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Optimization Fundamentals and Principles Sophisticated optimization mathematical techniques are developing that enable environmental and chemical engineers to consider, in more quantitative ways, problems concerned with technology, society, and the environment. The traditional approach, classical calculus, to determine the best values for optimal variables is still employed, but linear programming is being used to an increasing extent, although this method is used only for limited application by most practicing engineers. The presentation here is to simply acquaint engineers with the fact that other simpler optimization methods also exist. Further study is required for the reader to attain proficiency in advanced optimization theory and practice. Finally, the age of computers has, to a certain extent, allowed many professionals to become comfortable working in an area that was once considered to be beyond their abilities. Seven introductory optimization related chapters complement the presentation for Part I. The titles are as follows. Chapter 1: Optimization Overview Chapter 2: Mathematical Operations Chapter 3: Perturbation Techniques Chapter 4: Search Methods Chapter 5: Graphical Approaches Chapter 6: Analytical Methods Chapter 7: Linear Programming

1 Optimization Overview As one might suspect, the term optimization has come to mean different things to different people. It has also come to mean different things for different applications; for example, it could involve a simple two-step calculation or one that requires the use of a detailed numerical method. To take this a step further, the authors were initially undecided on how to include optimization subject matter in this text. After much deliberation and meditation, it was decided to present introductory material in the first Part of the book. This decision was primarily influenced by the desire for this book to be a book that both introduces optimization mathematical methods and addresses applications that appear in Parts II and III for environmental and chemical engineers, respectively, as well as in Part IV. The optimization problem has been described succinctly by Aris1 as “getting the best you can out of a given situation.” Problems amenable to solution by mathematical optimization techniques generally have one or more independent variables whose values must be chosen to yield a viable solution to distinguish between the many viable solutions generated by the different choices of these variables. Mathematical optimization techniques are generally used for guiding the problem solver to the choice of variables that maximizes the “goodness” measure (e.g., profit) or that minimizes some “badness” measure (e.g., cost). In addition to the optimization definition by Aris presented previously, one might offer the following generic definition for many engineers: “Optimization is concerned with determining the ‘best’ solution to a given problem.”2 Alternatively, a dictionary would offer something to this effect: “to make the most of … develop or realize to the utmost extent … often the most efficient or optimum use of.” The process of optimization in practice is required in the solution of many problems and can involve the maximization or minimization of a mathematical function. As one might suppose, many of the applications involve economic considerations. One of the most important areas for the application of mathematical optimization techniques is in chemical engineering design. Topics can include the following:

1. Generation of best functional representations (e.g., curve fitting) 2. Design of optimal control systems 3. Determining the optimal height (or length) of pollutant mass transfer control units 4. Determining the optimal diameter of units 5. Finding the best equipment materials of construction 6. Generating operating schedules 7. Selecting operating conditions

This first chapter in Part I serves to provide a broad overview of the general subject of optimization. Section headings and titles are listed here: 1.1 History of Optimization 1.2 The Computer Age 3

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1.3 The Scope of Optimization 1.4 Conventional/Established Optimization Procedures 1.5 Contemporary Optimization: Linear Programming The first three sections need no introduction. The fourth section addresses standard (some might say traditional) optimization procedures employed by engineers today and in the past. These include such topics as brute force methods, perturbation schemes, elementary search methods, graphical approaches, and analytical methods. A separate chapter is devoted to each of these topics later in this Part. The last section is called Contemporary Optimization (some might replace contemporary with complex) and primarily keys on linear programming. These approaches will be emphasized in the solution to the environmental and chemical engineering illustrative examples in Parts II and III, respectively, as well as in Part IV. Illustrative Example 1.1 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 environmental and chemical engineer to better solve a problem. This brief answer provides a qualitative look at optimization. As noted previously, 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, and so on. In engineering design, once a particular subject or process scheme has been selected for study, it is common practice to optimize the process from both a capital cost and a O&M (operation and maintenance) standpoint. 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. Although the use of these techniques in the majority of environmental and chemical engineering applications was not warranted in the past, more and more real-world problems are requiring the use of optimization techniques.

1.1 History of Optimization The subject of mathematics encompasses the study of relationships among quantities, magnitudes, and properties and of logical operations by which unknown quantities, magnitudes, and properties may be deduced. In the past, mathematics was essentially regarded as a science of magnitudes and numbers. Toward the middle of the 19th century, however, mathematics came to be regarded increasingly as the science of relations, or as

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the science that draws necessary conclusions. This latter view encompasses mathematical or symbolic logic, the science of using symbols to provide an exact theory of logical deduction and references based on definitions, axioms, postulates, and rules for combining and transforming primitive elements into more complex relations and theorems. Enter optimization. This brief survey of the history of optimization traces the evolution of mathematical ideas and concepts, beginning in prehistory. Indeed, mathematics is nearly as old as humanity itself; evidence of a sense of geometry and interest in geometrical patterns has been found in the designs of prehistoric pottery and textiles, and in cave paintings. Primitive counting systems were almost certainly based on using the fingers of both hands, as evidenced by the predominance of the number 10 as the base for many number systems employed today. Interestingly, mathematics of the late 19th century and the 20th century is characterized by an interest in unifying elements across numerous fields of mathematical endeavor, especially in logic. For example, group theory has proved a particularly effective unifier. The amount of new math and the particular topics arising at that time were numerous and varied: unified set theory, intuitive geometry, the development of the number systems – including methods of numeration, binary and other bases of notation, and modular arithmetic and measurement, with attention to accuracy, precision, and error. It also included studies of algebraic systems, linear algebra, modern algebra, vectors, matrices, logic, truth tables, the nature of proofs, Venn and Euler diagrams, relations, functions, probability and statistics, linear programming, and (of course) computer programming. As to the origin of optimization, it depends on who provides the response, because there are so many aspects of optimization of interest to the practitioner. For example, some claim it was Thomas Edison when he developed a long-lasting, high-quality light bulb in the 1870s. His success was primarily the result of an extensive trial-and-error search for the optimum filament material. A few now refer to it as the Edisonian approach. One of the authors refers to it as the perturbation approach; on occasion, he has modestly termed it the Theodore approach. Advances in computing have not only reduced the cost and time required to perform any iteration calculation but also provide a better understanding of how this method works. However, the success of this approach still relies heavily on the limitations of the user’s intuition and experience since one often cannot predict or even comprehend the effects of changing numerous variables in a complex system. Despite this barrier, the computer has expanded and improved the employment of the classical methods of optimization. With the promise of reducing design time and cost while improving product quality, automated design optimization held tremendous potential. Starting with a sub-optimal design, a numerical optimization algorithm could be used to iteratively adjust a set of preselected design parameters in an attempt to achieve a set of design targets. This new class of optimization technology enables broader, more comprehensive, and faster searches for innovative designs than was possible using previous generations of tools. Moreover, it requires no expertise in optimization theory, so it is easier to use for non-experts and experts alike. By leveraging an engineer’s potential to discover new design concepts, this new class of optimization technology overcomes the limits of human intuition and extends the designer’s professional capability to achieve breakthrough designs and accelerated innovation. As noted earlier, today’s computers now allow an optimization procedure (or mathematics, computer science, operations research, mathematical optimization, or mathematical programming) to select the best element (with regard to some criteria) from

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Introduction to Optimization for Chemical and Environmental Engineers

some set of available alternatives. In the simplest case, an optimization problem consists of optimizing (maximizing or minimizing) a real function by systematically choosing input values from within an allowed set of variables in order to compute the value of the function. Mathematically speaking, optimization allows one to find the best available value(s) of some (often referred to as the objective) function given a defined domain (or a set of constraints), including a variety of different types of objective functions and different types of domains. These various options are introduced in the last section of this chapter and treated in more detail in Chapter 7. A feasible solution to the objective function that minimizes (or maximizes), if that is the goal, is called an optimal solution.

1.2 The Computer Age Although digital computers are often viewed as fast calculators or special slide rulers, they are, in fact, rather general devices for manipulating symbolic information. As noted, the symbols being manipulated are numbers or digits (hence the name digital computer), and the operations being performed on the symbols are the standard arithmetical operations, such as addition and subtraction. Famularo offered the following thoughts in the late 1960s: “The digital computer can be viewed as a high-speed calculator that, with the availability of subroutines and a compiling language, is able to perform many mathematical operations such as add, multiply, generate analytic functions and logic decisions, etc.”3 Although the details of coding are not in the scope of this text, there is an intermediate step between the equations and the coding program, and that is to arrange the computing procedure in block diagram or information flow form. The operation of each block is described inside the block, and the computer performs the instructions around the loop until the condition in the decision block is satisfied, at which time the most recently computed values were once punched out on tape or card. For a large computer, this entire sequence, including a half-dozen cycles around the loop, would be accomplished within a millisecond or less. Of course, as the number of equations to be solved increases, the time required to obtain a solution increases. For cases where the solution time becomes excessive, mathematics or other more sophisticated programming is used to accelerate the convergence and decrease the total computation time required. High-speed computing equipment (today’s computers) 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 the mathematical operations described previously at tremendous speeds permits the analysis of many more applications and more engineering variables than could possibly be handled on the aforementioned slide ruler – the trademark of engineers of yesteryear. Scientific calculations previously estimated in lifetimes of computation time are currently generated in seconds and, on many occasions, microseconds, and in some rare instances, nanoseconds.4 Increased growth is expected in the microcomputer field as inexpensive informationprocessing devices continue to be developed. The cost will continue to decrease, while the cost of the associated software will probably tend to increase; computerized processes will, therefore, be quite inexpensive unless customized programs must be written.

Optimization Overview

7

1.3 The Scope of Optimization One can conclude from the aforementioned that the theory and application of optimization is mathematical in nature, and it typically involves the maximization or minimization of a function (usually known), which represents the “performance” of some “system.” This is carried out by the finding of values for those variables, which cause the function to yield an optimal value. There are a large number of traditional mathematical optimization methods available in practice. For example, problems might simply involve selecting the “best” choice between two alternatives – in effect a yes/no option. Or, it could involve three or more alternatives. A number of others are available in the mathematics literature. Some of these methods are reviewed in the chapters to follow in this Part I. There are a large number of “contemporary” optimization approaches available to the practicing environmental and chemical engineer. Most of the these methods involve finding the optimizer of a function (or functions) by a trial-and-error method until a satisfactory optimizer is achieved. The various methods involve a different procedure for making educated guesses. The overall process institutes the first guess, and changes this guess until a satisfactory solution is obtained.

1.4 Conventional/Established Optimization Procedures On to traditional optimization. One of the authors4 has often employed the phrase “…has come to mean different things to different people.” No technical phrase better fits the description of optimization. For many, including the authors, standard (or conventional, traditional, established, typical, etc.) optimization keys on four topics:

1. Perturbation methods 2. Search methods 3. Graphical approaches 4. Analytical methods

Introductory details on these topics follow, noting that linear programming briefly receives an introduction in Section 1.5. Topic (1) basically involves a guessing game when attempting to solve an optimization problem. High-powered mathematics is usually not involved. The same can be said for any trial-and-error method whether it be systematic or not. Numerous elementary search methods (2) are available but only two will be discussed in Chapter 4. Graphical approaches (3), as one might suppose, involve graphing available data and calculations in an attempt to arrive at an “optimization” solution. Analytical methods (4) often employ ordinary and partial derivatives in obtaining a solution. Topics (1)–(4) have been defined by most mathematics as direct approaches, while linear programming is referred to as an indirect approach. As one might assume, indirect methods are generally preferred. There are, however, numerous other methods – in addition to those mentioned previously – available for solving optimization problems. These can

8

Introduction to Optimization for Chemical and Environmental Engineers

involve the aforementioned procedures that simply require a yes or no answer, selecting the best option of the two alternatives, selecting the best of more than two options, and so on. The selection of the independent variable(s) is often set in academic problems involving standard optimization calculations. The choice of these variables is usually based on past experience or sound engineering judgment. For example, the performance of an absorber – for control of a pollutant in an environmental engineering application or for recovery in a chemical engineering application – is usually assumed to be a function of temperature and pressure. Other effects, such as the time of day, or lunch details, or … are usually and understandably neglected. This often reduces the problem to one that does not involve the need to resort to any sophisticated optimization mathematical methods. One should also note that these optimization problems, like many engineering problems, usually involve a simple two-step procedure:

1. Developing a describing equation (or equations) for the system to be optimized 2. Solving the describing equation or equations

Step (1) falls in the domain of the engineer. Step (2) traditionally fell in the domain of both the engineer and mathematician. However, the second step is essentially no longer a concern to many engineers with the advent of computers, particularly as it applies to contemporary optimization – a topic that is addressed in the next section. Illustrative Example 1.2 Which of the four traditional optimization methods discussed previously, that is,

1. Perturbation methods 2. search methods 3. graphical approaches 4. analytical methods

are most often employed by the environmental and chemical engineer? SOLUTION: This is a difficult question to answer since it is most probably a function of the experience, interest, and capabilities of the individual being questioned. The answer for one of the authors (2) earlier in his career would probably be (1) and (3). In more recent times, both authors would definitely select (4).

1.5 Contemporary Optimization: Linear Programming Perhaps the most important tool employed in detailed optimization studies by practicing engineers is linear programming. Linear programming consists of a linear, multivariable function, which is to be optimized (maximized or minimized), subject to a particular number of constraints. The constraints are normally expressed in linear form. Integer linear programming refers to optimization problems in which at least some of the

Optimization Overview

9

variables must assume integer values. If the assumption of linearity cannot be made, as in the case of linear programming, there exist some general procedures for nonlinear problems. The collection of techniques developed for these problems is called nonlinear programming. The application of linear programming usually involves a constrained optimization problem. The objective is to locate the values of the decision variable(s) that either maximize or minimize a linear objective function, where the decision variables are subject to linear constraints. These mathematical exercises involve locating a value (the feasibility value) that minimizes the objective function while satisfying the constraints. Thus, linear programming provides a method of obtaining the optimum feasible value from a near infinite number of values. Some optimization problems can be divided into parts, for which each part is then optimized. In some instances, it is possible to attain the optimum for the original problem by simply realizing how to optimize these constituent parts. This process is very powerful, as it allows one to solve a series of smaller, easier problems rather than one large one. One of the best-known techniques to attack such problems is dynamic programming. This approach is characterized by a process that is performed in stages, such as manufacturing processes. The possible states of any stage of the process must include enough information regarding conclusions and/or decisions made at the preceding stage (or stages) to allow an informed decision at the present stage. Rather than solving the problem as a whole, dynamic programming thus optimizes one stage at a time to produce an optimal set of decisions for the whole process–in effect reformatting the optimization problem as a multistage one. Although dynamic programming has applicability in some environmental and chemical engineering systems/processes, it is not reviewed in this book. One of the major responsibilities in optimization is to construct the correct objective function to be maximized (or minimized). For example, in a simple environmental problem, the objective function might be the profit associated with the operation of a landfill that accepts two categories of waste: “inert” and “non-inert.” One optimization problem might be to maximize the profit (from handling and treating the waste) by employing an operating treatment schedule subject to the landfill’s capacity and environmental regulatory requirements. This problem might also be rephrased in the following manner: What combination of the quantities of the two wastes will produce the maximum profit subject to the constraints connected with the landfill’s processing capabilities and the different regulatory requirements imposed on each waste? The objective function may be quite simple and easy to calculate in some environmental and chemical engineering applications, or it may be complicated and difficult to not only calculate but also specify and/or describe. The objective function may also be very elusive due to the presence of conflicting or dimensionally incompatible objectives; for example, one might be asked to optimize the profit for the aforementioned landfill that not only minimizes air and water contamination and/or emissions but also is aesthetically appealing. Thus, it may not be always possible to quantify an objective function and hence be able to use any of the mathematical optimization procedures available. Illustrative Example 1.3 The dependent variable (function) y is equal to four times x1 minus three times x2, where x1 and x2 are positive independent variables. The sum of x1 and x2 are both constrained by zero and 10, and their cross product is limited to the range of zero and 36. Describe this constrained mathematical system.

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Introduction to Optimization for Chemical and Environmental Engineers

SOLUTION: The (objective) function is y = 4 x1 − 3 x2

The constraints are

0 ≤ x1 + x2 ≤ 10

and 0 ≤ x1x2 ≤ 36

plus

0 ≤ x1 and 0 ≤ x2

The remainder of Part I addresses six additional topics related to optimization. Chapter numbers and subject matter are again presented here: Chapter 2: Mathematical Operations Chapter 3: Perturbation Techniques Chapter 4: Search Methods Chapter 5: Graphical Approaches Chapter 6: Analytical Methods Chapter 7: Linear Programming This material provides optimization fundamentals and principles ranging from the optimization (or minimization) of a single variable function to more detailed nonlinear constrained problems. Although the optimization methods are provided with computers in mind, the reader is again reminded that programming and coding details are not addressed. Also note that the performance criteria employed can vary from classical economic choices to technological factors, such as energy conservation, the degree of purity, and so on. Finally, most engineering optimization applications involve economics: maximizing profit, minimizing cost, or both. In addition to the traditional engineering applications discussed earlier, environmental applications can involve minimizing toxic emissions, maximizing energy usage and conservation, minimizing health and safety risks, standard economic concerns, and so on. The reader should not lose sight of the fact that most realworld industry applications involving optimization usually require simple solutions. Here is a comment from a retired engineer:5 “Generally, the consumer goods industry where I worked does not rely on sophisticated models. Basically, optimization was conducted to determine overall consumer preference or liking, to determine how to use minimum resources, and lastly, to determine how to minimize costs subject to constraints. In the determination of consumer preference, we typically used design of experiments6 to examine the broad range of attributes that led to maximum product liking or decision to purchase. Occasionally, we modeled processes and used regression models.6 In that case, we could search or compute the desired maximum. The last set of techniques used is what the

Optimization Overview

11

business world calls ‘Operations Research.’ Typically, linear or dynamic programming was used to evaluate an ‘objective function subject to constraints.’ For example, we used linear programming in hot dog manufacturing. The goal was to come up with a mixture of meat cuts that minimized costs while meeting governmental requirements for protein, fat, and water content.”

References

1. R. Aris, Discrete Dynamic Programming, Blaisdell, Boston, MA, 1964. 2. L. Theodore, personal notes, East Williston, NY, 2011. 3. J. Famularo, personal communication to L. Theodore (with permission), Englewood Cliffs, NJ, 1973. 4. L. Theodore, personal notes, East Williston, NY, 2005. 5. R. Altomare, communication to L. Theodore, Bronx, NY, 2016. 6. S. Shaefer and L. Theodore, Probability and Statistics Applications for Environmental Science, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2007.

2 Mathematical Operations The natural numbers, or so-called counting numbers, are the positive integers: 1, 2, 3, … and the negative integers: −1, −2, −3, … The following applies to real numbers:

a > b means that a – b is a positive real number (2.1)



If a < b and b < c, then a < c (2.2)



If a < b and c > 0, then ac < bc (2.3)



If a < b and c < d then a + c < b + d (2.4)



If a < b and ab > 0, then

1 1 > (2.5) a b



a + b = b + a; a ⋅ b = b ⋅ a = ( a )(b ) (2.6)



a + (b + c ) = ( a + b ) + c; a ⋅ (bc ) = ( ab ) ⋅ c (2.7)



a (b + c ) = a ⋅ b + a ⋅ c; with a + 0 = a; a ⋅ 1 = a (2.8)



If x = a, where a > 0, then x = a, or x = − a (2.9)



If x < c, then − c < x < c, where c > 0 (2.10)



a− n =



a ≠ 0 (2.11)

( ab )n = anb n (2.12)



1 , an

(a )

n m

= a nm , a n a m = a n+ m (2.13) a0 = 1, a ≠ 0 (2.14)

log ab = log a + log b ,

a > 0, b > 0 (2.15)



log a n = n log a (2.16)



 a log   = log a − log b (2.17)  b 13

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Introduction to Optimization for Chemical and Environmental Engineers



 1 log n a =   log a (2.18)  n



ln a = (log e 10 ) common log a = ( 2.3026 ) common log a = 2.3026 log a (2.19)

Based on the previous, one may write

( 4)(9) = (3.60)(10)1 (6)3 = (2.16)(10)2 3

3375 = (1.50 )(10 ) = 1

(

3

3.375

)(

3

1000

)

log ( 245) = 2.3892 log ( 24.5) = 1.3892

log ( 2.45) = 0.3892



log (0.245) = 9.3892 − 10 = −0.6108 ln ( 245) = ( 2.3892)( 2.3026 ) = 5.501 ln ( 24.5) = 3.199 ln ( 2.45) = (0.3892)( 2.3026 ) = 0.896 ln (0.245) = −11.4065 The authors have assumed that the reader is not only familiar with that described previously but also other more advanced topics that can include simultaneous linear algebraic equations, nonlinear equations, differentiation, integration, the solution of both ordinary differential equations and partial differential equations, and the Monte Carlo method. Only the numerical solution approaches for the previously mentioned are reviewed in this chapter. Elementary statistics and least squares approximation are not reviewed, but details are available in the literature.1 Ten sections complement this mathematical operations chapter: 2.1 The Quadratic Equation 2.2 Interpolation and Extrapolation 2.3 Significant Figures and Approximate Numbers 2.4 Errors 2.5 Differentiation 2.6 Numerical Integration 2.7 Simultaneous Linear Algebraic Equations 2.8 Nonlinear Algebraic Equations 2.9 Ordinary Differential Equations 2.10 Partial Differential Equations

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Mathematical Operations

The first four sections may be viewed as elementary material. The last six sections are more quantitative and can be viewed as fitting in the domain of numerical methods. Since a detailed treatment of each of these topics is beyond the scope of this text, the reader is referred to the literature2–4 for more extensive analysis and additional information. The remainder of this chapter briefly examines the topics listed previously.

2.1 The Quadratic Equation Given any quadratic equation of the general form

ax 2 + bx + c = 0 (2.20)

a number of methods of solution are possible depending on the specific nature of the equation in question. If the equation can be factored, then the solution is straightforward. For instance, consider

x 2 − 3 x = 10 (2.21)

First, the equation is put into the standard form,

x 2 − 3 x − 10 = 0 (2.22)

which can be factored as follows:

( x − 5) ( x + 2) = 0 (2.23)

This condition can be met, however, only when the individual factors in parentheses are zero, i.e., when x = 5 and/or x = −2. That these are indeed the solutions to the equation may be verified by substitution. If, upon inspection, no obvious means of factoring the above equation can be found, an alternative approach may exist. For example, in the equation

4 x 2 + 12x = 7 (2.24)

the expression

4 x 2 + 12x (2.25)

could be factored as a perfect square if it were

4 x 2 + 12x + 9 (2.26)

which equals

(2x + 3)2 (2.27)

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Introduction to Optimization for Chemical and Environmental Engineers

This can easily be achieved by adding 9 to the left side of the equation. The same amount must then, of course, be added to the right side as well, resulting in:

4 x 2 + 12x + 9 = 7 + 9 (2.28)



(2x + 3)2 = 16 (2.29)

This can be reduced to

(2x + 3 ) =



16 (2.30)

or 2x + 3 = +4 (2.31)

and

2x + 3 = −4 (2.32)



Since 16 has two solutions, the first equation leads to the solution x = 1/2 or x = 0.5, while the second equation leads to the solution x = −7/2, or x = −3.5. If the methods of factoring or completing the square are not possible, any quadratic equation can always be solved by the quadratic formula. This provides a method for determining the solution of the equation if it is once again in the form ax 2 + bx + c = 0 (2.33)



In all cases, the two solutions of x are given by the formula x=



−b ± b 2 − 4 ac (2.34) 2a

For example, to find the roots of x 2 − 4 x = −3 (2.35)



the equation is first put into the standard form x 2 − 4 x + 3 = 0 (2.36)



As a result, a = 1, b = −4 and c = 3. These terms are then substituted into the quadratic formula presented in Equation 2.34.



x=

− ( −4 ) ±

=

( −4)2 − 4 (1)(3) 4 ± = 2 (1)

16 − 12 (2.37) 2

4± 4 4±2 = = 3, and 1 (2.38) 2 2

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Mathematical Operations

The practicing environmental and chemical engineer occasionally has to solve not just a single equation but several at the same time. The problem is to find the set of all solutions satisfying all equations. These are called simultaneous equations, and specific algebraic techniques may be used to solve them. For example, a simple solution exists for the following two linear equations with two unknowns:

3 x + 4 y = 10 (2.39)



2x + y = 5 (2.40)

The variable y in Equation 2.40 is isolated (y = 5 − 2x), and then this value of y is substituted into Equation 2.39. The result is Equation 2.41.

3 x + 4 ( 5 − 2x ) = 10 (2.41)

This reduces the problem to one involving the single unknown x, and it follows that

−5x = −10 (2.42)

so that

x = 2 (2.43)

When this value is substituted into either previous equation, it follows that

y = 1 (2.44)

A faster method of solving simultaneous equations, however, is obtained by observing that if both sides of Equation 2.40 are multiplied by 4, then

8 x + 4 y = 20 (2.45)



3 x + 4 y = 10 (2.39)

If Equation 2.39 is subtracted from Equation 2.45, 5x = 10 results, or x = 2. This procedure leads to another development in mathematics, matrices, which can help to produce solutions for any set of linear equations with a corresponding number of unknowns (see also Section 2.7).

2.2 Interpolation and Extrapolation Experimental data (and data in general) in environmental and chemical engineering may be presented using a table, a graph, or an equation. Tabular presentation permits retention of all significant figures of the original numerical data. Therefore, it is the most numerically accurate way of reporting data. However, it is often difficult to interpolate between data points within tables.

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Introduction to Optimization for Chemical and Environmental Engineers

Graphical or tabular presentation of data is usually used if no theoretical or empirical equations can be developed to fit the data. This type of presentation of data is one method of reporting experimental results. For example, heat capacities of benzene might be tabulated at various temperatures. Data may also be presented graphically. One should note that graphs are inherently less accurate than numerical tabulations. However, they are useful for visualizing variations in data and for interpolation and extrapolation. Interpolation is of practical importance to the practitioner because of the occasional necessity of referring to sources of information expressed in the form of a table. Logarithms, trigonometric functions, water properties of steam, liquid water and ice, vapor pressures, and other physical and chemical data are commonly given in the form of tables in the literature. Although these tables are sometimes given in sufficient detail so that interpolation may not be necessary, it is important to be able to interpolate properly when the need arises. Assuming that a series of values of the dependent variable y for corresponding tabulated values of the independent variable x is provided, the problem of interpolation is to obtain the correct value of yet other values of x. (Extrapolation refers to a value of x lying outside the range of tabulated values of x.) Clearly, interpolation or extrapolation may be accomplished by using data for x and y to develop a linear relationship between the two variables. The general method would be to fit two points (y1, x1) and (y2, x 2) by means of y = a + bx (2.46)



and then employ this equation to calculate y for some value of x lying between x1 and x 2. Most practitioners do this mentally when reading values from the (steam) tables. If a number of points are used, a polynomial of a correspondingly higher degree may also be employed.1 Interpolation is necessary to find y when x is some value not given in the table. Thus, interpolation may be viewed as the process of finding the value of a function at some arbitrary point where the function is not known but is represented over a given range as a table of discrete points. (See also Table 2.1 where y represents a reservoir’s height as a function of time in days during a rainy season.) For instance, one may be interested in finding y when x = 7. (The process of finding x when y is known is referred to as inverse interpolation). Given a table with (x, y) data (Table 2.1), one can draw a picture (Figure 2.1) and write the equation of the straight line through the points (x1, y1) and (x 2, y2) for y. TABLE 2.1 Reservoir Height x, Time 0 3 6 9 12 15 18

y, Height 30 31 33 35 39 46 52

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Mathematical Operations

35 30

y

25 20

y2

15

y1

10 5 0

x1 1

2

3

x2

4

5

x

6

7

8

9

10

FIGURE 2.1 Interpolation procedure.

 y − y1  y − y1 =  2 ( x − x1 ) (2.47)  x2 − x1 



Equation 2.47 can be solved for y in terms of x y = y1 +



( y 2 − y1 ) ( x − x1 ) (2.48) x2 − x1

or (for points 0 and 1) y=



y 0 ( x − x 0 ) + ( y1 − y 0 ) ( x − x 0 ) (2.49) x1 − x0

Illustrative Example 2.1 Refer to Table 2.1. Find y at x = 11. SOLUTION: Proceed as follows, keying (obviously) on the values for x = 9 and x = 12. Data Point i

xi

yi

x i − x

1 2

9 12

35 39

2 1

Apply Equation 2.48 with x = 11. Therefore,

y = 35 +

( 4)( 2) = 35 + 8 = 37.67 3

3

Inverse interpolation involves estimating x which corresponds to a given value of y, and (as noted previously) extrapolation involves estimating values of y outside the interval in which the data x0, … , xn fall. It is generally unwise to extrapolate any empirical relation significantly beyond the last data point or before the first point. If, however, a certain form of equation is predicted by theory and substantiated by (other) available data, reasonable extrapolation is ordinarily justified.

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Introduction to Optimization for Chemical and Environmental Engineers

2.3 Significant Figures and Approximate Numbers1 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 nm (nanometers) implies that the actual quantity is somewhere between 8.315 and 8.325 nm. This applies to calculated and measured quantities; quantities that are known exactly (e.g., pure integers) have an infinite number of significant figures. Note, however, that there is an upper limit to the accuracy with which physical measurements can be made. The method for counting the significant digits of a number follows one of two rules depending on whether or not a decimal point is present. The significant digits of a number always start from the first nonzero digit on the left to either

1. the last digit (whether it is nonzero or zero) on the right, if there is a decimal point present, or 2. the last nonzero digit on the right of the number, if there is no decimal point present. For example, 370 370. 370.0 28,070 0.037 0.0370 0.02807

Has 2 significant figures Has 3 significant figures Has 4 significant figures Has 4 significant figures Has 2 significant figures Has 3 significant figures Has 4 significant 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 4113 (without a decimal). The least precise of the five numbers is 370, which has its last significant digit in the tens position. Therefore, the answer should also have its last significant digit in its tens position. Unfortunately, environmental and chemical engineers rarely concern themselves with significant figures in their calculations. However, it is recommended that the reader attempt to follow the calculation procedure set forth in this section. In the process of performing engineering scientific calculations, very large and very small numbers are often encountered. A convenient way to represent these numbers is to use scientific notation (see also the introduction to this chapter). Generally, a number represented in scientific notation is the product of a number and 10 raised to an integer power. For example,

21

Mathematical Operations

28, 070, 000, 000 = 2.807 × 1010 = ( 2.807 )(10 )

10



0.000002807 = 2.807 × 10 −6 = ( 2.807 )(10 )

−6



A positive feature of using scientific notation is that only the significant figures need to appear in the number. When approximate numbers are added, or subtracted, the results are also in terms of the least precise number. Since this is a relatively simple rule to master, note that the answer follows the aforementioned rule of precision. For example,

6.04 L + 2.8 L − 4.173 L = 4.7 L (2.50)

(The complete result is 4.667 L.) The terms in Equation 2.50 have two, one, and three decimal places, respectively. The least precise previous number (least decimal places) is 2.8, a value carried only to the tenths position. Therefore, the answer must be reported to the tenths position only. Thus, the correct answer is 4.7 L, with the last 6 and the 7 dropped from the 4.667 L, and the first 6 is rounded up to provide 4.7 L. In multiplication and division of approximate numbers, finding the number of significant digits is used to determine how many digits to keep (i.e., where to truncate). One must first understand significant digits in order to determine the correct number of digits to keep or remove in multiplication and division problems. As noted earlier in this section, the digits 1 through 9 are considered to be significant. Thus, the numbers 123, 53, 7492, and 5 contain three, two, four and one significant digits, respectively. The digit zero must be considered separately. Zeroes are significant when they occur between significant digits. In the following examples, all zeroes are significant: 10001, 402, 1.1001, 500.09 has five, three, five, and four significant figures, respectively. Zeroes are not significant when they are used as placeholders. When used as a placeholder, a zero simply identifies where a decimal is located. For example, each of the following numbers has only one significant digit: 1000, 500, 60, 0.09, 0.0002. In the number 1200, 540, and 0.0032 there are two significant digits, and the zeroes are not significant. When zeroes follow a decimal and are preceded by a significant digit, the zeroes are significant. In the following examples, all zeroes are significant: 1.00, 15.0, 4.100, 1.90, 10.002, 10.0400. For 10.002, the zeroes are significant because they fall between two significant digits. For 10.0400, the first two zeroes are also significant because they fall between two significant digits; the last two zeroes are significant because they follow a decimal and are preceded by a significant digit. Thus, when approximate numbers are multiplied or divided in a problem, the result is expressed as a number having the same number of significant digits as the expression in the problem having the least number of significant digits. When truncating (removing final, unwanted digits), rounding is normally applied to the last digit to be kept. Thus, if the value of the first digit to be discarded is less than 5, one should retain the last kept digit with no change. If the value of the first digit to be discarded is 5 or greater, one should increase the last kept digit’s value by one. Assume, for example, only the first two decimal places are to be kept for 25.0847 (the 4 and 7 are to be dropped). The number is then 25.08. Since the first digit to be discarded (4) is less than 5, the 8 is not rounded up. If only the first two decimal places are to be kept for 25.0867 (the 6 and the 7 are to be dropped), it should be rounded to 25.09. Since the first digit to be discarded (6) is 5 or more, the 8 is rounded up to 9.

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Introduction to Optimization for Chemical and Environmental Engineers

When adding or subtracting approximate numbers, a rule based upon precision determines how many digits are kept. In general, precision relates to the decimal significance of a number. When a measurement is reported as 1.005 cm, one can say that the number is precise to the thousandth of a centimeter. If the decimal is removed (1005 cm), the number is precise to thousands of centimeters. For example, in some water pollution studies, a measurement in gallons or liters may be required. Although a gallon or liter may represent an exact quantity, the measuring instruments that are used are only capable of producing approximations. Using a standard graduated flask in liters as an example, can one determine whether there is exactly one liter? Not likely. In fact, one would be pressed to verify that there was a liter to within ±1/10 of a liter. Therefore, depending upon the instruments used, the precision of a given measurement may vary. If a measurement is given as 16.0 L, the zero after the decimal indicates that the measurement is precise to within 1/10 L, that is, 0.1 L. Given a measurement of 16.00 L, one has precision to the 1/100 L. As noted, the digits following the decimal indicate how precise the measurement is. Thus, precision is used to determine where to truncate when approximate numbers are added or subtracted.

2.4 Errors This section introduces the various classes of errors that arise in technical calculations while the next section demonstrates the propagation of some of these errors. As one might suppose, numerous books have been written on the general subject of “errors.” Different definitions for errors appear in the literature, but what follows is the authors’ attempt to clarify the problem.1 Any discussion of errors would be incomplete without providing a clear and concise definition of two terms: the aforementioned precision and accuracy. The term precision is used to describe a state or system or measurement for which the word precise implies little to no variation; some refer to this as reliability. Alternatively, accuracy is used to describe something free from the matter of errors. The accuracy of a value, which may be represented in either absolute or relative terms, is the degree of agreement between the measured value and the true value. All measurements and calculations are subject to two broad classes of errors: determinate and indeterminate. The error is known as a “determinate error” if an error’s magnitude and sign are discovered and accounted for in the form of a correction. All errors that either cannot be or are not properly allowed for in magnitude and sign are known as “indeterminate errors.” A particularly important class of indeterminate errors is that of accidental errors. To illustrate the nature of these, consider the very simple and direct measurement of temperature. Suppose that several independent readings are made and that temperatures are read to 0.1°F. When the results of the different readings are compared, it may be found that even though they have been performed very carefully, they may differ from each other by several tenths of a degree. Experience has shown that such deviations are inevitable in all measurements and that these result from small unavoidable errors of observation due to the sensitivity of measuring instruments and the keenness of the sense of perception. Such errors are due to the combined effect of a large number of undetermined causes, and they can be defined as “accidental errors.”

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Regarding the words precision and accuracy, it is also important to note that a result may be extremely precise and at the same time inaccurate. For instance, the temperature readings just mentioned might all agree within 1°F. From this, it would not be permissible to conclude that the temperature is accurate to 1°F until it can be definitively shown that the combined effects of uncorrected constant errors and known errors are negligible compared with 1°F. It is quite conceivable that the calibration of the thermometer might be grossly incorrect. Errors such as these are almost always present and can never be detected individually. Such errors can be detected only by obtaining the readings with several different thermometers and, if possible, several independent methods and observers. It should also be understood once again that most numerical calculations are by their very nature inexact. The errors are primarily due to one of three sources: inaccuracies in the original data, lack of precision in carrying out calculations, or inaccuracies introduced by approximate or incorrect methods of solution. Of particular significance are the aforementioned errors due to “round-off” and the inability to carry more than a certain number of significant figures. The errors associated with the method of solution are usually the area of greatest concern.3 These usually arise as a result of approximations and assumptions made in the development of an equation used to calculate a desired result and should not be neglected in any error analysis. Finally, many list the following three errors associated with a computer (calculator): 1. Truncation error: With the truncation of a series after only a few terms, one is committing a generally known error. This error is not machine-caused but is due to the method. 2. Round-off error: This is machine-made and is caused by the limitations of the particular computer. If one computes using eight digits in multiplication, one would expect 16 digits by hand calculation. The computer retains only the most significant eight of these. If the number was rounded, this error would be minimized, but rounding is generally not done. 3. Propagation or inherited error: This is caused by sequential calculations that include points previously calculated by the computer that already are erroneous owing to the two previously mentioned errors. Since the result is already off the solution curve, one cannot expect any new points computed to be on the correct solution curve. Adding the round-off errors and truncation errors into the calculation causes further errors to propagate adding more error at each step.

2.5 Differentiation Several differentiation methods are available for generating expressions for a derivative. Consider the problem of determining the benzene concentration – time gradient dC/dt5 at t = 4.0 s; refer to Table 2.2. Six differentiating methods are presented subsequently.5 Method 1: This method involves the selection of any three data points and calculating the slope m of the two extreme points. This calculated slope is approximately equal to the slope at the point lying in the middle. The value obtained is the “equivalent” of the derivative at that point 4. Using data points from 3.0 to 5.0, one obtains

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Introduction to Optimization for Chemical and Environmental Engineers

TABLE 2.2 Time-concentration Data Time (s)

Concentration of Benzene (mg/L), C

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0

7.46 5.41 3.80 2.70 2.01 1.63 1.34 1.17

Slope = m =

C5 − C3 t5 − t3

1.63 − 2.70 = = −0.535 5.0 − 3.0 Method 2: This method involves determining the average of two slopes. Using the same points chosen previously, two slopes can be calculated, one for points 3 and 4 and the other for points 4 and 5. Adding the two results and dividing them by 2 will provide an approximation of the derivative at point 4. For the points used in this method, the results are:



m1 = slope1 =





C 4 − C3 t 4 − t3

2.01 − 2.70 = = −0.69 4.0 − 3.0 C − C4 m2 = slope2 = 5 t5 − t 4 1.63 − 2.01 = −0.38 = 5.0 − 4.0 mavg = slopeavg =





−0.69 + ( −0.38 ) = −0.535 2

Method 3: This method consists of selecting 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 following equation results:

C = 0.155t 2 − 1.775t + 6.63

The analytical derivative of the equation is calculated and may then be used to evaluate the slope at any point. Here, point 4 is used:

dC = 0.31t − 1.775 dt

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Evaluated at t = 4.0 s dC = 0.31 ( 4.0 ) − 1.775 = −0.535 dt



Method 4: This method uses the method of least squares.1 In this case, all the 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 is once again analytically differentiated.

C = 0.1626t 2 − 1.9905t + 7.3108



dC = 0.3252t − 1.9905 dt

Evaluated at t = 4.0 s:

dC = 0.3252 ( 4.0 ) − 1.9905 = −0.6897 dt

Methods 5 and 6: These two methods are somewhat similar. 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 literature6), respectively. Method 5 employs five data points to generate a five coefficient (fourth-order) model using an equation of the form f = A + Bx+ Cx2+ Dx3+ Ex4. This method is known as interpolating. A set of equations is used to evaluate numerical derivatives from the interpolating polynomial. The describing equations are listed here:

mo = f ′ ( x0 ) =

−25 f 0 + 48 f1 − 36 f 2 + 16 f 3 − 3 f 4 (2.51) 12h

m1 = f ′ ( x1 ) =

−3 f 0 − 10 f1 + 18 f 2 − 6 f 3 + 3 f 4 (2.52) 12h

mi = f ′ ( xi ) = mn−1 = f ′ ( xn−1 ) = mn = f ′ ( xn ) =

f i − 2 − 8 f i −1 + 8 f i + 1 − f i − 2 (2.53) 12h

− f n− 4 + 6 f n−3 − 18 f n−2 + f n−1 + f 3 (2.54) 12h

−3 f n− 4 − 16 f n−3 + 36 f n−2 − 48 f n−1 + 25 f 3 (2.55) 12h

where: h = xi + 1 − xi fi is the function evaluated at i For example, the equation obtained for “the five-data set” from 1.0 to 5.0 s, that is, t = 1.0, 2.0, 3.0, 4.0, and 5.0 s, using the equations given previously is

f ( x ) = −0.0012x 4 + 0.002x 3 + 0.2616 x 2 − 2.34 x + 7.467

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All these equations are evaluated for each value of x and f(x). The value obtained for point 4.0 is −0.5448. Method 6 also employs five data points, but only three coefficients are generated for a second-order polynomial equation of the form f = A + Bx+ Cx2. Another set of equations is used to evaluate the derivative at each point using this method. The describing equations are provided here:6

f ′ ( x0 ) =

( −54 f0 + 13 f1 − 40 f2 + 27 f3 − 26 f 4 ) (2.56)

f ′ ( x1 ) =

70 h

( −34 f0 + 3 f1 + 20 f2 + 17 f3 − 6 f 4 ) (2.57) 70 h

f ′ ( xi ) =



f ′ ( x n −1 ) =



f ′ ( xn ) =

( −2 fi−2 − fi−1 + fi+1 + 2 fi−2 ) (2.58) 70 h

(6 fn−4 − 17 fn−3 − 20 fn−2 − 3 fn−1 + 34 f3 ) (2.59) 70 h

(26 fn−4 − 27 fn−3 − 40 fn−2 − 13 fn−1 + 54 f3 ) (2.60) 70 h

At point 4.0, the solution for the derivative using the method is −0.6897. Comparing all six values obtained for the derivative at t = 4.0 s, one can conclude that the answers are in close proximity to each other. It is important to note that these are approximate values and that they vary depending on the approach and the number of data points used to generate the equations. Some useful analytical derivatives in engineering calculations are provided in the literature.4

2.6 Numerical Integration Numerous engineering and science problems require the solution of integral equations. In a general sense, the problem is to evaluate the function on the right-hand side (RHS) of Equation 2.61. b

∫

I = f ( x ) dx (2.61)



a

where I is the value of the integral. There are two key methods employed in their solution: analytical and numerical. If f(x) is a simple function, it may be integrated analytically. For example, if f(x) = x2, then b



∫

I = x 2dx = a

(

)

1 3 b − a 3 (2.62) 3

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If, however, f(x) is a function too complex to integrate analytically, for example,

( ) (2.63)

I = log  tanh e x 



2

−2

one may resort to any of the many numerical methods available. Two simple numerical integration methods that are commonly employed in environmental and chemical engineering practice are the trapezoidal rule and Simpson’s rule. These are described subsequently. 2.6.1 Trapezoidal Rule In order to use the trapezoidal rule to evaluate the integral I given by Equation 2.61 as b

∫

I = f ( x ) dx (2.61)



a

one may use the equation

I=

h  y0 + 2 y1 + 2 y 2 +  + 2 y n−1 + y n  (2.64) 2

where: h is the incremental change in x; that is, ∆x yi are the values of f(x) at xi, that is, f(xi) In addition,

y0 = f ( x0 ) = f ( x = a ) (2.65)



y n = f ( xn ) = f ( x = b ) (2.66)

b a (2.67) n This method is known as the trapezoidal rule because it approximates the area under the function f(x) – which is generally curved – with a two-point trapezoidal rule calculation. The error associated with this rule is illustrated in Figure 2.2. h=



f(x)

f(x)

f(a)

f(b)

x = Error FIGURE 2.2 Trapezoidal rule error.

= Calculated value

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There is an alternative for improving the accuracy of this calculation – the interval (a, b) can be subdivided into smaller intervals. The trapezoidal rule can then be applied repeatedly in turn over each subdivision. 2.6.2 Simpson’s Rule A higher-degree interpolating polynomial scheme can be employed for more accurate results. One of the more popular integration approaches is Simpson’s rule. For Simpson’s 3-point (or one-third) rule, one may use the equation

I=



h  y a + 4 yb+ a/2 + yb  (2.68) 3

The equation for the general form of Simpson’s rule (where n is an even integer) is



I=

h ( y0 + 4 y1 + 4 y2 +  + 4 yn −1 + yn ) (2.69) 3

This method also generates an error, although it is usually smaller than that associated with the trapezoidal rule. A diagrammatic representation of the error for a 3-point calculation is provided in Figure 2.3. The reader should note that 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 – subject to round-off error – to the analytical solution. The results of each numerical integration must be added together to obtain the final answer for smaller step sizes. Some useful analytical integrals in engineering calculations are provided in the literature.4

f(x)

f(x) Error

f(x0) x0

Simpson

f(x1) x1 = Error

FIGURE 2.3 Simpson’s rule error.

f(x2) x2

x

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2.7 Simultaneous Linear Algebraic Equations The environmental and chemical engineer often encounters problems that contain not only more than two or three simultaneous algebraic equations but also those that can be nonlinear. There is, therefore, an obvious need for systematic methods of solving simultaneous linear and simultaneous nonlinear equations.7 This section addresses the linear sets of equations; information on nonlinear sets is available in the literature.8 Consider the following set of n equations: a11x1 + a12 x2 + … a1n xn = y1 a21x1 + a22 x2 + … a2 n xn = y 2





(2.70)

an1x1 + an 2 x2 + … ann xn = y n where: a is the coefficient of the variable x y is a constant This set is considered to be linear as long as none of the x terms are nonlinear, for exam2 ple, x2 or ln x1; thus, a linear system requires that all terms in x be linear. The previously mentioned system of linear algebraic equations may be set in matrix forms:  a11 a12 … a1n   x1   y1   a a … a  x  y  2n  21 22   2  =  2  (2.71) … … … …  …  …        an1 an 2 … ann   xn   y n  However, it is often more convenient to represent Equation 2.71 in the augmented matrix form provided in the following equation:

 a11 a12 … a1n y1  a a …a y  2n 2  21 22  (2.72) … … … … …    an1 an 2 … ann y n 



Methods of solution available for solving these linear sets of equations include

1. Gauss–Jordan reduction 2. Gauss elimination 3. Gauss–Seidel approach 4. Cramer’s rule 5. Cholesky’s methods

Only methods 1 to 3 are discussed in this section.

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2.7.1 Gauss–Jordan Reduction Carnahan and Wilkes2 provide an example that solved the following two simultaneous equations using the Gauss–Jordan reduction method:

3 x1 + 4 x2 = 29 (2.73)



6 x1 + 10 x2 = 68 (2.74) The four-step procedure is provided here:



1. Divide Equation 2.73 through by the coefficient of x1: x1 +



4 29 x2 = (2.75) 3 3

2. Subtract a suitable multiple, in this case 6, of Equation 2.74 from Equation 2.73, so that x1 is eliminated. Equation 2.75 remains untouched, leaving 34 5 4 29 − x2 + x2 = (2.76) 3 3 3 3



with the corresponding solution 2x2 = 10 (2.77)



3. Divide Equation 2.77 by the coefficient of x2, that is, solve Equation 2.77.



x2 = 5 (2.78)

4. Subtract a suitable factor of Equation 2.78 from Equation 2.75 so that x2 is eliminated. When (4/3)x2 = 20/3 is subtracted from Equation 2.75, one obtains

x1 = 3 (2.79)

2.7.2 Gauss Elimination Gauss elimination is another method used to solve linear sets of equations. This method utilizes the augmented matrix as presented in Equation 2.72. The goal with Gauss elimination is to rearrange the augmented matrix into a triangle form, where all the elements below the diagonal are zero. This is accomplished in much the same way as in Gauss–Jordan reduction. The procedure employed follows. Start with the first equation in the set. This is known as the pivot equation and will not change throughout the procedure. Once the matrix is in triangle form, back substitution can be used to solve for variables.7 Gauss elimination is useful for systems that contain fewer than 30 equations. Systems larger than 30 equations become subject to round-off error where numbers are truncated by computers performing the calculations. This method is also effective to calculate the rank, determinant, and inverse of an invertible matrix.

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2.7.3 Gauss–Seidel Approach Another approach to solving an equation or series/sets of equations is to make an informed or educated guess. If the first assumed value(s) does not work, the value is updated. By carefully noting the influence of these guesses on each variable, one can approach these answers or correct set of values for a system of equations.7 The reader should note that when this type of iterative procedure is employed, a poor initial guess does not prevent the correct solution from ultimately being obtained. Ketter and Prawler3 provide several excellent illustrative examples.

2.8 Nonlinear Algebraic Equations The subject of the solution to a nonlinear algebraic equation is considered in this section. Although several algorithms are available in the literature, the presentation will focus on the Newton–Raphson (NR) method of evaluating the root(s) of a nonlinear algebraic equation. The solution to the equation f ( x ) = 0 (2.80)



is obtained by guessing a value for each x, for example (xold), that will satisfy this equation. This value is continuously updated (xnew) using the equation (the prime represents a derivative) xnew = xold −



f ( xold ) (2.81) f ′ ( xnew )

until either little or no change in (xnew − xold) is obtained. One can also express this operation graphically (see Figure 2.4). Noting that f(x)

Slope – f(xold)

xnew Exact solution since f(x) = 0 FIGURE 2.4 NR method.

xold

df (x) – f ¢ (xold) dx x = xold

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Introduction to Optimization for Chemical and Environmental Engineers

f ′ ( xold ) =

df ( x ) ∆f ( x ) f ( xold ) − 0 = ≈ (2.82) ∆x xold − xnew dx

one may rearrange Equation 2.82 to yield Equation 2.83 below. The xnew then becomes xold in the next calculation. This method is also referred to as Newton’s method of tangents (NMT) and is a widely used method for improving a first approximation to a root to the aforementioned equation of the form f(x) = 0. This development can be rewritten in subscripted form to (perhaps) better accommodate a computer calculation. Thus

f ′ ( xn ) =

f ( xn ) (2.83) x n − x n +1

from which

x n +1 = x n −

f ( xn ) (2.84) f ′ ( xn )

The term xn + 1is again the improved estimate of xn – the previous guess for a solution to the equation f(x) = 0. The value of the function and the value of the derivative of the function are determined at x = x n, and using the proposed procedure, and the new approximation to the root xn + 1 is obtained. The same procedure is repeated, with the new value, 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 ε, which controls the allowable error (or tolerance) in the root. Relative to the previous estimate, ε is given by

ε=

x n +1 − x n (2.85) xn

Despite its popularity, the method suffers for two reasons. First, an analytical expression for the derivative, specifically, f′(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 all cases. However, as is common with some numerical methods, it may fail occasionally in certain instances. A possible initial oscillation followed by a displacement away from a root can occur. Note, however, that the method would have converged if the initial guess had been somewhat closer to the exact root. Thus, the first guess may be critical to the success of the calculation.8,9

2.9 Ordinary Differential Equations The Runge–Kutta (RK) method is one of the most widely used techniques in engineering practice for solving first-order differential equations. For the equation

dy = f ( x , y ) (2.86) dx

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the solution takes the form

y n +1 = y n +

h (D1 + 2D2 + 2D3 + D4 ) (2.87) 6

where

D1 = hf ( x , y )



h D   D2 = hf  xn + , y n + 1   2 2



h D   D3 = hf  xn + , y n + 2   2 2 



D4 = hf ( xn + h, y n + D3 )

(2.88)

The term h represents the increment in x. The term yn is the solution to the equation at yn + 1, and yn + 1 is the solution to the equation at xn + 1 where xn + 1 = xn + h. Thus, the RK method provides a straightforward means for developing expressions for ∆y, namely, xn + 1 −yn, in terms of the function f(x,y) at various “locations” along the interval in question. Consider a simple equation of the form dC = a + bC (2.89) dt

where: t = 0 C = C0

The RK algorithm given previously becomes (for t = h)

C1 = C0 +

h (D1 + 2D2 + 2D3 + D4 ) (2.90) 6

where

D1 = hf ( x , y ) = h ( a + bC0 )



 h D  D    D2 = hf  xn + , y n + 1  = h  a + b  C0 + 1     2 2 2  



 h D    C + D2   D3 = hf  xn + , y n + 2  = h  a + b  0    2 2  2   



D4 = hf ( xn + h, y n + D3 ) = h  a + b (C0 + D3 )

(2.91)

The same procedure is repeated to obtain values for C2 at t = 2h, C3 at t = 3h, and so on.

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The RK method can also be used if the function in question also contains the independent variable.8,9 Consider the following equation: dC = f (C , t ) (2.92) dt

For this situation, one obtains

C1 = C0 +

h (D1 + 2D2 + 2D3 + D4 ) (2.93) 6

with

D1 = hf (C , t )



D h  D2 = hf  C0 + 1 , t0 +  (2.94)  2 2



D  D3 = hf  C0 + 2 , t0 +  2



h  2

D4 = hf (C0 + D3 , t0 + h )

For example, if

dC = 5C − e − Ct (2.95) dt

then

  D1   h     −  C0 + 2   t0 + 2    dC D1     = h  5  C0 +  − e  (2.96) 2 dt   

Situations may arise when there is a need to simultaneously solve more than one ordinary differential (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 = f 1 ( x , y1 , y 2 , … , y n ) dx

dy 2 = f 2 ( x , y1 , y 2 , … , y n ) dx (2.97)  dy n = f n ( x , y1 , y 2 , … , y n ) dx

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Note that the equations in Equation 2.97 are interrelated, that is, they are dependent on each other. This is illustrated in the following two equations:8

dC = Ae − E/RT C = f (C , t ) (2.98) dt



dT ∆H = − kC = g (C , t ) (2.99) dt ρCp

or in a more general sense



dy = f (x, y , z) ; dx dz = g (x, y , z) ; dt

(e.g., xyz ) (2.100)

(e.g., x y e ) (2.101) 2

2 −z

The RK algorithm for Equations 2.100 and 2.101 is

y1 = y 0 +

1 (RY1 + 2RY2 + 3RY3 + RY4 ) (2.102) 6



z1 = z0 +

1 (RZ1 + 2RZ2 + 3RZ3 + RZ4 ) (2.103) 6

where y1 − y0 = ∆y , z1 − z0 = ∆z , h = ∆x and

RY1 = h × f ( x0 , y0 , z0 )



RZ1 = h × f ( x0 , y0 , z0 )



h RY1 RZ1   RY2 = h × f  x0 + , y0 + , z0 +   2 2 2 



h RY1 RZ1   RZ2 = h × f  x0 + , y0 + , z0 +   2 2 2 



h RY2 RZ2   RY3 = h × f  x0 + , y0 + , z0 +   2 2 2 



h RY2 RZ2   RZ3 = h × f  x0 + , y0 + , z0 +   2 2 2 



RY4 = h × f ( x0 + h, y0 + RY3 , z0 + RZ3 )



RZ4 = h × f ( x0 + h, y0 + RY3 , z0 + RZ3 )

(2.104)

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Although the RK approach (and other similar methods) has traditionally been employed to solve first-order ODEs, it can also treat higher ODEs. The procedure requires reducing an nth-order ODE to n first-order ODE. For example, if the equation is of the form8 d2 y = f ( y , x ) (2.105) dx 2

set

z=



dy (2.106) dx

so that dz d 2 y = (2.107) dx dx 2



The second-order equation in Equation 2.107 has now been reduced to the two first-order ODEs provided in Equation 2.108: d 2 y dz = = f ( y , x ) (2.108) dx 2 dx



dy = z dx



The procedure expressed in Equation 2.102 and 2.103 can be applied to generate a solution to Equation 2.105. Note, however, that the first derivative (i.e., dy/dx or its estimate) is required at the start of the calculation. Extending the procedure to higher-order equations is left as an exercise for the reader. The selection of increment size remains a variable to the practicing engineer. Few numerical analysis methods provided in the literature are concerned with error analysis. In general, round-off and numerical errors appear as demonstrated earlier in Figure 2.5. In the limit, when the increment nears 0, one approaches an analytical solution. However, the number of calculations correspondingly increases the error Ɛ, which increases exponentially as the increment nears 0. Note that selecting the increment size that will minimize the error is rarely a problem in practice; in addition, computing it is also rarely a concern.

Roundoff ε Numerical

Increment FIGURE 2.5 Roundoff and numerical error.

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2.10 Partial Differential Equations Many practical problems in engineering practice involve at least two independent variables; thus, 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 differential equations (PDEs). 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 provided with a formal method that will yield exact solutions for all PDEs.9 Fortunately, numerical methods for solving these equations were developed during the mid-to late 20th century, and they are routinely used today. The three main PDEs encountered in practice are briefly introduced here, employing T (e.g., the temperature is the dependent variable) with t (time) and x, y, z (position) as the independent variables (note that any dependent variable, such as pressure or concentration, could have been selected). The parabolic equation ∂ 2T ∂T = α 2 (2.109) ∂t ∂z

The elliptical equation:

∂ 2T ∂ 2T = 0 (2.110) + ∂x 2 ∂y 2

The hyperbolic equation:

∂ 2T ∂ 2T = α (2.111) ∂t 2 ∂x 2



The preferred numerical method of solution involves finite differencing. Only the parabolic and elliptical equations are considered subsequently. 2.10.1 Parabolic PDE Examples of parabolic PDEs include

∂ 2T ∂T = α 2 (2.112) ∂t ∂x

and (the two-dimensional)

 ∂ 2T ∂ 2T  ∂T = α  2 + 2  (2.113) ∂t ∂y   ∂x

Ketter and Prawler,3 as well as many others, have reviewed the finite difference approach to solving Equation 2.112. This is detailed subsequently.

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Introduction to Optimization for Chemical and Environmental Engineers

Consider the (t, x) grid provided in Figure 2.6. The partial derivatives may be replaced by ∂T ∆T −T4 + T2 −T4 + T2 ≅ = ; ∆t = k (2.114) = ∂t ∆t 2 ( ∆t ) 2k

and

∂T 2 ∆  ∆T  T3 − 2T0 − T1 ; ∆x = h (2.115)  = 2 ≅ ∆x  ∆x  h2 ∂x



Substituting Equations 2.114 and 2.115 into Equation 2.112 leads to −T4 + T2 T3 − 2T0 − T1 = (2.116) 2k h2

Solving for T2:

T2 = T4 + 2r (T3 − 2T0 − T1 ) (2.117)



where r = k/h2. Thus, one may calculate T2 if T0, T1, T3, and T4 are known. Unfortunately, stability and error problems arise in employing the previously mentioned approach. These can be removed by replacing the central difference term in Equation 2.114 by a forward difference term: ∂T ∆T −T0 + T2 −T0 + T2 ≅ = = (2.118) ∂t ∆t k ( ∆t )



With this substitution, Equation 2.117 becomes T2 = T0 + r (T3 − 2T0 + T1 ) (2.119)



It can be shown that the problem associated with the central difference derivative is removed if r ≤ 0.5. T2 ∆t = k T3

T0

T1

T4 ∆x = h FIGURE 2.6 Two variable grid.

t

x

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Mathematical Operations

2.10.2 Elliptical PDE For this equation, examine the grid in Figure 2.7. Using finite differences to replace the derivatives in Equation 2.110 ultimately leads to 1 (T1 + T2 + T3 + T4 ) ; ∆x = ∆y (2.120) 4 In effect, each T value calculated reduces to the average of its four nearest neighbors in the square grid. This difference equation may then be written at each interior grid point, resulting in a linear system of N equations, where N is the number of grid points. The system can then be solved by one of several methods provided in the literature.9,10 Another solution method involves applying the Monte Carlo approach, requiring the use of random numbers.9,10 Consider the squares shown in Figure 2.8. If the describing equation for the variation of T within the grid structure is T0 =



∂ 2T ∂ 2T + = 0 (2.110) ∂x 2 ∂y 2



T2 y

∆y T3

T0

x

T1

T4 ∆x FIGURE 2.7 Elliptical grid. y y= a

y= 0 x= 0 FIGURE 2.8 Monte Carlo approach.

7

8

9

6

5

4

1

2

3

x =a

x

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Introduction to Optimization for Chemical and Environmental Engineers

with specified boundary conditions (BCs) for T(x, y) of T(0, y), T(a, y), T(x, 0), and T(x, a), one may employ the following approach:





1. Proceed to calculate T at point 1 (i.e., T1). 2. Generate a random number between 00 and 99. 3. If the number is between 00 and 24, move to the left. For 25 to 49, 50 to 74, 75 to 99, move upward, to the right, and downward, respectively. 4. If the move in step 3 results in a new position that is at the outer surface (boundary), terminate the first calculation for point 1 and record the T value of the boundary at that new position. However, if the move results in a new position that is not at a boundary and is still at one of the other eight interval grid points, repeat steps 2 and 3. This process is continued until an outer surface boundary is reached. 5. Repeat steps 2 to 4 numerous times, for example, 1000 times. 6. After completing step 5, sum all the T values obtained and divide this value by the number of times steps (2 to 4) have been repeated. The resulting value provides a reasonable estimate of T1. 7. Return to step 1 and repeat the calculation for the remaining eight grid points.

This method of solution is not limited to square systems. In addition, one of the authors11 has applied this method of solution to numerous real-world applications.

References

1. S. Schaefer and L. Theodore, Probability and Statistics Applications for Environmental Science, CRC press/Taylor & Francis Group, Boca Raton, FL, 2007. 2. B. Carnahan and J. Wilkes, Digital Computing and Numerical Methods, John Wiley & Sons, Hoboken, NJ, 1973. 3. R. Ketter and S. Prawler, Modern Methods of Engineering Computations, McGraw‑Hill, New York City, NY, 1973. 4. J. Reynolds, J. Jeris, and L. Theodore, Handbook of Chemical and Environmental Engineering Calculations, John Wiley & Sons, Hoboken, NJ, 2004. 5. L. Perez, homework assignment submitted to L. Theodore, Manhattan College, Bronx, NY, 2003. 6. F. Lavery, The Perils of Differentiating Engineering Data Numerically, Chem Eng, New York City, NY, 1979 7. L. Theodore and C. Prochaska, Introduction to Mathematical Methods for Environmental Engineers and Scientists, Wiley‑Scrivener, Salem, MA, 2018. 8. L. Theodore, class notes, Manhattan College, Bronx, NY, 1971. 9. L. Theodore, personal notes, East Williston, NY, 1965. 10. L. Theodore, Heat Transfer for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2011. 11. L. Theodore, personal notes, East Williston, NY, 1985.

3 Perturbation Techniques A significant number of optimization problems face the environmental and chemical engineer. The optimal design of industrial processes, as well as process equipment, has long been of concern to the practicing engineer, and indeed, for some, might be regarded as a definition of the function and goal of applied engineering. The attainment of an optimum design is generally a result of factors that include mathematical analysis, empirical information, and both the subjective and objective experience of the environmental and/or chemical engineer. Regarding design, formal optimization techniques have as their goal the development of procedures for the attainment of an optimum in a system that can be characterized mathematically. The mathematical characterization may be (1) partial or complete, (2) approximate or exact, and/or (3) empirical or theoretical. The resulting optimum may be a final implementable design or a guide to practical design and a criterion for judging practical designs. In either case, the optimization techniques should serve as an important part of the total effort in the design of the units, structures, and control of not only equipment but also industrial system processes. One simple procedure for solving optimization problems that is recommended by the authors is a perturbation scheme. This involves a systematic change of variables, one by one, in an attempt to locate an optimum. To be practical, this can mean that one must limit the number of variables by assuming values to those (process) variables that are known beforehand to play an insignificant role. Reasonable guesses or simple or shortcut mathematical methods can further simplify the procedure. Much information can be gathered from this type of study since it usually identifies those variables that significantly impact the solution, such as on the overall performance of equipment or process, and it helps identify the major contributors affecting the optimization calculations.1 What does the previous paragraph mean? If the relationship between the independent variables and the dependent variable cannot be solved graphically or analytically, some form of trial-and-error calculation may be employed to arrive at the optimum. For example, a sequence of values over a range of the independent variables may be calculated to completely map out the region of interest. Another scheme is to vary one factor at a time, holding the other(s) constant. These two schemes are outlined in this chapter in the three sections that follow. Finally, some in the technical community might malign this approach of solving optimization problems. Certainly, linear programming, which is addressed in the last two chapters in this Part, has its place in mathematical methods, but so does the perturbation approach. The readers should keep this in mind when there is a need to solve an engineering problem that involves optimization. In fact, one of the illustrative example applications in Part III involves the application of the Monte Carlo method,1 which some might argue requires a trial-and-error perturbation approach.

41

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This chapter contains three sections that address an optimization problem using the perturbation approach when there is

3.1 One independent variable 3.2 Two independent variables 3.3 Three independent variables

3.1 One Independent Variable This first section introduces the reader to the perturbation method and how it applies to an equation involving one dependent variable, for example, y, and one independent variable, for example, x. One could therefore write y = f ( x ) (3.1)

Three simple examples follow. Illustrative Example 3.1

Suppose one is interested in obtaining the maximum and/or minimum of the formula y = 100 (1 − x ) (3.2) 2

over the range (constraint)

0 ≤ x ≤ 1 (3.3)

SOLUTION:

To apply the perturbation method of solution, one could calculate y at 0.1 intervals of x between 0 and 1. The results are provided in Table 3.1. TABLE 3.1 Perturbation Method for y = 100(1 −x)2 y 100 81 64 49 36 25 16 9 4 1 0.0

x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

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Perturbation Techniques

There is both a maximum and a minimum over the 0.0–0.1 range of x. As expected, the maximum occurs at x = 0 with y = 100

and the minimum occurs at

x = 1.0 with y = 0.0



Illustrative Example 3.2 Consider the following function f ( x ) = x 3 − 2x 2 − 5x + 6



Calculate the minimum of the function over the range 0 ≤ x ≤ 4 using a perturbation technique. SOLUTION: Calculate f(x) in the range (0, 4) in increments of 0.4 for x. Results are presented in Table 3.2. Table 3.2 suggests the minimum is located approximately at x = 2.0 with a value of 4.0. TABLE 3.2 Calculations for Illustrative Example 3.2. f(x)

x

6.0 3.74 1.23 −1.15 −3.02 −4.0 −3.7 −1.73 2.29 8.74 18.0

0.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0

Illustrative Example 3.3 Consider the following function



y = 162 − 2.33 x −

11, 900 x

Locate the value of x that corresponds to the minimum value of y. Once again employ the perturbation technique.

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TABLE 3.3 Calculations for Illustrative Example 3.3 y

x

60 62 64 66 68 70 72 74 76 78 80

−176.1 −174.4 −173.1 −172.1 −171.4 −171.1 −171.1 −171.2 −171.7 −172.3 −173.2

SOLUTION: The following results if one were to neglect the last term in y: y = 162 − 2.33 x

Arbitrarily, set y = 0 and solve for x.

x=



162 ≈ 70 2.33

At this value of x, y = 162 − 2.33 x −



11, 900 70

y = 171.1



Generate values for x that straddle the value of 70 over the range 60–80 in increments of 2. Results are provided in Table 3.3. It appears the minimum is indeed around 70. Additional calculations could further refine these results. For example, for x = 71, y = −171.0. This example will be revisited in Chapter 6. Illustrative Example 3.4 Obtain a value for x2 that will minimize the function



 k −1   k −1         x2   k   x3   k   y = f ( x2 ) =   +  − 2 ;  x1    x2   

if , x1 = 1, x3 = 10, and k = 1.4.

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Perturbation Techniques

TABLE 3.4 Results of Illustrative Example 3.4 y

x2

2 3 4 5 6 7 8 9

0.800 0.777 0.780 0.800 0.814 0.814 0.875 0.901

SOLUTION: Set x1 = 1, x3 = 10, and k = 1.4 in the previous equation.

 x 0.285   10 0.285   y ( x2 ) =  2 + − 2     x2   1 

Since 1 ≤ x ≤ 10, select value of x2 in the interval (2, 9) in increments of 1.0 and calculate y. The results are provided in Table 3.4. The results of Table 3.4 suggest that the minimum of y is approximately 0.777 and occurs at x2 ≅ 3.0. The reader may check this for x2 = 3.5 where y = 0.777.

3.2 Two Independent Variables This section presents four examples that illustrate the method of solution when there are two independent variables. These examples introduce the reader to the concept of constraints. Illustrative Example 3.5 Obtain the maximum and/or minimum for the following function

y = 3 x1 + 5x2



0 ≤ x1 ≤ 3, 0 ≤ x2 ≤ 4

SOLUTION: Proceed as illustrated below. The function can be calculated at the 20 points shown in parenthesis in Table 3.5. The value of the Table 3.5 point locations for Illustrative Example 3.5 are shown immediately to the right of the point in parenthesis in Table 3.6. As expected, the optimum y is 29 and occurs at x1 = 3 and x2 = 4, and the minimum y is located at x1 = 0 and x2 = 0 with a minimum value of 0.

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Introduction to Optimization for Chemical and Environmental Engineers

TABLE 3.5 Optimization Calculations for Illustrative Example 3.5 (0, 0) (0, 1) (0, 2) (0, 3) (0, 4)

(1, 0) (1, 1) (1, 2) (1, 3) (1, 4)

(2, 0) (2, 1) (2, 2) (2, 3) (2, 4)

(3, 0) (3, 1) (3, 2) (3, 3) (3, 4)

TABLE 3.6 Function Values for Illustrative Example 3.5 (0, 0) (0, 1) (0, 2) (0, 3) (0, 4)

0 5 10 15 20

(1, 0) (1, 1) (1, 2) (1, 3) (1, 4)

3 8 13 18 23

(2, 0) (2, 1) (2, 2) (2, 3) (2, 4)

6 11 16 21 26

(3, 0) (3, 1) (3, 2) (3, 3) (3, 4)

9 14 19 24 29

A modified form of the previous equation will be revisited in Chapter 7 (Illustrative Example 7.2) when a graphical solution is provided that includes a plot of x1 vs.x2 for various values of y. Illustrative Example 3.6 Obtain the maximum and minimum for the following objective function with the same constraints employed in the previous example. E = 3 x1 + 5x2 + x12 + 2x2 2 − 5x1x2

SOLUTION:

The values for 20 points appear in Table 3.7. The above function exhibits both a maximum and a minimum over the specified range for x1 and x2. For this example, the minimum is located at (0, 0) with a value of 0, and the maximum is located at the point (0, 4) with a value of 52. This problem could also have been solved analytically; the solution is provided in Chapter 6. Illustrative Example 3.72 Employing the perturbation approach, obtain the maximum value of the function y = 162 − 2.33 x1 −



11, 900 − 1.86 x1 x1x2

TABLE 3.7 Calculations for Illustrative Example 3.6 (0, 0) (0, 1) (0, 2) (0, 3) (0, 4)

0 7 18 33 52

(1, 0) (1, 1) (1, 2) (1, 3) (1, 4)

4 6 12 22 36

(2, 0) (2, 1) (2, 2) (2, 3) (2, 4)

10 7 8 13 22

(3, 0) (3, 1) (3, 2) (3, 3) (3, 4)

18 10 6 6 10

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Perturbation Techniques

Earlier calculations suggest that 0 ≤ x1 ≤ 30, 0 ≤ x2 ≤ 30

SOLUTION:

The calculations for this example are presented in Table 3.8. It appears that the maximum value of y is approximately 80.4 and is located at x1 = 10 and x2 = 30. This problem will also be revisited in Chapter 6. Illustrative Example 3.8 Obtain values for x1 and x2 that minimize the subsequent function below employing a perturbation approach. Both x1 and x2 are constrained in the range (1, 10) with x2 > x1.  x  0.285  x2  0.285  10  0.285  y ( x1 , x2 ) =  1  +  −  − 3  x1   x2   1  



SOLUTION: Since x2 > x1, initially assume increments of 1 for x1 and x2 in the range of (2, 5) and (4, 9), respectively. Calculated values for y for various combinations of x1 and x2 are provided in Table 3.9. The results in Table 3.9 suggest the minimum occurs in the neighborhood of x1 = 2.0 and x2 = 4.0. What further calculation would the reader suggest? One of the authors followed the previous calculations with two additional calculations, the results of which are presented here:

x1 = 1.5, x2 = 4.0; y = 0.743



x1 = 2.0, x2 = 3.5; y = 0.739

These two latter values for y further implies that the minimum is indeed in the range of x1 = 2.0 and x2 = 4.0. The reader is referred to both Chapters 4, 5, 6, and 7 and Part III – Fluid Flow, Chapters 13. The reader will soon discover that the analytical answer is x1 = 2.154, x2 = 4.642; y = 0.644



TABLE 3.8 Calculation Chart for Illustrative Example 3.7 x2 0 5 10 15 20 25 30

x1 0

5

10

15

20

25

30

−∞ −∞ −∞ −∞ −∞ −∞ −∞

−∞ −335.0 −97.0 −17.6 +22.1 +45.9 +61.7

−∞ −117.9 +1.1 +40.8 +60.6 +72.5 +80.4

−∞ −59.5 +19.8 +46.3 +59.5 +67.4 +72.7

−∞ −40.8 +18.7 +38.5 +48.5 +54.4 +58.4

−∞ −38.0 +9.7 +25.5 +33.5 +38.2 +41.4

−∞ −43.0 −3.4 +9.9 +16.5 +20.4 +23.1

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Introduction to Optimization for Chemical and Environmental Engineers

TABLE 3.9 Calculation Chart for y(x1, x2) for Illustrative Example 3.8 x2 2 3 4 5

x1 4

5

6

7

8

9

0.654 0.671 − −

0.734 0.742 0.768 −

0.743 0.742 0.762 0.792

0.754 0.748 0.765 0.790

0.810 0.798 0.810 0.791

0.784 0.766 0.775 0.794

3.3 Three Independent Variables This section presents one illustrative example involving three independent variables. Illustrative Example 3.9 Consider the following function: y = 3 x1 − 5x2 − 8 x3



Assume the same constraints as the previous two examples for x1 and x2; the constraint for x3 is 1 ≤ x3 ≤ 2



Obtain the maximum and minimum of y. SOLUTION: Apply the perturbation method again as shown below. See Table 3.10 for x3 = 1: TABLE 3.10 Calculations for x3 = 1 for Illustrative Example 3.9 (0, 0) (0, 1) (0, 2) (0, 3) (0, 4)

−8 −13 −18 −23 −28

(1, 0) (1, 1) (1, 2) (1, 3) (1, 4)

−5 −10 −15 −20 −25

(2, 0) (2, 1) (2, 2) (2, 3) (2, 4)

−2 −7 −12 −17 −22

(3, 0) (3, 1) (3, 2) (3, 3) (3, 4)

1 −4 −9 −14 −19

(3, 0) (3, 1) (3, 2) (3, 3) (3, 4)

−1 −6 −11 −16 −21

For x3 = 2, refer to Table 3.11: TABLE 3.11 Calculations for x3 = 2 for Illustrative Example 3.9 (0, 0) (0, 1) (0, 2) (0, 3) (0, 4)

−16 −21 −26 −31 −36

(1, 0) (1, 1) (1, 2) (1, 3) (1, 4)

−13 −18 −23 −28 −33

(2, 0) (2, 1) (2, 2) (2, 3) (2, 4)

−10 −15 −20 −25 −30

Perturbation Techniques

49

Based on the these two trials for x3, the maximum is located at (3, 0) and the minimum value is located at (0, 4) with corresponding y values of 1 and −36, respectively. The previous solution could be further refined by performing calculations at x3 = 1.25, 1.5, and 1.75.

References

1. L. Theodore, Chemical Engineering: The Essential Reference, McGraw-Hill, New York City, NY, 2014. 2. L. Theodore, personal notes, Theodore Tutorials, East Williston, NY, 2015.

4 Search Methods The mathematicians’ literature abounds with search methods, that is, methods that allow one to search for either a solution to an equation or the optimum value for an equation. There are both indirect and direct approaches. However, direct methods – those that produce an exact answer – are generally preferred by environmental and chemical engineers. These methods have also been defined in the literature as seek methods, and usually apply to single-independent variable unimodal equations; they provide results in intervals (or limits) of uncertainty for the variable in a (minimum) number of trials during the search process. The unimodal function referred to for single-independent variable equations contains a single optimization (maximum or minimum) point over a specified interval/range. Two of the procedures that are reviewed in this chapter are the interval halving (or bisection) and golden section search methods. These two methods can also be employed to obtain the solution to some equations. For example, the interval halving method depends on finding an approximation to a solution of the form f(x) = 0. Initial guesses x0 and x1 are sequenced provided that f(x) is continuous for x0 ≤ x ≤ x1 and the product [(  f(x0))(  f(x1))]      2 2  ∂x2   ∂x3  ∂x2∂x3

Therefore, the previously calculated y is a minimum.



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Analytical Methods

6.5 Three Independent Variables This section presents one illustrative example involving three independent variables. Illustrative Example 6.8 Consider now a traditional mathematician’s exercise to determine the maximum or minimum of a function dependent on three variables.

P = 8 x + 6 y + 8 z + xy + xz + yz − 2x 2 − 3 y 2 − 4 z 2

SOLUTION:

∂P = 8 + y + z − 4x = 0 ∂x



∂P = 6 + x + z − 6y = 0 ∂y



∂P = 8 + y + x − 8z = 0 ∂z

The solution to the above three linear simultaneous equations is

x = −2.900, y = 1.794, z = 1.806

for which

P = 78.07

To determine whether the above P is a maximum or minimum, calculate the following three second-order partial derivatives.

∂ 2P = −4 ∂x 2



∂ 2P = −6 ∂y 2



∂P = −8 ∂z 2

Since the three derivatives are negative, the value of P of 78.07 is a maximum; that is, any other calculated P for a different x, y, and z would produce a value of P less than 78.07. The “proof” is left as an exercise for the reader. However, if one were to select values of x, y, and z straddling the maximum set of values, P should be below 78.07. For example, the value of P for x = 2, y = 2, and z = 2 is −20. For x = y = z = 10, the value of P is −380.

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References

1. L. Theodore, Chemical Engineering: The Essential Reference, McGraw-Hill, New York City, NY, 2014. 2. J. Happel, Chemical Process Economics, John Wiley & Sons, Hoboken, NJ, 1958.

7 Linear Programming Linear programming is one of several techniques for optimization that has increased in usefulness because of the availability of digital computers. The title results from the assumptions in the method that linear relationships describe the problem/system/relationship under consideration. For example, each production element increase per day in a process could produce a corresponding linear profit increase per day. However, there are many instances where a linear model does not accurately describe the totality of the problem/system/relationship. It may then be necessary to resort to other procedures such as nonlinear programming; some brief details will be presented for handling the nonlinear situation. The definitions associated with some of these terms appear earlier in this Part. The reader should note that the terms linear programming and nonlinear programming are essentially similar from an applications perspective. There were problems around the middle of the last century because of some difficulties that arose in attempting to solve nonlinear programming problems. However, the arrival of the modern-day computer (see Chapter 1) and sophisticated software (e.g., Excel) have removed these problems. Unless a solution is presented graphically or algebraically, it will be obtained directly from Excel so that the reader need not be concerned with whether a system’s pertinent describing equations and constraints are linear or nonlinear. Thus, details regarding the solution methodology for both linear programming and nonlinear programming problems as well as for some of the illustrative examples will not be presented. They are simply not necessary and beyond the scope of this introductory optimization text. However, illustrative examples involving traditional optimization will address both the development of the describing equations, that is, the model, and the solution to the problem. Other than linear programming, dynamic programming is another often used optimization method by the practicing environmental and chemical engineer. As described earlier, it is a process involving decisions made at multiple stages. Just as an optimization model is defined by its variables, constraints, and objective function, a multistage decision process is defined by its stages and decisions. At any stage, there is often enough information about decisions made at preceding stages to make informed decisions at the present stage. Thus, some optimization problems can be reformulated as multistage decision process problems. One then proceeds through subsequent stages until the final stage of the process is reached. Some additional details are provided in the next section. As one might suppose, there are numerous other optimization methods available to the practicing environmental engineer and scientist. Several of these are detailed in Perry’s Handbook:1

1. Mixed integer 2. Mixed integer linear programming 3. Mixed integer nonlinear programming 4. Quadrative programming 5. Differential/nondifferentiable systems 6. Convex/nonconvex 95

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The previously mentioned topics are not reviewed in this text. The reader is referred to the optimization literature2–6 for details. Five sections compliment the presentation for this chapter:

7.1 Definitions 7.2 Basic concepts of optimization 7.3 Applied mathematical concepts in linear programming 7.4 Applied engineering concepts in linear programming 7.5 Other real-world applied concepts in linear programming

7.1 Definitions Seven key definitions – as they apply to optimization and linear programming are provided below (in alphabetical order). 1. Algebra: A branch of mathematics in which letters are used to represent basic arithmetic relations. As in arithmetic, the basic operations of algebra are addition, subtraction, multiplication, and division. Arithmetic, however, cannot generalize such mathematical relations as the Pythagorean theorem (which states that the sum of the squares of the sides of any right triangle is also a square). Arithmetic can produce specific instances of this relations (e.g., 3, 4, and 5, where 32 + 42 = 52). Algebra can make a purely general statement that fulfills the conditions of the theorem, that is, a2 + b2 = c2. Any number multiplied by itself is termed squared and is indicated by a superscript number 2. For example, 3 × 3 is noted as 32 and is termed three squared; similarly, a × a is equivalent to a2. Classical algebra, which is concerned with solving equations, uses symbols instead of specific numbers and employs arithmetic operations to establish procedures for manipulating symbols, while modern algebra has evolved from classical algebra by placing more attention to the structures within mathematics. 2. Constraint: “A constraining or being constrained…confinement or restrictions… forced…something that constrains.” The term constraint is also defined as “to bond together, draw together…to force into or hold in, close bounds, confine…restrain.” 3. Dynamic programming: The dimensionality of a particular operation is reduced from (m)(n) variables to m repeated n times. Thus, if it is not practical or possible to solve a large complex system of equations, the system is reduced in complexity in a manner that requires the solution of a smaller subset of the original problem, noting that for each n (e.g., each stage in a mass transfer operation)7, 8 only m variables are involved in the optimization process. One can then work backward and combine the solutions generated at each n. 4. Linear algebra: A branch of mathematics that is concerned with systems of linear equations, linear transformations, vectors, vector spaces, and related topics. The major interest is with linear equations. 5. Linear programming: A mathematical and operations research technique used in engineering, science, administrative engineering, and economic planning to

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maximize the linear function(s) of a large number of variable(s) subject to certain constraint(s). It is basically used to find a set of values, chosen from a prescribed set of numbers, that will maximize or minimize, that is, optimize, a given equation. 6. Operations research: A confusing term that most often has been used to describe linear programming. 7. Optimization: As noted earlier in this Part, optimization has come to mean different things to different people. Its definition has also varied in the literature. In its most elementary and basic form, optimization is concerned with the determination of the optimum solution to a given problem. This process is required in the solution of many problems in environmental and chemical engineering that involve the maximization or minimization of a given function(s). Some specific examples include a. Environmental engineering research b. Applied science c. In the selection of control variables to accomplish a desired operation d. In the scheduling of a series of operations or events to control dates of a given project e. The development of optimal layouts of organizational units within a given design space, and so on.7

7.2 Basic Concepts of Optimization The term linear programming defines a particular class of optimization problems in which the constraints of the system can be expressed as linear equations or inequalities and the function is linear with respect to the independent design variable(s). These techniques are widely used to solve a number of engineering, environmental, chemical, structural, economic, and societal problems. The primary reasons for its wide use today are the availability of a host of commercial software to solve complex problems and the ease with which any data variation that could involve sensitivity analysis can be handled. As noted in the earlier chapters in this Part, the optimization problem in which the performance measure is a function of one variable is the most elementary type of optimization problem since it is a type of problem that the environmental and chemical engineer commonly encounters in practice. It also allows single-variable optimization to be employed as a sub-problem within iterative procedures for solving multivariable optimization problems. Constraints, constraints, constraints. Constraints continue to be imposed on individuals, organizations, groups by governments, religions, relatives, friends, and so on. It seems to be getting worse each year. And, one of the hallmarks of linear programming is the inclusion of constraints in the problem, that is, the objective function. Linear programming involves linear constraints. For example, the function

f ( x ) = x 4 + 3 x 2 − 2x + 5; all values of x (7.1)

is an unconstrained function. However, the function

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f ( x ) = x 6 − 4 x + 6; − 3 ≤ x ≤ 6 (7.2)

is a constrained function. When constraints are present, the function being optimized is usually referred to as the objective function to distinguish it from any other functions that may be used to define the constraints or the problem. Although the emphasis to follow is on linear constraints, many real-world applications also involve nonlinear constraints and nonlinear objective functions. To further complicate the problem, the constraints can be of an equality or monoequality/inequality nature. From a basic perspective, the problem could be to optimize the objective function

y = f ( x1 , x2 ) (7.3)

The constraints can include

g ( x1 , x2 ) < 0 (7.4)



x1 ≥ 0 (7.5)



x2 ≥ 0 (7.6)

This is a simple yet typical problem in optimization – the maximization or minimization of a real-value function of a number of real variables (occasionally just a single variable) subject to a number of constraints (occasionally the number is zero). However, the mathematical model normally contains several (n, e.g.) variables known as the independent design variables denoted as x1, x2, x3, … xn. This set of variables has been described by some as a vector, x. One may view each x as one dimension in the set (space) of variables, and any particular set of variables, and any particular set of values for these variables in this system as a solution. One of the more important recent areas for the application of mathematical optimization techniques is in engineering and science. Many problems require an iterative and/or trial-and-error solution and the problem of concern is often stated in an ambiguous and open-ended way. The first task is to decide what the problem is and to identify the requirements or specifications for any proposed and/or feasible solution. One then generates a concept, perhaps in the form of a rough configuration, for the object, system, or process to be optimized. This phase of the problem is usually the most difficult and can require the greatest ingenuity and engineering judgment. The next step attempts to model mathematically the object, system, or process. This model may assume a wide variety of forms, but the usually preferred ones are mathematical in nature. One may then use many mathematical, scientific, and engineering tools. These can involve not only the most empirical of data correlation but also the most rigorous mathematical procedures to the least formal cut-and-try methods for assisting in the analysis of this aspect of the solution. As noted above, it may happen that certain functional equality constraints also apply, for example, f1 ( x1 , x2 , x3 ,… , xn ) = 0

 f k ( x1 , x2 , x3 ,… , xn ) = 0

There may also be certain inequality constraints such as

(7.7)

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Linear Programming

g1 ( x1 , x2 , x3 ,… , xn ) < 0



(7.8)

g m ( x1 , x2 , x3 ,… , xn ) < 0 that are specified on the space so that only points in a portion of the total space may be considered as acceptable values for the solution of the problem. This region is defined as the feasible space and points in the region are defined as feasible solutions and/or values. The previously mentioned equality constraints – Equation 7.7 – come from functional relationships that must be satisfied among the aforementioned n variables. If there is to be a choice in the values for which at least some of the variables may be assumed, then k must be smaller than n. The inequality constraints in Equation 7.8 usually arise from some specified limitations (maximum permissible stress, minimum emissions, minimum allowable temperatures, etc.). Note that there is no upper limit on the value of m. Additional information is provided in Section 7.4. Summarizing, the optimization problem to be considered is that of minimizing (minimizing should be italicized) f(x) subject to certain constraints. The term f(x), is a real-valued function called the objective function (or cost function). A vector x is an n-vector of independent variables x1, x2, …, xn. These variables are often referred to as decision variables. The problems require finding the “best” values of the vector x over all the possible values of x. The “best” value refers to the one that results in the smallest value of the aforementioned objective function. Optimization problems can also involve maximization of the objective function. This can be represented similarly to the minimization presentation above since maximizing f(x) is equivalent to minimizing f(x). Therefore, one can apply the presented to the minimization problem without concern. Finally, the following comments are provided on the objective function. It may be quite simple and easy to determine, or complicated and difficult to calculate. The objective function may also be very elusive due to the presence of conflicting or dimensionally incompatible sub-objectives; for example, one might be asked to operate an absorber7 at minimum cost and that it is aesthetically appealing. In some cases, it may not be possible to find a quantitative objective function. In this case, one can assume that an alternative suitable objective function can be formulated.

7.3 Applied Mathematical Concepts in Linear Programming As noted earlier, linear programming deals with the determination of an optimum solution of a problem expressed in linear relationships where there may be a large number of possible solutions. In general, the methods of solution employed are of a trial-and-error nature, where the procedure systemically follows a mathematical treatment that minimizes one’s labor involved and ensures that a correct result has been obtained. The following graphical approach to a linear programming problem illustrates some of the mathematics employed in the solution. Consider now an example that involves minimizing an objective variable E (e.g., expenditure or emission)

E = a1x1 + a2 x2 (7.9)

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subject to two constraints

B ≤ b1x1 + b2 x2 (7.10)



C ≤ c1x1 + c2 x2 (7.11)

In addition,

x1 ≥ 0 (7.12)



x2 ≥ 0 (7.13)

In terms of a solution, first examine the previous system of equations analytically. The problem can be viewed in Figures (7.1) and (7.2). First, note that only the first quadrant need be considered since Equations 7.12 and 7.13 must hold. Equations 7.10 and 7.11 are rewritten in the following form:

x2 ≤

B  b1  − x1 (7.14) b2  b2 

x2 ≤

C  c1  − x1 (7.15) c2  c2 

and x2

x2

x2

Equation (7.14)

Equation (7.15)

x1

(a)

x3

(b)

x1

x

(c) FIGURE  7.1 Two-variable minimization problems.

x2

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Linear Programming

x2

E*

x2 a2 x2 + a2 x1 a 1 x 1+ = E 1 =a 1 E2

x2

=

E

3

=

a

1x 1

a

2x 2

x1

(a)

x2

E1 E 2 E* = E3

x2

+

(b)

x1

x2

FIGURE  7.2 Solution to a two-variable minimization problem.

It is apparent that the only values of x1 and x 2 that will satisfy Equations 7.14 and 7.15 lie above the solid lines (the cross-hashed area) in Figure 7.1. This shaded area is termed the constraint set. Now place in Figure 7.2 a series of straight lines (or contours) corresponding to Equation 7.9 with a different value of E used for each line (see Figure 7.2a). Of all these lines, the smallest E for which x1 and x 2 remain in the specified region will be that minimum E (see Figure 7.2b), i.e., E3. This will yield that proportion of x1 and x 2 that minimizes the objective function. Stated in other words, one desires the lowestvalue contour having some point in common with the constraint set. For the particular value of slope chosen for Equation 7.9, the minimum point is seen to correspond to E where

E3 = Es* (7.16)

The point that represents the minimum is therefore determined by the constraints a1 and a2 of the problem. Further, if the constants c1 and c2 in Equation 7.11 are changed, it is possible that the minimum point would occur elsewhere. The situation is essentially reserved for a maximization problem, that is, for a profit P. This is demonstrated in Figure (7.3). It should be pointed out that if the objective function – Equation 7.9 – was nonlinear, the contour lines would be curved, and if the constraint equations were nonlinear, the constraint lines would not be straight. It should be intuitively obvious that locating the minimum point graphically is more complicated in this case.

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x2

=P P3 *

x2 *

P3 P2 P1

x1

x1

FIGURE  7.3 Possible solution to a two-variable maximization problem.

The previously mentioned methodology can be extended to more than two variables and two constraints. The following would apply for three variables and four constraints:

y = f ( x1 , x2 , x3 ) = a1x1 + a2 x2 + a3 x3 (7.17)

and

B = b1x1 + b2 x2 + b3 x3 (7.18)



C = c1x1 + c2 x2 + c3 x3 (7.19)



D = d1x1 + d2 x2 + d3 x3 (7.20)



E = e1x1 + e2 x2 + e3 x3 (7.21)

and

x1 ≥ 0 (7.22)



x2 ≥ 0 (7.23)



x3 ≥ 0 (7.24)

Naturally, the previous can be rewritten for n objective functions and m constraints.

7.4 Applied Engineering Concepts in Linear Programming An engineer’s first task in problem-solving is to decide what the problem is and to identify the requirements or specifications for any proposed and/or feasible solution. One then generates a concept, perhaps in the form of a rough configuration, for the object, or process

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Pressure, P

to be optimized. As noted previously, this phase of the problem is often the most difficult and requires the greatest ingenuity and engineering judgment. The next step attempts to model the object, system, or process. This model may assume a wide variety of forms, but the usually preferred ones are mathematical in nature. One may then use many mathematical, scientific, and engineering tools. As noted earlier, these can involve not only the most empirical of data correlations but also the most rigorous mathematical procedures or formal cut-and-try methods for assisting in the analysis of this aspect of the solution. Regarding constraints, Figure (7.4) illustrates a simple situation involving the operation of a control equipment absorber7–10 with temperature and pressure limits for which each depends on the magnitude of the other. The former might come from limitations due to equilibrium considerations, since “the higher the temperature in the absorber the lower the recovery absorptive capacity of the absorbing liquid while the higher the pressure in the absorber increases the absorption.”8–10 The region of feasible solutions for the absorber problem is the region enclosed by the constraints. The optimization problem is to find the one set of variables or point in the feasible solution space that corresponds to the best solution. In order to locate the best feasible solution of the potentially infinite number of feasible solutions using a mathematical optimization technique, it is essential that a meaningful objective function f(x) of the independent design variables be specified for this operation concerned with the control and/or recovery of a particular component of gaseous emission. For simplicity, assume that one must choose for the aforementioned two independent variables – temperature T and pressure P. The space is, therefore, two-dimensional, with temperature and pressure as the coordinates. In practice, there usually are certain inequality constraints placed upon the range on both T and P. As shown in Figure 7.4, there could be lower limits for the temperature and pressure, and an upper limit for each that depends on the magnitude of the other. Once again, the optimization problem is to find the one set of variables or point, x, in the feasible solution space that corresponds to the best solution. In order to locate the best

Lower constraint on T

Feasible solution

Upper constraint on both T and P

Lower constraint on P

Temperature, T FIGURE  7.4 Feasible solution for operation of an absorber.

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of the potentially infinite number of feasible solutions using a mathematical optimization technique, it is essential that the aforementioned objective function f(x) – see Equation 7.25, for example – of the design variables be specified to provide for a comparison of alternative solutions. Perhaps Happel11 best described linear programming from a practicing engineer’s perspective. His edited writing follows. The general problem of finding a maximum usually assumes that some partial derivative of the function y = f (x1, x2, x3) vanishes within the range of interest of the variables. In practical problems, it often happens, however, that the optimum solutions will be obtained at the boundary or limiting values of the variables. This is, of course, always true where the function to be optimized is linear. In such cases, explicit analytical solution is not possible, and the optimum must be obtained by some method or iteration. In general, even the simplest type of formulations may require a great deal of computation in practical applications, which is facilitated by the availability of high-speed computers. Linear programming, discussed in greater detail subsequently, is beginning to be used to a substantial extent by practicing engineers. As noted earlier, linear programming deals with the determination of an optimum solution of a problem expressed in linear relationships where there are a large number of possible solutions. In general, the methods employed are trial-and-error, where the procedure systematically follows a mathematical treatment that minimizes the labor involved and ensures that a correct result has been obtained. In order to apply linear programming to a problem, the relationships between variables must be expressed as a set of linear equations or inequalities. The variables must be otherwise independent and must exceed the number of equations. Mathematically expressed,

∑ a x = b ( i = 1, 2,…, m; j = 1,…, n; n > m) (7.25) i

ij i

i

where the sum of the coefficient aij times the value of each variable xi equals a requirement bi in the ith equation. The number of variables must exceed the number of equations, so that an infinite number of solutions is possible. The best one is selected. The relationships in Equation 7.25 are often expressed as inequalities. “Dummy” additional variables can then be introduced in order to convert these inequalities into appropriate equations. Another requirement is that no quantity can be negative. Thus, mathematically xi ≥ 0 and bi ≥ 0 (7.26)



This requirement will ensure that only useful answers are obtained. The solution of a problem by linear programming also requires a profit function or objective to be maximized (or minimized). This requirement is stated as

P=

∑ c x = maximum ( or minimum ) (7.27) i

i i

where ci is the profit (or cost) per unit of xi used. The solution is carried out by an iterative process, which at each stage of calculation assigns either zero or positive values to all xi variables. As the calculation proceeds, the values selected, while meeting the constraints imposed, will tend to increase the summation to be maximized at each stage. If a solution exists, a maximum profit value will finally be realized.”

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7.5 Other Real-World Applied Concepts in Linear Programming This last section attempts to clarify the ideas presented previously via three illustrative examples. An attempt will also be made to relate the variables to real-world quantitative factors. That will occur in the next part of the text. In fact, the first illustrative example will be revisited as it applies to pollutant emission concerns from a coal-fired boiler in Part II. Illustrative Example 7.1 You have been requested to obtain the maximum value of P subject to the following conditions: Objective function: P = 0.3 x1 + 5x2 (1)

Constraints:

x2 ≤ −0.5x1 + 3 (2)



x2 ≤ −3 x1 + 6 (3)



x1 ≤ 3 (4)



x2 ≤ 4 (5)

Once again, the solution is first provided in graphical form. Refer to Figure 7.5. Conditions (4) and (5) require that the solution, that is, the values x1 and x2 that will maximize P, lie within the rectangle provided in Figure 7.5a. Conditions (2) and (3) require that the solution be within the area ABCD in Figure 7.5b. When the objective function equation is superimposed on Figure 7.5b, Figure 7.5c results. The maximum P is located at x1 = 1.2 and x2 = 2.4 for which P = 15.6. Excel provides the same solution. Illustrative Example 7.2 A company manufactures two products, A and B, with an accompanying unit profit of PA and PB, respectively. A minimum number of N products can be produced on a daily basis. The daily production ratio of A and B is xA and xB, respectively. In addition, the fractional defective rate of producing A and B are a and b, respectively. Each defective product reduces the profit associated with product A and B by c and d, respectively. Determine the describing equations (the model for the production rate of A and B) that will maximize daily profits P. SOLUTION: As noted, xA and xB are the daily production rates of A and B respectively. Therefore,

P = x A PA + xB PB − x A ac − xBbd



x A + xB ≥ N



x A , xB ≥ 0 This problem will be evaluated later in the text.

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(5)

4.0

(3) B

3.0 x2

(4)

2.0

C

x2

(2)

1.0

(a)

1.0

3.0

x1

(b)

A

D

x1

5.00 4.80 3.12 Objective function x2 = 2.4

(c)

P*

5.2 6.7 8.2

x1 = 1.2

FIGURE  7.5 Graphical solution for Illustrative Example 7.1.

Illustrative Example 7.3 A batch chemical reactor12 can produce either product A, B, or C. The daily production levels of each are represented by xA, xB, and xC with an accompanying unit profit of PA, PB, and PC, respectively. The two-unit expressions of concern for each product are manpower (M) and energy costs (E), neither of which can exceed Y and Z, respectively. Develop the linear programming model to optimize the profit P for this system. SOLUTION: Based on the problem statement one concludes:

P = PA x A + PB xB + PC xC



M A x A + M B x B + MC x C ≤ Y



EA x A + EB xB + EC xC ≤ Z



x A , x B , xC ≥ 0 Numerical problems of this nature will also be revisited later.

Linear Programming

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References 1. D. Green and R. Perry (editors), Perry’s Chemical Engineers’ Handbook, 8th edn, McGraw-Hill, New York City, NY, 2007. 2. R. Ketter and S. Prawel, Modern Methods of Engineering Computation, adapted from, McGrawHill, New York City, NY, 1969. 3. L. Lapidus, Digital Computations for Chemical Engineers, adapted from, McGraw-Hill, New York City, NY, 1962. 4. E. Chong and S. Zak, An Introduction to Optimization, 4th edn, John Wiley & Sons, Hoboken, NJ, 2013. 5. G. Reklaitis, A. Ravindran, and K. Ragsdell, Engineering Optimization: Methods and Applications, John Wiley & Sons, Hoboken, NJ, 1983. 6. A. Levy, The Basics of Practical Optimization, Society for Industrial and Applied Mathematics, Philadelphia PA, 2009. 7. H. Mickley, T. Sherwood, and C. Reed, Applied Mathematics in Chemical Engineering, McGrawHill, New York City, NY, 1957. 8. L. Theodore and F. Ricci, Mass Transfer Operations for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2010. 9. L. Theodore, Chemical Engineering: The Essential Reference, McGraw-Hill, New York City, NY, 2014. 10. L. Theodore, F. Ricci, and T. VanVilet, Thermodynamics for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2009. 11. J. Happel, Chemical Process Economics, John Wiley & Sons, Hoboken, NJ, 1958. 12. L. Theodore, Chemical Rector Analysis and Application for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2008.

Part II

Environmental Engineering Applications Five chapters compliment the presentation in Part II. The titles follow: Chapter 8: Air Pollution Management Chapter 9: Water Pollution Management Chapter 10: Solid Waste Management Chapter 11: Health Risk Assessment Chapter 12: Hazard Risk Assessment Optimization provides important methods for decision-making in environmental engineering. It has evolved from a methodology of academic interest into a technology that continues to have a significant impact on both engineering research and practice. Optimization algorithms include tools for experimental design, model development, and statistical analysis; process synthesis analysis, design, and retrofit; model predictive control and real-time optimization; and, planning, scheduling, and the integration of process operations. Part II keys on optimization applications in environmental engineering. The five topics addressed are: air pollution management (Chapter 8), water pollution management (Chapter 9), solid waste management (Chapter 10), health risk assessment (Chapter 11), and hazard risk assessment (Chapter 12). Each chapter contains a brief introduction to the subject in question which is followed by a host of illustrative examples keying on industrial applications that employ some—if not all—of the optimization methods introduced in Part I. The reader should note that illustrative examples involving traditional optimization will address both the development of the describing equation(s); that is, the model and the solution to the problem. Linear programming illustrative examples will key solely on the modeling aspect of the problem and not the solution, which more often than not will be “obtained” from Excel. This statement particularly applies to some of the illustrative examples at the end of each chapter.

8 Air Pollution Management The first chapter in Part II contains six illustrative examples. Example titles are listed here:

8.1 Outdated air pollution control device 8.2 Coal-fired power plant options 8.3 Particulate control equipment options 8.4 Recovering dust 8.5 Incinerating mercury-contaminated waste 8.6 Optimizing incinerator performance Seven select references for suggested reading for Chapter 8 are provided below:



1. L. Theodore, Chemical Engineering: The Essential Reference, McGraw-Hill, New York City, NY, 2014. 2. J. Reynolds, J. Jeris, and L. Theodore, Handbook of Chemical and Environmental Engineering Calculations, John Wiley & Sons, Hoboken, NJ, 2004. 3. L. Theodore, R. Dupont, and K. Ganesan, Unit Operations in Environmental Engineering, Wiley‑Scrivener, Salem, MA, 2018. 4. M.K. Theodore and L. Theodore, Introduction to Environmental Management, CRC Press/ Taylor & Francis Group, Boca Raton, FL, 2009. 5. G. Burke, B. Singh, and L. Theodore, Handbook of Environmental Management and Technology, 2nd edn, John Wiley & Sons, Hoboken, NJ, 2000. 6. L. Stander and L. Theodore, Environmental Regulatory Calculations Handbook, John Wiley & Sons, Hoboken, NJ, 2007. 7. L. Theodore, Air Pollution Control Equipment Calculations, John Wiley & Sons, Hoboken, NJ, 2008.

8.1 Outdated Air Pollution Control Device Air pollution control technology is self-defeating if it creates undesirable side effects in meeting (limited) air pollution control objectives. Air pollution control should be considered in terms of both the total technological system and ecological consequences. The former considers the technology that can be brought to bear on not only individual pieces of equipment but also the entire technological system. Consideration of ecological side effects must also be taken into account; for example, the problem of disposal of possibly unmanageable accumulations of contaminants. These may be concentrated in the collection process, such as groundwater pollution resulting from landfill practices or pollution of streams from the discharges of air pollution control systems. 111

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Gaseous and particulate pollutants discharged into the atmosphere can be controlled. The five generic devices available for particulate control include gravity settlers, cyclones (centrifugal separators), electrostatic precipitators, wet scrubbers, and baghouses (fabric filtration). The four generic devices for gases include condensers, absorbers, adsorbers, and incinerators. Detailed information on these control devices are available in the literature.1 There are a number of factors to be considered prior to selecting a particular piece of air pollution control hardware. In general, they can be grouped into three categories: environmental, engineering, and economic. The latter category is discussed later since economics are considered in the applications to follow. Key economic factors include1

1. Capital cost (equipment, installation, engineering, etc.) 2. Operating cost (utilities, maintenance, etc.) 3. Expected equipment lifetime and salvage value Illustrative Example 8.1 The annual total operation cost of an outdated air pollution control device is $75,000. Under a proposed air toxin emission reduction plan, the installation of a new processing system will require an initial cost of $150,000, but it will reduce annual air toxin pollution control costs to $15,000 in each of the first five years and $25,000 in the latter years. Determine the annualized cost for the new processing system by assuming the system only has five years operational life. The interest rate is 7%. Compare the costs for both operations and select the optimum option. Note: The capital recovery factor (CRF) or annual payment of a capital investment can be calculated as follows: CRF =



i (1 + i )

(1 + i)

n

n

−1

(8.1)

where: i = interest, fractional value n = equipment lifetime, years SOLUTION: The annualized cost for a new process is determined based on the following input data: Capital cost = $150, 000



Interest , i = 7%



Term , n = 5 yr



With i = 7% and n = 5 yr, the CRF from Equation 8.1 is:

CRF =

i (1 + i )

n

(1 + i)n − 1

=

0.07 (1 + 0.07 )

5

(1 + 0.07 )5 − 1

= 0.2439

The total annualized cost for the new process is then calculated by Equation 8.2:

Annualized cost = (CRF)(AIC) + AOC = Annualized installation + Annualized operation cost



Annualized cost = (0.2439)($150, 000 ) + $15, 000 = $36, 585 + $15, 000 = $51, 585

(8.2)

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Since this annual cost is lower than the original annual cost of $75,000 for the old process, select the new process.

8.2 Coal-Fired Power Plant Options An environmental standard of performance is a standard for emissions of air pollutants that reflects the degree of emission limitation achievable through the application of, for example, best available control technology (BACT) or the best system of emission reduction that (taking into account the cost of achieving such a reduction and any non-air-quality health and environmental impact plus energy requirements) the EPA (Environmental Protection Agency) determines has been adequately demonstrated. Four source categories (as they relate to coal) for which performance standards have been established are

1. Fossil fuel-fired steam generators 2. Electric utility steam-generating units 3. Industrial/commercial/institutional steam-generating units 4. Small industrial/commercial steam-generating units

Coal combustion products (CCPs) can play an important role in its usage. CCPs are the by-products generated from burning coal in coal-fired power plants. These by-products include2

1. Fly ash 2. Bottom ash 3. Boiler slag 4. Flue gas desulfurization gypsum 5. Other types of material, such as fluidized-bed combustion ash, cenospheres, and scrubber residues

The key air pollution control option for coal-fired boilers today is the baghouse, details of which will be provided in the next section (Section 8.3). Illustrative Example 8.2 Consider the following application. A baghouse is needed at a small coal-fired power plant for a design operating period of 20 years. If the unit fails at any time (bag meltdown), a 45% (of the initial cost) reinvestment cost will result. Two companies submit bids for this particulate control device with the following cost and operating characteristics data (see Table 8.1): TABLE 8.1 Data for Coal-Fired Power Plant Options Company A B

Initial Cost

Time to Failure (yr)

Annual Operation Cost

Salvage Value in Year 20

$15,000 $22,000

10 45

$2,300 $1,100

0 0

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If each alternative has an annual probability of failure that is inversely proportional to its time to failure, choose the optimum option for particulate control for this facility assuming an annual interest rate of 4%. SOLUTION: The annual probability of failure of the baghouse provided by Company A is 1/10 = 0.10, while that of Company B is 1/45 = 0.022. With these results, the estimated annual cost of each of the baghouse options can be determined as follows: Annual cost = Annualized capital cost + Reinvestment cost + Annual O/M cost (8.3) The value of the CRF for 20 years at 4% is 0.0736 (see Section 8.1). For Company A, the estimated annual cost is: Annual cost = ($15, 000 )(0.0736 ) + $15, 000 (0.45)(0.10 ) + $2, 300

= $1,, 104 + $675 + $2, 300



= $4, 079 / yr For Company B, the estimated annual cost is Annual cost = ($22, 000 )(0.0736 ) + $22, 000 (0.45)(0.022) + $1, 100

= $1, 620 + $218 + $1, 100



= $2, 938 / yr From this present worth evaluation, Company B’s bid for the baghouse installation should be selected since the annual cost is minimized.

8.3 Particulate Control Equipment Options 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 affect the separation is considerably less than for other types of gas cleaning apparatuses. 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 100 inches of water for scrubbers and 10 inches of water for baghouses. The 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.1 As the name scrubber implies, wet collectors or wet scrubbers are devices that use a liquid for removing particles or polluted gases from an exhaust 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 remove particulate matter from process exhaust gases. They can also effectively remove gases such as HCl or SO2, but removal conditions must be right. In addition, gas-liquid contact can bring about gas conditioning, and to a lesser extent, liquid conditioning. In effect, these air pollution control devices can remove both particulate

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matter and gases from effluent gas streams. The evaluation and selection of a scrubber system is somewhat simplified by the observation that collection efficiency is a function of the amount of energy required to operate the scrubber. This means, that independent of design and generally speaking, the more power put into the system, the greater the collection efficiency. Therefore, for systems of nearly equivalent power inputs, the collection efficiencies should be nearly equivalent. The selection or evaluation procedure between two systems could then concentrate on ease of operation, potential maintenance problems, and comparative costs.1 One of the oldest, simplest, and most efficient methods for removing solid particulate containments from gas streams is by filtration through fabric media. The fabric filter is capable of providing high collection efficiencies for particles as small as 0.1 µm and will remove a substantial quantity of those particles as small as 0.01 µm. 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 nonexistent, and the primary filtering mechanisms are a combination of inertial forces, impingement, and so on, on the fabric. 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.1 Proper selection of a particular system for a specific application can be extremely difficult and complicated. In view of the multitude of complex and often ambiguous pollution control regulations, it is in the best interest of the prospective user (as noted above) to work closely with regulatory officials as early as possible in the process. The final choice in equipment selection is usually dictated by that piece of equipment capable of achieving compliance with regulatory codes at the lowest uniform annual cost (amortized capital investment plus operation and maintenance costs). More recently, there have been attempts to include liability problems, neighbor/consumer goodwill, employee concerns, and so on, in the economic analysis, but these effects – although important – are often extremely difficult to quantify.1 In order to compare specific control equipment alternatives, knowledge of the particular application and site is also essential. A preliminary screening, however, may be performed by reviewing the advantages and disadvantages of each type of air pollution control equipment. For example, if water or a waste treatment system is not available at the site, this may preclude the use of a wet scrubber system and instead focus particulate removal on dry systems, such as cyclones, baghouses, and/or electrostatic precipitators. If auxiliary fuel is unavailable on a continuous basis, it may not be possible to combust the organic pollutant vapors in an incineration system. If the particulate-size distribution in the gas stream is relatively fine, gravity settlers and cyclone collectors most probably would not be considered. If the pollutant vapors can be reused in the process, control efforts may be directed to adsorption systems. There are many other situations where knowledge of the capabilities of the various control options, combined with common sense, will simplify the selection process.1 Illustrative Example 8.3 A stream of 100,000 acfm of particulate-contaminated gas from a waste facility is to be cleaned according to environmental regulations. You have been requested to install the optimum pollution control device at the site. A reputable vendor – Behan Doyle

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Environmental Inc. – has provided information on the cost of the equipment, as well as installation, operating and maintenance costs, according to the conditions specified for it to be treated. Given in Table 8.2 is all the collected data. Determine what equipment you would select for this work in order to minimize costs on an annualized basis. Costs are based on comparable overall collection efficiencies. The interest rate is 10% and there is zero salvage value. SOLUTION: The first step is to convert the equipment, installation, and operating costs to total costs by multiplying each by the total gas flow, 100,000 acfm. Hence, for the electrostatic precipitator, the total costs are

 $3.1  Equipment cost = 100, 000 acfm  = $310, 000  acfm 



 $0.80  Installation cost = 100, 000 acfm  = $80, 000  acfm 



 $0.06  Operating cost = 100, 000 acfm   = $6, 000/yr  acfm − yr 

Note that the operating costs are on an annualized basis. The equipment cost and the installation cost must then be converted to an annual basis using Equation 8.1 with n = 20 and i − 0.10.3

CRF =

(0.1)(1 + .10)20 = 0.1175 (1 + .10 )20 − 1  

The annual costs for the equipment and the installation is the product of the CRF and the total costs of each.



Equipment Annual Cost = $310, 000 (0.1175) = $36, 412/yr Installation Annual Cost = $80, 000 (0.1175) = $9, 400/yr





The calculations for the scrubber and the baghouse are performed in the same manner. The three devices can be compared after all the annual costs are added. The tabulated results are provided in Table 8.3. TABLE 8.2 Precipitator, Scrubber, and Baghouse Cost; Illustrative Example 8.3 Equipment cost Installation cost Operating cost Maintenance cost Lifetime of equipment

Electrostatic Precipitator

Venturi Scrubber

Baghouse

$3.1/acfm $0.80/acfm $0.06/acfm-yr $14,000/yr 20 yr

$1.9/acfm $1.4/acfm $0.06/acfm-yr $28,000/yr 15 yr

$2.5/acfm $1.0/acfm $0.095/acfm-yr $9,500/yr 20 yr

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TABLE 8.3 Calculation for Air Pollution Equipment; Illustrative Example 8.3 Electrostatic Precipitator

Venturi Scrubber

Baghouse

$310,000 $80,000 0.1175 $36,412 $9,400 $6,000 $14,000 $65,812

$190,000 $140,000 0.1315 $24,985 $18,410 $6,000 $28,000 $77,395

$250,000 $100,000 0.1175 $29,375 $11,750 $9,500 $9,500 $60,125

Equipment cost Installation cost CRF Annual equipment Annual installation Annual operating Annual maintenance Total annual cost

According to the analysis, the baghouse is the most economically attractive, that is, the optimum device, since the annual cost is the lowest.

8.4 Recovering Dust 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 (P) is, therefore, P = I − C (8.4)

At break-even, the profit is zero. Illustrative Example 8.4

A process emits 50,000 acfm of gas containing a dust (it may be considered ash and/ or metal) at a loading of 2.0 gr/ft3. A particulate control device is employed for particle capture and the dust captured from the unit is worth $0.03/lb of dust. Experimental data have shown that the collection efficiency, E, is related to the system pressure drop, ∆P, by the formula: E=



∆P (8.5) ∆P + 15.0

where: E = fractional collection efficiency ∆P = 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 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 dust loading w, and the value of the dust (DV):

RV = ( q) ( w )( DV )(E) (8.6)

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Substituting in Equation 8.6 yields:

 50, 000 ft 3   2.0 gr   1 lb   0.03$  RV =    (E) = 0.429E$/ min  3   min   ft   7000 gr   lb 

The recovered value can be expressed in terms of pressure drop, that is, replace E by Equation 8.5:

RV =

(0.429)( ∆P) $/min (8.7) ∆P + 15.0

The cost of power (CP) in terms of ∆P, q, the cost of electricity (CE) and the fan efficiency, Ef, is

( )

CP = ( q) ( ∆P )( CE) / E f (8.8)

Substitution yields  50, 000ft 3   ∆P lbf   0.18$   1 min ⋅ kW   1   1h  CP =  = 0.006 ∆P$/min       min   ft 2   kWh   44, 200 ft ⋅ lbf   0.55   60 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 = CP (8.9)

Substituting Equations 8.7 and 8.8 for RV and CP, respectively ultimately leads to

∆P = 56.5 lbf /ft 2 = 10.9 H 2 O



Figure 8.1 shows the variation of RV, CP, and profit with pressure drop. The collection efficiency corresponding to the previously calculated ∆P from Equation 8.5 is: E=



=

∆P ∆P + 15.0

(8.5)

56.5 56.5 + 15.0

= 0.79 = 79%

The reader should note that operating below this efficiency (or the corresponding pressure drop) will produce a profit; operating above the value leads to a loss. The operating condition for maximum profit can also be estimated from the graph. Calculating this value is left as an exercise for the reader. [Hint: Set the first derivative of the profit (i.e., RV-CP) with respect to ∆P equal to zero. The answer is 13.9 lbf/ft2.] However, this application will be revisited later in the book. The reader should realize that this problem was addressed earlier in the text. One is left the option of revisiting Illustrative Example 5.1 and 6.2.

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0.5 0.4

CP

Value, $/min

0.3

RV

0.2 Profit

0.1

Break-even

0 –0.1 –0.2

0

10

20

30

40

50

60

70

80

Pressure drop, lbf /ft2 FIGURE 8.1 Profit as a function of pressure drop; illustrative example 8.4.

8.5 Incinerating Mercury-Contaminated Waste Combustion and/or incineration is often used to control the emissions of organic compounds from process industries. At a sufficiently high temperature and adequate residence time, any hydrocarbon can be oxidized to carbon dioxide and water by the combustion process. Combustion systems are often relatively simple devices capable of achieving very high removal destruction efficiencies. They consist of burners, which ignite the fuel and organic material, and a chamber, which provides appropriate residence (or detention/retention) time for the oxidation process. Because of the high cost and decreasing supply of fuels, combustion systems may be designed to include some type of heat recovery. Combustion is also used for the more serious emission problems that require high destruction efficiencies, such as emission of toxic or hazardous gases. There are, however, some problems that may occur when using combustion. Incomplete combustion of many organic compounds results in the formation of aldehydes and organic acids as well as metals, such as mercury, lead, calcium, and so on, which may create additional problems. Oxidizing organic compounds containing sulfur or halogens produce unwanted pollutants such as sulfur dioxide, hydrochloric acid, hydrofluoric acid, or phosgene. If present, these pollutants would require a scrubber to remove them prior to release into the atmosphere. Several basic combustion systems are in use. Although these devices are physically similar, the conditions under which they operate may be different. Choosing the proper device depends on many factors, including the type of hazardous contaminants in the waste stream, concentration of combustibles in the steam, process flow rate, control requirements, and an economic evaluation.4 Combustion is a chemical process occurring from the rapid combination of oxygen with various elements or chemical compounds resulting in the release of heat. The process of combustion is also referred to as oxidation or incineration. Most fuels used for combustion along with the waste are composed essentially of carbon and hydrogen, but can include other elements, such as sulfur, nitrogen, and chlorine. Although combustion seems to be a very simple process that is well understood, in reality it is not. The exact manner in which

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a fuel or waste is oxidized occurs in a series of complex, free radical chain reactions. The precise set of reactions by which combustion occurs is termed the mechanism of combustion. By analyzing the mechanism of combustion, the rate at which the reaction proceeds and the variables affecting the rate can be estimated. For most combustion devices, the rate of reaction proceeds extremely fast compared to the mechanical operation of the device. Maintaining efficient and complete combustion is somewhat of an art rather than a science. Therefore, this section focuses on the factors that influence the completeness of combustion rather than analyzing the mechanism involved.4,5 To achieve complete combustion once the air (oxygen), waste, and fuel have been brought into contact, the following conditions must be provided: a temperature high enough to ignite the waste/fuel mixture, turbulent mixing of the air and waste/fuel; and, sufficient residence time for the reaction to occur. These three conditions are referred to as the “three T’s of combustion.” Time, temperature, and turbulence govern the speed and completeness of reaction. They are not independent variables since changing one can affect the other two.4 For combustion processes, ignition is accomplished by adding heat to speed up the oxidation process. Heat is needed to combust any mixture of air and fuel until the ignition temperature of the mixture is reached. By gradually heating a mixture of fuel and air, the rate of reaction and energy released will gradually increase until the reaction no longer depends on the outside heat source. In effect, more heat is being generated than is lost to surroundings. The ignition temperature must be reached or exceeded to ensure complete combustion. To maintain combustion of the waste, the amount of energy released by the combusted waste must be sufficient to heat the incoming waste (and air) up to its ignition temperature; otherwise, a fuel must be added. The ignition temperature of various fuels and compounds can be found in the literature.4,5 The dangers of human exposure to many metallic chemical elements have long been recognized. Metals such as mercury, lead, cadmium, and arsenic are toxic. Health effects range from retardation and brain damage, especially in children from lead poisoning, to the impairment of the central nervous system as a result of mercury exposure. Since these toxic metals are chemical elements, they cannot be broken down by any chemical or biological process. As a result of this, it is not uncommon to have metallic buildup or bioaccumulation.6 Mercury and most of its compounds are toxic substances. Nineteenth-century hat makers, for example, developed a characteristic shaking and slurring of speech from occupational exposure to large quantities of inorganic mercury during the manufacturing process – symptoms that gave rise to the phrase “mad as a hatter.” Such problems result from impairment of the central nervous system. In addition, high levels of mercury can cause kidney damage, birth defects, and in extreme cases, death. The most common source of the silvery liquid can be found in the bulbs of thermometers and thermostats. Large quantities of mercury can also be found in one type of barometer or another. Small quantities are also found in fluorescent lights, mercury switches, batteries, hearing aids, smoke detectors, watches, and cameras.6 Consider the following application. Illustrative Example 8.5 Two hospital wastes are received and stored in separate tanks at an incineration facility. The first, a sludge waste with a net heating value (NHV) of 6000 Btu/lb contains 2% Hg by weight. The second, a mercury-contaminated plating waste with an NHV of 8000 Btu/ lb, contains 8% Hg by weight. A minimum of 1000 lb/h of each waste is to be incinerated. Because of pump limitations, no more than 5000 lb/h of each waste can be utilized.

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The incinerator can operate with a heat rate between 25 and 40 million Btu/h. These specifications provide two additional constraints on the operation of the incinerator.

msludge = 3890 −

3890 mplate ; < 40 MMBtu/h (8.10) 5000



msludge = 2431 −

2431 mplate ; > 25 MMBtu/h (8.11) 3125

Determine the maximum amount of mercury-contaminated plating waste that can be incinerated employing a graphical approach involving the plotting of the two principal variables, msludge and mplate (plating waste). SOLUTION: Let msludge and mplate represent the flowrates of the two wastes to the waste incinerator in lb/h. The restrictions on the available blending options are provided in Table 8.4 and shown graphically in Figure 8.2 using numbered lines for the constraints. The permissible area in Figure 8.2 is the quadrilateral area (ABCD) bounded by the lines 2, 4, 5, and 6. The point farthest to the right (point A), represents the maximum rate at which plating waste can be incinerated. This occurs where constraints 2 and 5 intersect. The describing equation for a minimum sludge rate of 1000 lb/h is then (employing Equation 8.12)

1000

lb lb  3890 lb/h  = 3890 −  mplate h h  5000 lb/h 

so that mplate = 3715 lb/h



Can the reader determine the maximum rate at which the sludge can be incinerated?

8.6 Optimizing Incinerator Performance The performance of an incinerator is strongly influenced by temperature T. The effect pressure, P, is normally assumed to be negligible. However, some innovative and/or specially designed incinerators can be adversely affected by pressure. TABLE 8.4 Flowrate Restrictions Related to Illustrative Example 8.5 Line 1 Line 2 Line 3 Line 4 Line 5

Maximum of 5000 lb/h of sludge Minimum of 1000 lb/h of sludge Maximum of 5000 lb/h of plating waste Minimum of 1000 lb/h of plating waste Maximum heat rate of 40 MMBtu/h5

Line 6

Minimum heat rate of 25 MMBtu/h5

msludge  1000 mplate  1000 msludge = 3890 −

3890 mplate 5000

msludge = 2431 −

2431 mplate 3125

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6000 Line 4

Line 3 Line 1

5000

msludge lb/h

4000 3000

B

Line 5

2000 D

C

1000 0

0

1000

Line 2

A

Line 6 2000

3000

4000

5000

6000

mplate lb/h FIGURE 8.2 Diagram showing six constraints for mercury incinerator application.

Illustrative Example 8.6 The emissions rate, E, of a toxic pollutant as a function of both temperature and pressure from one such unit has been described by the following equation: E (lb/h ) = 0.23T + 0.19P +



12, 000 + 4.31; T = °C, P = psi (8.12) PT

Concerned about the worst-case scenario of emissions, you have been hired to determine the maximum emission rate from the previously mentioned incinerator. Earlier tests suggest that the allowable operating range of the unit is in the 25°C–50°C temperature and 14.7–50.0 psi pressure ranges. Obtain the solution to this application employing a brute force perturbation approaches (The reader may choose to revisit Illustrative Example 6.6). SOLUTION: Generate emission rates from Equation 8.12 for a number of combinations of pressure and temperature. Some of the calculated results are provided in Table 8.5. TABLE 8.5 Incineration Emission Rates as a Function of Temperature and Pressure; Illustrative Example 8.6 P, psi

14.7 20.0 25.0 30.0 40.0 50.0

T,°C 25.0

30.0

35.0

40.0

45.0

50.0

45.51 37.86 34.01 31.76 29.66 29.16

41.21 35.01 31.96 30.24 28.81 28.71

38.27 33.30 30.82 29.49 28.53 28.72

36.71 32.31 30.26 29.21 28.61 29.01

33.59 31.79 30.08 29.25 28.93 29.49

34.93 31.61 30.16 29.51 29.41 30.11

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The reader is left the exercise of refining (expanding) the table. The results suggest that the minimum emission rate occurs with T ≈ 35–40°C and P ≈ 40 psi with E ≈ 28.5 lb/h. The reader should also realize that the application could have been solved analytically by employing an approach by requiring partial derivatives.7,8

∂E 12, 000 = 0.23 − (8.13) ∂T T 2P



∂E 12, 000 = 0.19 − (8.14) ∂P TP 2

Setting both Equations 8.13 and 8.14 equal to zero and solving simultaneously leads to

T = 35.06°C



P = 42.44 psi

The corresponding value of E is E = 0.23 ( 35.06 ) + 0.19 ( 42.44 ) +

12, 000 + 4.31 ( 42.44)( 35.06)

= 8.06 + 8.06 + 8.06 + 4.31



= 28.49 lb/h Evaluating the second partial derivatives is necessary to determine if E is a maximum or minimum.

(12, 000)( 2) ∂ 2E =+ ∂T 2 T 3P



(12, 000)( 2) ∂ 2E =+ 2 ∂P TP 3

Since both second partial derivatives are positive, the calculated value for the previously mentioned E is indeed a minimum.

References

1. L. Theodore, Air Pollution Control Equipment Calculations, John Wiley & Sons, Hoboken, NJ, 2008. 2. K. Skipka and L. Theodore, Energy Resources: Availability, Management, and Environmental Impacts, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2016. 3. S. Shaefer and L. Theodore, Probability and Statistics for Environmental Science, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2007. 4. J. Santoleri, J. Reynolds, and L. Theodore, Introduction to Hazardous Waste Incineration, 2nd edn, John Wiley & Sons, Hoboken, NJ, 2004. 5. L. Theodore, “Chemical Reactor Analysis and Application for the Practicing Engineer”, John Wiley & Son, Hoboken, NJ, 2012.

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6. L. Theodore and R. Dupont, “Environmental Health and Hazard Risk Assessment: Principles and Calculations”, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2012. 7. L. Theodore, personal notes, East Williston, NY, 2000. 8. L. Theodore and C. Prochaska, Introduction to Mathematical Methods for Environmental Engineers and Scientists, Wiley‑Scrivener, Salem, MA, 2018.

9 Water Pollution Management The second chapter in Part II contains six illustrative examples. Example titles are listed here:

9.1 Break-even point operation 9.2 Wastewater treatment equipment options 9.3 Water disinfection options 9.4 Water plant operation 9.5 Generating waste revenues from waste streams 9.6 Membrane construction costs Eight select references for Chapter 9 are provided here:



1. L. Theodore, Chemical Engineering: The Essential Reference, McGraw-Hill, New York City, NY, 2014. 2. J. Reynolds, J. Jeris, and L. Theodore, Handbook of Chemical and Environmental Engineering Calculations, John Wiley & Sons, Hoboken, NJ, 2004. 3. L. Theodore, R. Dupont, and K. Ganesan, Unit Operations in Environmental Engineering, Wiley‑Scrivener, Salem, MA, 2018. 4. M. K. Theodore and L. Theodore, Introduction to Environmental Management, CRC Press/Taylor & Francis, Boca Raton, FL. 5. G. Burke, B. Singh, and L. Theodore, Handbook of Environmental Management and Technology, 2nd edn, John Wiley & Sons, Hoboken, NJ, 2000. 6. L. Stander and L. Theodore, Environmental Regulatory Calculations Handbook, John Wiley & Sons, Hoboken, NJ, 2007. 7. R. Thomann and J. Muller, Principles of Surface Water Quality Modeling and Control, Harper & Row, New York City, NY, 1987. 8. R. Dupont, K. Ganesan, and L. Theodore, Pollution Prevention, Sustainability, Industrial Ecology, and Green Science and Engineering, CRC Press/Taylor & Francis, Boca Raton, FL, 2017.

9.1 Break-Even Point Operation From an economic point of view, the break-even point of a process operation is defined as that condition when the costs (C) exactly balances the income (I). As noted in the previous chapter, the profit (P) is, therefore, P = I − C. At break-even, the profit is zero. 125

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Illustrative Example 9.1 The cost and income (in dollars) for a particular operation are given by the following equations:

I = $60, 000 + 0.021N



C = $78, 000 + 0.008 N

where N is the yearly production of the item 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 = 0. I =C

Substitute for C and I in terms of N:

$60, 000 + 0.021N = $78, 000 + 0.008 N

Solving for N at the break-even point

N ≈ 1, 384, 600

Calculating the cost at the break-even point: C = $78, 000 + 0.008 N

= $78, 000 + 0.008 (1, 384, 600 ) ≈ $89, 077

The reader should note that as N decreases below 1,384,600 items, P is negative (there is a cost). Higher values of N lead to profits.

9.2 Wastewater Treatment Equipment Options Numerous technologies exist for treating industrial wastewater. These technologies range from simple clarification in a settling pond to a complex system of advanced technologies requiring sophisticated equipment and skilled operators. Finding the proper technology or combination of technologies adequate to treat a particular wastewater to meet federal and local requirements and yet still be cost-effective can be a challenging task.1 Treatment technologies can be divided into three broad categories: physical, chemical, and biological. Many treatment processes combine two or all three categories to provide the most economical treatment. There are a multitude of technologies for each of these categories. Therefore, although the technologies selected for subsequent discussion are among the most widespread in the environmental field, they represent only a fraction of the available technologies.

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Physical treatment processes include clarification (sedimentation), floatation, and oilwater separation. Chemical treatment processes include coagulation, precipitation, and neutralization. Biological processes include actuated sludge, aerobic attached growth, aerobic lagoons (stabilization ponds), and anaerobic lagoons.1 Illustrative Example 9.2 Consider the following application. Three different water pollution control devices are available for the removal of a toxic contaminant from a wastewater stream. The service life is 10 years for each device. The capital and annual operating costs are as follows (see Table 9.1): Which is the most economical unit (that will minimize cost)? Employ a straight-line depreciation method of analysis. SOLUTION: To select the most economical control device, a comparison can be performed among the three units based on the total annualized cost (TAC). Table 9.2 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.

9.3 Water Disinfection Options The exploration of an individual water quality problem begins with an investigation of the impact of bacteria and the organisms that may cause communicable diseases. This is appropriate because:

1. The kinetics of bacterial die-away are usually considered to be first order. 2. It is the oldest of the water pollution problems in the discovery of the links between contaminated water and communicable disease. TABLE 9.1 Economic Cost Data for Waste Pollution Control Options Device A B C

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

TABLE 9.2 Total Annual Cost Calculations Equipment Capital investment Depreciation Operating costs Total annual costs

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

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3. The problem is still quite relevant today and is manifested in the continuing high levels of waterborne diseases in some countries, closing of bathing beaches, and restrictions on water consumption.

The transmission of waterborne diseases (e.g., gastroenteritis, amoebic dysentery, cholera, and typhoid fever) has been a matter of concern for many years. The impact of high concentrations of disease-producing organisms on waste uses can be significant. Bathing beaches may be closed permanently or intermittently during rainfall conditions when high concentrations of pathogenic bacteria and discharge from urban runoff and combined sewer overflows. Diseases associated with water used for drinking purposes continue to occur. The modes of transmission of pathogens are (1) ingestion of contaminated water and food, and (2) exposure to infected persons or animals. Infections of the skin, eyes, ears, nose, and throat may result from immersion in the water while bathing. The water uses impacted by pathogenic bacteria, viruses, and parasites are2

1. Drinking water: municipal, domestic, industrial, and individual supplies 2. Primary contact recreation, such as bathing and water skiing 3. Secondary contact recreation, such as boating and floating 4. Shellfishing, such as harvesting for clams and oysters

There are three broad categories for control of pathogenic bacteria, viruses, and parasites: control at the input source of the microorganism, control at the area of water use, and control of the product that is affected by contamination. Point sources of municipal wastes are the usual principal inputs of communicable disease organisms, and such inputs can be reduced by treatment of wastes without disinfection, and disinfection by chlorination, ultraviolet (UV) radiation, ozonation, and chlorine dioxide. Controls of the area of water use would include bathing restrictions on a transient basis (i.e., during and after storms, resulting in combined sewer overflows), construction of dikes, and diversion structures to protect a given area. Controls of the product would include chlorination or other forms of disinfection of water used for municipal water supply, treatment plants for contaminated shellfish to allow depuration of bacteria prior to marketing, and distribution of high-quality bottled water during emergencies.1 Illustrative Example 9.3 Consider the following application. A large wastewater treatment facility currently uses chlorine for disinfection. The average chlorine dosage is 6.0 mg/L at an average flow rate of 70 MGD. Chlorine (Cl2) costs $1.00/lb. Sulfur dioxide (SO2) is used to dechlorinate the effluent before it is discharged (a requirement placed on the facility to protect the fish inhabiting the receiving water) and is consumed at an average dose of 2.0 mg/L. Its cost is $1.20/lb. Concerned about operating costs and the risks associated with Cl2 usage, the utility has completed a pilot study showing that UV light disinfection could be used to replace Cl2 and SO2. The UV system has a capital cost of $12,000,000 and will cost $500,000 a year to operate with a 20-year service life. The utility will draw from bank savings accounts currently earning 3.0% simple interest per year to pay for the UV system’s capital cost. Does the savings in operating costs justify the capital cost? Assume straight-line depreciation of the UV system, with $0 salvage value at the end of its service life.

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ADDITIONAL INFORMATION: Conversion factor: A dose of 1 mg/L = 8.34 lb/106 gal (MG) Straight-line depreciation equation: Depreciation Rate in $ yr =



Initial Value − Salvage Value (9.1) Service Life

Rate of ROI equation in %yr:

ROI =

Scenario A operating costs − Scenario B operating costs (9.2) Investmeent

where: Scenario A = estimated cost for continuing the status quo Scenario B = estimated cost for proposed alternatives SOLUTION: Calculate the daily Cl2 and SO2 dosage rates as follows:





Cl 2 dosage rate = ( 70 MGD )( 6.0 mg L ) ( 8.34 L ⋅ lb MG ⋅ mg ) = 3503 lb day

SO 2 dosage rate = (70 MGD )( 2.0 mg L ) ( 8.34 L ⋅ lb MG ⋅ mg ) = 1168l l b MG ⋅ mg





These daily rates are then converted to annual rates: Annual Cl 2 dosage rate = ( daily Cl 2 rate )( day yr )

= ( 3503 lb day ) ( 365 day yr ) = 1, 278, 595 lb yr Annual SO 2 dosage rate = ( daily SO 2 rate )( day yr )



= ( 1168 lb day ) ( 365 day yr ) = 426, 320 lb yr

The annual costs for Cl2 and SO2 addition are calculated as follows: Annual Cl 2 cost = ( annual Cl 2 dosage rate)( Cl 2 cost in $ lb)

= (1, 278, 595 lb yr )( $1.00 lb)

(9.3)

= $1, 278, 595 yr Annual SO 2 cost = ( annual SO 2 dosage rate) ( SO 2 cost in $ lb)



= ( 426, 320 lb yr )( $1.00 lb) = $511, 584 yr

(9.4)

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Total operating cost = Annual Cl 2 cost + Annual SO 2 cost = $1, 278, 595 yr + $511, 584 yr



(9.5)

= $1, 790, 179 yr The straight-line depreciation of the system is then calculated as follows: Depreciation =

initial value − salvage value service life

= ( $12, 000, 000 − $0 ) ( 20 yr )



(9.6)

= $600, 000 yr The total annual cost of the UV system is then calculated as: Total annual cost = Straight line depreciation + Operating cost

= $6000, 000 yr + $500, 000 yr

(9.7)

= $1, 100, 000 yr The annual cost savings for switching to the UV system is determined as follows:

Annual cost savings = Scenario A operating cost − Scenario B operating cost (9.8)

where scenario A is the total annual Cl2 cost and scenario B is the total annual UV cost. Annual cost savings = Annual Cl 2 cost − Annual UV cost = $1, 790, 000 yr − $1, 100, 000 yr





= $690, 000 yr The expected ROI on the investment in the UV system is determined as follows: ROI = 100 ( Annual cost savings ) ( investment )

= 100 ( $690, 000 yr ) ( 2, 000, 000 ) = 100 ( 0.0575 )

(9.2)

= 5.8% This is more than the 3.0% simple interest the utility could make from cash in its bank account (at the time of preparation of this problem). Thus, based on operating cost considerations alone, the utility should invest in the UV system. Other considerations may make UV even more attractive. There may be hidden costs to continued Cl2 and SO2 usage, such as maintaining safety equipment and training, hazardous materials planning, worker and public concerns, stockholder support, and, perhaps, capital costs to maintain or upgrade the Cl2 and SO2 storage and dosing systems. Additional calculations should be made that include these as well as potentially other hidden costs before a final decision is made.

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9.4 Water Plant Operation Illustrative Example 9.4 The Canadian Rocky Mountain Water Plant is located in the foothills of the Canadian Rockies near Calgary. The plant “produces” bottled water and is the hub of their operation. The bottled water is delivered to two local municipalities – Edmonton (1) and Winnipeg (2). Locality (1) and (2) are approximately 200 and 650 miles from the plant. Henry Hydroxic was recently assigned the task of maximizing profits for the plant’s operation. The following information is available. The plant can only produce a maximum of 17,300 bottles/day. Income derived from the sale of water to (1) and (2) is 27 cents and 38 cents per bottle respectably. However, Harry is confronted with two constraints – one associated with energy costs and one associated with environmental regulations. The energy constraint is based on company policy that energy operating costs – 0.015 and 0.035 $/bottle for (1) and (2) – be limited to $1700/day. The environmental constraints arise because the water contains 0.3 ppm dissolved SO2; local policy at (1) and (2) require a “tax” of 1.1 cents and 1.9 cents/bottle, respectively. Company policy limits this charge to $1000/day. Unable to arrive at a solution to the company’s request, Harry has requested that you provide a weekly delivery schedule that will maximize profits, P. SOLUTION: Set x1 and x2 equal to the daily delivery schedule of bottled water for Edmonton1 and Winnipeg2, respectively. The profit objective function is then P = 0.27 x1 + 0.38 x2 $ day (9.9)

The two constraints are

Energy: 0.015x1 + 0.035x2 ≤ 1700 $ day (9.10)



Environmental: 0.011x1 + 0.019x2 ≤ 1000 $ day (9.11)

In addition, based on physical grounds,

x1 ≥ 0 (9.12)



x2 ≥ 0 (9.13)

Employing the previously mentioned three equations, Excel provides the following solution:

x1 = 27 , 000 bottles day



x2 = 37 , 000 bottles day



P = 21, 350 $ day

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9.5 Generating Waste Revenues from Waste Streams The waste from one industry is often an acceptable raw material for another. In response to this common interest, material and waste exchanges were established. Material and waste exchanges were organized to enable industrial process wastes, by-products, surpluses, or materials that do not meet specifications to be transferred from one company to another company where they can be used as a process input.3 Because many of these materials are typically of low or negative value, it usually does not pay to transport them long distances. As a result, these exchanges are generally regional ventures servicing a wide variety of industries. Many of these exchanges have been operating in the United States since the early 1970s. Initially, a major stumbling block to widespread adoption of this method of waste recycling in this country has been the lack of an intermediary or an information bank to help match supply and demand. Although more waste exchanges have been set up in this country in recent years, they have not been fully accepted by industry because of liability concerns, particularly for wastes that can be considered hazardous under the Resource Conservation and Recovery Act (RCRA).4 Waste exchanges are usually nonprofit operations; however, some waste exchanges charge for access to their database services or for copies of their catalogs. A partial list of waste exchanges is available in the literature.5 Illustrative Example 9.5 As noted previously, a waste stream may be one company’s problem, but another company’s source of revenue. A chemical company has found an environmental waste exchange that is prepared to pay $50/ton for Waste A that contains 40% chromium (Cr), 15% nickel (Ni), and 45% water; $20/ton for Waste B that contains 10% chromium, 80% nickel, and 10% water; and, $35/ton for Waste C that contains 5% chromium, 20% nickel, and 75% water. How many tons of waste A, B, and C should be purchased that will minimize the cost ($) of the waste and yet ensure that the mixture contains at least 20 tons of chromium, 12 tons of nickel, and 15 tons of water? Develop the describing equations for this operation. SOLUTION: Set x1, x2, and x3 as the tons of waste A, B, and C, respectively. The objective function for the cost is given by

$ = 50 x1 + 20 x2 + 35x3 (9.14)

The constraints are:

0.40 x1 + 0.15x2 + 0.45x3 ≥ 20 (9.15)



0.10 x1 + 0.80 x2 + 0.10 x3 ≥ 12 (9.16)



0.05x1 + 0.20 x2 + 0.75x3 ≥ 15 (9.17)

and

x1 ≥ 0 (9.18)

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x2 ≥ 0 (9.19)



x3 ≥ 0 (9.20)

To minimize the waste cost, 29.0 tons of Waste A, 9.4 tons of Waste B and 15.6 tons of Waste C should be purchased. This will produce a minimum cost of $2,181. The calculations are left as an exercise for the reader. Hint: Employ Excel.

9.6 Membrane Construction Costs Membrane processes today 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 manufacturing, 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 or extraction. Membrane processes are generally more efficient and effective since they can simultaneously concentrate and purify, and they can perform separations at ambient conditions.6 There are four major membrane processes of interest to the practicing engineer:

1. Reverse osmosis (hyperfiltration) 2. Ultrafiltration 3. Microfiltration 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/ concentrated are generally solids or colloids rather than molecules in solution. 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.6 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 the selective action.6

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Illustrative Example 9.6 Consider the following application. Theodore International manufactures four different membranes (A, B, C, and D) for water purification systems. Four input variables affect the production process labor requirements: individual days, raw membrane material, lb, processed membrane material, lb; and labor costs. Each membrane has different input requirements. Current information on the production process for the four membranes is TABLE 9.3 Membrane Data; Illustrative Example 9.6 Membrane

Individual days Lbs of raw membrane Lbs of preferred membrane Labor costs

A

B

C

D

Maximum Input

3 5 3 2

3 5 3 5

1 7 6 4

2 6 10 3

40 80 65 50

provided in Table 9.3. If the A, B, C, and D membranes produced sell for $5, $5, $8, and $20, respectively, determine the maximum production revenue subject to the constraints listed in Table 9.3. SOLUTION: Set x1, x2, x3, and x4 as the production numbers for membranes A, B, C, and D. The objective function describing the production membrane number for maximizing the total revenue (R) is presented here:

R = f ( x1 , x2 , x3 , x 4 ) = 5x1 + 5x2 + 8 x3 + 20 x 4 (9.21)

The constraints are

3 x1 + 3 x2 + x3 + 2x 4 ≤ 40 (9.22)



5x1 + 5x2 + 7 x3 + 6 x 4 ≤ 175 (9.23)



3 x1 + 3 x2 + 6 x3 + 10 x 4 ≤ 65 (9.24)



2x1 + 5x2 + 4 x3 + 3 x 4 ≤ 50 (9.25)

with

x1 ≥ 0 (9.26)



x2 ≥ 0 (9.27)



x3 ≥ 0 (9.28)



x 4 ≥ 0 (9.29)

The solution to this problem, subject to the information provided is from the previous equation,

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x1 = 6.71



x2 = 3.71



x3 = 1.47



x 4 = 1.82



P = $100

The calculations are left as an exercise for the reader. Hint: employ Excel.

References 1. G. Burke, B. Singh, and L. Theodore, Handbook of Environmental Management and Technology, 2nd edn, John Wiley & Sons, Hoboken, NJ, 2000. 2. M. K. Theodore and L. Theodore, Introduction to Environmental Management, CRC Press/Taylor & Francis Group, Boca Raton, FL. 3. L. Theodore and F. Ricci, Mass Transfer Operations for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2010. 4. R. Dupont, K. Ganesan, and L. Theodore, Pollution Prevention: The Waste Management Approach for the 21st Century, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2000. 5. R. Dupont, K. Ganesan, and L. Theodore, Pollution Prevention, Sustainability, Industrial Engineering, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2017. 6. S. Slater, Membrane Technology, NSF Workshop Notes, Manhattan College, Bronx, NY, 1991 (adapted with permission).

10 Solid Waste Management The third chapter in Part II contains six illustrative examples. Example titles are listed below:

10.1 Solid waste treatment 10.2 Cement kiln versus rotary kiln incinerator 10.3 Sludge waste receiving tank size 10.4 Discounted cash flow application 10.5 Maximum mercury waste emission constraint 10.6 Structural considerations of concrete mix Eight select references for Chapter 10 are provided here:



1. L. Theodore, Chemical Engineering: The Essential Reference, McGraw-Hill, New York City, NY, 2014. 2. J. Reynolds, J. Jeris, and L. Theodore, Handbook of Chemical and Environmental Engineering Calculations, John Wiley & Sons, Hoboken, NJ, 2004. 3. L. Theodore, R. Dupont, and K. Ganesan, Unit Operations in Environmental Engineering, Wiley‑Scrivener, Salem, MA, 2018. 4. M. K. Theodore and L. Theodore, Introduction to Environmental Management, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2008. 5. G. Burke, B. Singh, and L. Theodore, Handbook of Environmental Management and Technology, 2nd edn, John Wiley & Sons, Hoboken, NJ, 2000. 6. L. Stander and L. Theodore, Environmental Regulatory Calculations Handbook, John Wiley & Sons, Hoboken, NJ, 2007. 7. J. Santoleri, J. Reynolds, and L. Theodore, Introduction to Hazardous Waste Incineration, 2nd edn, John Wiley & Sons, Hoboken, NJ, 2000. 8. R. Dupont, K. Ganesan, and L. Theodore, Pollution Prevention, Sustainability, Industrial Ecology, and Green Science and Engineering, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2017.

10.1 Solid Waste Treatment Technical considerations involving solid waste treatment include waste characteristics, technical sustainability of the available treatments, and treatment objectives. The waste must be analyzed so that the proper form of treatment may be selected. Should the waste 137

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contain more than one type (e.g., a mixture of heavy metals and organics), more than one method of treatment may be needed. Waste treatment may be divided into five categories: physical, chemical, biological, thermal (incineration), and solidification or encapsulation (a subcategory of physical treatment). Physical treatments may be used alone or with the other types of treatments; their function is to concentrate wastes, reduce waste volume, and separate different waste components for continued treatment or disposal. Chemical treatments are used to convert hazardous wastes into other less hazardous forms; they are generally used on one substance or similar substances because the reactions involved are specific. Biological treatments are more specific than the other treatment types and are used mainly on organic wastes. Thermal treatment is usually concerned with incineration (see later paragraph). Solidification and encapsulation are also treatments for specific types of waste; these methods work well on inorganic wastes, but physical and/or chemical pretreatment is usually necessary. Incineration is one of the most popular forms of thermal waste treatment, especially when organic wastes are involved. It destroys most organic components in the waste, and the resulting sludge and ash are almost always less hazardous and easier to dispose of than the original waste. It is difficult to obtain a permit for incineration, however, and many stringent regulations govern its use.1,2 Processes to perform various waste treatment tasks include:

1. Separation of heavy metals from liquid waste streams 2. Destruction of organics 3. Separation of toxic anions from any liquid waste stream 4. Processes that can handle slurries or sludges 5. Processes the can handle tars or solids

Not all of these processes are currently feasible for hazardous wastes. The objective of the treatment is another important factor that influences treatment selection. It may be desired, for example, to destroy the hazardous components in the waste, reduce the hazardous components, and then isolate them, or merely to separate the hazardous components. The end product of the waste treatment must be compatible with the type of ultimate disposal to be used for the waste: sewage discharge, landfilling, deep-well injection, and so on. For example, if the waste is to be disposed of by landfarming, the end product of the waste treatment should consist of biodegradable organic components. Economic feasibility and environmental factors are obviously also important factors to be considered in treatment selection. Cost is a strong function of the amount of waste to be treated and the simplicity and degree of commercial usage of the treatment process. The various costs include capital investment, operating costs, utility costs, and expenses associated with the final disposal of the waste product. Environmental factors, in the form of state and local environmental regulations, also influence waste treatment selection. The Environmental Protection Agency (EPA) Resource Conservation and Recovery Act (RCRA) regulations especially control the type of waste treatment that may be used and the characteristics of the waste that is to finally be disposed.1 Illustrative Example 10.1 A copper solid waste treatment facility can produce three grades of copper – pure grade (99.99%), medium grade (99.95%), and poor grade (99.90%). Because of industry’s copper demand, the production of the three classes of copper is limited. The profit after

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producing the three copper wastes, A, B, and C, is 2.20, 1.94, and 3.10, respectively, $/lb. Copper Waste A generates 43%, 14%, and 43% of pure, medium, and poor grade copper, respectively. Copper Waste B generates 35%, 35%, and 30% of pure, medium, and poor grade copper, respectively. Copper Waste C generates 75%, 7%, and 18% of pure, medium, and poor grade copper, respectively. However, the maximum production of pure, medium, and poor grade copper is limited to 340,000, 66,000, and 145,000 lb/mo. Calculate the daily processing rate of the three wastes to maximize the profits. SOLUTION: Set x1, x2, and x3, as the daily processing rate of wastes A, B, and C, respectively. The objective function for the income I is then I = 2.20 x A + 1.94 xB + 3.10 xC (10.1)

There are three constraints:

0.43 x1 + 0.35x2 + 0.75x3 ≤ 340, 000 (10.2)



0.14 x1 + 0.35x2 + 0.07 x3 ≤ 66, 000 (10.3)



0.43 x1 + 0.30 x2 + 0.18 x3 ≤ 145, 000 (10.4)

with

x1 ≥ 0 (10.5)



x2 ≥ 0 (10.6)



x3 ≥ 0 (10.7) Excel provides the following solution:



x1 = 154, 838



x2 = 59, 255



x3 = 336, 907



I = $1, 263, 052

10.2 Cement Kiln versus Rotary Kiln Incinerator In the field of waste incineration, there is more experience with liquid injection (LI) incinerators than all other types combined. However, various other types of incinerators are used to handle solid wastes as well as wastes in other forms, that is, liquid sludges, slurries, and fumes. The subsequent discussion is confined to the LI type of incinerator, which, as the name implies, is applicable almost exclusively to pumpable liquid waste. The waste is burned directly in a burner (combustor) or injected into the flame zone or combustion zone of the incinerator chamber (furnace) via nozzles. The heating value of the waste is the primary determining factor for nozzle location.

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Waste liquids in the process industries are numerous in kind and therefore defy easy definition. Disposal of these waste liquids has become a serious problem for plant operators. Regulations today cover wastes once considered wastewater streams, as well as the organic streams known to be made up of hazardous compounds. LI incinerators are available today to handle the various liquid streams that are being generated by the process industries. These units may be on-site (at the location of the generator) or off-site (at a sister facility or at a commercial disposal operation). LI incinerators are usually refractory-lined chambers (horizontal, vertical with orientation up or down), generally cylindrical, and equipped with a primary combustor (waste and auxiliary fuel-fired) and often secondary combustors, or injection nozzle for low-calorific-value materials (aqueous wastes containing either organic or inorganic compounds, or both). These units operate at temperature levels from 950°C (1742°F) to 1700°C (3092°F). Residence time in the chamber may vary from milliseconds to 2.5 s. The viscosity determines whether the material being incinerated is considered a liquid, slurry, or sludge. Liquid units today are capable of burning high viscosity materials of 4500 Saybolt seconds universal (SSU) or less.1,2 Critical to the operation of the unit would be the atomizing nozzle used to convert the liquid stream into finely atomized droplets.2 When considering the LI incinerator, one must be concerned about the ability of the system, that is, the design, components, controls, and air pollution control system, to meet the requirements of the federal and state regulations. The physical, chemical, and thermodynamic properties of the waste must be considered in the basic design requirements of the total incinerator system. These include storage tanks, mixers, pumps, control valves, piping, atomizers, combustors, refractory, heat recovery, quench system(s), and the air pollution equipment. The types of data needed by the designer to properly engineer the total system and the individual components are available in the literature.1 Rotary kiln (RK) process incinerators were originally designed for lime processing. Recently the RK has been utilized to process hazardous waste because it has the unique capability to stabilize the combustion process for many different materials that may be fed simultaneously. These kilns have demonstrated the ability to handle a wide variety of materials with minimal impact on their performance. The result is minimal preprocessing of the wastes and is one of the main reasons for the RK popularity. The processing of raw materials has been a historic application of the RK. The basic building materials of a community – cement, lime, coal, iron ore, aggregates, and other bulk materials – require heating to high temperatures. These materials historically have been produced in an RK. This high-temperature treatment and the continual mixing and blending of the raw materials for cement are constantly exposed to the heat of the kiln. Lime, lightweight aggregate, cement, and other process kilns may operate at temperatures up to 1925°C (1350°F). Hazardous waste incinerators operate in the range of 700−1315°C (1300−2400°F) depending on the waste types and shapes fed. The RK is a cylindrical refractory-lined shell that is mounted at a slight incline from the horizontal plane to facilitate mixing the waste materials and exposing the surface to the auxiliary burner flame plus the waste fuel flames and flames generated over the burning surface of solids. The kiln accepts all types of solid waste materials with heating values between 555 and 8333 kcal/kg (1000 and 15,000 Btu/lb), and even higher. Solid wastes and drummed wastes are usually fed by a pack-and-drum feed system, which may consist of a bucket elevator for loose solids and a conveyor system for drummed wastes. Pumpable sludges and slurries are injected into the kiln through nozzles. Temperatures for burning vary from 700°C to 1315°C (1300°F to 2400°F) depending upon the waste types and shape.

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RKs are classed as slagging [>1000°C (1832°F)] or non-slagging [ 101) = P  > (10.8) 0.5   0.5



 X − 100 101 − 100  X − 100 P ( X > 101) = P  > ; Z=  0.5  0.5  0.5



P ( X > 101) = P(Z > 2) TABLE 10.2 Incinerator Cost Comparisons for Illustrative Example 10.2 Total installed ($/yr) Operation ($/yr) Maintenance ($/yr) Total annual cost ($/yr) Income credit ($/yr) Profit ($/yr)

Cement Kiln

Rotary Kiln

678,000 400,000 650,000 1,728,000 2,000,000 272,000

755,000 550,000 775,000 2,080,000 2,500,000 420,000

(10.9)

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From a standard normal distribution table,3,4 P = 0.023



However, if X is not normally distributed, then X, the mean of a sample of n observations of X, is still approximately normally distributed with mean µ and standard deviation σ n, provided the sample size n is large (>30). This result is based on an important theorem in probability called central limit theorem.3,4 Illustrative Example 10.3 Consider the following application. Tests indicate that a sludge waste arriving in 55-gal drums to a treatment facility has a mean lead content of 11 ppm with a standard deviation of 10 ppm. The drums are unloaded into a 250-gal receiving tank. The facility is required to keep the lead concentration entering the incinerator at or below 15 ppm in order to meet the required regulatory particulate emission levels. Assume that the lead contents from one drum to the next are not correlated and that the tank is nearly full. What size should the receiving tank be to ensure, with 98% confidence, that the facility treats a waste with a mean lead concentration below 15 ppm? SOLUTION: For this condition, the probability that the lead (Pb) concentration in the receiving tanks exceeds 15 ppm is

1.0 − 0.98 = 0.02 or 2.0%

The value of z0 from the standard normal table3 is then 2.05, which corresponds to the number of standard deviations above the mean tank concentration. According to the aforementioned central limit theorem, the standard deviation of the mean lead concentration is given by

10 ppm σ = (10.10) n n

where n is the number of drums. Therefore, for a 2% probability that the mean concentration in the tank exceeds 15 ppm, the number of drums can be found by solving

2.05 =

15 − 11  10     n

or

n = 26.3

The tank volume V must then be (approximately)



V = 26.3 drums ( 55 gal/drum ) = 1446 gal



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10.4 Discounted Cash Flow Application Illustrative Example 10.4 Two small commercial solid waste facility designs are under consideration. The first design involves fixation and the second involves encapsulation. (In fixation, a solidifying agent is used to solidify the waste. Encapsulation is a process in which the waste is surrounded by a bundle; it has often been solidified by a chemical agent.) For the fixation system, the total capital cost (TCC) is $2.5 million, the annual operating costs (AOC) are $1.2 million, and the annual revenue generated from the facility (R) is $3.6 million. For the encapsulation system, TOC, AOC, and R are $3.5, 1.4, and 5.3 million, respectively. Using straight-line depreciation and the discounted cash flow method, which design is more attractive? Assume a 10-year facility lifetime and a two-year construction period. Note that the solution involves the calculation of the rate of return for each of the two proposals.2,5 An after-tax rate of return on an initial investment of at least 30% is usually desirable. The method used to arrive at a rate of return is discussed below. An annual after-tax cash flow can be computed as the annual revenues (R) less the annual operating costs (AOC) and less total income taxes (IT). Total IT can be estimated at 50% (this number could be significantly lower subject to the passage of the new tax laws at the time of the preparation of this manuscript) of taxable income (TI). IT = 0.5 ( TI ) (10.11)



The TI is obtained by subtracting the AOC and the depreciation of the plant (D) from the revenues generated (R) or

TI = R − AOC − D (10.12)

If straight-line depreciation is assumed, the plant will depreciate uniformly over the life of the plant. Thus, for a 10-year lifetime, the facility will depreciate 10% each year.

D = 0.1( TCC ) (10.13)

The annual after-cash flow (A) is then

A = R − AOC − IT (10.14)

This procedure involves a trial-and-error solution. There are both positive and negative cash flows. The positive cash flows consist of A and the recoverable working capital in year 10. Both should be discounted backward to time = 0, the year the facility begins operation. The negative cash flows consist of the TCC and the initial working capital (WC). In actuality, the TCC is assumed to be spent evenly over the two-year construction period. Therefore, one-half of this flow is adjusted forward from after the first construction year, (time = −1 yr) to the year the facility begins operating (time = 0). The other half, plus the WC, is assumed to be expended at time = 0. Forward adjustment of the 50% TCC is accomplished by multiplying by an economic parameter known as the singlepayment compound amount factor F/P, given by

m F = (1 + i ) (10.15) P

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where: i = rate of return (fraction) n = the number of years (in this case, 1 yr) For the positive cash flows, the annual after-tax cash flow (A) is discounted backward by using a parameter known as capital recovering factor (CRF) or the uniform series present worth factor (P/A). This factor is dependent on both interest rate (rate of return) and the lifetime of the facility and is defined by

(1 + i ) − 1 (10.16) n i (1 + i ) n

P/A =



where n is the lifetime of the facility (in this case, 10 years). The recoverable working capital at year 10 is discounted backward by multiplying WC by the single payment present worth factor (P/F) P/F =



1

(1 + i )

n

(10.17)

where n is the lifetime of the facility (in this case, 10 years). The positive and negative cash flows are now equated and the value of i, the rate of return, may be determined by trial-and-error from Equation 10.18. Term 1 = Term 2 − Term 3 + Term 4 (10.18)

where:

 (1 + i )10 − 1   A; worth at year 0 of annual after − tax cash flows (10.19) Term 1 =  10  i (1 + i ) 



 1   Term 2 =  10 WC ; worth at year 0 of recoverable WC after 10 year (10.20)  (1 + i ) 



1   Term 3 =  WC + TCC  ; assumed exoenditures at year = 0 (10.21) 2  

Term 4 =

1 1 ( TCC ) (1 + i ) ; worth at year = 0 of assumed expenditures at year = −1 (10.22) 2

SOLUTION: For the fixation system, calculate, D, WC, TI, IT, and A. The depreciation is D = 0.1( TCC )

= ( 0.1)( $2, 500, 000 ) (10.23) = $250, 000

The WC is set at 10% of the TCC.

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WC = 0.1( TCC ) = ( 0.1)( $2, 500, 000 ) (10.24)



= $250, 000 In addition, TI = R − AOC − D = $3, 600, 000 − $1, 200, 000 − $250, 000 (10.25)



= $2, 150, 000 and IT = ( 0.5 ) TI = ( 0.5 )( $2, 150, 000 ) (10.26)



= $1, 075, 000 The after-tax cash flow is calculated using A = R − AOC − IT = $3, 600, 000 − $1, 200, 000 − $1, 075, 000 (10.27)



= $1, 325, 000 The rate of return, i, for the fixation unit is also calculated. The rate of return can be computed by solving the following equation:

 (1 + i )10 − 1   1  1  A+  10 10 WC = WC + ( 0.5 ) TCC + ( 0.5 ) TCC ( 1 + i ) (10.28)  I (1 + i )   (1 + i ) 

or (on substituting)  (1 + i )10 − 1   1    1.325 × 10 6 +   0.250 × 10 6 10 10  I (1 + i )   (1 + i ) 

(





(

)

)

(

) (

(

(

= 0.250 × 10 6 + ( 0.5 ) 1.250 × 10 6 +  ( 0.5 ) 1.250 × 10 6 

)

)) (1 + i )  1

By trial-and-error, i = 39.6%

For the encapsulation system,

WC = D = (0.1)($3, 500, 000 ) = $350, 000



(10.29)

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TI = $5, 300, 000 − $1, 400, 000 − $350, 000



= $3, 550, 000 IT = (0.5) ($3, 550,, 000 ) = $1, 775, 000 The annual after-tax cash flow is A = $5, 300, 000 − $1, 400, 000 − $1, 775, 000





= $2, 125, 000

The rate of return equation for the encapsulation unit becomes



 (1 + i )10 − 1   1    2.125 × 10 6 +   0.3650 × 10 6 10 10  I (1 + i )   (1 + i ) 

(

(

)

(

(

)

) )



= 0.350 × 10 6 + ( 0.5 ) 1.750 × 10 6 + ( 0.5 ) 1.750 × 10 6 (1 + i )   1

By trial-and-error,

i = 44.8%

Hence, by the discounted cash flow method, the rate of return on the initial capital investment is approximately 5% greater for the encapsulation system than the fixation system. From a purely financial standpoint, the encapsulation system is more attractive.

10.5 Maximum Mercury Waste Emission Constraints Are mercury emissions important? This question was briefly discussed in Chapter 8, Section 8.5. This application involves an extension of that problem. One of the most important issues is what environmental factors influence the formation of methyl mercury, the element’s most toxic chemical compound and the one most easily incorporated into an aquatic food chain. More than 95% of the mercury in fish is in the form of methyl mercury. Most atmospheric mercury occurs simply in its elemental form, which must be oxidized before being carried to the earth in rain. In a lake, the transformation of the oxidized form to methyl mercury, a process called methylation, takes place mainly through the action of bacteria. The amount of methyl mercury available in the water at any time depends on a balance between methylation and its opposite reaction, demethylation. Researchers are particularly interested in determining what factors can tip this balance one way or the other. Fish can absorb the methyl mercury directly through their gills or ingest it by eating smaller organisms. Microscopic plants (phytoplankton) absorb considerable amounts of methyl mercury from the water, so the small fish that eat them receive an already concentrated dose. Large predatory fish then eat the smaller ones, and methyl mercury builds up in their bodies because it is difficult to excrete. Eventually, the magnitude of such

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bioconcentration becomes staggering. Fish at the top of a food chain may have levels of methyl mercury a million times the level in the surrounding water.6,7 Illustrative Example 10.5 Consider the following application. Refer to Chapter 8, Section 8.5. If no more than 150 lb/h of mercury can be incinerated, what is the maximum amount of mercurycontaminated plating waste that can be incinerated in order to operate at maximum profit? SOLUTION: A seventh constraint is added to the earlier application (See also Table 8.4 and Figure 8.1). It is required to keep the Hg  f1

a21x1 + a22 x2 ,… + a2 n xn > f 2 

(11.9)

am1x1 + am 2 x2 ,… + amn xn > f m with

xn ≥ 0 xn ≥ 0

 xn ≥ 0

(11.10)

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11.3 Reducing Insurance Costs Illustrative Example 11.3 A structural engineering summer intern at NAHEB Construction Associates was assigned a project concerned with reducing a company’s insurance costs by installing induced draft fans on every floor of the company’s eight-story building. The purpose of the fans is to provide the ventilation needed to reduce the indoor air pollutants that reportedly have affected the health and performance of several of the company’s personnel. After several months of reviewing the literature and numerous discussions with Lou Theodore, the nation’s foremost authority on air pollution, the intern developed the following economic model: IR = A +



Bx − Dx ; $ yr (11.11) C+x

( 1) ( 2 ) ( 3 )





where: IR is the annual insurance reduction Term1 is the savings factor concerned with making the employees aware of the problem Term2 is the savings factor concerned with the installation of x fans Term3 is the savings factor (a cost) concerned with the purchase of x fans Determine the number of fans that should be installed to maximize the annual insurance reduction. SOLUTION: In order to reduce insurance costs (a maximization operation), the intern decided to solve the insurance reduction (IR) model analytically. The calculation involved generating d(IR)/dx and setting the result equal to zero, that is, d ( IR ) = 0 (11.12) dx



For the equation provided in the problem statement, d ( IR ) B Bx = − − D = 0 (11.13) dx C + x ( C + x )2



Equation 11.13 is a form similar to that for a quadratic equation, that is,

(

)

B ( C + x ) − Bx − D C 2 + 2Cx + x 2 = 0 (11.14)

or

Dx 2 + 2CDx + C ( CD − B ) = 0 (11.15)

The solution to Equation 11.15 is given by

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x=



−b ± b 2 − 4 ac (11.16) 2a

where: a = D b = 2CD c = C(CD−B) Substitution yields

x=

−2CD ± 4C 2D2 − 4CD2 − 4BCD (11.17) 2D

The intern was notified that she would be hired next summer to generate numerical values for coefficient A, B, C, and D.

11.4 Reducing Health Concerns Associated with Flue Gas Emissions As noted earlier, there are many definitions of the word “risk.” Some of the definitions provided are as follows: It is a combination of uncertainty and damage; a ratio of hazards to safeguards; a triplet combination of event, probability, and consequences; or even 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 every day, 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 health risk assessments (HRAs). The usual objective of HRA and the accompanying calculations are to evaluate the potential for adverse health effects from the release of chemicals into the environment. Unfortunately, the environment is very complex since there is a vast array of potential receptors present. The task of testing and evaluating all of the enormous number of chemicals on the market for their impact on human populations and ecosystems becomes extremely difficult. To further complicate this problem, health is a concept that has come to mean different things to different people. Some have defined it as follows: “…a state of complete physical, mental and social well-being and not merely the absence of disease or infirmary.” Many other definitions and concepts have been proposed and appear in the literature. Since 1970, the field of HRA has received widespread attention within both the scientific and the regulatory communities. It has also attracted the attention of the public. Properly conducted risk assessments and risk assessment calculations have received fairly broad acceptance, in part because they put into perspective the terms toxic, health, risk, and hazard (see also Chapter 13). Toxicity is an inherent property of all substances. It states that all chemical and physical agents can produce adverse health effects at some specific exposure conditions. In contrast, exposure to a chemical that has the capacity to produce a particular type of adverse effect represents a health problem or “hazard.” Health risk (in a general sense), 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 the

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hazardous agent. Health risk is therefore a function of exposure and dose. Consequently, HRA 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. Corporations that manufacture, sell, and purchase chemicals now realize that the chemicals they handle present health risks to their employees, their customers, and/or the public. The assessment process is often an enormous task since the health risks of scores of chemicals may have to be assessed from a risk perspective. Each individual HRA is a multistep process consisting of

1. Health problem identification 2. Toxicological concerns 3. Exposure characterization 4. Risk characterization Illustrative Example 11.4 In an attempt to minimize the reduction of a host of gaseous pollutants, a state-run utility has proposed to convert some of its coal-fired boilers to oil-fired in order to reduce health concerns associated with flue gas emissions. DuPont Associates was hired to generate a model (objective function) that would minimize the cost of the replacement relative to the quantity of oil to be combusted. Two oils – (1) and (2) – are available to supply the necessary energy for the new boilers, x1 and x2. The proposal also requested that an attempt be made to salvage as many of the old boilers (x3) currently burning coal.6 The cost associated with each oil varies with its content, regulatory requirements, heating value, and so on. The engineers at Dupont Associates proposed the following cost model, DA:

DA = 3 x1 + 4 x2 − x3 ; 10 −4 $ (11.18)

(Salvaging the existing boilers resulted in a credit.) The constraints, primarily due to emission rules for NOX, SOX, and ROX were:

x2 + x3 ≤ 100 ( nitrous oxide, NOX ) (11.19)



x1 + x3 ≤ 50 ( sulfur dioxide, SOX ) (11.20)



x1 + x2 x3 + x32 ≤ 1000 ( particulates, ROX ) (11.21)

with

x1 < x2 < x3 ≤ 50 (11.22)

The state-run facility has been informed that capital cost expenditures would have to be below $500,000 for this cost minimization project. SOLUTION: Excel provides the following solution:

x1 = 4

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x2 = 10



x3 = 26

with

DA = $26 × 10 4 = $260, 000

Thus, this minimization project should go forward. The solution requires the purchase of four new boilers that burn oil (1), 10 new boilers that will burn oil (2), and the reassignment of 26 of the old coal-fired boilers.

11.5 Vitamin Caloric and Protein Requirements A protein is any of a large number of organic compounds that makes up living organisms and is essential to their functioning. Protein molecules range from the long, insoluble fibers that make up connective tissue and hair to the compact, soluble globules that can pass through cell membranes and set off metabolic reactions. They are all large molecules, ranging in molecular weight from a few thousand to more than a million, and they are specific for each species and for each organ of each species. Humans have an estimated 30,000 different proteins, of which only about 2% have been adequately described. Proteins in the diet primarily serve to build and maintain cells, but their chemical breakdown also provides energy, yielding close to the same four calories per gram as do carbohydrates. Illustrative Example 11.5 Vitamin (x1) costs $0.20 per mg and contains 50 cal/vitamin and 20 mg. protein/ vitamin. Vitamin  (x 2) costs $0.15 per mg and contains 40 cal/vitamin and 10 mg. protein/vitamin. What amounts of vitamins x1 and x 2, in mg would meet the minimum regulatory requirement while minimizing the cost, C. The US Department of Agriculture (USDA) provided the following objective function for C:

C($) = 0.20 x1 + 0.15x2

(11.23)

along with the constraints

50 x1 + 40 x2 ≥ 300 (11.24)



20 x1 + 10 x2 ≥ 120 (11.25)

and

x1 ≤ 6 (11.26)



x2 ≤ 6 (11.27)

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SOLUTION: Excel provides the following solution for the number of vitamins:

x1 = 6



x2 = 0

and

C = $1.20

11.6 Pharmaceutical Cancer Drug Cancer is defined as new growth of tissue resulting from a continuous proliferation of abnormal cells that have the ability to invade and destroy other tissues. Cancer is known throughout the world as a fearsome cause of illness and death. Not all cancers are the same, however, nor are all persons equally at risk. The disease can occur in any tissue where cells are growing and dividing, including such widely different tissues as those of the lungs, skin, brain, stomach, and bones. Each kind of cell gives rise to its own kinds of cancer, with its own causes, frequency, symptoms, treatment, and outcome. Some of the most common kinds of cancer are those of the lung, prostate, and large bowel (for older men) and of the breast, large bowel, and uterus (for older women). Cancer is less common in children and young adults; when it occurs, it is likely to arise in the bone marrow (the forms of cancer called leukemia), lymphatic tissue (including Hodgkin’s disease), brain, kidneys, and bones. These generalizations exclude the common forms of skin cancer, which – unique among cancers – are easily diagnosed and treated in their early stages and that rarely cause serious disability unless long neglected. Cancer is often discovered because it causes symptoms that lead the patient to seek medical help, or it may be found by special screening examinations beforehand. Good reasons exist to believe that the earlier a cancer is detected and treated, the higher the chance of cure. Cancer is a disease of individual cells, and even one cell can grow and divide many times to produce a new cancer mass. Thus, the cure can be accomplished only by removing or killing every single malignant cell. The three accepted ways of doing this are surgery, radiation, and drugs, which are often used in combination. Several dozen drugs are used at present to treat cancer, and research has shown just how the drugs should be used for best effect. The size and timing of each dose may have profound effects on drug action, as may the drug’s integration with surgical or radiation treatment. Moreover, because some of the drugs attack cancer cells in different ways, the use of as many as four or five such drugs in a precisely determined combination or sequence is commonly more effective than the use of just one. Illustrative Example 11.6 ABC Drug Laboratory is working to produce four variations of a new cancer drug. Initial estimates of the predicted hourly profit from the production of drugs A, B, C, and D were $100, $150, $200, and $250 respectively. Table 11.1 shows how much time

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TABLE 11.1 Drug Laboratory Production Time Expenditure (Hours) for Illustrative Example 11.6 Lab 1 2 3

Drug A

Drug B

Drug C

Drug D

10 6 3

4 5 9

8 3 7

9 9 12

is required in each of the three research labs to make one unit of each of the four products. Laboratory 1 has no more than 300 man-hours available per day. Laboratory 2 is limited to no more than 400 man-hours per day. Similarly, Laboratory 3 is limited to 425 man-hours per day. As a recent biomedical engineering graduate, you have been hired to determine the number of A, B, C, and D units that should be produced to maximize daily profit, P. The following information is provided

P = 100 x1 + 150 x2 + 200 x3 + 250 x 4 ; $ h (11.28)

where x1, x2, x3, and x4 represent the number of drugs A, B, C, and D, respectively, produced per hour. The man-hour constraints are

10 x1 + 4 x2 + 8 x3 + 9 x 4 ≤ 300 (11.29)



6 x1 + 5x2 + 3 x3 + 9x 4 ≤ 400 (11.30)



3 x1 + 9x2 + 7 x3 + 12x 4 ≤ 425 (11.31)

and

x1 ≥ 0 (11.32)



x2 ≥ 0 (11.33)



x3 ≥ 0 (11.34)



x 4 ≥ 0 (11.35)

SOLUTION: Excel provides the following solution:

x1 = 0



x2 = 29



x3 = 23



x4 = 0

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with a profit of



P = $8, 950 h = $214, 900 day ( round the clock operation )



References 1. L. Theodore and R. Dupont: Environmental Health and Hazard Risk Assessment: Principles and Calculations, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2012. 2. J. Reynolds, J. Jeris, and L. Theodore, Handbook of Chemical and Environmental Engineering Calculations, John Wiley & Sons, Hoboken, NJ, 2004. 3. L. Theodore, R. Dupont, and K. Ganesan, Unit Operations in Environmental Engineering, Wiley‑Scrivener, Salem, MA, 2018. 4. L. Theodore, J. Reynolds, and K. Morris, Concise Dictionary of Environmental Terms, Gorden and Breach Science Publishers/CRC Press/Taylor & Francis Group, Boca Raton, FL, 1997. 5. USEPA, Guidelines for Carcinogen Risk Assessment, EPA/630/R-98/002, Risk Assessment Forum, Washington, DC, 2005. 6. K. Skipka and L. Theodore, “Energy Resources: Availability Management, and Environmental Impacts”, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2014.

12 Hazard Risk Assessment The last chapter in Part II contains six illustrative examples. Example titles are listed here:

12.1 Safety expenditures versus return of safety investments 12.2 Plant safety proposal comparisons 12.3 Effect of failure distribution rate 12.4 Estimating the effect of worst-case explosions 12.5 Economics of fire protection equipment 12.6 Optimizing the operation of a cement incinerator Seven references are recommended for suggested reading:





1. L. Theodore, Chemical Engineering: The Essential Reference, McGraw-Hill, New York City, NY, 2014. 2. J. Reynolds, J. Jeris, and L. Theodore, Handbook of Chemical and Environmental Engineering Calculations, John Wiley & Sons, Hoboken, NJ, 2004. 3. L. Theodore, R. Dupont, and K. Ganesan, Unit Operations in Environmental Engineering, Wiley‑Scrivener, Salem, MA, 2018. 4. M. K. Theodore and L. Theodore, Introduction to Environmental Management, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2008. 5. G. Burke, B. Singh, and L. Theodore, Handbook of Environmental Management and Technology, 2nd edn, John Wiley & Sons, Hoboken, NJ, 2000. 6. L. Theodore and R. Dupont, Environmental Health and Hazard Risk Assessment: Principles and Calculations, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2012. 7. L. Theodore, Environmental Risk Analysis: Probability Distribution Calculations, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2016.

12.1 Safety Expenditures versus Return of Safety Investments Protective systems are employed to reduce/eliminate hazards and risks and may be viewed as a special category of process equipment. The management of the protective system’s operation and maintenance and testing information should occur within the overall management approach. Provisions for protection and safety equipment should be incorporated in the original plant design. The size of the plant, nature of the

165

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hazards, and potential exposures will determine the amount, kind, and location of this equipment. Regarding fires, water is the primary extinguishing agent, and it should be available in adequate supply and at adequate pressure at all locations in the plant. The layout for various types of fire protection installations and the appropriate recommendations for their design are found in the standards of the National Fire Protection Association. Fire hydrants, hose lines, and automatic sprinklers and water spray systems should all be a part of the permanent equipment facilities of the plant. Fire extinguishing systems can include foam, carbon dioxide, and dry chemicals. Wetting agents and high-expansion foam have been used in some plant protection systems. All fire extinguishing systems should be evaluated for their potential health risks as well as their overall effectiveness before incorporation into a plant. Since many chemical plants have severe potential health and hazard problems, it is essential to provide medical facilities and first aid stations in case rapid medical response is required. In addition, showers and eyewash stations are necessary in certain hazardous areas. Also, guards and covers should be provided for all moving equipment. Ten of the key protective equipment elements in an environmental or chemical process plant or manufacturing facility are listed here:1 1. Pressure relief/vent collection 2. Release devices (flares, scrubbers, etc.) 3. Plant equipment isolation 4. Critical alarms/interlocks 5. Fire detection/protection 6. Hazardous/toxic gas detection 7. Flame arrestors 8. Emergency system services 9. Appropriate grounding and bonding 10. Personal protection equipment Illustrative Example 12.1 BETH Construction Company has decided to implement a new proposal from Doyle Engineers in the construction of a 57-story high-rise building in San Francisco that will house offices on floors 1–6 and apartments on the floors above. The new proposal from the structural engineer at Doyle recommends that their recent patented quakebuster columns be inserted at strategic locations in the building in order to allow the structure to survive earthquakes up to 7.5 on the Richter scale. Billy B engineers have estimated that the value (worth) of the augmented high rise, including the column costs, is given by



V = 25.6 +

x ( 2.56 ) − 0.015x ; MM$ (12.1) x + 3.5

where x = number of quake-buster columns Determine the optimum number of columns that BETH should install in the high rise.

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Hazard Risk Assessment

SOLUTION: This illustrative example requires the optimization of V. To employ an analytical solution, proceed as follows:

x ( 2.56 ) dV 2.56 = − − 0.015 (12.2) dx x + 3.5 x + 3.52

Setting this equation equal to zero, that is, dV/dx = 0 and solving for x (positive/negative) yield x ≅ 20. Therefore, 20 quake-buster columns should be employed in the new highrise to maximize the value of the structure, increasing it from 25.6 to 27.5 MM$.

12.2 Plant Safety Proposal Comparisons The academic training of engineers with respect to plant equipment has traditionally focused on design and predicting performance. Optimizing operation from a safety perspective is often an overlooked consideration. Little to no effort has historically been expended in attempting to answer the question

WHAT CAN GO WRONG?

This question requires answers in today’s high-tech environment. Failure to properly and realistically address and answer this question can result/lead to an

ACCIDENT !

Once an industrial accident has occurred, the responsible individual (usually the operator or engineer) should react as quickly as possible in a responsible manner (exercise good judgment) to detect and correct the situation, and hopefully reduce and/or eliminate potential danger(s). Industry now knows that equipment sometimes partially fails and sometimes catastrophically fails. In any event, it behooves the engineer to carefully and thoroughly examine the various pieces of plant equipment that can fail. A partial list of some of the more common equipment that can be expected to fail in a plant with some regularity is provided subsequently. Details are available in the literature.1,2

1. Foundations 2. Structural steel 3. Vessels and tanks 4. Pumps 5. Compressors 6. Fans 7. Heat exchangers 8. Turbines 9. Electrical systems 10. Instrumentation and controls

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11. Piping systems 12. Valves and joints 13. Mass transfer operations 14. Heaters and furnaces 15. Chemical reactors 16. Fire protection equipment 17. Safety equipment

Deviations, that is, abnormal conditions that can lead to a failure and/or an accident generally occur because of

1. Abnormal temperatures 2. Abnormal pressures 3. Material flow stoppage 4. Equipment leaks 5. Equipment spills 6. Materials failure due to wear 7. Materials failure due to imperfections 8. Materials failure due to poor maintenance 9. Materials failure due to corrosion

Some specific operational failures include

1. Blocked outlets 2. Opening/closing valves 3. Cooling water failure 4. Power failure 5. Instrument air failure 6. Thermal expansion 7. Vacuum problems Illustrative Example 12.2 Consider the following application. Four different plant safety proposals are under consideration at the KBLT Power Plant. The life of each project is 20 years. The individual capital and annual operating costs are provided in Table 12.1. The salvage value at the termination of the project is zero and there are no demolition costs. Which is the least expensive project? Assume the straight-line depreciation method applies. SOLUTION: The total annualized cost calculations (TAC) are provided in Table 12.2. Comparing projects 1, 2, 3, and 4 indicates that unit 3 has the lowest TAC and should be selected as the most economical project of the four being evaluated.

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TABLE 12.1 Cost Data for Four Safety Purposes; Illustrative Example 12.1 Project 1 2 3 4

Initial Cost, $

Operating Cost, $/yr

2,600,000 2,400,000 2,450,000 1,850,000

200,000 250,000 275,000 350,000

TABLE 12.2 Cost Analysis for Four Safety Purposes; Illustrative Example 12.1 Project Capital cost ($) Depreciation ($) Operating costs ($) TAC ($)

1

2

3

4

2,600,000 130,000 200,000 330,000

2,400,000 120,000 150,000 270,000

2,450,000 125,000 275,000 400,000

1,850,000 92,250 350,000 442,250

12.3 Effect of Failure Distribution Rate The probability distribution of a random variable concerns the distribution of probability over the range of the random variable. The distribution probabilities are specified by the probability distribution function (pdf). This introduction is devoted to providing general properties of the pdf in the case of discrete and continuous random variables, as well as an introduction to the cumulative distribution function (cdf). Special pdfs finding extensive application in hazard and risk analysis are available in the literature.3 The pdf of a discrete random variable X is specified by f(x), where f(x) has the following essential properties: 1. f ( x ) = P ( X = x ) = probability assigned to the outcome



corresponding to the number x in the range of X ; (12.3) i.e., X is a specifically designated value of x 2. f ( x ) ≥ 0 (12.4)



3.

∑ f (x) = 1 (12.5) x

Property 1 indicates that the pdf of a discrete random variable generates probability by substitution. Property 2 and Property 3 restrict values of f(x) to non-negative real numbers and numbers whose sum is 1, respectively. The pdf of a continuous random variable x has the following properties: b



∫ f ( x ) dx = p(a < x < b) (12.6) a

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f ( x ) ≥ 0 (12.7)

∞

∫ f ( x ) dx = 1 (12.8)



−∞

Equation 12.6 indicates that the pdf of a continuous random variable generates probability by integration of the pdf over the interval whose probability is required. When this interval contracts to a single value, the integral over the interval becomes zero. Therefore, the probability associated with any particular value of a continuous random variable is zero. Consequently, if x is continuous P ( a ≤ x ≤ b ) = P( a < x ≤ b) = P( a < x < b) (12.9)



= P( a ≤ x < b) Equation 12.7 restricts the values of f(x) to non-negative numbers. Equation 12.8 follows from the fact that P ( −∞ < x < ∞ ) = 1 (12.10)



The following properties of the cdf of a random variable, x, can be deduced directly from the definition of F(x).

F ( b ) − F ( a ) = P( a < x ≤ b) (12.11)



F ( +∞ ) = 1 (12.12)



F ( −∞ ) = 1 (12.13)



F ( x ) is a nondecreasing function of x (12.14)

These four properties apply to the cases of both discrete and continuous random variables. Illustrative Example 12.3 Consider the following application. You have been assigned to run ten 80-day corrosion tests. Each testing device can be used for only one test, after which it must be discarded. Device A costs $400 each. Data from past tests indicate these devices have a failure rate distribution curve shown in Figure 12.1.

1. How many devices would you expect to use? 2. Competing device, Device B, with the uniform failure rate distribution provided in Figure 12.2, is available for $500 each. To minimize costs, is it economically advantageous to purchase the more expensive device?

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Hazard Risk Assessment

0.015 0.0125 f(t)

0.010 0.075 0.005 0.025 0.0

8 16

32

64

128

Time (d) FIGURE 12.1 Failure rate, f(t), as a function of time for Device A; Illustrative Example 12.3. 0.015 0.010 f(t)

f(t)= 0.008

0.005 0.0

8 16

32

64

128

Time (d) FIGURE 12.2 Failure rate, f(t) as a function of time for Device B; Illustrative Example 12.3.

SOLUTION:

1. Calculate the probability of the device lasting at least 80 days using the following equation:

80



∫

F ( t ) = f ( t ) dt = 0

8

80

0

8

∫ (0.0125) dt + ∫ (0.005) dt



= 0.10 + 0.36 = 0.46 Number of A devices most likely required =

number of tests required 10 = = 21.7 probability of success 0.46

Therefore, 22 are needed at a total cost of $8,800.





2. The probability of success for the B device devices is

F (t ) =

80

∫ (0.008 ) dt = 0.64 0

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Therefore, the number of B devices needed = 10/0.64 = 15.6 = 16 for a total cost = $8,000. The more expensive devices appear to be justified. Since this question relates to “most likely” outcomes, it would also be acceptable to base the calculations on factional devices needed. Doing so yields

 $ 400  = $8, 680 Device A: 21.7   device 



 $500  = $7 , 800 Device B: 15.6   device 

So, the conclusion remains the same.

12.4  Estimating the Effect of Worst-Case Explosions1,4 One of the most hazardous situations encountered in a process plant is the atmospheric detonation of hydrocarbon vapors that have been accidentally released. Such explosions, referred to as “unconfined vapor cloud explosions,” have been responsible for some of the industry’s worst accidents, such as the one at Flixborough, England, in 1974.1 An explosion at the Nypro Ltd. caprolactam factory in Flixborough, England, on June 1, 1974, was one of the most serious in the history of the chemical industry and the most serious in the history of the United Kingdom. Of those working on the site, 28 were killed and 36 others injured. Outside the plant, 53 people were reported injured, while 1821 houses and 167 shops suffered damage. The estimated cost of the damage was well over $100 million. The cyclohexane oxidation plant in Flixborough, which stored this solvent at about 120 psig and 145°C, contained six reactors in series. One reactor had been removed for repairs, and the resulting gap was bridged by a temporary 20-in pipe, connected by a bellows at each end and supported on temporary scaffolding. This pipe collapsed, and in the minute or so that elapsed before ignition, about 50 tons of cyclohexane escaped. The extensive damage that occurred could have resulted from the deflagration of as little as 10 to 20 tons of the chemicals. It was later determined that no calculations had been done to ascertain whether the bellows or pipe could withstand the strain. Only the capacity of the assembly to carry the required flow was calculated. No reference was made to any accepted standard, nor to the designer’s guide issued by the manufacturer of the bellows. Neither the pipe nor the complete assembly unit was pressure tested before being fitted. Apparently, no one realized that the pressurized assembly would be subject to a turning moment, imposing shear forces on the bellows for which they were not designed. Nor did anyone observe that the hydraulic thrust on the bellows would tend to make the pipe buckle at the joints. The temporary pipe connection functioned satisfactorily after the initial installation. It was never closely inspected, however, even though when at operating temperature and pressure, it was observed to lift off the scaffolding that had been put in to support it. What happened on the final shift will never be known because all those in the control room were killed, and all instrumentation and records were destroyed. The equivalent force of the explosion was estimated at 15 tons of TNT, minimum.

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Hazard Risk Assessment

Explosions are frequently calibrated in terms of the aforementioned equivalent mass of TNT. One lb of TNT releases 1983 Btu (500 Kcal) on detonation; therefore, any explosion that releases 1983 Btu is said to be a “1 lb of TNT” explosion. The overpressure at any distance from an explosion can be estimated using the following method. The overpressure from one kg of TNT, as a function of distance, is given by P ( psi ) =



1737 1.875 + − 0.01556 (12.15) D2 D

with D in feet, for the range from 3 to 400 ft. The distance for equal overpressures then scales as the 1/3 power of the energy of the explosion (E): 1

 D1   E1  3  D  =  E  (12.16) 2 2



Illustrative Example 12.4 Consider the following hypothesis. The hot hydrocarbon vapor from a catcracker at Petro Unlimited suddenly releases vapors to the atmosphere, mixes with the air, and explodes.

1. Estimate the maximum possible energy release (Emax) and the TNT equivalency. 2. Estimate the radius from the center of the explosion at which overpressures of 1 psi (sufficient to injure many people and cause damage to buildings), and 5 psi (sufficient to break wooden telephone poles) would occur. 3. Based on the results of (1) and (2), discuss how the hazard risks from the catcracker can be minimized.

The cracker has a volume of 800 ft3. The vapors at the time of release consist of light hydrocarbons with an average molecular weight of 55, at 25 psi and 350°F. The average heat of combustion in air of the hydrocarbons is 19,000 Btu/lb. SOLUTION:

1. Calculate the moles of hydrocarbon released and convert to lbs: N hc =

=



PV RT

( 25)(8000) (10.73)(810)

= 2.3 lbmol

Mhc = ( 2.30 )( 55 ) = 127 lbs    

Determine the maximum possible combustion energy release and convert it to lb TNT equivalent: Emax = (127 )(19, 000 ) = 2, 400, 000 Btu

=

2, 400, 000 = 1210 lb TNT 1983



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2. Calculate the distance for an overpressure of 5 psi. Employ Equation 12.15. For 5 psi overpressure with 1 kg TNT:

5=



1737 1.875 + − 0.01156 D2 D

Solving for D: D = 18.8 ft



  Convert the energy of the blast to kg TNT: E=



1210 lb TNT = 550 kg TNT 2.2 lb / kg

Scale the distance for 550 kg TNT. Employ equation 12.16: 1

1

 E 3  550  3 D1 = D2  1  = (18.8 )   = 154 ft E  1   2



  Finally, calculate the distance for an overpressure of 1 psi. The distance for 1 psi overpressure with 1 kg TNT is given by: 1=



1737 1.875 + − 0.01156 D2 D

Solving for D:

D = 42.1 ft

Scale the distance for 550 kg TNT employing Equation 12.16. 1



1

 E 3  550  3 D1 = ( D2 )  1  = ( 42.1)   = 345 ft E  1  1   3. It has been reported that in most hydrocarbon vapor explosions, only 10 to 25% of the maximum possible combustion energy is released. This results from the vapor/air mixture being ignited when the mixture is still “rich”, that is, before the amount of air for complete combustion has had a chance to mix with the hydrocarbon vapors. Thus, efforts should be directed to reducing the amount of energy released. In addition, consideration should be given to surrounding the catcracker with barriers.

12.5 Economics of Fire Protection Equipment The most common cause of fire accidents in environmental and chemical process plants is equipment failure. This is primarily a result of poor equipment maintenance or poor equipment layout and design. Maintenance performed according to a detailed and well-structured schedule will significantly reduce the occurrence of fire accidents. The second largest cause

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175

of fire accidents is ignorance of the properties of a specific chemical or chemical process. Proper training of employees will increase their knowledge of the properties of a specific chemical and chemical process, and it can prevent many of these chemical fire accidents. Leaks and spills resulting from an equipment failure will frequently cause fires. Large leaks can occur when a vessel, pipe, or pump fails. Failure of a pump seal, pipe flange, or bore connection will cause a somewhat smaller leak. Since fires at pumps and/or at flanges are highly possible, it is essential to assess and minimize the effect of such fires on the equipment above or near pumps or flanges. Incidence of pump fires can be minimized by using double mechanical seals. Pipe flanges also can be protected against fires by placing an insulating pad between the joint ring and pipe bore and by specifying more massive flanges to minimize the temperature gradient. Thermal insulation (or lagging) on plant equipment may become soaked or impregnated with oils and other flammable liquids. When the lagging gets hot, spontaneous combustion can occur. Lagging fires are affected by oil leaks, insulation material, and temperature. Spontaneous combustion occurs only when the oil is non-volatile, since volatile oil evaporates more easily, thus delaying the accumulation of oil that reaches its auto-ignition temperature (AIT).1 A continuous leak may actually inhibit lagging fires since the lagging fire can be over-saturated. A good insulating material with a low thermal conductivity is favored by the lagging fire. A low density, porous structure provides surface area for the air to diffuse in. A thick lagging gives good insulation; therefore, spontaneous combustion is more likely on a thicker lagging. To prevent lagging fires, the following precautions are recommended:

1. Prevent leaks. 2. Use appropriate sealings (cement finish or aluminum foil). 3. Use special insulation materials such as foam glass or crimped aluminum sheeting.

The potential for fire hazards is rather high in the chemical industry. However, this potential is generally judged to be less than that of an explosion or toxic release. The scale of a fire hazard can be determined by assessing the following factors: 1,2 1. Inventory. The larger the inventory of material, the greater the loss potential. 2. Energy. The more energy available for a release, such as the stored energy in a material state or chemical reaction, the greater the potential. 3. Time. A higher rate of release of a hazard and the distance over which it may cause injury or damage also directly affect the potential. 4. Exposure. The intensity of the hazard and the distance over which it may cause injury or damage also directly affect the potential. Most fire accidents involve a large loss of life and extensive property damage. Frequently, a fire is preceded by an explosion. Explosions are usually more lethal than fires.1 Some major fires in industrial plants have been described in detail in the literature.1,2 The thermal radiation intensity and the time duration of fires often are used to estimate injury and damage due to a fire. The best way to fight a fire is to remove any one of three essential conditions required to sustain the fire:

1. The fuel source (e.g., by eliminating the leaks in a process plant) 2. Heat (e.g., by dousing the fire with water)

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3. The supply of oxygen (e.g., by applying foams or inert gases). Various fire extinguishing systems can be used. These include a. Water systems i. Automatic sprinklers ii. Fire hoses b. Foam systems i. Chemical foams: formed by the chemical reaction in which bubbles of CO2 gas and a foaming agent combine to produce and expand froth. ii. Mechanical foams: bubbles of air, which are produced when air and water are mechanically agitated with a foam-making agent. iii. High-expansion foams: tiny foam bubbles filled with air, created by a fan blowing the air through a mesh over which a detergent base solution is flowing. High-expansion foam is used where water damage is a problem or access to an area is not feasible. For example, such a foam would be suitable if a flammable liquid held in an atmospheric storage tank were to spill as a result of overfilling, causing the formation of an ignitable vapor cloud. c. Carbon dioxide systems. d. Dry chemical systems (e.g., sodium bicarbonate). These are not toxic and do not conduct electricity or freeze. e. Water spray systems. These are used for explosion protection of buildings, tanks, and control of flammable liquid fires. f. Steam jet systems. These are used to smother some fires in closed containers or in confined spaces. Illustrative Example 12.5 Consider the following application. Ricci Consultants have provided Behan and Theodore (BAT) Associates with the following model to determine a company’s profits (P) gained from the installation of two different fire protection devices. Assume the number of devices installed are x1 and x2. The model is given by

P = 2000 x1 + 2750 x2 (1)



2x1 + 3 x2 ≤ 30 (2)



x1 + 10 x2 ≤ 25 (3)

with

x1 ≥ 0 (4)



x2 ≥ 0 (5)

Qualitatively explain the previously mentioned profit mathematical model; that is, what number of the two devices will maximize the profits.

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SOLUTION: Equation (1) is the objective function and describes the profit that will be produced from the installation of the safety devices. Equations (2) and (3) are part of the constraints on the limitation of the number of devices that can be installed because of space and regulatory issues. Equation (4) and (5) require that the number of devices (naturally) be a positive number. The calculations are left as an exercise for the reader. (Hint x1 = 14 devices and x2 = 2 devices)

12.6 Optimizing the Operation of a Cement Incinerator Monte Carlo simulation is a procedure for mimicking observations on a random variable that permits verification of results that would ordinarily require difficult mathematical calculations or extensive experimentation. The method normally uses computer programs called random number generators. A random number is a number selected from the interval (0,1) in such a way that the possibilities that the number comes from any two sub intervals of equal “length” are equal. For example, the probability that the number is in the sub-interval (0.1,0.3) is the same as the probability that the number is in the subinterval (0.5,0.7). Thus, random numbers are observations on a random variable x having a uniform distribution on the interval (0,1). This means that the probability distribution function (pdf) of x is specified by

f ( x ) = 1; 0 < x < 1 (12.17)



= 0; elsewhere (12.18)

The previously mentioned pdf assigns equal probability to subintervals of equal length in the interval (0,1). Using random number generators, Monte Carlo simulation can generate observed values of a random variable having any specified pfd. For example, to generate observed values of T, the time to failure, when T is assumed to have a pdf specified by f(t), one first uses the random number generator to generate the value of T between 0 and 1. The solution is an observed value of the random variable T having a pdf specified by f(t). Monte Carlo methods find application in a host of applied mathematical and engineering areas. Most of these involve the solution of problems that cannot be solved by analytical or ordinary numerical methods. Solving these types of problems using Monte Carlo methods has recently been employed in optimization. In terms of introduction, consider the system pictured in Figure 12.3. The system is a solid and the variable T is the temperature; the base temperature is maintained at 100°C and the other three surfaces at 0°C. For this system, the temperature at each location can be described by the average temperature of its four neighboring points. The Monte Carlo procedure provided previously can be used to generate the temperature profile in the solid in the following manner.

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y y = b 0°C

0°C

0°C

0°C

y= 0

100°C x= 0

0°C

0°C 0°C

0°C

1

2

3

4

5

6

7

8

9

100°C

100°C 100°C

100°C

0°C

0°C

0°C

0°C

100°C x =a

x

a =b FIGURE 12.3 Monte Carlo gird approach.

With specified boundary conditions (BCs) for T(x, y) of T(0, y), T(a, y), T(x, 0), and T(x, a), one may employ the following approach:





1. Proceed to calculate T at point 1 (i.e., T1). 2. Generate a random number between 00 and 99. 3. If the number is between 00 and 24, move to the left. For 25 to 49, 50 to 74, 75 to 99, move upward, to the right, and downward, respectively. 4. If the move in step 3 results in a new position that is at the outer surface (boundary), terminate the first calculation for point 1 and record the T value of the boundary at that new position. However, if the move results in a new position that is not at a boundary and is still at one of the other eight interval grid points, repeat steps 2 and 3. This process is continued until an outer surface boundary is reached. 5. Repeat steps 2 to 4 numerous times, for example, 1000 times. 6. After completing step 5, sum all the T values obtained and divide this value by the number of times steps (2 to 4) have been repeated. The resulting value provides a reasonable estimate of T1. 7. Return to step 1 and repeat the calculation for the remaining eight grid points. The results?



T1 = 7.07°C



T2 = 9.80°C

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T3 = 7.16°C



T4 = 18.67°C



T5 = 25.12°C



T6 = 18.77°C



T7 = 42.93°C



T8 = 52.57°C



T9 = 42.80°C

This procedure can be extended to rectangles; it is not limited to squares. Illustrative Example 12.6 Maximize the following profit objective function, MC 1, it is clear  ∂ 2E  is greater than zero, a positive quantity. Similarly, by differentiation of that,   ∂P2 2  0 Equation 13.21 with respect to P3 and substitution of values from Equation 13.24 and 13.25

 1+ 3 k   1− 3 k   ∂ 2E   k − 1    3 k  −  3 k   = 2 NRT P P   (13.28) 1 1 4    ∂P 2    k   3 0

This, by similar reasoning, is positive. The product of the second derivatives lead to  NRT1 ( k − 1)    ∂ 2E   ∂ 2E  = 4   P1  ∂P 2   ∂P 2  k 2 3   0 0 2



2−8k   3k 

P4

 2+ 4k  −  3 k 

(13.29)

Similarly, the cross derivative is given by  NRT1 ( k − 1)    ∂ 2E  =   P1  ∂P ∂P  k 2 3 0   2



2−8k   3k 

P4

 2+ 4k  −  3 k 

(13.30)

Thus, it has also been demonstrated that 2

 ∂ 2E   ∂ 2E   ∂ 2E   ∂P 2   ∂P 2  >  ∂P ∂P  (13.31) 2 3  2 3 Finally, it should be noted that considerable accuracy in computation is often required to establish the optimum. This frequently occurs in applying other methods so that it is necessary to use an analytical expression to establish the optimum accurately. Another limitation of the procedure, using the method of steepest descent/ascent previously discussed and the general analytical differentiation procedure, is that they do not take into account external constraints imposed on the system. For example, the gradient may lead into an area in which operations are not possible.

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References

1. I. Farag, Fluid Flow, A Theodore Tutorial, Theodore Tutorials, East Williston, NY, originally published by the USEPA/APTI, RTP, NC, 1995. 2. P. Abulencia and L. Theodore, Fluid Flow for the Practicing Chemical Engineer, John Wiley & Sons, Hoboken, NJ, 2009. 3. L. Theodore, Heat Transfer Applications for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2011. 4. L. Theodore, F. Ricci, and T. VanViliet, Thermodynamics for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2009. 5. R. Heinsohn, Industrial Ventilation: Engineering Principle, John Wiley & Sons, Hoboken, NJ, 1991. 6. R. Dupont, T. Baxter, and L. Theodore, Environmental Management: Problems and Solutions, CRCLewis Publishers, Boca Raton, 1998. 7. K. Ganesan, L. Theodore, and R. Dupont, Air Toxins: Problems and Solutions, Gordon and Breach, New York City, NY, 1996. 8. L. Theodore, “Nanotechnology: Basic Calculations for Engineers and Scientists”, John Wiley & Sons, Hoboken, NJ, 2006.

14 Chemical Reactor Applications The second chapter in Part III contains six illustrative examples concerned with the general subject of chemical reactors (CR). The title and one-line descriptions of each example are provided here: 14.1 Comparing continuous stirred tank reactor (CSTR) to tubular flow (TF) reactor performance 14.2 Optimizing fluidized-bed reactor performance 14.3 Two reactors in series: achieving maximum conversion 14.4 Maximizing selectivity 14.5 Optimizing batch reactor performance 14.6 Optimizing operating schedules The six references that follow are suggested reading for chemical reactors:

1. R. Perry and D. Green, Perry’s Chemical Engineers’ Handbook, 8th edn, McGraw‑Hill, New York City, NY, 2008. 2. L. Theodore, Chemical Engineering: The Essential Reference, McGraw‑Hill, New York City, NY, 2014. 3. J. Reynolds, J. Jeris, and L. Theodore, Handbook of Chemical and Environmental Engineering Calculations, John Wiley & Sons, Hoboken, NJ, 2004. 4. L. Theodore, R. Dupont, and K. Ganesan, Unit Operations in Environmental Engineering, Wiley‑Scrivener, Salem, MA, 2018. 5. L. Theodore, Chemical Reactor Analysis and Applications for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2012. 6. L. Theodore, Chemical Reaction Kinetics, A Theodore Tutorial, Theodore Tutorials, East Williston, NY, originally published by USEPA/APTI, RTP, NC, 1992.

14.1 Comparing a Continuous Stirred Tank Reactor to a Tubular Flow Reactor Performance There are certain inherent advantages and disadvantages to using each of the three categories of reactors: batch reactors (B), CSTRs (C), and (TF) reactors. The advantages and disadvantages of these three classes of reactor are provided in the literature.1,2 These reactors are sometimes used in combination for any one of a variety of reasons.

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A reaction where mixing is important is the tank flow or (CSTR); it is also referred to as a “back-mix” reactor. This type of reactor, like the batch reactor discussed in a later section, essentially consists of a tank or kettle equipped with an agitator. As with the batch reactor, perfect mixing is normally assumed so that both the concentration and temperature within the reactor is the same, that is, there is no spatial variation so that both terms are not a function of position within the reactor. This perfect mixing assumption also requires that both the concentration and temperature at the discharge point (the outlet pipe) is identical to that in the reactor. Fogler3 notes that for perfect mixing: “The idea that the composition is identical everywhere in the reactor and in the exit pipe requires some thought. It might seem that, since the concentration changes instantly at the entrance where mixing occurs, reaction occurs there and nothing else happens in the reactor because nothing is changing. However, reaction occurs throughout the reactor, but mixing is so rapid that nothing appears to change with time and position.” One can extend this reasoning to also conclude the same within the reactor and the discharge port. This class of reactor finds application in industry when

1. The production rate (or quantity) is high. 2. The reaction time is relatively low.

The bottom line is that in addition to producing greater quantities with smaller equipment, most industrial reactors are operated in a continuous rather than batch mode. They also require less labor and maintenance; however, flow processes are more difficult to start and stop than their batch counterparts. As noted previously, the CSTR essentially consists of a tank or kettle equipped with an agitator. It may be operated under steady or transient conditions. Reactants are fed continuously, and the product(s) are withdrawn continuously. The reactants and products may be liquid, gas, or solid, or a combination of them all. If the contents are perfectly mixed, the reactor design problem is greatly simplified for steady conditions because the mixing results in the aforementioned uniform concentration, temperature, and so on, throughout the reactor. This means that the rate of reaction is also constant and the describing equations are not differential and, therefore, do not require integration. In general, CSTRs are used for liquid phase reactions. High reactant concentrations can be employed with low flow rates so that conversions approaching 100% can be achieved. However, the overall economics of the system may be reduced because of low throughput rates. CSTRs (as well as TF reactors to be addressed shortly) are often connected in series, in such a manner that the exit stream of one reactor is the feed stream for another reactor. Under these conditions, it is convenient to define the conversion at any point downstream in the battery of CSTR reactors in terms of inlet conditions, rather than with respect to any one of the reactors in the series. The conversion X (or X A) is then the total moles of A that have reacted up to that point per mole of A fed to the first reactor. However, this definition should only be employed if there are no side stream withdrawals and the only feed stream enters the first reactor in the series. The conversion from reactors 1, 2, 3,… in the series are usually defined as X1, X2, X3, …, respectively, and effectively represent the overall conversion for that reactor relative to the feed stream to the first reactor. As noted, the CSTR has certain advantages because of the near uniform temperature, concentration, and so on, that results because of mixing. However, for perfect mixing, the reaction occurs at a rate determined by the concentration of the discharge (or exit) stream from the reactor. Since the rate usually decreases with the extent of reaction (conversion), the CSTR operates in the range between the high reaction rate corresponding to

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201

the concentration in the feed stream and the (normally) lower rate corresponding to the concentration leaving the reactor. The CSTR therefore requires a larger volume than a tubular reactor to accomplish a given degree of conversion. This statement applies to all positive reaction orders. The reaction/residence time ratio for a zero order reaction is one; that is, both perform identically. However, the previous analysis occasionally does not apply, particularly if the reaction mechanism is complex and/or non-elementary. The second reactor to be examined is the TF reactor. The most common type involves the use of the single-pass cylindrical tube; another type is one that consists of a number of tubes in parallel. Both will be reviewed in more detail subsequently. The TF reactor can be conceptualized by two different views. The first is to consider a CSTR (see previous development) of given volume V, and feed rate q (or F) and divide the volume into a number of much smaller CSTRs in series. In the limit, an infinite number of these CSTRs with a total volume V reduces to a TF reactor. In other words, the time of reaction in a flow system may be visualized as the time required for an infinitesimal parcel of mass to pass through the reactor in an imaginary compartment (a CSTR of differential volume) that moves with the flowing stream and reacts, producing changes to the temperature, pressure, and composition of the stream. The second approach is to envision a pipe (or tube) of uniform diameter with reactants and products flowing axially along the pipe. This simpler perspective is also a TF reactor. In terms of kinetic behavior, the TF reactor may be viewed as “identical” to a batch reactor. The time t or reaction that can be defined for a batch reactor is now expressed in terms of τ (the residence time) and represents the length of time necessary for any given increment of feed to travel the entire length of the reactor. TF reactors are characterized by a continuous and decreasing concentration of reactants in the direction of flow. Either horizontal or vertical orientation is common. The reactants are charged continuously at one end, and the products are removed continuously at the other end. The unit almost always operates in a steady-state mode. This greatly simplifies design and predictive calculations. It is a unit that is amendable to automatic control and to experimental work. As noted, the most common type of TF reactor is the single-pass cylindrical tube. Another type is one that consists of a number of tubes in parallel. These reactor(s) may be vertical or horizontal. The feed is charged continuously at the inlet of the tube, and the products are continuously removed at the outlet. If heat exchange with surroundings is required, the reactor setup includes a jacketed tube and effectively acts as either a double pipe or a single pass heat exchanger.1 If the reactor is “empty,” a homogeneous reaction – one phase present – usually occurs. If the reactor contains catalyst particles, the reaction is said to be heterogeneous. This type will be considered later in this chapter. TF reactors are usually operated under steady conditions so that physical and chemical properties do not vary with time. Unlike the batch and tank flow reactors, there is no mechanical mixing. Thus, and as indicated earlier, the state of the reacting fluid will vary from point to point in the system, and this variation may be in both the radial r and axial z-direction. The describing equations are then differential, with position as the independent variable. In the describing equations to follow, the reacting system is assumed to move through the reactor in plug flow. It is further assumed that there is no mixing in the axial direction and complete mixing in the radial direction so that the concentration, temperature, and so on, do not vary through the cross section of the tube. Thus, the reacting fluid flows through the reactor in an undisturbed plug of mass. Note once again that time for this hypothetical plug to flow through this type of reactor is the same as the contact or reaction time in a batch reactor. Under these conditions, the form of the describing equations for batch reactors will also apply to TF reactors.1,2 In a plug flow reactor, the entire feed

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stream moves with the same radially uniform axial velocity along parallel streamlines. The entire feed stream, therefore, has the same residence time since there is no mixing in the axial direction but (effectively) complete mixing radially. From a qualitative point of view, as the length of the reactor approaches infinity, the concentration of a (single) reactant approaches zero for irreversible reactions (except zero order) and the equilibrium concentration for reversible reactions. In actual practice, TF reactors deviate from the plug flow model because of the velocity variations in the radial direction. For any of these conditions, the residence time for annular elements of fluid within the reactor will vary from some minimal value at a point where the velocity is a maximum, to a maximum value near the wall where the velocity approaches zero. The concentration and temperature profiles, as well as the velocity profile, are therefore also not constant across the reactor. The describing equations based on the plug flow assumption are then not applicable. The design equation for tubular reactors operating in a steady-state mode will be differential in form, unlike the algebraic design equation for CSTRs. The describing equation is of an ordinary differential form for variations along the axial (longitudinal) length of the reactor. A partial differential equation arises if there are variations along both the axial and radial directions. The key assumption with a CTSR is that it is completely mixed. The key assumption with a TF unit is that plug flow is present, that is, the fluid is completely unmixed along the reactor length and flows down the tube as a plug since it is perfectly mixed radially. Even though the plug-flow approximation may not apply rigorously in some applications, the simplicity of solutions in the limit of perfect plug flow makes it a very useful model from both an academic and industrial perspective. Ultimately, the performance of the three different classes of reactor is based on a host of considerations. They include

1. Conversion 2. Production rate 3. Volume requirement(s) 4. Selectivity (and the effect of unwanted by-products)

Energy considerations, space, power requirements, and so on, also play a role but superimposed on all of the previously mentioned is the economics associated with the choice. Walas4, in an earlier article, provides some general “rules of thumb” regarding reactor selection. Some of his (unedited) suggestions are noted here:

1. The rate of reaction in every instance must be established in the laboratory, and the residence time or space velocity and product distribution must be found in a pilot plant. 2. The optimum proportions of stirred tank reactors are a liquid level equal to tank diameter; at high pressures, slimmer proportions are more economical. 3. Power input to a homogenous-reaction stirred tank is 0.5–1.5 hp/1000 gal, but input is three times this amount when heat is to be transferred. 4. CSTR behavior is approached when the mean residence time is 5–10 times the time needed to achieve homogeneity, which is accomplished by 500–2000 revolutions of a properly designed stirrer.

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5. Batch reactions are conducted in stirred tanks for small, daily production rates, or when the reaction times are long, or when some condition, such as feed rate or temperature, must be programmed in some way. 6. Relatively slow reactions of liquids and slurries are conducted in continuous stirred tanks. A battery of four or five units in series is most economical. 7. TF reactors are suited to high production rates at short residence times (seconds or minutes) and when substantial heat transfer is needed. Jacketed tubes or shelland-tube construction are used. 8. For conversions under about 95% of equilibrium, the performance of a five-stage CSTR unit approaches plug flow. Illustrative Example 14.1 Consider the following applications. The liquid reaction A → products

follows the rate law

−rA =



k1C1A/2 (14.1) 1 + k 2CA2

where: k1 = 5 (gmol/L)1/2/h k2 = 10(L/gmol)2 Initial concentration of A = 0.5 gmol/L Feed rate of A = 200 gmol/h

1. Find the volume necessary to achieve 60% conversion in a CSTR. 2. Would a larger or smaller TF reactor be required to achieve the same degree of conversion? 3. Explain the difference in the results of (2).

SOLUTION:

1. The rate of reaction, −rA, in terms of the conversion variable X is C A = C A 0 (1 − X ) ; X = X A (14.2)

so that

k1C A1/02 (1 − X )

1/2

−rA =



1 + k 2C A2 0 (1 − X )

2

(14.3)

The design equation if the reaction is conducted in a CSTR is1,2 V=

FA 0 X ; X = 0.6 − rA

 1 + (10 )(0.5)2 (1 − 0.6)2   = FA 0 X  1/2  ( 5)(0.5) (1 − 0.6)1/2 

(14.4)

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Further substitution yields



  2 2  1 + (10 )( 0.5 ) (1 − 0.6)  V = ( 200 )( 0.6 )  1  ; X = 0.6 1  ( 5 )( 0.5 ) 2 (1 − 0.6) 2    = 75.1 liters



2. The design equation if the reaction is conducted in a TF reactor is1,2 X

dX rA

∫

V = FA 0 − 0



F = A10/2 k1C A 0

X

∫

1 + k 2C A2 0 (1 − X )

(1 − X )

1/2

0

(14.5)

2

dX

The integral I may be defined as 0.6

I=

∫ 0



1 + k 2C A2 0 (1 − X )

(1 − X )1/2

2

dX (14.6)

0.6

=

∫ f (X ) dX 0

Simpson’s three-point rule is used to evaluate the integral:1,2  h I =    f (0.0 ) + ( 4 ) f (0.3 ) + f (0.6 )  3

h = 0.3

 0.3  ( 3.5) + ( 4 )( 2.659) + 2.214  =  3  



= 1.635 The volume is then



V=

FA 0 I k1C A1/02

(14.7)

= 92.5 liters Note that the integral could have been evaluated by any one of several methods. 3. Surprisingly, the CSTR, is the optimum choice since it requires a smaller volume. This result arises because of the unique nature of the rate expression.

14.2 Optimizing Fluidized-Bed Reactor Performance Fluidization is the process in which fine solid particles are transformed into a fluid-like state through contact with either a gas or liquid, or both. Fluidization is normally carried out in a vessel filled with catalyst solids. The fluid is normally introduced through the

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bottom of the vessel and forced through the bed. At a low flow rate, the fluid (liquid or gas) moves through the void spaces between the stationary and solid catalyst particles and the bed is referred to as fixed. As the flow rate increases, the catalysts begin to vibrate and move about slightly, resulting in the onset of an expanded bed. When the flow of fluid reaches a certain velocity, the solid catalysts become suspended because the upward frictional force between the catalyst and the fluid balances the gravity force associated with the weight of the catalyst. This point is termed minimum fluidization or incipient fluidization and the velocity at this point is defined as the minimum or incipient fluidization velocity. Beyond this stage, the bed enters the fluidization state where bubbles of fluid rise through the solid catalysts, thereby producing a circulatory and/or mixing pattern.5 From a force balance perspective, as the flow rate upward through a packed bed is increased, a point is reached at which the frictional drag and buoyant force is enough to overcome the downward force exerted on the bed by gravity. Although the bed is usually supported at the bottom by a screen, it is free to expand upward, as it will if the velocity is increased above the aforementioned minimum fluidization velocity. At this point, the catalysts are no longer supported by the screen, but rather are suspended in the fluid and act and behave as the fluid. The bed is then said to be fluidized. From a momentum or force balance perspective, the sum of the drag, buoyancy, and gravity forces must be equal to zero. The terminal settling velocity can be evaluated for the case of flow past one catalyst particle in the bed. By superimposition, this case is equivalent to that of the terminal velocity that a catalyst particle would attain flowing through a fluid. Once again, a force balance can be applied, and empirical data can be used to evaluate a friction (drag) coefficient.1,2 At intermediate velocities between the minimum fluidization velocity and the terminal velocity, the bed is expanded above the volume that it would occupy at the minimum value. Note also that above the minimum fluidization velocity, the pressure drop stays essentially constant. One of the novel characteristics of fluidized beds is the uniformity of temperature found throughout the system. Essentially constant conditions are known to exist in both the horizontal and vertical directions in both short and long beds. This homogeneity is due to the turbulent motion and rapid circulation rate of the solid catalyst particles within the fluid stream described previously. In effect, excellent fluid-particle contact results. Temperature variations can occur in some beds in regions where quantities of relatively hot or cold catalyst particles are present but these effects can generally be neglected. Consequently, fluidized beds find wide application in industry, that is, oil cracking, zinc coating, coal combustion, gas desulfurization, heat exchangers, plastics cooling, and fine powder granulation. Illustrative Example 14.2 Consider the following application. A fluidized-bed reactor is to be designed to destroy 99.99% of a unique liquid hazardous waste. Based on laboratory and pilot plant studies, researchers have described the waste reaction by a first-order reversible mechanism. Their preliminary findings are given here:





k A↔R k′

A = waste

k = 1.0e −10 ,000/T

T = °R

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k′ = 9.89e −35 ,000/T

T = °R

Calculate the bed reaction operating temperature that will minimize the volume of the reactor and achieve the optimum desired degree of waste conversion. SOLUTION: A fluid-bed reactor is best described by a CSTR model.1,2 The rate of reaction, −rA, in terms of the concentration of A, CA, is1,2

−rA = kC A1 − k′CR1 (14.8)

Since a fluidized-bed incinerator is best described by a CSTR,1,2



V C A1 − C A1 = ; q = volumetric flow rate (14.9) −rA q =

C A 0 − C A1 (14.10) kC A1 − k′CR1

where: V = reactor volume q = volumetric flowrate of the waste CA1 and CR1, the outlet concentrations of A and R, respectively, can now be expressed in terms of CA0. Note that for 99.99% destruction of the waste, A, the conversion variable X A becomes

X A = 0.9999

Thus,

C A1 = 0.0001C A 0



CR1 = 0.9999C A 0

The design equation for V may now be rewritten in terms of CA0 and the k’s



V C A 0 − 0.0001(C A 0 ) = q ( k )( 0.0001) C A 0 − ( k′ )( 0.9999 ) CA 0 0.9999 0.9999(CA 0 ) = = ′ ( 0.0001k − 0.9999k )( CA0 ) 0.0002k − 0.9999k



The reaction velocity constants, k and k′, are described by the Arrhenius equation given in the problem statement. Thus,

k = Ae − E/RT (14.11)



k′ = A′e − E′/RT (14.12)



A = 1.0

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A’ = 9.89



E = −10, 000 R



E = −35, 000 R

To calculate the operating temperature that will require the minimum volume to accomplish a conversion of 99.99%, minimize the volume by setting dV/dT = 0 and solving for the temperature:



d (V / q ) dT

  dk   dk′     dT  − ( 9999 )  dT        (14.13) = ( 0.9999 )  2   ′ ( k − 9999k )    

Setting d (V / q )



dT

= 0 (14.14)

dk dk′ = ( 9999 ) dT dT



( )

≈ 10 4

dk′ dT



Since

dk AE − E/RT = e (14.15) dT RT 2



dk′ A′E′ − E′/RT = e (14.16) dT RT 2

rearranging the previous equation yields  A E 10 4 =     e  A ′   E′ 

E′ − E RT

 −35 , 000 ′ − −10 , 000  ) ( ) ( 

 1.0   −10, 000   10 =  e  9.89   −35, 000  4



346, 150 = e

−

25 , 000 T

T



T = 1960 °R = 1500°F This represents the optimum operating temperature to minimize the volume of the reactor.

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14.3 Two Reactors in Series: Achieving Maximum Conversion The sequence in which reactors are placed can have an impact on the design and/or the final conversion. Illustrative Example 14.3 Consider the following application. Your company has two reactors of equal volume, which it would like to use in the production of a specified product formed by a firstorder irreversible liquid reaction. One reactor is a CSTR and the other is a TF reactor. How should these reactors be hooked up to achieve maximum conversion (see Figure  14.1)? Justify your answer using concentration (not conversion) terms in the design equations. SOLUTION: Write the design equation for a CSTR in terms of the concentration (not the conversion) and the residence time τ.1,2

V = FA0 ( X − 0 ) / ( −rA ) (14.17)



= qC A0 ( X − 0 ) / kC A1 (14.18)

Since C A0 X = C A0 − C A1 and V/q = τ.

τ = (C A0 − C A1 )/ kC A1 (14.19)



C A1 = C A0 / (1 + k τ ) (14.20)

Write the design equation for a TF reactor in terms of the concentration (not the conversion) and the residence time τ. X

 1 V = FA0  −  rA 0

∫



X

  dX (14.21) 

 1 V/q = C A0  −  dX (14.22)  rA  0

∫



(a) q or (b) q FIGURE 14.1 Physical arrangements of CSTR and TF.

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Since C A0 dX = −dC A τ=−



1 k

C A1

 1   dC A A 

∫  C

CA0

C  −k τ = ln  A1   CA0 

(14.23)



CA1 = exp ( −k τ ) CA0

C A1 = C A0 exp ( −k τ ) (14.24)

Consider the first case (a) – a CSTR followed by a TF reactor – and determine the outlet concentration C A1 from the CSTR in terms of τ1.

C A1 = C A0 / (1 + k τ1 ) (14.25)

Determine the outlet concentration, C A2 , for the TF reactor. Set τ = τ2 for the TF reactor

C A2 = C A1 exp ( − k τ 2 ) = C A0 exp ( − k τ 2 ) / (1 + τ 2 )

(14.26)

Since τ = τ1 = τ2,

C A2 = C A0 exp ( −k τ ) / (1 + k τ ) (14.27)

Consider the second case (b) – a TF reactor followed by a CSTR – and determine C A1 and C A2. C A1 = C A0 exp ( −k τ )

C A2 = C A0 / (1 + k τ )

(14.28)

= C A0 exp ( −k τ ) / (1 + k τ ) The previous results are identical; that is, both cases provided the same outlet conversion. How would the analysis in the previous example be affected if the volumes of the two reactors are not equal? One would repeat the previous calculation if the two volumes are not equal by employing τT and τc, rather than τ. For the first case,

C A2 = C A0 exp ( −k τT ) / (1 + k τC ) (14.29)

For the second case,

C A2 = C A0 exp ( −k τT ) / (1 + k τC ) (14.29)

The results are once again identical. How would the results of the previous two examples be affected if the reaction was not first order? The reader should not interpret the previous two results too loosely. For any order reaction other than first order, the sequence in which the reactors are placed

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may affect the performance. The reader is left the exercise of verifying this statement; however, for a zero order or ∞ order reaction, that is, for n equal to either zero or ∞, the “performance” or degree of conversion would still be identical.

14.4 Maximizing Selectivity When multiple reactions (either parallel or consecutive) occur, the side or undesired reactions compete with the main or desired reaction; the less predominant the main reaction is, the smaller the amount of desired product is for a given amount of reactant. Suppose the following hypothetical parallel reactions occur: A→B A → C (14.30)



A→D Suppose also that, with a starting amount of 10 mol of A, four form the desired product B, two form the undesired product C, one forms the undesired product D, and three remain unreacted. While this process produces 4 mol of valuable product, it could have produced 10 if everything went the way a practicing engineer would want it to; that is, if all 10 mol of A reacted to form 10 mol of B. The ratio of the 4 mol of B actually produced, to the 10 it potentially could have produced, is called the yield (0.40 or 40%). By definition, the yield of a reaction is a measure of how much of the desired product is produced relative to how much would have been produced if only the desired reaction occurred and if that reaction went to completion. Obviously, the fractional yield must be a number between zero and infinity. Another definition that has been used by industry is the “ratio of product formed to the initial (inlet) reactant.” An additional term used in conjunction with multiple reactions is selectivity. Selectivity is a measure of how predominant the desired reaction is relative to one of the undesired products produced by a side reaction. Obviously, selectivity becomes more important than conversion if “undesired” side reaction(s) take precedence over “desired” reaction(s). Other definitions of selectivity that have appeared in the literature include the popular “ratio of the amount of desired products formed to the amount of all products formed.” Definitions have also included “rate of production of one product relative to another product (point selectivity)” and “amount of one product formed relative to another product (overall selectivity).” In the previously mentioned example, the selectivity of B over C is 2.0 and that of B over D is 4.0. These definitions are employed in the illustrative example to follow. Illustrative Example 14.4 Consider the following reactions3 kD



A → D ( desirable ) (14.31)



A → U (undesirable) (14.32)

kU

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with

rD = kDC Aα1 (14.33)



rU = kU C Aα2 (14.34)

and

−rA = rD + rU (14.35)



= kDC Aα1 + kU C Aα 2 (14.36)

Suggest procedures that can be employed to maximize the point selected, S, where S = rD/rU. SOLUTION: If one desires rD ≫ rU, then the point (or local) selectivity is

S=

rD kD α1 −α2 = CA (14.37) rU kU

If α1 − α2 = a, then

k S= D  kU

 a  CA (14.38) 

To maximize S, the need to maintain the product of both terms in Equation 14.38 is high. If a is positive, one needs to maximize CA. This generally means no inert(s), no product(s), and for gas phase reactions to operate at low temperature and high pressure. If a is negative, the reverse applies. The effect of temperature is handled through the Arrhenius equation: 1



kD AD − RT ( ED −EU ) = (14.39) e kU AU

If ED > EU, the reaction should be run at the highest temperature economically possible. If ED  rc. The radius at which this occurs is denoted at r*. The term r* may be obtained by solving the equation

2πL (T1 − T0 ) 2πL (T1 − T0 ) = (16.16) r0 r0 r* ln ln ln 1 1 1 1 r1 r1 r0 + + + + + ki kw r0 h0 ri hi ri hi ki r* h 0

( )

( ) ( )

One may solve for r* using a suitable trial-and-error procedure.

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Resistance, R

Heat Transfer Applications

ri

r0

rc

r*

Radius, r FIGURE 16.3 Resistance associated with the critical insulation thickness for a bare surface.

The previously mentioned development applies where r is less than rc. If r0 is larger than rc, the previous analysis again applies, but only to the results presented for r > rc; that is, it will decrease indefinitely as r increases. Note that there is no maximum/minimum (inflection) for this case since values of ln(r/r0) are indeterminate for r > r0. Once again, it approaches zero in the limit as r approaches infinity. As noted earlier, as the thickness of the insulation is increased, the cost associated with heat loss decreases, but the insulation cost increases. The optimum thickness is determined by the minimum of the total costs. Thus, as the thickness of the insulation is increased, the heat loss reaches a maximum value and then decreases with further increases in insulation. Reducing this effect can be accomplished by using an insulation of low conductivity. Illustrative Example 16.4 With reference to the previously mentioned development, calculate the outer critical radius of insulation (k = 0.44 Btu/h · ft · °F) on a 2.0-inch outside diameter (OD) pipe. Assume the air flow heat transfer coefficient to be 1.32 Btu/h · ft2 · °F. Comment on the effect of insulation on the heat rate lost from the pipe. SOLUTION: Assume the resistances are approximately equal, i.e., ki = horc and employ Equation (16.13). Since BL ≈ 1.0, set rc =

=

ki h0 0.44 = 0.333 ft 1.32

= 4.0 inches The critical insulation thickness is therefore

 2.0  L1 = ∆xi = 4.0 −   = 3.0 inches  2 

Since r0 = 1.0 inches, r0  rc.

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16.5 Recovering Quality Energy The law of conservation of energy is defined by many as the first law of thermodynamics. Its application allows calculations of energy relationships associated with various processes. The second law of thermodynamics is referred to as the “limiting law”. Historically, the study of the second law was developed by individuals, such as Carnot, Clausius, and Kelvin in the middle of the 19th century. This development was made purely on a macroscopic scale and is referred to as the “classical approach” to the second law. Concerns involving conservation of energy issues gained increasing prominence during and immediately after the OPEC oil embargo of 1973. In addition, global population growth has led to an increasing demand for energy. Although the use of energy has resulted in great benefits, the environmental and human health impact of this energy use has become a concern. One of the keys to reducing and/or eliminating this problem will be achieved through what has come to be referred to as meaningful energy conservation. Illustrative Example 16.5 Discuss optimizing the recovery of quality energy from an environmental perspective. SOLUTION: The first law of thermodynamics is a conservation law concerned with energy transformation. Regardless of the types of energy involved in processes – thermal, mechanical, electrical, elastic, magnetic, and so on – the change in the energy of a system also allows free convertibility from one form of energy to another, as long as the overall quantity is conserved. Thus, this law places no restriction on the conversion of work into heat, or on its counterpart – the conversion of heat into work. However, the second law is another matter. One of the areas where the meaningful energy conservation measures can be realized is in the design and specification of process (operating) conditions for heat exchangers. This can be best accomplished by the inclusion of second law principles in the analysis. The quantity of heat recovered in an exchanger is not alone in influencing size and cost. As the temperature difference driving force (LMTD) in the exchanger approaches zero, the “quality” heat recovered increases.5,6 Any discussion of energy conservation regarding heat exchangers leads to an important second law consideration – energy has quality as well as quantity. Because work is 100% convertible to heat, whereas the reverse situation is not true, work is a more valuable form of energy than heat. Although it is not obvious, it can also be shown through second law arguments that heat also has “quality” in terms of the temperature at which it is discharged from a system. The higher the temperature, the greater the possible energy transformation into work. Thus, thermal energy stored at high temperatures is generally more useful to society than that available at lower temperatures. While there is an immense quantity of energy stored in rivers and oceans for example, its present availability to society for performing useful tasks is its “quality” since it is degraded when it is transferred by means of heat transfer from one temperature to a lower one. Other forms of energy degradation include energy transformation due to frictional effects and electrical resistance. Such effects are highly undesirable if the use of energy for practical purposes is to be maximized.5,6 The second law provides the means of measuring this energy degradation through a thermodynamic term referred to as entropy, and it is the second law (of thermodynamics) that serves to define this important property. It is normally designated as S with units of energy per absolute temperature (e.g., Btu/R or cal/K). Furthermore, entropy calculations can provide quantitative information on the “quality” of energy and energy degradation.5,6

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Heat Transfer Applications

There are a number of other phenomena that cannot be explained by the first law of conservation of energy. It is the second law of thermodynamics that provides an understanding and analysis of these adverse effects. However, among these considerations, it is the second law that can allow the measuring of the aforementioned “quality” of energy, including its effect on the design and performance of heat exchange.

16.6 Maximizing Profit through Energy Recovery Illustrative Example 16.6 Two hot liquid (essentially water) process streams are discharged from sources A and B in a plant. A young engineer has proposed to recover energy from the two streams by employing two heat exchangers that are currently available on-site: a double pipe (DP) exchanger and a shell-and-tube (ST) exchanger. Due to diameter differences in the two lines that discharge the hot liquid from source A, there is a 60/40 split (mass basis) of the flowrate. There are also two lines discharging liquid from source B with a 75/25 split to the ST/TB exchangers. Due to the location of the exchangers relative to the two sources, the 40% flow from A and 75% flow from B can be diverted (fed) to the ST exchanger. The remaining discharge can only be sent to the DP exchanger. The young engineer has obtained additional information. The maximum flowrate that exchanger ST and DP can accommodate is 12,000 ft3/day and 6000 ft3/day respectively. In addition, the maximum flow that be drawn from sources A and B is 8000 ft3/ day and 6000 ft3/day respectively. The profit value of the recovered energy from sources A and B are $1.70/ft3 and $2.00/ 3 ft , respectively. Develop the describing equations that will provide values of the (volumetric) flowrates from A and B, that is, qA and qB, and that will maximize the profit P for the proposed processes. SOLUTION: The line diagram of the system is provided in Figure 16.4. The describing equations are as follows. The objective function for the profit is P = (1.70 ) qA + ( 2.00 ) qB (16.17)



0.75 qB A

qA ≤ 8000

Shell and tube (ST)

qST ≤ 12,000

0.40 qA

0.60 qA Double pipe (DP) 0.25 qB qB ≤ 6000 B FIGURE 16.4 Shell-and-tube double pipe exchangers; Illustrative Example 16.6.

qDP ≤ 6000

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The constraints are

qA ≤ 8, 000 (16.18)



qB ≤ 6, 000 (16.19)



0.75qA + 0.40qB = qST ≤ 12, 000 (16.20)



0.60qA + 0.25qB = qDP ≤ 6, 000 (16.21)



qA ≥ 0; qB ≥ 0 (16.22)

SOLUTION: One may also calculate the annual (365-day basis) profit based on this condition. Employing a suitable optimization program gives

qA = 7 , 500 ft 3 /day



qB = 6, 000 ft 3 /day



P = 24, 750 $/day = $8, 910, 000 $/year



Note that operating, maintenance, depreciation, and so on, costs have not been included in the analysis.

References

1. R. Perry and D. Green, Perry’s Chemical Engineers’ Handbook, 8th edn, McGraw-Hill, New York City, NY, 2008. 2. L. Theodore, Chemical Engineering: The Essential Reference, McGraw-Hill, New York City, NY, 2014. 3. L. Theodore, Heat Transfer Applications for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2011. 4. Webster’s New World Dictionary, Second College Edition, Prentice-Hill, Upper Saddle River, NJ, 1971. 5. L. Theodore, F. Ricci, and T. VanVliet, Thermodynamics for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2009. 6. L. Theodore and J. Reynolds, Thermodynamics, A Theodore Tutorial, Theodore Tutorials, East Williston, NY, originally published by the USEPA/APTI, RTP, NC, 1991.

17 Plant Design Applications The last chapter in Part III contains six illustrative examples concerned with the general subject of plant design (PD). The title and one-line description of each example are provided here:

17.1 Shipping facilities 17.2 Tank farms 17.3 Cyclone selection and design 17.4 Minimizing the cost of a batch plant operation 17.5 Ventilation models with system variables 17.6 Plant structure design The seven references that follow are for suggested reading for plant design applications:



1. R. Perry and D. Green, Perry’s Chemical Engineers’ Handbook, 8th edn, McGraw‑Hill, New York City, NY, 2008. 2. L. Theodore, Chemical Engineering: The Essential Reference, McGraw‑Hill, New York City, NY, 2014. 3. J. Reynolds, J. Jeris, and L. Theodore, Handbook of Chemical and Environmental Engineering Calculations, John Wiley & Sons, Hoboken, NJ, 2004. 4. L. Theodore, R. Dupont, and K. Ganesan, Unit Operations in Environmental Engineering, Wiley‑Scrivener, Salem, MA, 2018. 5. J. Happel, Chemical Process Economics, John Wiley & Sons, Hoboken, NJ, 1958. 6. F. Vilbrandt and C. Dryden, Chemical Engineering Plant Design, McGraw-Hill, New York City, NY, 1959. 7. M. Peters, Plant Design and Economics for Chemical Engineers, McGraw-Hill, New York City, NY, 1958.

17.1 Shipping Facilities A large chemical plant complex is likely to have facilities for shipping by pipeline, sea, inland waterway, rail, and road. Smaller plants may have facilities for only one or two of these modes. In any case, they must be designed for safe loading and unloading. In addition, there is often a major economic incentive to load and unload ships, barges, rail cars, and trucks as quickly as possible.

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The cost of shipping materials, both raw materials and products, is often a very significant factor in the chemical industry. The energy cost associated with shipping can range from 0.001 gallons of fuel per ton-mile of cargo moved for supertankers to 0.2 for cargo jet aircraft. Illustrative Example 17.11 A new process plant on the Ohio River expects to receive 500,000 gallons per day of a single raw material and ship out 100,000 gallons per day of each of two liquid products and 500,000 lbs each of four solid products. Estimate the total shipping cost per day for this plant’s operations if all shipments are by barge and if all shipments are by rail. Select the optimum mode of shipping. Data is provided in Tables 17.1 and 17.2. SOLUTION: The number of tons per day in transit is given by the volume per day times the density. The density of water is 8.33 (specific gravity = 1.00) in lb/gal, and 2000 converts from lbs to tons. To calculate the shipping cost, the tons are multiplied by the number of miles shipped and the ton-mile shipping cost. Liquid products are calculated in the same manner. Solid products are calculated similarly, except that the number of pounds per day is already given. Calculate the shipping cost of the feed by barge and rail: Barge

( 500, 000)  2000  (800)(0.015) = 25, 000 $ day 8.33

Rail

( 500, 000)  2000  (800)(0.08) = 133, 300 $ day 8.33



TABLE 17.1 Shipping Distance Data; Illustrative Example 17.1 Distance Shipped (miles)

Product Feed Liquid product A Liquid product B Solid product C Solid product D Solid product E Solid product F

800 250 200 40 300 45 120

TABLE 17.2 Shipping Cost Data; Illustrative Example 17.1 Transportation Type Barge

Rail

Distance (miles) 20–80 80–300 300–1000 10–100 100–1000

Cost ($/ton-mile) 0.03 0.02 0.015 0.08 0.04

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Calculate the shipping cost of liquid products by barge and rail. Barge, Product A:

(100, 000)  2000  (800)( 250)(0.02) = 2, 080 $ day 8.33

Barge, Product B:

(100, 000)  2000  (800)( 200)(0.02) = 1, 670 $ day 8.33

Rail, Product A:

(100, 000)  2000  (800)( 250)(0.04) = 4, 170 $ day 8.33

Rail, Product B:

(100, 000)  2000  (800)( 200)(0.04) = 3, 330 $ day 8.33



Calculate the shipping cost of solid products by barge and rail. Barge, Product C:  500, 000    ( 40 )(0.03 ) = 3, 000 $ day 2000 

Barge, Product D:

 500, 000    ( 300 )(0.02) = 1, 500 $ day 2000 

Barge, Product E:

 500, 000    ( 45)(0.03 ) = 340 $ day 2000 

Barge, Product F:

 500, 000    (120 )(0.02) = 600 $ day 2000 

Rail, Product C:

 500, 000    ( 40 )( 0.03 ) = 800 $ day  2000 

Rail, Product D:

 500, 000    ( 300 )(0.04 ) = 3, 000 $ day 2000 

Rail, Product E:

 500, 000    ( 45)(0.08 ) = 900 $ day 2000 

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Rail, Product F:

 500, 000    (120 )(0.04 ) = 1, 200 $ day 2000 

The sum of the costs for barge transport are shown in Table 17.3. The sum of the costs for rail transport are shown in Table 17.4. Conclusion: Transport by ship and/or barge is usually much cheaper than transport by rail or truck. It often requires longer travel times, however, and thus more inventory can be tied up in shipping.

17.2 Tank Farms Liquid feeds, products, intermediates, and fuels at chemical and petrochemical plants are stored in tanks, which are usually located in a “tank farm” adjacent to the process plant area. The tank capacities are most often expressed in gallons or barrels (one barrel = 42 gallons). The individual tanks may range in size from a thousand gallons or less to several million gallons. TABLE 17.3 Costs for Barge Transport Product Feed Product A Product B Product C Product D Product E Product F

Cost ($) 25,000 2,080 1,670 300 1,500 340 1,200 Total: 31,490

TABLE 17.4 Costs for Rail Transport Product Feed Product A Product B Product C Product D Product E Product F

Cost ($) 133,300 4,170 3,330 800 3,000 900 600 Total: 146,100

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Whenever possible, liquids are stored at ambient temperature and pressure. Volatile liquids may have to be stored under pressure. In addition, they often require vapor recovery systems2 or other devices to prevent releases to the atmosphere as the tanks are filled and emptied. Liquids that are very viscous at ambient temperature or that would solidify at ambient temperature are kept in heated tanks. Standard practice calls for each tank to be surrounded by a dike or berm sufficiently high to contain all the liquid stored in a tank in case it should rupture. Firefighting equipment are normally located near tanks containing flammable liquids. Illustrative Example 17.2 Consider the same process plant described in Illustrative Example 17.1. The new process plant expects to receive 500,000 gallons per day of a single raw material and ship out 100,000 gallons per day of each of two liquid products and 500,000 lbs each of four solid products. Estimate the storage tank requirements for the liquid feed and products. If shipments are all made by barge, the plant will require tankage for a 15-day supply of feed and a 10-day supply of each product. If shipments are by rail, which can be scheduled more reliably, the plant will require a seven-day supply of feed and a five-day supply of each product. For ease of maintenance, there should be at least three tanks for each liquid, with one tank being “off-line” at any given time. Which mode of shipment should be employed for optimum operation? Available standard tank sizes are provided in Table 17.5. SOLUTION: Calculate the storage requirement for feed and select an appropriate number of tanks: Barge shipping:

Storage needed = (15)( 500, 000 ) = 7 , 500, 000 gallons



Use : three 120 ft × 48 ft tanks or five 100 ft × 36 ft tanks Rail shipping:



Storage needed = (7 )( 500, 000 ) = 3, 500, 000 gallons



Use : three 100 ft × 36 ft tanks or five 70 ft × 36 ft tanks

Calculate the storage requirement for one of the liquid products. Barge shipping:

Storage needed = (10 )(100, 000 ) = 1, 000, 000 gallons



Use : four 45 ft × 36 ft tanks or six 35 ft × 30 ft tanks TABLE 17.5 Available Standard Tank Sizes; Illustrative Example 17.2 Capacity (gallons) 216,000 429,000 1,040,000 2,110,000 4,060,000

Height (ft)

Diameter (ft)

30 36 36 36 48

35 45 70 100 120

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Rail shipping

Storage needed = ( 5)(100, 000 ) = 500, 000 gallons



Use : four 35 ft × 30 ft tanks

Rail shipping 45 ft × 36 ft tanks would also be satisfactory. Using the 35 ft × 30 ft tanks allows a little more flexibility since there would be a total of eight tanks the of the same size. Specify a single set: Barge shipping Feed: five 100 ft × 36 ft tanks Product A: four 35 ft × 36 ft tanks Product B: four 45 ft × 36 ft tanks Rail shipping Feed: three 100 ft × 36 ft tanks Product A: four 35 ft × 36 ft tanks Product B: four 35 ft × 36 ft tanks It should be mentioned that many different combinations would be acceptable. It appears either modes of shipment would be acceptable. Storage tanks can present major risks both in terms of fire and environmental pollution. In both cases, this is primarily due to the large inventories of materials that could be involved in any accident. There has been a major effort in industry in the last several decades to reduce the amount of storage at process plants, especially for flammable and/ or toxic materials.

17.3 Cyclone Selection and Design 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 force is known as centrifugal force. The whirling motion of the gas in a cyclone, that is, centrifugal separator, causes particulate matter in the gas to sense this force and move out to the walls and ultimately be collected.2 Cyclones provide a relatively low-cost method of recovering or 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 better than that of settling chambers. However, cyclones are not as efficient as electrostatic precipitators, baghouses, and venturi scrubbers, but they are often installed as pre-cleaners or pre-recovery units before these more effective devices.2 Cyclones come in many sizes and shapes and have no moving parts. From the small 1 and 2 cm diameter source sampling cyclones used for particle size analysis, to the large 5 m diameter cyclone separators used after wet scrubbers, the basic separation principle remains the same. As noted previously, particles enter the device with the flowing gas; the

Plant Design Applications

255

gas stream is forced to turn, but the larger particles have more momentum and cannot turn with the gas. These larger particles impact and, under the influence of gravity, fall down the cyclone wall and are 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. The removal efficiency of a cyclone for a given size particle is very dependent on the cyclone dimensions. The efficiency at a given volumetric flow rate is most affected by the diameter. The overall length determines the number of turns of the vortex. The greater the number of turns, the greater the efficiency. The length and width of the inlet are also important, since the smaller the inlet, the greater the inlet velocity becomes. A greater inlet velocity gives greater efficiency but also increases the pressure drop. 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 going into the cyclone. Multiple-cyclone collectors (multiclones) are high-efficiency devices that consist of a number of small-diameter cyclones operating in parallel with a common gas inlet and outlet. The flow pattern differs from a conventional cyclone in that instead of bringing the gas in at the side to initiate the swirling action, it is imparted by a stationary vane positioned in the path of the incoming gas. The diameters of the collecting tubes usually range from 6 to 24 in. with pressure drops in the 2–6 in range.2 Properly designed units can be constructed and operated with collection efficiency as high as 90% for particulates in the 5–10 µm range. The most serious problems encountered with these systems involve plugging and flow equalization. Since the gas flow to a multicone is axial (usually from the top), the cross-sectional area available for flow inlet conditions is given by the annual area between the outlet tubes and cyclone body. The outlet tube diameter is usually one-half the body diameter.2 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 recovery/control device. With cyclones, an increase in pressure drop usually means that there will be an improvement in collection efficiency (one exception to 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 idea is that with having such an equation, one could work back and optimize the design of new cyclones.2 Illustrative Example 17.3 Consider the following application. A recently hired plant engineer has been assigned the job of selecting and specifying a cyclone unit to be used to reduce an inlet fly ash loading (with the particle size distribution given in Table 17.6) from 3.1 gr/ft3 to an outlet value of 0.06 gr/ft3. The flowrate from the coal-fired plant’s boiler is 100,000 acfm. Fractional efficiency data provided by a vendor is presented subsequently (see Figure  17.1) for three different types of cyclones (multiclones). Which type and how many cyclones are required to meet the previously mentioned specifications? The optimum operating pressure drop is 3.0 in. H2O; at this condition, the average inlet velocity may be assumed to be 60 ft/s.

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TABLE 17.6 Particle Size Distribution; Illustrative Example 17.3 Particle Diameter Range (µm)

Weight Fraction (w i)

5–35 35–50 50–70 70–110 110–150 150–200 200–400 400–700

Collection efficiency, %

100

0.05 0.05 0.10 0.20 0.20 0.20 0.10 0.10

6" diameter 24" diameter

80 12" diameter

60 40 20 0

0

20

40

60

80

100

Particle diameter, microns FIGURE 17.1 Fractional efficiency data; Illustrative Example 17.3.

SOLUTION: Calculate the required collection efficiency, ER:



 3.1 − 0.06  ER =  (100) 3.1   = 98%

= 0.98 Calculate the average particle size associated with each size range (see Table 17.7). Table 17.8 provides the overall efficiency, E6, for the 6-in tubes. Since E6 > ER, the 6-in tubes will do the job. Table 17.9 is generated for the 12-in tubes. Since the overall efficiency, E12  0 (18.8)



Failure rate

where α and β are constants, referred to by some as shape parameters or curve-fitting parameters. For β  1, the failure rate increases with time. Using Equation 18.8 to translate the assumption about failure rate into a corresponding assumption about the probability distribution function (pdf)12 of t, time of failure, one obtains Equation 18.9, which primarily defines the pdf of the Weibull distribution. (The exponential distribution is a special case of the Weibull distribution with β = 1.)



 t f ( t ) = αβtβ−1 exp  αβtβ−1dt    0 

∫

(

(18.9)

)

β>0 = αβtβ−1 exp −αtβ ; t > 0; α > 0,β The variety of assumptions about failure rate and the probability distribution of time to failure that can be accommodated by the Weibull distribution make it especially attractive in describing failure-time distributions in industrial and process plant applications. Estimating the coefficients in the Weibull distribution can be accomplished using a graphing procedure developed by Bury.13 The Weibull distribution has played an important role in helping to reduce and/or eliminate accidents and other hazards from occurring in both the workplace and in everyday life. Shaefer and Theodore11, and Theodore and Dupont7 have provided both development material and numerous illustrative examples that are concerned with this distribution. It has also served as the basis for a host of technical papers prepared by one of the authors. In recent years, the Weibull distribution has come under fire, due primarily to the efforts of one of the authors. The last 6 years has provided an opportunity for Dupont and Theodore14, along with Ricci and others14–17, to carefully analyze the merits and limitations of the Weibull distribution. This has led to the development of the DRaT models. Details are given subsequently. As noted in the previous section, the general two-coefficient Weibull model is represented by an equation that can be applied to three failure rate periods representing three failure mode stages. As such, the model consists of six coefficients – two for each of the three failure mode stages. Theodore and Dupont14 viewed this six-coefficient relationship as both cumbersome and unnecessary. After some deliberation, it was decided to employ a new, simpler approach to represent failure behavior – specifically the failure-time (as opposed to failure rate-time) relationship depicted in Figure 18.8. After even more deliberation and analysis, these authors settled on four models that are based on the melding of either a power function (P) or an exponential relationship (E) for the BI period with either a power function or exponential relationship for the WO period – the sum of which results in a curve as shown in Figure 18.8. There are therefore four combinations of the power function and exponential term equations: BI(P) – WO(P), BI(E) – WO(E), BI(E) – WO(P), and BI(P) – WO(E), and the corresponding equations resulting from these combinations are defined as the DRaT II, DRaT III, DRaT IV, and DRaT V models, respectively. Note that the general relationship for the power function for either the BI or WO period takes the form

P ( t ) = P1 + P2t + P3t 2 + P4t 3 (18.10)

The general form of the exponential function takes the form of either

(

)

E ( t ) = A1 1 − e − A2t (18.11)

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Number of failures, N

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Time, t FIGURE 18.8 Failure-time relationship for the Weibull distribution.

for the BI period, and

(

)

E ( t ) = B1 e B2t − 1 (18.12)



for the WO stage. The failure-time functional relationships for BI and WO periods are represented by g(t)and h(t), respectively, for the development to follow, with the sum of the two resulting in the DRaT model, f(t), that is, f ( t ) = g ( t ) + h ( t ) (18.13)



For example, if both the BI and WO periods are represented by an exponential function (see also Figure 18.9) it would appear as

(

)

(

)

g ( t ) = A1 1 − e − A2t (18.14)

and

h ( t ) = B1 e B2t − 1 (18.15)



The sum of Equations 18.14 and 18.15, presented in Figure 18.9, represents the failure model previously defined as the DRaT III model, that is,

(

)

(

)

N ( III ) = g ( t ) + h ( t ) = A1 1 − e − A2t + B1 e B2t − 1 ; N = number of failures (18.16)

Specific details on each of the remaining DRaT models are provided in the next three subsections. 18.5.1 DRaT II Model This model assumes that BI(P) and WO(P) relationships apply. Thus,

N ( III ) = N ( BI ) + N ( WO ) (18.17)

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g(t) = A1(1 – e–A2t)

(a)

h(t) = B1(eB2t – 1)

(b) f(t) = g(t) + h(t) g(t) h(t)

(c) FIGURE 18.9 DRaT model II representation. (a) BI period, (b) WO period, and (c) combined BI and WO period.

and

N ( II ) = A1 + A2t + A3t 2 + A4t 3 (18.18)

Interestingly, this model only contains three coefficients. 18.5.2 DRaT IV Model This model assumes that BI(E) and WO(P) relationships apply; thus,

(

)

N ( BI ) = A1 1 − e − A2t (18.19)

and

N ( WO ) = B1 + B2t + B3t 2 ; B1 = 0 (18.20)

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so that

(

)

N ( IV ) = A1 1 − e − A2t + B2t + B3t 2 (18.21)



This model contains four coefficients. 18.5.3 DRaT V Model The last model assumes BI(P) and WO(E) relationships apply. Thus, N ( BI ) = A1 + A2t + A3t 2 ; A1 = 0 (18.22)

and

(

)

N ( WO ) = B e B2t − 1 (18.23)

so that

(

)

N ( V ) = A2t + A3t 2 + B e B2t − 1 (18.24)



Similar to the last two models, this model also contains four coefficients. Three improvements in the proposed DRaT models relative to the Weibull model become immediately apparent: 1. The DRaT model requires either three of four (not six) coefficients to be estimated. 2. The DRaT model is continuous over the entire time range, as opposed to the Weibull model that is evaluated separately over three compartmentalized failure stages. 3. The requirement of a constant failure rate period in the Weibull model has been replaced by a more realistic failure rate that can continue to increase slightly with time during the supposed constant failure rate period. The reader should note once again that the failure rate for most applications is a calculated quantity obtained from the number of failures (N) versus time (t) data. The failure rate at a specified time is thus approximately equal to the slope (or derivative) of N versus t at the time point in question. Theodore and Ricci18 provide six different numerical differentiation procedures to calculate this derivative, that is the failure rate, and the reader is directed to that reference for more details.

References

1. L. Theodore, Air Pollution Control Equipment Calculations, John Wiley & Sons, Hoboken, NJ, 2009. 2. L. Theodore and F. Ricci, Mass Transfer Operations for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2010. 3. D. Kaufmann, Plant Design, A Theodore Tutorial, Theodore Tutorials, East Williston, NY, originally published by the USEPA/APTI, RTP, NC, 1997.

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4. K. Skipka and L. Theodore, Energy Resources: Availability, Management, and Environmental Impacts, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2014. 5. J. Happel, Chemical Engineering Economics, John Wiley & Sons, Hoboken, NJ, 1958. 6. C. Tyler, Chemical Engineering Economics, 3rd edn, McGraw-Hill, New York City, NY 1948. 7. L. Theodore and R. Dupont, Environmental Health and Hazard Risk Assessment: Principles and Calculations, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2002. 8. D. Hendershot, A simple example problem illustrating the methodology of chemical process quantitative risk assessment, Paper presented at AIChE Mid-Atlantic Region Day in Industry for Chemical Engineering Faculty, April 15, 1988, Bristol, PA. 9. L. Theodore and F. Taylor, Probability and Statistics, A Theodore Tutorial, Theodore Tutorials, East Williston, NY, originally published by the USEPA/APTI, RTP, NC, 1999. 10. J. Reynolds, J. Jeris, and L. Theodore, Handbook of Chemical and Environmental Engineering Calculations, John Wiley & Sons, Hoboken, NJ, 2002. 11. S. Shaefer and L. Theodore, Probability and Statistics for Environmental Science, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2007. 12. L. Theodore, Environmental Risk Analysis: Probability Distribution Calculations, CRC Press/Taylor & Francis Group, Boca Raton, FL, 2016. 13. K. Bury, Statistical Methods in Applied Science, John Wiley & Sons, Hoboken, NJ, 1975. 14. L. Theodore and R. Dupont, Calculating hazard probabilities using the Weibull distribution, Paper #125, AWMA Conference, Orlando, FL, 2011. 15. R. Dupont, J. McKenna, and L. Theodore, Baghouse failures: applying the Weibull distribution to estimate bag failure emissions as a function of time, Paper #112, AWMA Conference, Chicago, IL, 2013. 16. R. Dupont, F. Ricci, and L. Theodore, An improved failure rate model applied to baghouse failures, AWMA Conference, Long Beach, CA, 2014. 17. R. Dupont, F. Ricci, and L. Theodore, Replacing the Weibull distribution failure model, AWMA Conference, Raleigh, NC, 2015. 18. L. Theodore and F. Ricci, Mass Transfer Operations for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2011.

19 Select Chemical Engineering Applications Five sections complement the presentation in this final chapter:

19.1 Fluid flow 19.2 Chemical reactors 19.3 Heat exchangers 19.4 Mass transfer operation 19.5 Plant design

19.1 Fluid Flow Fluids are usually transported in pipes or tubes. Generally speaking, pipes are heavywalled and have a relatively large diameter. Tubes are thin-walled and often come in coils. Pipes are specified in terms of their diameter and wall thickness. The nominal diameters range from 1/8 to 30 inches for steel pipe. Standard dimensions of steel pipe are provided in the literature and are known as IPS (iron pipe size) or NPS (nominal pipe size). The wall thickness of the pipe is indicated by the schedule number, which can be approximated from 1000(P/S) where P is the maximum internal service pressure (psi) and S is the allowable bursting stress in the pipe material (psi). (The S value varies by material, grade of material and temperature; allowable S values may be found in piping handbooks).1–3 Tube sizes are indicated by the outside diameter. The wall thickness is usually given a BWG (Birmingham Wire Gauge) number. The smaller the BWG, the heavier the tube. The literature lists the sizes and wall thicknesses of condenser and heat exchanger tubes. For example, a ¾ inch 16 BWG tube has an outside diameter (OD) of 0.75 inches, an inside diameter (ID) of 0.62 inches, a wall thickness of 0.065 inches, and a weight of 0.476 lb/ft. When designing a piping system for an industrial process, savings can be effected by determining the optimum pipe diameter and hence the economic optimum condition by means of an economic balance. As mentioned previously, to obtain an economic balance, one type of cost must increase while the other decreases with an increase in the independent variable. For the case under consideration, the primary independent variable is usually the diameter of the pipe. One set of costs, namely, the fixed charges (cost of the pipe, fittings, and installation) will increase with an increase in the pipe diameter. On the other hand, for a given flow rate, the power costs due to pressure drop decrease as the pipe size increases. In other words, if the pipe is oversized, the fixed charges will be too large, and if the pipe diameter is too small, the pressure drop will be excessive, resulting in large power costs. The economic balance therefore is made between the fixed costs and the power costs.

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TABLE 19.1 Pipe Cost Data; Illustrative Example 19.1 Pipe Size, I.D. (in)

Installed Cost ($/ft)

1.0 1.5 2.0 2.5 3.0

1.80 1.89 2.19 3.18 3.99

Illustrative Example 19.1 Consider the following application. It is desired to design a pipeline to operate 8700 h/yr in which water at 60°F is to be fed to a cooling tower at a rate of 30 gal/min. The overall length of the horizontal piping system is to be 1000 ft. If the annual fixed charges (depreciation, taxes, insurance, etc.) are 30%/yr of the installed cost of the pipe and power costs $0.06 kW‑h, calculate the economic optimum pipe diameter based solely on operating costs. Assume that a pump for the installation is already on hand. SOLUTION: First, perform the calculations for the 1.0-in pipe assuming turbulent flow and a friction factor of approximately 0.005.1,2 Initial cost of pipe = (1000)(1.80 ) = $1, 800



Annual initial charges = (1800 )(0.30 ) = $540

Annual power costs will depend on the pressure drop. First, calculate the velocity: v=



( 30 )( 4 ) 2 ( 60 )( 7.48 )(1 / 12 )



= 38.5 ft / s Apply Fanning’s equation.

1–3



∆P =

4fLv 2 2g c D

∆P =

( 4 )(1000 )( 0.005 )( 38.5 ) ( 2 )( 32.2 )(1 / 12 )

(19.1) 2



∆P = 5, 523 ft H 2 O = 28, 732 lbf / ft 2 The annual power cost is

PC =

( 28, 732 )( 30 )( 60 )( 8700 )( 0.06 ) ( 7.48 )( 778 )( 3412 )

Annual power costs = $1359

Total annual cost for the 1.0-in pipe = 540 + 1359 = 1899.

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Table 19.2 provides the calculated results for the remaining pipe sizes. Plotting both the annual fixed costs and power costs versus the pipe size will result in a graph similar to that in Figure 19.1, as presented earlier in Part III, Figure 13.1. The optimum pipe diameter is approximately 1.75 inches. Performing the calculations for pipe sizes of 1.25 inches and 1.75 inches is left as an exercise for the reader. The reader should note that this topic will be revisited in Part IV, Chapter 19.

19.2 Chemical Reactors A reactor where mixing is important is the tank flow or continuous stirred tank reactor (CSTR). It is also referred to as a “back-mix” reactor. This type of reactor, like the batch reactor,4,5 essentially consists of a tank or kettle equipped with an agitator. As with the batch reactor discussed previously, perfect mixing is normally assumed so that both the concentration and temperature within the reactor is the same, that is, there is no spatial variation so that both terms are not a function of position within the reactor. This perfect mixing assumption also requires that both the concentration and temperature at the discharge point (the outlet pipe) is identical to that in the reactor. As noted in Chapter 16 in Part III, Fogler6 noted that for perfect mixing: “The idea that the composition is identical everywhere in the reactor and in the exit pipes requires some thought. It might seem that, since the concentration changes instantly at the entrance where mixing occurs, the reaction occurs there and nothing else happens in the reactor because nothing is changing. However, reaction occurs throughout the reactor, but mixing is sufficiently rapid that TABLE 19.2 Summary of Pipe Diameter Calculations; Illustrative Example 19.1 Pipe Size (in)

Initial Charges ($/yr)

Power Cost ($/yr)

Total Costs ($/yr)

540 567 657 954

1359 179 42 14

1899 746 700 968

1.0 1.5 2.0 2.5

Cost $/yr

Total cost

Operating cost Capital cost

Pipe diameter FIGURE 19.1 Economic balance plot.

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nothing appears to change with time or position.” One can extend this reasoning to also conclude that the rate of reaction within the reactor is also uniform so that the extent of conversion of reactant is the same within the reactor and the discharge port. The CSTR essentially consists of a tank or kettle equipped with an agitator (see Figure 19.2) and it may be operated under steady or transient conditions. Reactants are fed continuously, and the product(s) are withdrawn continuously. The reactant(s) and product(s) may be liquid, gas, or solid, or a combination of them all. If the contents are perfectly mixed, the reactor design problem is greatly simplified for steady conditions because the mixing results in the aforementioned uniform concentration, temperature, and so on, throughout the reactor. This means that the rate of reaction is also constant and the describing equations are not differential and, therefore, do not require integration. Consider the elementary irreversible reaction between benzoquinone (A) and cyclopentadiene (B): A + B → products (19.2)



If one employs a feed containing equimolar concentrations of reactants, the reaction rate expression can be written as

−rB = kC A C B = kC 2B ; k = k B (19.3)

Proceed to calculate the reactor size requirements for one CSTR. Also, calculate the volume requirements for a cascade composed of two identical CSTRs. Assume isothermal operation at 25°C where the reaction rate constant is equal to 9.92 m3/(kgmol/ks). Reactant concentrations in the feed are each equal to 0.08 kgmol/m3, and the liquid feed rate is equal to 0.278 m3/ks. The desired degree of conversion is 87.5%. The rate equation, −rB, in terms of conversion variable X is4,5 −rB = −r = kC B2 (19.4)



Motor

Feed line

Impeller

Cooling or heating jacket

Product FIGURE 19.2 Typical continuous stirred tank flow reactor.

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C B = C B0 ( 1 − X ) (19.5)



−r = kC 2B ( 1 − X ) (19.6) 2

If only one reactor is employed,4,5 FB0 X −r

V=

FB0 X

=

kC B2 0 ( 1 − X )

(19.7) 2

Since FB0 = C B0 q = (0.08 )(0.278 )



= 0.02224 kgmol / ks

(19.8)

= 0.02224 gmol / s

The volume may be calculated employing Equation (19.7).



2 2 V = ( 0.02224 )( 0.875 ) / ( 9.92 )( 0.08 ) ( 1 − 0.875 )    3 = 19.6 m

The previous design equation may also be applied to two reactors of equal volume in series, noting that X1 − 0 = X1.3 V1 =



V2 =

FB0 X1 (19.9) kC 2B0 ( 1 − X1 )2

FB0 X1 ; ∆X = X 2 − X1 (19.10) kC 2B0 ( 1 − X 2 )2

For reactor 1,

kC 2B0 V1 X1 = 2 FB0 1 − ( X1 )

For reactor 2,

kC 2B0 V2 X − X1 = 2 2 FB0 (1 − X 2 )

Since the LHS of both of the previous equations are equal, (V1 = V2),

X1

(1 − X1 )

2

=

X 2 − X1

(1 − X 2 )

2



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Solving with X2 = 0.875 yields X1 = 0.7251

Thus,



V1 =

( 0.278 )( 0.08 )( 0.7251) 2 2 ( 9.92 )( 0.08 ) (1 − 0.7251)

= 3.36 m 3 In addition, VT = V1 + V2

= ( 2 )( 3.36 ) (19.11) = 6.72 m 3

The previously mentioned results – in terms of the volume requirements – for the two cases may be compared. Using one reactor requires a volume of

V1 = 19.6 m 3

Use of two equal volume reactors in series produced a volume requirement of

V2 = 3.36 + 3.36 = 6.72 m 3

As expected, the latter case provided the smaller volume requirement. The previously mentioned calculations may be further extended to determine the effect of using a cascade of two CSTRs, that differ in size, on the volume requirements for the reactor network. If the same feed composition and flow rate as previously mentioned are employed, and if the reactors are operated isothermally at 25°C, the minimum total volume required and the manner in which the volume should be distributed between the two reactors can also be calculated if the fractional overall conversion of 0.875 is to be achieved again. Set up an equation to determine the total volume requirement if the two reactor volumes are not equal.4,5

VT = V1 + V2 ; V1 ≠ V2 (19.12)



 F   X1 X 2 − X1  (19.13) VT =  B02   2 + 2  ; X 2 = 0.875  kC B0   (1 − X 1 ) (1 − X 2 ) 

To solve the previous equation for the intermediate conversion X1 under minimum volume conditions, one needs to minimize VT, by setting

dVT = 0; VT = VT ( X 1 ) (19.14) dX 1

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291

and then solve for X1. Analytical differentiating yields dVT 1 2X 1 = + − 64 = 0 (19.15) dX1 ( 1 − X1 )2 ( 1 − X 2 )2



Solving by trial-and-error gives X1 = 0.702



The volumes of the two individual reactors and the total minimum volume requirement can now be calculated:



V1 =

( 0.02224 gmol / s ) ( 0.702 ) ( 9.92 L / gmol ⋅ s ) (0.08 gmol / L)2 (1 − 0.875 )2

= 2.77 L



V2 =

( 0.02224 gmol / s ) ( 0.875 − 0.702 ) ( 9.92 L / gmol ⋅ s ) (0.08 gmol / L)2 (1 − 0.875 )2

(19.16)

= 3.88 L VT = V1 + V2

= 2.77 L + 3.88 L = 6.65 L

The previous result is slightly lower than that calculated earlier (VT = 6.72 L). The effect of using a cascade of three CSTRs that differ in size on the volume requirements for the reactor network may also be investigated. If the same feed composition and flow are employed and the reactors are operated isothermally at 25°C, the minimum total volume required and the manner in which the volume should be distributed between the three reactors is another optimization problem. An overall conversion of 0.875 is to be achieved again. Proceeding as before, write the equation for the volume of each reactor and total volume requirement for three reactor volumes that are not equal. For each reactor

 F   X1  (19.17) V1 =  B02   2  kC B0   (1 − X1 ) 



 F   X − X1  (19.18) V2 =  B02   2 2  kC B0   (1 − X 2 ) 

with

 F   X − X2  (19.19) V3 =  B02   3 2  kC B0   (1 − X 3 ) 

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The total volume requirement is

VT = V1 + V2 + V3 (19.20)



 F   X 1   FB0   X 2 − X 1   FB0   X 3 − X 2  (19.21) VT =  B02     2 + 2 2 +  kC B0   (1 − X 1 )   kC 2B0   (1 − X 2 )   kC 2B0   (1 − X 3 ) 

Take the partial derivative of the equation obtained previously with respect to X1. Note that the partial derivative must now be employed since the total volume, VT, depends on two variables, X1 and X2, that is, VT = VT(X1, X2).7

 ∂VT  FB0   1 2X 1 1 (19.22) = 2 + 3 − 2 2   ∂X1  kC B0   (1 − X 1 ) (1 − X 2 ) (1 − X 3 ) 

Set this derivative equal to zero.

 ∂VT  FB0   1 2X 1 1 (19.23) = 2 + 3 − 2 = 0 2   ∂X 1  kC B0   (1 − X 1 ) (1 − X 2 ) (1 − X 3 ) 

Take the partial derivative for VT with respect to X2.

 ∂VT  FB0   1 2(X 2 − X 1 ) 1 (19.24) = 2 + 3 − 2 2   ∂X 2  kC B0   (1 − X 1 ) (1 − X 2 ) (1 − X 3 ) 

Also, set this derivative equal to zero.

 ∂VT  FB0   1 2(X 2 − X 1 ) 1 (19.25) = 2 + 3 − 2 = 0 2   ∂X 2  kC B0   (1 − X 1 ) (1 − X 2 ) (1 − X 3 ) 

Solve the resulting two algebraic equations obtained from the partial derivative equations simultaneously. For X3 = 0.875, X2 is obtained as follows by trial-and-error (see Table 19.3). Therefore, X1 = 0.585, X2 = 0.7879. Finally, calculate the volume of the three individual reactors and the total minimum volume requirement.  F   X1  V1 =  B02   2  kC B0   (1 − X1 ) 

   0.585  0.02224 gmol / s (19.26) = 2 2   (9.92 L / gmol ⋅ s ) (0.08 gmol / L)   (1 − 0.585)  = 1.19 L  F   X − X1  V2 =  B02   2 2  kC B0   (1 − X 2 ) 



  0.02224 gmol / s = 2  (9.92 L / gmol ⋅ s ) (0.08 gmol / L)  = 1.58 L

 0.7879 − 0.585  (19.27)  2   (1 − 0.7879) 

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TABLE 19.3 Trial-and-Error Calculation for Three Reactor Set X1 0.6 0.59 0.58 0.585

X2

Residual

0.8 0.7918 0.7835 0.7879

11 4 −2.5 0.8

 F   X − X2  V3 =  B02   3 2  kC B0   (1 − X 3 ) 

   0.875 − 0.7879  0.02224 gmol / s (19.28) = 2  2   (9.92 L / gmol ⋅ s ) (0.08 gmol / L)   (1 − 0.875)  = 1.95 L VT = V1 + V2 + V3



= 1.19 L + 1.58 L + 1.95 L (19.29) = 4.72 L

Note the reduction in the volume requirement when three CSTRs in series are employed. The reader is left the exercise of showing that the total volume requirement for three CSTRs of equal volume is

V1 = V2 = V3 = 2.17 m 3



VT = V1 + V2 + V3 = 6.51 m 3

with

X1 = 0.6710



X 2 = 0.7782

Theodore4,5 also calculated the reactor size requirement for a cascade composed of two CSTRs with the first inlet reactor THRICE the size the second, the reactor size requirements for a cascade composed of two CSTRs with the first inlet reactor ONE-THIRD the size of the second, if a very large number (approaching infinity) of CSTRs are employed, and if the reaction is reversible for the previously mentioned conclusions.

19.3 Mass Transfer Operation One of the more important mass transfer operations involves the transfer of gases – including those that are toxic – within a plant. Exposure to contaminants in a workplace can be reduced by proper ventilation. Ventilation can be provided either by dilution ventilation

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or by a local exhaust system. In dilution ventilation, air is brought into the work area to dilute the contaminant sufficiently to minimize its concentration and subsequently reduce worker exposure. In a local exhaust system, the contaminant itself is removed from the source through hoods. A local exhaust is generally preferred over a dilution ventilation system for health problems because a local exhaust system removes the contaminants directly from the source, whereas dilution ventilation merely mixes the contaminant with uncontaminated air to reduce the contaminant concentration. Dilution ventilation may be acceptable when the contaminant concentration has a low toxicity, and the rate of contaminant emission is constant and low enough that the quantity of required dilution air is not prohibitively large. However, dilution ventilation generally is not acceptable when the concentration is less than 100 ppm. In determining the quantity of dilution air required, one must also consider mixing characteristics of the work area in addition to the quantity (mass or volume) of contaminant to be diluted. Thus, the amount of air required in a dilution ventilation system is much higher than the amount required in a local exhaust system. In addition, if the replacement air requires heating or cooling to maintain an acceptable workplace temperature, then the operating cost of a dilution ventilation system may further exceed the cost of a local exhaust system. Illustrative Example 19.2 Discuss the optimization of a ventilation system. SOLUTION: The major components of an industrial ventilation system include the following:

1. Exhaust hood 2. Ductwork 3. Contaminant control device 4. Exhaust fan 5. Exhaust vent or stack

Several types of hoods are available. One must select the appropriate hood for a specific operation to effectively remove contaminants from a work area and transport them into the ductwork. The ductwork must be sized such that the contaminant is transported without being deposited within the duct; adequate velocity must be maintained in the duct to accomplish this. Selecting a control device that is appropriate for the contaminant removal is important to meet certain pollution control removal efficiency requirements. The exhaust fan is the workhorse of the ventilation system. The fan must provide the volumetric flow at the required static pressure, and it must be capable of handling contaminated air characteristics such as dustiness, corrosiveness, and moisture in the air stream. Properly venting the exhaust out of the building is equally necessary to avoid contaminant recirculation into the air intake or into the building through other openings. Such problems can be minimized by properly locating the vent pipe in relation to the aerodynamic characteristics of the building. In addition, all or a portion of the cleaned air may be recirculated to the workplace. Primary (outside) air may be added to the workplace and is referred to as makeup air; the temperature and humidity of the makeup air may have to be controlled. It also may be necessary to exhaust a portion of the room air. Each of these variables needs careful analysis if a ventilation system’s operation and performance is to be maximized.

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To stack

Duckwork

Particulate control equipment Makeup air

Return/ recirculation air Source

Room/workplace

Discharge/ exhaust air

FIGURE 19.3 Industrial ventilation systems.

A line diagram of a typical industrial ventilation system is provided in Figure 19.3. Note that either the control device or the fan (or both) can be located in the room/ workplace.

19.4 Heat Exchangers In many environmental, chemical, and petrochemical plants there are cold streams that must be heated and hot streams that must be cooled. Rather than use steam to do all the heating and cooling water to do all the cooling, it is often advantageous to have some of the hot streams heat the cold ones. In addition, the problem of optimizing the heat exchanger networks that would be required has been extensively studied, but it remains a daunting exercise for practicing engineers.8–9 What follows is one simple illustration of a heat exchanger network. Consider the following application.10 A plant has three streams to be heated and three streams to be cooled. Cooling water (90°F supply, 155°F return) and steam (saturated at 250 psia) are available. Devise a network of heat exchangers that will make full use of heating and cooling streams against each other, using utilities only if necessary. Information on the three streams to be heated are presented in Table 19.4. Information on the three streams to be cooled are presented in Table 19.5. Saturated stream at 250 psia has a temperature of 401°F. Here is the solution represented by Kaufman.10 The heating duties for all streams are first calculated. The results are tabulated in Table 19.6. The total heating and cooling duties are next compared:

Heating : 7 , 745, 000 + 6, 612, 000 + 9, 984, 000 = 24, 071, 000 Btu / h



Cooling : 12, 600, 000 + 4, 160, 000 + 3, 150, 000 = 19, 910, 000 Btu / h

As a minimum, 4,160,000 Btu/h might have to be supplied by steam.

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TABLE 19.4 Heated Network Streams Stream 1 2 3

Flowrate (lb/h)

cp [Btu/(lb⋅°F)]

Tin(°F)

Tout (°F)

50,000 60,000 80,000

0.65 0.58 0.78

70 120 90

300 310 250

Flowrate (lb/h)

cp [Btu/(lb⋅°F)]

Tin(°F)

Tout (°F)

60,000 45,000 35,000

0.70 0.52 0.60

420 300 240

120 100 90

TABLE 19.5 Cooled Network Streams Stream 4 5 6

TABLE 19.6 Network Heating and Cooling Duties Stream Number 1 2 3 4 5 6

Stream Type

Duty

Heating Heating Heating Cooling Cooling Cooling

7,745,000 6,612,000 9,984,000 12,600,000 4,160,000 3,150,000

Figure 19.4 represents a system of heat exchangers that will transfer heat from hot streams to the cold ones in the amounts desired. It is important to note that this is but one of many possible schemes. The optimum system would require a trial-and-error procedure that would examine a host of different schemes. Obviously, the economics would also come into play. Theodore9, and Abulencia and Theodore,6 discuss the 2 4 3

120° F 120° F

310° F

6,612,000 1,600,000

90° F

Steam

158.1° F 115.6° F

4,388,000

420° F

1,496,000 250° F

2,500,000

240° F

6 1 5

90° F 70° F

650,000

3 6

Steam

121° F 90° F

2 4

2,665,000 4,160,000

100° F

FIGURE 19.4 Flow diagram for a heat exchanger network.

218° F

300° F

300° F

1 5

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ramifications that arise if the heat exchanger cost is a function of the heat exchanger area; that is, the network should be such that the total cost of the resulting network is minimized. They also discuss how to solve the problem so that the network’s entropy increase11,12 and the network’s cost are minimized. The practicing engineer or scientist should consider how to arrive at a reasonable solution to this energy conservation problem. Finally, highly interconnected networks of exchangers can save a great deal of energy in a chemical plant. The more interconnected they are, however, the harder the plant is to control, startup, and shut down. Often auxiliary heat sources and cooling sources must be included in the plant design in order to ensure that the plant can operate smoothly. The  previously mentioned problem and considerations indicate how difficult it can be to solve some real-world industrial applications. The interested reader is encouraged to contact the authors if other methods of solution are available and possible.

19.5 Plant Design Plant design requires the use of engineering principles and theories combined with a practical realization of the limits imposed by industrial conditions. The development of a plant-design project involves a wide variety of talents. Research, market analyses, design of individual pieces of equipment, plant-location surveys, and many other problems are involved in the design of industrial plants.10 Although the environmental and chemical engineer must understand the principles related to the various design phases, the individual parts of the design are of little benefit until they are combined to give one complete unit. The complete plant, i.e., the complete unit – that is the design engineer’s goal. However, one major qualification must be added before the goal is attained. This is the qualification that the operation must be able to proceed economically, and this often requires that energy be conserved. Concerns and details regarding the conservation of energy follow. One of the areas where meaningful energy conservation measures can be realized is in the design and specification of process (operating) conditions for heat exchangers. This can be best accomplished by the inclusion of second law principles in the analysis. The quantity of heat recovered in an exchanger is not alone in influencing size and cost. As the temperature difference driving force (LMTD) in the exchanger approaches zero, the “quality” heat recovered increases.5 Most heat exchangers are designed with the requirements/specification that the temperature difference between the hot and cold fluid be at all times positive and be at least 20°F. This temperature difference or driving force is referred to by some as the approach temperature. However, the corresponding entropy change is also related to the driving force, with large temperature difference driving forces resulting in large irreversibilities and the associated large entropy changes.11,12 The individual designing a heat exchanger is faced with two choices. He/she may decide to design with a large LMTD that results in both a more compact (smaller area) design and a large entropy increase that is accompanied by the loss of “quality” energy. Alternatively, a design with a small driving force results in both a large heat exchanger and a small entropy change with a larger recovery of “quality” energy.9,11

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Illustrative Example 19.3 Consider the following application. Shannon O’Brien, a recent graduate from Manhattan College’s prestigious chemical engineering program was given an assignment to design the most cost-effective heat exchanger to recover the energy from a hot flue gas at 500°F. The design is to be based on pre-heating 100°F incoming air (to be employed in a boiler) to a temperature that will result in the maximum annual profit to the utility. A line diagram of the proposed countercurrent exchanger temperature profile is provided in Figure 19.5. Having just completed a heat transfer course with Dr. Annmarie Flynn and a thermodynamics course with the nefarious Dr. Theodore, Shannon realizes that there are two costs that need to be considered:

1. The heat exchanger employed for energy recovery 2. The “quality” (from an entropy perspective) of the recovered energy

She also notes that the higher the discharge temperature of the heated air, t, the smaller will be the temperature difference driving force, and the higher the area requirement of the exchanger and, correspondingly the higher the equipment cost. Alternatively, with a higher t, the “quality” of the recovered energy is higher, thus leading to an increase in recovered profits (by reducing fuel costs). Based on similar system designs, Francesco Ricci Consultants (FRC) has provided the following annual economic models for the heat exchanger,

Recovered energy profit : A ( t − tc ) ; A = $ / yr ⋅°F (19.30)



Recovered energy cost : B / ( TH − t ) ; B = $ / yr ⋅°F (19.31)

For the previously mentioned system, FRC suggests a value for the coefficients in the cost model be set at A = 10 B = 100,000 Employing the information provided, Shannon has been asked to calculate a t that will

TH = 500°F

t Hot stream T = 120°F Cold stream

FIGURE 19.5 Proposed countercurrent exchanger; Illustrative Example 19.3.

100°F = tc

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1. Provide breakeven operation 2. Maximize profits

She is also required to perform the calculation if A = 10, B = 4000, as well as A = 10, B = 400,000. Finally, an analysis of the results is requested. SOLUTION: 1. Since there are two contributing factors to the cost model, one may write the following equation for the profit, P

P = A ( t − tc ) −

B ; TH = 500 and tc = 100 (19.32) ( TH − t )

For breakeven (BE) operation, set P = 0 so that

( t − tc ) ( TH − t ) =



B (19.33) A

This may be rewritten as B  t 2 − ( TH + tc ) T +  + TH tc  = 0 (19.34) A 



The solution to this quadratic equation for A = 10 and B = 100,000 is

t=

=

600 ±

( 600 ) − ( 4 )(1) (10, 000 + 50, 000 ) 2

2 600 ± 346 2



= 473°F , 127°F 2. To maximize the profit, take the first derivative of P with respect to t and set it equal to zero, that is,

dP B = A− 2 = 0 (19.35) dt ( TH − t )

Solving,

( TH − t )

2

=

TH − t =

B A B A

= 10, 000 = 100 Since TH = 500,

t = 400°F



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B = 4,000

B = 100,000 Profit 100°F

t 500°F

B = 400,000

FIGURE 19.6 Heat exchanger results.

Upon analyzing the first derivative with t values greater than and less than 400°F, Shannon observes that the derivative changes sign from + → − about t = 400, indicating a relative maximum. Similarly, for A = 10, B = 4000, tBE = 499°F , 101°F tmax = 480°F

For A = 10, B = 400,000,

tBE = 300°F tmax = 300°F



Graphical results for the three scenarios are shown in Figure 19.6. Finally, there are a host of problems peripheral to this “quality energy” concern associated with heat exchangers that the reader may choose to consider. This includes topics that involve minimizing capital costs, minimizing operating costs, minimizing entropy change(s), minimizing area requirements, minimizing energy requirements, and so on, and/or any combination of these.

References 1. P. Abulencia and L. Theodore, Open-Ended Problems: A Future Chemical Engineering Education Approach, Wiley‑Scrivener, Salem, MA, 2015. 2. W. Baasel, Preliminary Chemical Engineering Plant Design, 2nd edn, Van Nostrand Reinhold, New York City, NY, 1990. 3. R. Huntington, High Pressure Pipe Line Research, Clark Bros., Olean, NY, 1947. 4. L. Theodore, Chemical Reactor Analysis and Applications for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2012.

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5. L. Theodore, Chemical Reactor Kinetics, A Theodore Tutorial, Theodore Tutorials, East Williston, NY, originally published by the USEPA/APTI, RTP, NC, 1995. 6. H. Fogler, Elements of Chemical Reaction Engineering, 4th edn, Prentice-Hall, Upper Saddle River, NJ, 2006. 7. H. Mickley, T. Sherwood, and C. Reed, Applied Mathematics in Chemical Engineering, McGrawHill, New York City, NY, 1957. 8. J. Reynolds, Heat Transfer, A Theodore Tutorial, Theodore Tutorials, East Williston, NY, originally published by the USEPA/APTI, RTP, NC, 1995. 9. L. Theodore, Heat Transfer Applications for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2011. 10. D. Kauffman, Process Synthesis and Design, A Theodore Tutorial, Theodore Tutorials, East Williston, NY, originally published by the USEPA/APTI, RTP, NC, 1992. 11. L. Theodore, F. Ricci, and T. VanVliet, Thermodynamics for the Practicing Engineer, John Wiley & Sons, Hoboken, NJ, 2009. 12. L. Theodore, Thermodynamics, A Theodore Tutorial, Theodore Tutorials, East Williston, NY, originally published by the USEPA/APTI, RTP, NC, 1995.

20 Advanced Optimization Term Projects Three sections complement the presentation in this final chapter:

20.1 Sulfuric acid process 20.2 Optimum indoor ventilation flow Rate 20.3 Fluid transportation system 20.4 Mass transfer application

The final chapter (20) in Part IV contains four illustrative examples that could best be described as advanced optimization term projects. Illustrative Examples 20.3 and 20.4 were drawn, with permission, from the classic work of Happel1 titled, Chemical Process Economics, John Wiley & Sons, Hoboken, NJ, 1957; one example is concerned with flow/ piping optimization and the other with a distillation process. The first example in this chapter is concerned with a sulfuric acid process and the second with optimizing indoor ventilation flowrates. Note that no introductory material for each example is provided in this final chapter.

20.1 Sulfuric Acid Process Illustrative Example 20.1 Two sulfuric acid streams (x1, x2) are processed at a small refinery in northern New Jersey. The first (1) contains a 2.0% (by volume) concentration of fine catalyst particles while the second (2) contains a concentration of 8.0%. The maximum and minimum flow rate of each stream that can be processed is 50 gal/hr and 250 gal/hr, respectively. An earlier economic study by Lake Engineers concluded that for maximum profit, the flow rate of the two acid streams should be constrained by the relationship

x1 ≤ 194.5 − 0.69x2 , gal/hr (20.1)



x1 ≥ 126.6 − 0.84 x2 , gal/hr (20.2)

In addition, Fitzmaurice Environmental (FE) introduced another constraint based on environmental concerns that the platinum laden catalyst discharge be limited to 7.5 gal/hr (approximately 50 lb/hr). Assuming a worst-case condition, FE concluded that the following constraint be met:

0.02x1 + 0.08 x2 ≤ 7.5; gal/hr (20.3)

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Rufus Hampton Jr., the plant engineer recently requested that the authors calculate the maximum and minimum hourly flow rate of each stream that can be processed that satisfies all of the aforementioned constraints and maximizes profit. SOLUTION: Employ Equations (20.1) and (20.2) in conjunction with the following two equations:

50 ≤ x1 ≤ 250 (20.4)



50 ≤ x2 ≤ 250 (20.5)

Excel provides the following flow rate (gal/hr) solutions:

x1 (max ) = 154.78 with x2 = 57.55



x1 (min ) = 57.91 with x2 = 81.77



x2 (max ) = 81.77 with x1 = 57.91



x2 (min ) = 57.55 with x1 = 81.77

The reader is left the exercise of solving the problem graphically.

20.2 Optimum Indoor Ventilation Flow Rates Indoor air quality (IAQ) is a major concern because indoor air pollution may present a greater risk of illness than exposure to outdoor pollutants. People spend nearly 90% of their time indoors. This situation is compounded as sensitive populations – for example, the very young, the very old, and sick people who are more vulnerable to disease – spend many more hours indoors than the average population. Indoor air quality problems have become more serious and of greater concern now than in the past because of a number of developments that are believed to have resulted in increased levels of harmful chemicals in indoor air. Some of these developments are the result of construction of more tightly sealed buildings to save on energy costs, the reduction of the ventilation rate standards to save still more energy, the increased use of synthetic building materials and synthetics in furniture and carpeting that can release harmful chemicals, and the widespread use of new pesticides, paints, and cleaners. Some of the immediate health effects of indoor air quality problems are irritation of the eyes, nose, and throat, headaches, dizziness, fatigue, asthma, pneumonitis, and “humidifier fever.” Some of the long-term health effects of indoor air quality problems are respiratory diseases and cancer. These are most often associated with radon, asbestos, and second-hand tobacco smoke. The US Environmental Protection Agency (EPA), in a report to Congress nearly three decades ago, estimated that the costs of IAQ problems were in the tens of billions of dollars per year. The major types of costs from IAQ problems are direct medical costs, lost productivity due to absence from the job because of illness, decreased efficiency on the job, and damage to materials and equipment.

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Illustrative Example 20.2 Refer to Illustrative Example 13.5.







1. Calculate minimum air ventilation flow rate containing 10 ng/m3 nanoparticles into the room to ensure that the nanoagent concentration does not exceed 35.0 ng/m3. The nanoagents are appearing in the laboratory at a rate of 250 ng/min. Assume steady-state conditions. 2. Calculate the steady-state concentration in the laboratory; the initial concentration of the nanochemical is 500 ng/m3. There is no additional source of nanoagents (generated) and the ventilation air is essentially pure, that is, there is no background nanoagent concentration. 3. If the room volume is 142 m3, the flow rate of the 10 ng/m3 ventilation air is 12.1 m3/min, and nanoparticles are being generated at a steady rate of 30 ng/min, calculate how long it would take for the concentration to reach 20.7 ng/m3. The initial concentration in the laboratory is 85 ng/m3. 4. Refer to part 3. How long would it take to achieve a concentration of 13.65 ng/m3? How long would it take to reach 12.2 ng/m3?

SOLUTION:

1. The applicable model for this case is (see also Part III, Illustrative Example 13.5): v0 (c0 − c ) + rV = V



dc (20.6) dt

Under steady-state, dc/dt = 0. Pertinent information includes rV = 250 ng / min c0 = 10 ng / m 3





c = 35 ng / m 3

Substituting gives v0 =

− rV c0 − c

=

rv c0 − c



(20.7)

250 = 35 − 10 = 10 m 3 / min = 353 ft 3 / min

2. The applicable model is v0 (c0 − c ) + rV = V





dc (20.8) dt

For steady-state conditions dc/dt = 0. In addition, based on the information provided, r = 0 and c0 = 0. Therefore, and as expected, c=0

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3. First, note that



τ = 142 / 12.1 = 11.73 min



k = r / V = 30 / 142 = 0.211 ng / m 3 ⋅ min

The applicable describing equation is c = ci e





 t  −   + (c0 + k τ ) 1 − e  τ   (20.9)  

Substituting gives 20.7 = 85e



 t −   τ

 t  −  11.73 

 t   −  + (10 + 2.48 ) 1 − e  11.73    

Solving by trial-and-error gives (approximately)



t = 29 min



4. First, calculate the steady-state concentration for this condition. The applicable model is obtained from part 2, after setting dc/dt = 0 c0 + k τ c − = 0 (20.10) τ τ



Solving and substituting gives c = c0 + k τ



= 10 + (0.211)(11.73 ) = 12.48 ng / m 3

Since this is the steady-state concentration, it will take an infinite period of times to reach this value. However, the steady-state concentration represents the minimum concentration achievable in the laboratory based on the conditions specified. Therefore, the concentration will never reach a value of 12.2 ng/m3.

20.3  Fluid Transportation System1 The general subject of fluid flow encompasses a host of topics. One topic of importance to both environmental and chemical engineers is the optimizing of a fluid transport system. Happel1 provides one such example that requires the use of optimization principles. His example and solution (employing his notation), in which the total yearly cost of operating a pipeline system in the year 1957 is formulated, is provided subsequently. First, fixed charges on compressors and pipes will be established. The installed cost of compressors is taken as X dollars/hp. Q cu ft/day of gas are to be pumped a distance of L miles. Power requirement will be aZ2 bhp/(MM cu ft)/day, where Z2 is the deviation from

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perfect-gas laws at the suction pressure, and a is the reference brake horsepower per million cubic feet of gas per day. The investment in compressors is then

 aQZ2  X ; dollars / mile (20.11)  L106 

The fixed charge on compressors, α, includes maintenance, depreciation, and allowance for profit on investment. The yearly fixed charge on the compressors is then

αXQZ2 ; dollars / ( yr ) (mile) (20.12) L106

Pipe costs Y dollars/ton. If W tons of pipe per mile are necessary for construction of the line, the investment in pipe is YW tons/mile. For steel pipe W = 28.2(D + t)t where D is the inside diameter of the pipe in inches, and t is the pipe wall thickness in inches. Barlow’s formula for the wall thickness of pipe in terms of maximum allowable pressure, P lbf/sq in., and maximum allowable stress, S lbf/sq in., is t = P1A / 2 (S − P1 ) (20.13)

The direct pipe investment is then

 P1D   P1D  YW = Y 28.2  D +   ; dollars / mile (20.14) 2 (S − P1 )   2 (S − P1 )  

A good designation of laying cost is H dollars/mile for right of way, surveying, clearing, supervision, engineering; G dollars/ton for unloading, hauling, aligning, placing, and welding or coupling; and, N dollars/(mile)(in. diam.) for ditching, painting, back filling, and laying equipment. The total pipe cost is then

YW + H + GW + ND ; dollars / mile (20.15)

B is the fractional fixed charge per year on the installed pipe. The yearly charges on the pipe will be

 P1D   P1D  B ( Y + G)( 28.2)  D +   + BND + BH dollars / ( yr ) (mile) (20.16) 2 (S − P1 )   2 (S − P1 )  

Next, operating expenses must be considered. Labor, supervision, and salaries are taken at $3850/month for a 4000-hp station. If direct proportionality is assumed for other stations for approximately the same size, this expense is

3850 × 12  QZ2   QZ2    = 11.6   ; dollars / ( yr ) (mile) (20.17) 4000  L106  L106 

Supplies for maintenance of compressors amount to $3.40/hp/yr. The total yearly cost for these items is thus,

 QZ2  3.40  ; dollars / ( yr ) (mile) (20.18)  L106 

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About 10 std cu ft of fuel/bhp are required at a cost of Cf dollars/M std cu ft. The yearly cost of fuel is 10 × 24 × 360  QZ2   QZ2  Cf  = 86.4Cf  ; dollars / ( yr ) (mile) (20.19) 6    L106  L10 1000



In addition, an allowance of $100/(yr)(mile) is taken to allow for loss of gas by leakage. The addition of all these fixed and operation costs, taking the constant a = kln (P1 / P2 ) + b (20.20)



gives the total yearly cost as follows:

Cy =

  P1 P12 KQZ2 2   kln P P + b + K D + /  ( ) 1 2 1 2  + BND + BH + 100 (20.21) 6   L10  2 (S − P1 ) 4 (S − P1 ) 

where:

K = αX + 86.4Cf + 15.0 (20.22)



K 1 = 28.2B ( Y + G) (20.23)

The following values of various constants mentioned previously are believed to be representative of 1957 costs: α = 5% maintenance plus 5-year payment = (0.05)(5) = 0.25/yr X = $250/bhp, compressor cost not installed Cf = $0.10/1000 cu ft a = 56.8 ln(P1/P2) + 5.1 k = 56.8 b = 5.1 B = 5.4% maintenance plus 5-yr payout = 0.27/yr Y = $240/ton, pipe cost not installed G = $30/ton, unloading pipe, and so on. S = $20,000 lb/sq in, allowable stress N = $500/(in. diam.)(mile), unloading pipe, and so on. H = $5900/mile, survey, and so on. With these constants provided, the cost Cy of operating a pipeline system for a specified capacity is a function only of the pressures P1 and P2, the distance between stations L, and diameter of pipe D. It is desired to compute the optimum values for these parameters for the economic transportation of 150 MM std ft3/day of natural gas. Additional data from various standard sources is provided by Huntington.2 The flow equation relating Q to the independent variables involved is

309

Advanced Optimization Term Projects

where: D = P1 = P2 = L = Zavg =

Q = K2

(

D 2.622 P12 − P2 2

(

L0.541 Zavg

)

)

0.541

0.541

(20.24)

inside diameter of pipe, in. discharge pressure from compressors, lb/sq in. abs suction pressure to compressors, lb/sq in. abs distance between compressor stations, miles average deviation from perfect-gas laws for terminal pressures P1 and P2

with

K2 =

80.8  M    ρ0µ 0.0814  T 

0.541

(20.25)

where for methane ρ0 = density, lb/cu ft at 60°F = 0.0422 μ = viscosity in ln/(ft)(sec) at 60°F = 7 × 10−6 M = molecular weight of methane = 16 T = absolute temperature, °F abs = 520°R From this data K2 = 2.15 × 105. The optimum can be established either by direct differentiation of the cost function with respect to each of the variables involved, setting the derivatives equal to zero, and solving the resulting simultaneous equations, or by Lagrange’s method. Instead of working the solution out for all four variables P1, P2, L, and D, Huntington chooses to assume successive values for P1, reducing the analytical solution to a problem involving three independent variables, P2, L, and D. Employment of Lagrange’s method3,4 results in the following relationships if Q = ψ(P2, L, D) is taken as the restrictive condition imposed on Cy in Equation (20.24).

∂C y ∂ψ +λ = 0 (20.26) ∂P2 ∂P2



∂C y ∂ψ +λ = 0 (20.27) ∂L ∂L



∂C y ∂ψ +λ = 0 (20.28) ∂D ∂D

whence

 −1.082P2Q KQZ2 ∂a KQZ2 ∂Z2 Q ∂Zavg  + + λ − 0.541 = 0 (20.29) 6 2 2 6 L10 ∂P2 L10 ∂P2 Zavg ∂P2   P1 − P2

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Introduction to Optimization for Chemical and Environmental Engineers

−



Q KQZ2a − λ 0.541 = 0 (20.30) L2 106 L

  P P12 Q K 1D  1 + = 0 (20.31) 2  + NB + λ 2.622 S P D − 2 (S − P1 )  1 



The previous three equations and Equation (20.24) constitute four independent equations with unknown P1, L, D, and λ. They are solved by elimination and on the assumption that Z2 and Zavg are sufficiently close to being constant so that their derivatives will equal zero. With these assumptions it is found that P1 = 1.33 (20.32) P2

and D 6.85 + =

NB

K 1 [ P1 /(S − P1 ) + P12 / 2 (S − P1 ) ] 2

(

K 12.85 P12 − P2 2

)

D 5.85

4.85KQ 2.85Z2Zavg a 2 K 1 P1 / (S − P1 ) + P12 / 2 (S − P1 )  106  

(20.33)

The method of solution then involves calculations of P2 from an assumed value of P1. D is then computed from Equation (20.31), and finally the flow Equation (20.24) is used to solve for L. In the present instance, a series of trials were made for a P1 variation between 300 and 800 lb/sq in. abs. Thus, assuming P1 P2 Z2 Zavg

= 600 = 451 = 0.933 = 0.923

P1 P2  =  0.0312 + S − P1 2 (S − P1 )2 N B K1 K

= 500 = 0.27 = 28.2 × 0.27(240 + 30) = 2,060 = 0.25(250) + 26.4(0.1) + 15 = 86.1

from which the right-hand side of Equation (20.31) equals

( 4.85)(86.1)( 2.01 × 1023 ) (0.933)(22.3) = 7.43 × 108 2 6 5 (2.15)(10 ) ([600] − [ 451] ) (2060)(0.0312)(10 )

311

Advanced Optimization Term Projects

The left-hand side of equation (20.31) equals D 6.85 +



(500)(0.27 ) D 5.85 (2060)(0.0312)

= D 6.85 + 2.10D 5.85

The value of D is obtained by solving the equation D 6.85 + 2.10D 5.85 = 7.43 × 108



The solution of the previous equation by trial-and-error gives D = 19.0 in. The value of L from the flow equation is

(2.15 × 10 )(1.75 × 10 )(3.6 × 10 ) × 0.435 = 48.7 miles (1.31 × 10 ) (0.923) 5

L=



6

5

15

whence the yearly cost, Cy corresponding to the assumed 600 lb/sq in. abs discharge pressure is 86.1 × 1.5 × 108 × 0.933 × 22.3 (19) × 0.0312 + 0.27 × 500 × 19 + 0.27 × 5900 + 100 + 2060 + C y = 48.7 × 106 2 2

C y = $21, 370



The results of the calculations at other discharge pressures for Q = 150 MM ft3/day are given in Table 20.1. It appears that the optimum lies close to 400 lb/sq in. abs discharge pressure for the conditions assumed. One notes that spacing between stations does not vary a great deal.

20.4  Mass Transfer Application1 As previously explained in Chapter 15, 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. This separation process is controlled by the properties TABLE 20.1 Calculation for Optimum Compressor Installation Discharge Pressure for Compressor lb/sq in. abs, P1 300 400 500 600 800

Inside Diameter of Pipe, in., D

Distance between Stations, miles, L

Annual Cost, $/yr, Cy

26 23 21 19 17

51.6 51.9 53.6 48.7 53.1

$21,640 $21,180 $21,250 $21,370 $21,560

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Introduction to Optimization for Chemical and Environmental Engineers

of the components involved, and the physical arrangement of the unit used for distillation. A common technique used for this separation is continuous distillation. This process removes and absorbs the less volatile (heavier) components from the vapor, thus efficiently enriching the vapor with the more volatile (lighter) components.6,7 It is often convenient to express the profitability equation for a mass transfer operation in terms of some dependent process variable other than the venture worth (W) itself. Simplification that is possible is based on the observation that unit costs of each process variable are usually expressible as simple functions of the independent variables. The modified equation to be differentiated graphically may then be expressed as follows: W = Aa + Uu + Vv +  + Zz (20.34)



where unit costs of each process variable A, U, V,⋯, Z are expressed as simple functions of the independent variables. The dependent variable a is taken as a desired process result, and the independent variables u, v,⋯, z as the operating conditions whereby it is obtained. Upon differentiation with respect to each of the independent variables, the following equations result:



∂W = ∂u ∂W = ∂v  ∂W = ∂z

∂A ∂a ∂U ∂V ∂Z +A +u +U+v ++ z =0 ∂u ∂u ∂u ∂u ∂u ∂A ∂a ∂U ∂V ∂Z a +A +u +v + V ++ z =0 (20.35) ∂v ∂v ∂v ∂v ∂v  a

a

∂A ∂a ∂U ∂V ∂Z +A +u +v ++ z +Z=0 ∂z ∂z ∂z ∂z ∂z

A system for employing Equation (20.35), using the plan of variation of one variable at a time is provided in the following mass transfer application1 developed by Harbert5 that illustrates its employment in a mass transfer problem involving two independent variables where process correlations are available for the full range of interest of the variables. Consider the following application. A chemical by-product is obtained at a constant rate of 28 mole % solution in water. It is desired to recover this by-product at a purity corresponding to 99.2 mole % as the overhead from a distillation column. One percent recovery of this chemical for one day’s operation is worth $4.20. Sufficient heat pickup is available to bring the feed to its boiling point. The relative volatility for dilute water in a chemical (top of column) is 1.58, and for a dilute chemical in water (bottom of column) is 1.75. The cost of one additional MM (million) Btu of reboiler heat to the column is $0.60. This figure includes the cost of heat itself, the cost of the condenser cooling to remove the heat, and the cost of additional reboiler and condenser surfaces, allowing for an appropriate profit for the investment involved for the additional surface. For the completely installed column, the cost of one additional plate together with the section of tower it occupies is estimated to be $96 plus $44/sq ft of area. With 18-in. plate spacing, 1 sq ft of plate area will handle a vapor load corresponding to 20 MM Btu of reboiler heat per day. Plate efficiency for these conditions is taken at 90%. It is estimated that maintenance will amount to 2%/yr of installed cost and that operating time will be 300 days/yr. Management expects a payout time at most of two years.

313

Advanced Optimization Term Projects

It is desired to determine the conditions for the most economical design. The operating cost may be expressed in terms of the variables, a = recovery of chemical, % of total u = reboiler heat duty, MM Btu/day v = number of equilibrium plates in tower The interrelations between these variables as calculated for given conditions are shown in Figure 19.3, which provides the product recovery versus the reboiler heat input for towers of various numbers of plates (a vs. u at constant v). These curves were obtained by a trial-and-error computation, using McCabe–Thiele diagrams.6,8 The first step in determining the required optimum is to express the costs of additional units of the variables (values of A, U, and V) on the basis of dollars per operating day. For the product recovery, the value given for one additional percent for one day is A = 4.20



For the reboiler heat variable, the cost per day for 1 MM Btu is $0.60 plus the cost of the additional plate area required. From the data given, the cost of this ($1/20 sq ft) is

( 44.0 / 20) (0.02 + 1 / 2) / (300)(0.9) = $0.00423 / day

where (44.0/20) is the first cost of the area, (0.02 + 1/2) the annual charge per year, 300 the operating days per year, and 0.9 the plate efficiency. Then, this cost becomes $0.00423 v for a tower of v plates, so that U = − (0.60 + 0.00423 v ) (20.36)



The negative sign denotes a cost instead of profit item. For the number of plates variable, the total area required per plate is (u/20) sq ft, so that the original cost of one additional plate is 96.0 + 44.0(u/20). The cost per day of an equilibrium plate then becomes

V=−

96.0 + 44.0(u / 20)   0.02 + 300 × 0.9

1  = − (0.185 + 0.00423u ) (20.37) 2

With these values, appropriate values for the partial derivatives in Equation (19.8) may be computed. ∂A/∂u, ∂A/∂v, ∂U/∂u, ∂V/∂v, all equal zero, and

∂U = −0.00423 (20.38) ∂v



∂V = −0.00423 (20.39) ∂u

Therefore, from Equation (20.35)



 ∂a  4.20   − (0.60 + 0.00423 v + 0.00423 v ) = 0  ∂u   ∂a  4.20   − (0.60 + 0.00846 v ) = 0  ∂u 

(20.40)

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Introduction to Optimization for Chemical and Environmental Engineers

where ∂a/∂u is the slope of the curves of Figure 20.1. Figure 20.2, the cross plot of the data of Figure 20.1, as percent recovery versus number of plates for constant amounts of reboiler heat (a vs. v at constant u) gives, as the slope of its curves, values for ∂a/∂v. Similarly,  ∂a  4.20   − (0.185 + 0.00846 v ) = 0 (20.41)  ∂u 



A set of reasonable values for the independent variables is next assumed: u = 180, v = 50. Substitution of v = 50 into Equation (20.40) gives ∂a 1.023 = = 0.244 (20.42) ∂u 4.20



The corresponding value of u is 165 MM Btu/day from Figure 20.2. Substitution of u = 180 into Equation (20.40) gives ∂a 1.705 = = 0.406 ∂v 4.20



a, product recovery, % of total

100

98

equ υ = n ilib o. o riu mt f ra ∞ ys

The slope ∂a/∂v = 4.06 corresponding to the line u = 180 on Figure 20.2 gives v = 39. As a second trial, u = 165 and v = 39 are assumed. This gives, first ∂a/∂v = 0.222 corresponding to u = 164, and second, ∂a/∂v = 0.382 corresponding to v = 38. Alternatively, one might have selected a series of values of v and plotted Equation (20.50) on Figure 20.1 (dashed line). Similarly, Equation (20.40) can be plotted on Figure 20.1 (dot-dash line). Values for Equation (20.35) could be cross-plotted to Figure 20.1 (dashed line). The two curves on Figure 20.2 would intersect as shown at approximately the same value as obtained by the two successive trails; namely u = 164, v = 38.

60 55 50 45

96

94

40 35 Eq. (3.15) 30

92 160

170

180

190

200

u, reboiler heat duty, MM Btu/day FIGURE 20.1 Recovery versus number of plates.

210

220

315

Advanced Optimization Term Projects

he 22 0 1

Eq. (3.16)

Eq. (3.15)

0

20

2

0

98

ay

d ler u/ oi Bt b re M u, y, M t du at 0

0 17

0

18

96

19

a, product recovery, % of total

100

94 0

16

92 30

35

40

45

50

55

60

υ, number of plates

FIGURE 20.2 Recovery versus number of plates.

The economic design thus corresponds to 38 equilibrium plates with a reboiler duty of 164 MM Btu/day. Resultant chemical recovery is 93.4%. The tower would have 38/0.9 or 42 actual trays with a vapor area of 164/20 = 8.4 sq ft. This design would result in a maximum overall profit for the operation.

References

1. J. Happel, Chemical Process Economics, John Wiley & Sons, Hoboken, NJ, 1958. 2. R. Huntington et al., High Pressure Pipe Line Research, Clark Bros., Olean, NY, 1947. 3. R. Ketter and S. Prawler, Modern Methods of Engineering Computations, McGraw-Hill, New York City, NY, 1969. 4. H. Mickley, T. Sherwood, and C. Reed, Applied Mathematics in Chemical Engineering, McGrawHill, New York City, NY, 1957. 5. W. Harbert, title unknown, Petroleum Refinery, 27, 185, location unknown, 1948. 6. L. Theodore and J. Barden, Mass Transfer, A Theodore Tutorial, Theodore Tutorials, East Williston, NY, originally published the USEPA/APTI, RTP, NC, 1995. 7. L. Theodore and F. Ricci, Mass Transfer Operations for the Practicing Engineer”, John Wiley & Sons, Hoboken, NJ, 2010. 8. L. Theodore, Chemical Engineering: The Essential Reference, McGraw-Hill, New York City, NY, originally published by USEPA/APTI, RTP, NC, 2014.

Index ABC Drug Laboratory, 161 Absorber vs. extractor, 227–232 Accidental errors, 22 Advanced optimization term projects, 303–315 fluid transportation system, 306–311 mass transfer application, 311–315 optimum indoor ventilation flow rates, 304–306 sulfuric acid process, 303–304 Air pollution management, 111–123, 265–268 coal-fired power plant options, 113–114 incinerating mercury-contaminated waste, 119–121 optimizing incinerator performance, 121–123 outdated air pollution control device, 111–112 particulate control equipment options, 114–116 recovering dust, 117–118 AIT, see Auto-ignition temperature (AIT) Algebra, 96 Analytical methods, 83–93 breakeven considerations, 83–85 formulation of optimum, 88–89 one independent variable, 85–88 three independent variables, 93 two independent variables, 89–92 Applied engineering concepts, in linear programming, 102–104 Applied mathematical concepts, in linear programming, 99–102 Approximate numbers, 20–22, 21 Aris, R., 1 Arithmetic relations, 96 Auto-ignition temperature (AIT), 175 Back-mix reactor, 200, 287 Backward curved/inclined fans, 184 BACT, see Best available control technology (BACT) BAT, see Behan and Theodore (BAT) Associates Batch plant operation cost, 258–259 Batch reactor performance, 212–214 Bathtub curve, 279 Behan and Theodore (BAT) Associates, 176 Behan Doyle Environmental Inc., 115–116 Behan Structural Engineers (BSE), 261 Best available control technology (BACT), 113 BETH Construction Company, 166

Biological treatment processes, 126–127 Biological waste treatments, 138 Birmingham Wire Gauge (BWG), 285 Bisection method, 56 Break-even point operation, 83–85, 117, 125–126 BSE, see Behan Structural Engineers (BSE) BWG, see Birmingham Wire Gauge (BWG) Canadian Rocky Mountain Water Plant, 131 Cancer, 155, 161 Cancer-fighting drugs cost, 155–156 Capital cost, 112 Capital recovery factor (CRF), 112, 145, 185 Carbon dioxide systems, 176 Carcinogen, 155 Carnahan, B., 30 Cartesian coordinates, see Rectangular coordinates CCPs, see Coal combustion products (CCPs) Cdf, see Cumulative distribution function (Cdf) Cement incinerator operation, 177–180 Cement kiln vs. rotary kiln incinerator, 139–141 Centrifugal compressors, 189 Centrifugal fan, 184 Centrifugal force, 254 Centrifugal pumps, 185, 186 Chemical engineering applications, 285–300 chemical reactors, 287–293 fluid flow, 285–287 heat exchangers, 295–297 mass transfer operations, 293–295 plant design, 297–300 Chemical foams, 176 Chemical Process Economics, 60, 264, 303 Chemical reactor (CR) applications, 199–215 achieving maximum conversion, 208–210 batch reactor performance, 212–214 continuous stirred tank reactor vs. tubular flow reactor, 199–204 fluidized-bed reactor performance, 204–207 maximizing selectivity, 210–212 operating schedules, 214–215 Chemical reactors (CR), 287–293 Chemical stability, 229 Chemical treatment processes, 126–127 Chemical waste treatments, 138 Classical algebra, 96 Cleaning bath life, 224

317

318

Coal, 271 Coal combustion products (CCPs), 113 Coal-fired power plant options, 113–114 Combustion systems, 119 Computer age, 6 Concrete mix, 149–150 Constraint, 96, 97 Contaminants, 191–192 Contemporary optimization, 8–11 Continuous distillation, 218, 312 Continuous stirred tank reactor (CSTR) vs. tubular flow (TF) reactor, 199–204, 208–209, 288–291, 293 Conventional optimization procedures, 7–8 Copper, 224 Corrosiveness, 228 Counting numbers, see Natural numbers CR, see Chemical reactor (CR) applications CRF, see Capital recovery factor (CRF) Critical insulation thickness, 242–245 Crude oil, 232–233 CSTR, see Continuous stirred tank reactor (CSTR) vs. tubular flow (TF) reactor Cumulative distribution function (Cdf), 169 Cyclone selection and design, 254–258 Determinate error, 22 Differentiation, 23–26 Digital computer, 6, 95 Dilutant and solute, 230 Dilution ventilation, 294 Direct approach, 51 Discounted cash flow application, 144–147 Distillation, 218, 232, 312 Double pipe (DP) heat exchangers, 239–240, 241, 247 Doyle Engineers, 166 Doyle Financial Consultation, 230 DP, see Double pipe (DP) heat exchangers Drag-out reduction, 224 DRaT II model, 280, 281–282 DRaT IV model, 280, 282–283 DRaT V model, 280, 283 Dry chemical systems, 176 Dummying process, 224 Dupont, R., 159, 274, 280 Dust recovering, 117–118 Dynamic programming, 9, 11, 95, 96 Economic lot size, 221 Economics and optimization applications, 10 Edison, Thomas, 5 Edisonian approach, 5

Index

Electrostatic precipitators (ESPs), 114 Elliptical PDE, 39–40 Encapsulation, 138, 144, 147 Energy quality, 246–247 Energy recovery, 247–248 Environmental engineering applications, 265–283 air management, 265–268 hazard risk assessment, 279–283 DRaT II model, 281–282 DRaT IV model, 282–283 DRaT V model, 283 health risk assessment, 273–277 solid waste management, 271–272 water management, 268–271 Environmental Protection Agency (EPA), 113, 138, 155, 156, 304 Equipment lifetime and salvage value, 112 Errors, 22–23 ESPs, see Electrostatic precipitators (ESPs) Extraction, 229 Extractor, absorber vs., 227–232 Extra expenses, 222 Extrapolation, 17–19 Fabric filter, 115 Failure distribution rate, 169–172 Failure rate curve, 279 Fans and blowers, 184 Fan selection, 183–185 FE, see Fitzmaurice Environmental (FE) Feasible solutions/values, 99 Filter press operation, 219–221 Filtration process, 219 Finite differences, 39 Fire accidents, 174–175 Fire extinguishing systems, 166 Fire protection equipment, 174–177 First-degree equation, 67 First law of thermodynamics, 246 First-order differential equations, 32 First-order ODEs, 34, 36 Fitzmaurice Environmental (FE), 303 Flue gas emissions, 158–160 Fluid flow, 285–287 Fluid flow applications, 183–196 fan selection, 183–185 pipe diameter selection, 188–189 pump selection, 185–187 three-stage compressor, 195–196 two-stage compressor, 189–191 ventilation models, 191–194 Fluidization, 204

319

Index

Fluidized-bed reactor performance, 204–207 Fluid transportation system, 306–311 Flynn, Annmarie, 298 Foam insulation, 238 Foam systems, 176 Fogler, S., 200 Forward curved fans, 184 Freezing point, 229 Full logarithmic graph, 68, 71 Gas absorption, 227, 265 Gaseous pollutants, 112, 265 Gas permeation (GP), 133 Gas phase reactions, 213 Gas solubility, 228 Gauss elimination, 30 Gauss–Jordan reduction, 30 Gauss–Seidel approach, 31 Geometrical patterns, 5 Golden section method, 57–60 GP, see Gas permeation (GP) Graphical approaches, 65–80 applications, 77–80 logarithmic-logarithmic (log-log) coordinates, 68–73 methods of plotting data, 76 rectangular coordinates, 66–68 intercepts, 68 linear vs. non-linear, 67–68 quadrants, 67 semi-logarithmic (semi-log) coordinates, 73–76 plotting straight lines, 73–76 triangular (trilinear) coordinates, 76 Graphical/tabular presentation, 17–18 Gravity filters, 220 Guidelines for Carcinogen Risk Assessment, 155 Hampton, Rufus, Jr., 304 Happel, J., 60, 88, 104, 264, 271, 303, 306 Hazard risk assessment, 165–180, 279–283 cement incinerator operation, 177–180 DRaT II model, 281–282 DRaT IV model, 282–283 DRaT V model, 283 failure distribution rate effect, 169–172 fire protection equipment, 174–177 plant safety proposal comparisons, 167–169 safety expenditures vs. return of safety investments, 165–167 worst-case explosions effect, 172–174 Hazard risk assessment of accidents (HZRA), 279

Health risk assessments (HRAs), 153–163, 158–159, 273, 273–277 cancer-fighting drugs cost, 155–156 health concerns associated with flue gas emissions, 158–160 insurance costs, 157–158 minimum purging time, 153–155 pharmaceutical cancer drug, 161–163 vitamin caloric and protein requirements, 160–161 Heat exchangers, 239, 295–297 Heat transfer applications, 235–248 critical insulation thickness, 242–245 maximizing profit through energy recovery, 247–248 optimum insulation thickness, 238–239 profitable exchanger, 239–242 recovering quality energy, 246–247 steam options, 235–237 Heinsohn, R., 192 High-expansion foams, 176 High-speed computing equipment, 6 HRAs, see Health risk assessments (HRAs) Hydrocarbon vapors, 172 Hydrocracker, 232 HZRA, see Hazard risk assessment of accidents (HZRA) IAQ, see Indoor air quality (IAQ) Incineration, 119, 138 Incinerator performance optimization, 121–123 Incipient fluidization, see Minimum fluidization Independent design variables, 98 Indeterminate errors, 22 Indirect approach, 51 Indoor air pollution, 191 Indoor air quality (IAQ), 304 Industrial ventilation, 191 Industrial wastewater, 126 Inert waste, 9 Inherited error, see Propagation error Insurance costs, 157–158 Integer linear programming, 8 Intercepts, 68 Interpolating method, 25 Interpolation, 17–19 Interval halving method, 53–55 Inverse interpolation, 18, 19 Kaufmann, D., 295 KBLT Power Plant, 168 Ketter, R., 31, 37

320

Latent enthalpy effects, 235 Law of conservation of energy, 246 Leaf filters, 220 Least squares, 25 LI, see Liquid injection (LI) incinerators Linear algebra, 96 Linear programming, 8–11, 95–106, 182 applied engineering concepts, 102–104 applied mathematical concepts, 99–102 definitions, 96–97 optimization concepts, 97–99 real-world applied concepts in linear programming, 105–106 Linear vs. non-linear equation, 67–68 Liquefied nitrogen gas (LNG), 84 Liquid injection (LI) incinerators, 139–140 Liquid–liquid extraction, 229 Liquid phase reactions, 213 Liquid–solid changes, 236 LNG, see Liquefied nitrogen gas (LNG) Logarithmic-logarithmic (log-log) coordinates, 68–73 Logarithmic-probability (log-normal) graphics, 76 Mass transfer applications, 217–233, 311–315 absorber vs. extractor, 227–232 filter press operation, 219–221 optimum reflux ratio, 217–219 process and equipment modifications, 223–227 processing crude oil, 232–233 production rates, 221–223 Mass transfer operations, 293–295 Material and waste exchanges, 132–133 Material substitution, 223 Mathematical ideas and concepts, 5 Mathematical operations, 13–40 approximate numbers, 20–22 differentiation, 23–26 errors, 22–23 extrapolation, 17–19 interpolation, 17–19 nonlinear algebraic equations, 31–32 numerical integration, 26–28 Simpson’s rule, 28 trapezoidal rule, 27–28 ordinary differential equations, 32–36 partial differential equations, 37–40 elliptical, 39–40 parabolic, 37–38 quadratic equation, 15–17 significant figures, 20–22

Index

simultaneous linear algebraic equations, 29–31 Gauss elimination, 30 Gauss–Jordan reduction, 30 Gauss–Seidel approach, 31 Mathematical optimization technique, 103, 104 McCabe–Thiele diagrams, 313 Mechanical foams, 176 Mechanism of combustion, 120 Membrane construction costs, 133–135 Membrane processes, 133 Mercury-contaminated waste incineration, 119–121 Mercury waste emissions, 147–149 Methylation, 147 Methyl mercury, 147 MF, see Microfiltration (MF) Microfiltration (MF), 133 Minimum expenses, 222 Minimum fluidization, 205 Mixed solvent, 230 Monte Carlo approach, 39, 177, 179 Multi-independent variable equations, 52 Multilayer insulation, 238 Multistage decision process, 95 Multivariable optimization problem, 97 NAHEB Construction Associates, 157 Nanoparticles, 192 National Fire Protection Association, 166 National Science Foundation (NSF), 192 Natural numbers, 13 Negative integers, 13 Newton–Raphson (NR) method, 31, 52 Newton’s method of tangents (NMT), 32 Non-inert waste, 9 Nonlinear algebraic equations, 31–32, 52 Non-linear equation, linear vs., 67–68 Nonlinear programming, 9, 95 NR, see Newton–Raphson (NR) method NSF, see National Science Foundation (NSF) Numerical error, 36 Numerical integration, 26–28 Simpson’s rule, 28 trapezoidal rule, 27–28 Numerical optimization algorithm, 5 Nypro Ltd. caprolactam factory, 172 Objective function, 98, 99 O’Brien, Shannon, 298 ODEs, see Ordinary differential equations (ODEs) One independent variable, 42–45, 85–88 One-independent variable search methods, 60 OPEC oil embargo (1973), 246

321

Index

Operating costs, 112, 222 Operating schedules, 214–215 Operations Research, 11, 97 Optimization, 1, 3–11, 109 computer age, 6 concepts, 97–99 contemporary, 8–11 definition, 1 history, 4–6 linear programming, 8–11 model, 95 procedures, 7–8 scope, 7 Optimum indoor ventilation flow rates, 304–306 Optimum insulation thickness, 238–239 Optimum reflux ratio, 217–219 Ordinary differential equations (ODEs), 32–36 Organic wastes, 138 Organization costs, 222 Outdated air pollution control device, 111–112 Oxidation, see Incineration Parabolic PDE, 37–38 Partial differential equations (PDEs), 37–40 elliptical, 39–40 parabolic, 37–38 Particulate control equipment options, 114–116 Particulate pollutants, 112, 265 Pasquill-Gifford model, 276 Pathogens, 128 PD, see Plant design (PD) applications PDEs, see Partial differential equations (PDEs) Pdf, see Probability distribution function (Pdf) Performance measure, 97 Perry’s Handbook, 95, 238 Perturbation techniques, 41–49 one independent variable, 42–45 three independent variables, 48–49 two independent variables, 45–48 Pharmaceutical cancer drug, 161–163 Physical treatment processes, 126–127 Physical waste treatments, 138 Pipe diameter selection, 188–189 Pivot equation, 30 Plant design (PD), 297–300 Plant design (PD) applications, 249–262 batch plant operation cost, 258–259 cyclone selection and design, 254–258 shipping facilities, 249–252 structure design, 261–262 tank farms, 252–254 ventilation models with system variables, 259–261

Plant safety proposal comparisons, 167–169 Plate-and-frame filters, 220 Plotting data methods, 76 Positive displacement pumps, 185 Positive integers, 13 Powder insulation, 238 Prawler, S., 31, 37 Precision and accuracy, 22 Primitive counting systems, 5 Probability distribution function (Pdf), 169 Propagation error, 23 Protective systems, 165 Protein molecules, 160 Pump selection, 185–187 Purging time, 153–155 Pythagorean theorem, 96 Quadrants, 67 Quadratic equation, 15–17 Quadratic formula, 16 Radial blade fans, 184 Raffinate, 230 Random number generators, 177 RCRA, see Resource Conservation and Recovery Act (RCRA) Real numbers, 13 Real-world applied concepts, in linear programming, 105–106 Reciprocating compressors, 189 Reciprocating pumps, 185, 186 Rectangular coordinates, 66–68 intercepts, 68 linear vs. non-linear, 67–68 quadrants, 67 Refractory materials, 238 Regression equation, 25 Resource Conservation and Recovery Act (RCRA), 132, 138 Return of safety investments, safety expenditures vs., 165–167 Reverse osmosis (RO), 133 Rezifp Pharmaceuticals and Chemicals Corp., 214 Ricci, F., 176, 280, 283, 298 Risk, 158, 274 RK, see Rotary kiln (RK) incinerator, cement kiln vs.; Runge–Kutta (RK) method RO, see Reverse osmosis (RO) Rotary compressors, 189 Rotary kiln (RK) incinerator, cement kiln vs., 139–141 Rotary pumps, 185, 186

322

Rotary vacuums filters, 220 Round-off error, 23, 30, 36 Runge–Kutta (RK) method, 32–4 Safety expenditures vs. return of safety investments, 165–167 SCC, see Secondary combustion chamber (SCC) Scientific notation, 20 Search methods, 51–63 bisection, 56 golden section, 57–60 interval halving, 53–55 steepest ascent/descent, 60–63 Secondary combustion chamber (SCC), 141 Second law of thermodynamics, 246, 247 Second-order polynomial equation, 25 Seek methods, see Direct approach Selectivity, 210–212 Semi-logarithmic (semi-log) coordinates, 73–76 plotting straight lines, 73–76 triangular (trilinear) coordinates, 76 SF, see Slope factor (SF) Shaefer, S, 280 Shell-and-tube (ST) heat exchangers, 239, 241 Shipping facilities, 249–252 Significant figures, 20–22 Simpson’s rule, 28 Simultaneous linear algebraic equations, 29–31 Gauss elimination, 30 Gauss–Jordan reduction, 30 Gauss–Seidel approach, 31 Single-pass cylindrical tube, 201 Single-payment compound amount factor, 144 Single-phase heat exchanger, 241 Single solvent, 230 Single-variable optimization, 97 Slope factor (SF), 155, 156 Slope-intercept method, 65, 70, 73 Slurry flows, 219 Solidification, 138 Solid–liquid extraction, 229 Solid waste management, 137–150, 271–272 cement kiln vs. rotary kiln incinerator, 139–141 discounted cash flow application, 144–147 maximum mercury waste emission constraints, 147–149 sludge waste receiving tank size, 142–143 solid waste treatment, 137–139 structural considerations of concrete mix, 149–150 Solid waste treatment, 137–139 Solvent, 230

Index

ST, see Shell-and-tube (ST) heat exchangers Statistical methods, 76 Steam jet systems, 176 Steam options, 235–237 Steam Tables, 236 Steepest ascent/descent method, 60–63 Straight blade fans, see Radial blade fans Structure design, plant, 261–262 Sulfuric acid process, 303–304 Superinsulation, 238 Superproduction costs, 222 Tank farms, 252–254 TF, see Tubular flow (TF) reactor, continuous stirred tank reactor (CSTR) vs. Theodore, Lou, 157, 243, 280, 283, 296, 298 Theodore approach, 5 Theohan Gas Company, 84 Thermal conductivity, 238 Thermal insulation, 238–239 Thermal waste treatment, 138 Three independent variables, 48–49, 93 Three-point equal-interval search method, 53 Three-stage compressor, 195–196 “Three T’s of combustion,” 120 TNT, 172–174 Toxicity, 229 Transient operation, 212 Trapezoidal rule, 27–28 Treybal, R., 229 Trial-and-error method, 5, 41, 60, 98, 99, 104, 296, 313 Triangular coordinates, 76 Truncation error, 23 Tubular flow (TF) reactor, continuous stirred tank reactor (CSTR) vs., 199–204, 208–209 Tumble burners, 141 Two independent variables, 45–48, 89–92 Two-phase heat exchanger, 241 Two-stage compressor, 189–191 Ultrafiltration (UF), 133 Unconfined vapor cloud explosions, 172 Unimodal function, 51, 60 Unsteady-state operation, see Transient operation US Department of Agriculture (USDA), 160 Vacuum insulation, 238 Vapor–liquid changes, 236 Vapor–solid changes, 236 Ventilation models, 191–194, 259–261, 293

323

Index

Viscosity, 229 Vitamin caloric and protein requirements, 160–161 Volatility, 228 Walas, S., 202 Wastewater treatment equipment options, 126–127 Waterborne diseases, 128 Water disinfection options, 127–130 Water management, 268–271 Water plant operation, 131 Water pollution management, 125–135 break-even point operation, 125–126 membrane construction costs, 133–135

waste revenues from waste streams, 132–133 wastewater treatment equipment options, 126–127 water disinfection options, 127–130 water plant operation, 131 Water quality problem, 127 Water sprays, 114, 176 Water systems, 176 Weibull distribution, 279–280 Wet collectors, 114, 115 Wet scrubbers, see Wet collectors Wilkes, J., 30 Worst-case explosions effect, 172–174 Zeroes, 21