Dimitrios P. Tassios Applied Chemical Engineering Thermodynamics extras.springer.com SpringerVerlag Berlin Heidelber
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Dimitrios P. Tassios
Applied Chemical Engineering Thermodynamics
extras.springer.com
SpringerVerlag Berlin Heidelberg GmbH
Prof. Dimitrios P. Tassios National Technical University of Athens Zographou Campus 15780 Zographos Athens, Greece Fonnerly: New Jersey Institute of Technology 323 Dr. Martin Luther King Jr. Blvp. Newark, N.J., 07102, USA
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ISBN 9783662016473 ISBN 9783662016459 (eBook) DOI 10.1007/9783662016459 This work is subject to copyright. All rights are reserved, wether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version , and permission for use must always be obtained from SpringerVerlag Berbn He idelberg GmbH . Violations are liable for prosecution under the German Coyright Jaw.
© SpringerVerlag Berlin Heidelberg 1993 Originally published by SpringerVerlag Berlin Heidelberg New York in 1993
The use of general descriptive names, registered names, trademarks. etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typesetting: Camera·ready by author; 6 1/3020
5 4 3 2 10
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To my students, Past, Present, and Future
The Hazards of Literary Life "The hardships endured by authors are typified by a letter bound inside the flyleaf of a volume in exhibition of annotated rare books on view at the Smithsonian Institution . .... . The book in question is "History and Progress of the Steam Engine", published in London in 1837. It was written by civil engineer Elijah Galloway, who addressed the letter inside to Charles Manby, a prominent English engineer. It reads: 'Dear Sir: I am just on the eve of completing a new work on "The Mechanic Powers and Their Application to Machinery". But literary work is only paid for when the Manuscript is completed and what with ill health and the all absorbing nature of this kind of labor I am in the awful predicament of wanting a bit of bread! I have by me an only copy of my well known "History of the Steam Engine", which Mrs Galloway will herewith hand you. Will you in charity purchase it of her? This is the only legitimate source by which I can avert the horror of hunger and cold for we have neither food,fire, nor a penny in the world! I am, Dear Sir, Your ever Obliged E. Galloway."'
(Chern. & Eng. News, Nov. 10, 1986, p.40.)
This is by way of expressing my appreciation to NJIT for granting me a Sabbatical leave for the 19851986academic year, during which most of this book was written.
Preface
There are many excellent books on Chemical Engineering Thermodynamics, but their use in teaching the subject in the classroom leads to two serious difficulties: First, most books tend to concentrate either on undergraduate or graduate level material only. This creates problems in the review of undergraduate material, which I have found essential for an effective graduate course. Second, their presentation of the subject is too extensive for the number of credits allocated to it (about six semester hours in the undergraduate level and three in the graduate one). As a result, Chapters  or Sections of them must be skipped, making the subject that is by its very nature complex  still more difficult for the students to comprehend. Thus, not only learning suffers, but a fear of Thermodynamics is embedded into the student's mind, which hinders his or her future learning and use of the subject. (This fear is apparent to anyone teaching the graduate level course; and felt by many practicing engineers and chemists, leading a well known thermodynamicist to state a few years ago: "The most underutilized subject in the chemical and petroleum industries is Thermodynamics.") These problems are, hopefully, avoided with this book for the material included in it covers both levels and can be presented in the aforementioned semester hours. Obviously, less material is thus presented, but the book does provide the undergraduate and graduate student with the material needed in industrial practice and for future study; in addition, through the extensive list of pertinent references, it provides for: * a more in depth study of the different subjects; * practical applications; * the development of awareness of data sources, and * making the student appreciate the importance of using the literature in problem solving. The emphasis is, of course, placed on covering the undergraduate level material, while that for the graduate one is presented in Sections of Chapters 3,5, 7, 8, and 12, with the Problems numbered over 50 assigned to it; and in Chapters 10, 16 and 17. The presentation is made in such a fashion, however, that minimizes any loss of continuity in the undergraduate material, while the latter's review in the graduate course is facili
X
tated with the assignment of some of the aforementioned Problems (51, 52 ... ). The short length of the book should also make it useful to the practicing engineer and chemist that needs a relatively quick review of the subject, as well as references for a more detailed study of items of his/her interest. The development of the subject is based on the Laws of Thermodynamics, i.e. it follows the historical approach, which seems easier for most students to understand than the postulatory one. (The latter is based on a small number of postulates, that cannot be proved but only disproved by showing that consequences resulting from them are in disagreement with experimental evidence, and from which the laws of thermodynamics can be derived.) We start the presentation of Chemical Engineering Thermodynamics by taking a Glimpse at the Subject and proceed with a discussion of its Laws, followed by a brief Historical Account of its development. (A Nation cannot exist without a sense of its history, why should a field of Science and Technology?) We continue with a discussion of Efficient Energy Utilisation  a subject of immense importance that became, unfortunately, apparent only after the energy crisis of th·e early 1970's  followed by a brief presentation of the Broader Implications of the Second Law. (We are educating the technological leaders of tomorrow, not just technologists.) We proceed with a limited discussion of Intermolecular Forces, which will help in understanding the Physical and Thermodynamic Properties of Pure Fluids which are considered next and, of course, the rest of the subject matter. Cubic Equations of State are then considered for they have become a major tool in describing quantitatively the properties of Pure Fluids, as well as those of Mixtures, which are discussed next. We continue with a discussion of Equilibrium and Stability, that prepares the ground for the ensuing presentation of the two topics of paramount importance to chemical engineers: Phase and Chemical Reaction Equilibria. We close with a brief discussion of Statistical Mechanics that should prepare the student for further ·study in this area, which is becoming progressively more and more important in applied thermodynamics. The inclusion of computer Programs in the enclosed diskette should help the students in developing familiarity with the determination of fluid properties and phase equilibria calculations. The presentation of the material has been influenced by the excellent books of Professors K. Denbigh (The Principles of Chemical Equilibrium, 4th Ed., Cambridge University Press, 1981), J.M. Prausnitz (with R.N.
XI Lichtenthaler and E.G. de Azevedo, Molecular Thermodynamics of Fluid Phase Equilibria, 2nd Ed., Prentice Hall, 1986) and H.C. Van Ness (with J .M. Smith, Introduction to Chemical Engineering Thermodynamics, 4th Ed., McGrawHill, 1987), that are often mentioned in the book. The philosophy of presentation, however, has been shaped by many students, who always want to know the why and the relevance of what they are taught, and to whom this book is dedicated; and from my own conviction, that we should strive to make the topic more attractive to students by exposing them to the broader world of Thermodynamics. If they like this book, the effort will have been more than worthwhile. In closing I would like to gratefully acknowledge: Professor D. Cardwell for permitting the use of material from his book: From Watt to Clausius; Professor H. Van Ness for providing the program that generates the Steam Tables; and Academic Press for permitting the use of material from W. Kenney's book Energy Conservation in the Process Industries; and, of course, to express my appreciation to NJIT for granting me a Sabbatical leave for the 19851986 academic year, during which most of this book was written.
Dimitrios P. Tassios Athens, 1992
Contents
1 A Glimpse at Thermodynamics 1.1 1.2 1.3 1.4 1.5
Introduction Objective and Approach The Vocabulary of Thermodynamics Pressure, Temperature and Volume Work 1.5.1 Definition 1.5.2 Example 1.1 1.5.3 Example 1.2
1.6 Energies 1.6.1 1.6.2 1.6.3 1.6.4 1.6.5
Potential Kinetic Internal Enthalpy Helmholtz and Gibbs Free Energies
1.7 Heat 1.8 System, Surroundings, and Related Terms 1.8.1 1.8.2 1.8.3 1.8.4 1.8.5
System and Surroundings Modes of Interaction Between Them State Process Equilibrium: Internal and External
1.9 Internal Energy and Molecular Energy 1. 10 Entropy 1.11 State Properties I . 12 Reversibility 1.13 Auxiliary Terms
1 1 2 2 4 5 5 5 6 7 7 8 8 8 8 9
10 10 10
10 11 11 11 12 12 13 15
Contents
XIV I .13.1 Heat Capacity 1.13.2 Ideal Gas 1.13.3 Real Fluid versus Ideal Gas 1. I 3.4 Ideal Solution 1. I3.5 Real versus Ideal Solution 1. I3.6 Activity Coefficient I. I 3. 7 Chemical Potential and Fugacity I. I 3. 8 Intensive and Extensive Properties
I .14 The Foundations of Thermodynamics 1.14.1 The Laws of Thermodynamics 1.14.2 The Thermophysical Data 1.15 The Purpose of Chemical Engineering Thermodynamics 1.15.1 Feasibility of Processes and Efficient Energy Utilization I. 15.2 Thermophysical Properties 1.15.3 Chemical and Phase Equilibrium 1.16 Examples 1.3 through 1.9
15 15 16 16 17 17 18 18 18 18 20 20 21 22 23 24
2 The Zeroth and First Laws of Thermodynamics
35
2.1 Introduction 2.2 Objective and Approach 2.3 The Zeroth Law and the Ideal Gas Temperature 2.3.1 The Statement 2.3.2 Is it Really a Law? 2.3.3 Temperature: the Common Property of Systems in Thermal
35 36 36 36 37
Equilibrium 2.3.4 The Ideal Gas Temperature Scale 2.3.5 Example 2.1
2.4 The First Law 2.4.1 Background and Statement 2.4.2 The Analytical Expression 2.4.3 Comments 2.4.4 Example 2.2 2.4.5 Example 2.3 2.5 The First Law Expression for a SteadyState Flow Process 2.5.1 The Problem 2.5.2 Development of the Expression 2.5.3 Comments 2.6 Thermodynamic Relationships of the Ideal Gas 2.6. 1 Importance of Ideal Gases
37 38 40 41 41 43 44 44 45 46 46 46 47 48 48
Contents
XV
2.6.2 Equation of State 2.6.3 Internal Energy and Enthalpy Changes Due to Heating or Cooling 2.6.4 The Difference in Heat Capacities 2.6.5 The PYTRelationship for an Adiabatic and Reversible Change 2.6.6 Comments 2.7 Examples 2.4 through 2.7
2.8 Concluding Remarks
3 The Second Law of Thermodynamics 3. 1 Introduction 3.2 Rationale for a Second Law 3.2.1 3.2.2 3.2.3 3.2.4
The Direction of Natural Phenomena The Feasibility of Processes The Big Question: Where is the Energy Gone? Summary
3. 3 Objective and Approach 3.4 Second Law: The Verbal Statement 3.4.1 Statements 3.4.2 Example 3.1 3.5 The Carnot Cycle 3.5.1 Description 3.5.2 The Expression for the Thermal Efficiency 3.5.3 The Maximum Thermal Efficiency 3.5.4 Comments 3.5.5 Example 3.2 3.5.6 Example 3.3 3.6 A Real Heat Engine: A Simple Power Plant 3.6.1 Description 3.6.2 The Expression for the Thermal Efficiency 3.6.3 Comments 3.7 Second Law: The Analytical Statement
3.8 Proposition I 3. 9 Proposition II 3. 9. I Approach 3.9.2 j(tpt2 ) = j(t 1)/f(t2 ) 3.9.3 The Thermodynamic Temperature 3. 10 Proposition III
48 49 49 49 50 51 56
61
61 62 62 62 63 63 64 65 65 65
66 66 67 69 70 72 72 73 73 74 74 75 76 77 77 77 79 79
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Contents
3 11 Proposition IV 3011.1 Objective 30110 2 Entropy Change of a System Undergoing a Spontaneous 0
Process
3011.3 The Total Entropy Change 3 012 Entropy Change Calculations 3 12 I The Problem and Approach 301202 Isothermal Removal or Addition of Heat 301203 Heating or Cooling of a Body from T1 to T2 3012.4 Entropy Changes of an Ideal Gas 301205 Entropy of Mixing 3013 A Molecular Interpretation of Entropy 3014 Examples 304 through 3012 3015 Steam Power Plants 3 015 01 Operating Conditions 3o15o2 Comments 301503 Example 30130 A look at Example 1.9 3 016 Concluding Remarks 0
0
4 Thermodynamics: A Historical Perspective 401 402 403 4.4
Introduction Objective and Approach The Power of Heat: Early Recognition The Steam Engine 4.401 James Watt 4.402 Richard Trevithick and the Railway Locomotive 405 The Science of Heat 40501 The Thermometer 40502 The Nature of Heat 40503 The Conservation of Fire 405.4 The Differentiation of Heat and Temperature 40505 The Conservation of Heat 40506 The Caloric Theory and its Problems 406 The Establishment of Thermodynamics 407 Sadi Carnot 40701 Motivation 40702 Arguments 40703 Conclusions 407.4 The Acceptance of his Work
81 81 81 82 82 82 83 83 83 84 85 88 97 97 99 101 102
107 107 108 109 110 112 113 114 115 116 117 118 119 120 121 123 124 124 125 126
Contents 4.8 Emile Clapeyron 4.8.1 Contributions 4.8.2 The Efficiency of the Camot Cycle 4.8.3 Example 4.1. The Clapeyron Equation 4.8.4 Example 4.2. Evaluation of the Camot Factor C 4.9 The Conversion of Work to Heat 4.9.1 J.R. Mayer 4.9.2 J.P. Joule 4.9.3 Early Experimental Work 4.9.4 The Expansion/Compression of Gases and Later Work
4.10 Kelvin, Clausius, and the Establishment of the First and Second Laws 4.10.1 Lord Kelvin 4.10.2 R.J .E. Clausius
XVll 127 128 129 130 131 132 132 133 134 135 137 137 139
4.11 The Absolute Temperature and the Analytical Expression
of the Second Law 4.11.1 The Absolute Temperature 4.11. 2 The Analytical Statement of the Second Law: Reversible Processes 4. 11.3 Real Life Processes and Entropy: Irreversible Processes
4.12 Thermodynamic Properties of Matter 4.12.1 J.W. Gibbs 4.12.2 EquationsofState 4.12.3 Activity Coefficients 4.12.4 The Third Law of Thermodynamics 4.12.5 OtherContributors
5 Efficient Energy Utilization: Energy Conservation 5. 1 Introduction 5.2 Objective and Approach 5.3 Ideal Work 5.3.1 Definition 5.3.2 Example 5.1 5.3.3 Evaluation of Ideal Work: the General Expression 5.3.4 Example 5.2 5.3.5 Example 5.3
5. 4 Useful Energy 5.4.1 Definition
142 142 142 143 144 145 147 149 150 151
155 155 156 158 158 159 160 162 162 163 163
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Contents
5.4.2 Example 5.4 5.5 Exergy 5.5.1 Maximum Obtainable Useful Energy from a System 5.5.2 Definition 5.6 Evaluation of Exergy 5.6.1 Exergy of a System at T and P 5.6.2 Example 5.5 5.6.3 Example 5.6 5.6.4 Exergy of Thermal Energy Available at a Temperature T 5.6.5 Exergy of Any Energy Except Thermal 5.6.6 Summary 5. 7 First and Second Law Efficiencies 5. 7.1 Definitions 5. 7.2 Example 5. 7 5.7.3 Example5.8 5. 7.4 Comments 5.8 Cogeneration 5.8.1 Rationale 5.8.2 Example 5.9 5.8.3 Example5.10 5. 9 Limitations of Cogeneration
164 165 165 166 166 166 167 167 168 168 170 171 171 172 173 173 174 174 175 179 181
5.10 Energy Conservation: The Favorable Economics and the Institutional Barriers 5.11 Concluding Remarks
185
6 Second Law and Entropy: A Broader View 6.1 6.2 6.3 6.4
Introduction Objective and Approach Material Dissipation and the 'Fourth' Law Second Law and World History 6.4.1 Common Threads in World History 6.4.2 Entropy Watersheds and the Main Transitions in World History
6.5 Energy Demand and Supply in the Future 6.5.1 The Energy Crisis of the Early 1970's: Was it a Warning? 6.5.2 The U.S. Oil Demand and Supply to the Year 2000 6.5.3 Global Energy Demand and·Supply in the Future 6.6 Meeting the Energy Demand in the Future: Problems 6.6.1 Cost
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191 191
192 192 193 193 194 197 197 198 200 203 203
Contents
XIX
60602 Environmental 60 7 So, Is There a Third Entropy Watershed Coming? 60 7 01 The Magnitude of the Problem 60702 Life After This Entropy Watershed 6o 1o3 Comments 608 Concluding Remarks
204 208 209 209 211 211
7 Intermolecular Forces 7 01 702 703 7.4
Introduction Objective and Approach Intermolecular Forces and Potential Energy Origin and Types of Intermolecular Forces 7.401 Attractive Forces 70402 Repulsive Forces
705 Physical Attractive Forces 70501 DipoleDipoleForces 70502 DipoleInduced Dipole Forces 70503 Dispersion Forces 70504 Example 7010 Comparison of Attractive Forces 70505 Comments 706 Intermolecular Potentials 70601 The Mie Potential 70602 The Lennardlones Potential 70603 Comments 706.4 Example 7020 The Origin of the 'Geometric Mean' Rule 707 Chemical Forces 70 701 Hydrogen Bonding 70 702 Weaker Forces 7 07 03 Applications 707.4 Example 703 708 Concluding Remarks
8 Physical Properties of Pure Fluids 801 Introduction 802 Objective and Approach 803 The Volumetric Behavior of Pure Gases 80301 The Origin of Fluid Pressure
217 217 217 218 219 219 220 221 221 221 222 223 224 225 225 226 227 228 229 229 231 231 231 233
237 237 238 239 239
XX
Contents 8.3.2 PressureVolume Isotherms
8.4 The Volumetric Behavior of Pure Vapors and Liquids 8.4.1 PressureVolume Isotherms 8.4.2 Terminology 8.4.3 Phase Diagrams
8.5 Volumetric Behavior Common to All Pure Fluids 8.6 The Volumetric Behavior of Pure Fluids: Quantitative Treatment 8.6.1 The Compressibility Factor 8.6.2 Determination of the PVT Behavior
8. 7 The Corresponding States Principle 8. 7.1 8.7.2 8.7.3 8. 7. 4 8. 7.5
The TwoParameter Version The ThreeParameter Version Example 8.1 Comments Example 8.2. The Molecular Theory of the Corresponding States Principle
8.8 Equations of State: General Remarks 8. 8. 1 The Purpose of an Equation of State 8.8.2 Classification of Equations of State
8.9 The Virial Equation of State 8.9.1 The Expression 8.9.2 Evaluation of Virial Coefficients 8.9.3 Example 8.3. The Relationship Among the Coefficients of the Volume and Pressure Series Expansions 8.9.4 Example 8.4 8.9.5 Example 8.5. Estimation of Second Virial Coefficients: Nonpolar Compounds 8.9.6 Example 8.6. Estimation of Second Virial Coefficients: Polar Compounds 8. 9. 7 Comments
8.10 Cubic Equations of State 8. 10.1 8.10.2 8.10.3 8.10.4 8.10.5 8.10.6 8.10.7 8.10.8
Development of a Cubic Equation of State Example 8. 7 Example 8.8 The RedlichKwong Equation of State Example 8.9 The SRK, PR, and vdW711 Equations of State Example8.10 Comments
8.11 The BenedictWebbRubin (BWR) Equation of State
240 240 240 241 243 244 245 245 246 246 246 248 250 250 251 251 251 252 252 252 253 255 255 257 260 260 261 261 262 262 263 264 264 265 267 270
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XXI
8.12 Recommendations for the Estimation of the Volumetric Behavior of Pure Fluids 8.13 Vapor Pressures 8. 13. 1 Correlation of Vapor Pressure Data 8.13.2 Estimation of Vapor Pressures 8.13.3 Comments
8.14 Enthalpies of Vaporization 8.14.1 Determination and Prediction 8.14.2 Example 8.11
8.15 Heat Capacities 8.15.1 Determination 8.15.2 Example 8.12
9 Thermodynamic Properties of Pure Fluids 9. 1 Introduction 9.2 Objective and Approach 9.3 The Fundamental Equations: Closed Systems 9.3.1 The Fundamental Equation for the Internal Energy 9.3.2 The Fundamental Equation for H, A, and G 9.3.3 Example 9.1
9.4 The Fundamental Equations: Open Systems 9.5 Evaluation of Thermodynamic Properties of Pure Fluids 9.5.1 The Problem 9.5.2 The Approach
9.6 The Maxwell Relations 9. 7 Departure Functions 9. 7.1 9.7.2 9.7.3 9. 7.4 9.7.5 9. 7.6
The Approach Example 9.2 Example 9.3 Evaluation of Departure Functions Evaluation of Enthalpy and Entropy Differences Development of Enthalpy and Entropy Tables and Charts
9. 8 Estimation of Departure Functions 9.8.1 9.8.2 9.8.3 9.8.4 9.8.5
The Methods Example 9.4 Example 9.5 Example 9.6 Comments on the Estimation of Entropy and Enthalpy Departures
270 271 2 71 273 273 274 275 276 276 277
278
285 285 286 286 287 288 290 290 293 293 293 296 297 297 298 300
301 302 303 303 303
304 305
305 306
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Contents
9. 9 The Chemical Potential 9. I0 Fugacity 9.10.1 Definition 9.10.2 A Physical Interpretation
9. 11 Evaluation of Fugacities from Experimental Data 9.11.1 The Mathematical Formulation 9.11.2 Evaluation of Fugacities with an Equation of State 9.11.3 Example 9.7 9.11.4Example 9.8 9.11.5 Liquid Phase Fugacities 9.11.6 Example 9.9
9. 12 Estimation of Fugacities 9.12.1 The Methods 9.12.2Example9.10 9.12.3 Comments
9.13 Summary
10 Cubic Equations of State 10.1 Introduction 10.2 Objective and Approach 10.3 Evaluation of Vapor Pressures with an Equation of State 10.3.1 The Problem 10.3.2 Example 10.1 10.3.3 Comments
10.4 Development of a Cubic Equation of State for Vapor Pressure Prediction 10.4.1 The Approach 10.4.2 The Dependency of Alpha on Temperature 10.4.3 Generalization of the EoS
10.5 Prediction of Saturated Liquid Volumes 10.5.1 The Problem 10.5.2 Methodology and Results
10.6 Comments 10.7 Concluding Remarks
11 Properties of Mixtures 11.1 Introduction
307 308 308 309 310 310 311 312 312 313 314 315 315 315 316 317
323 323 324 324 324 325 326 326 326 327 328 328 328 329 331 336
339 339
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11.2 Objective and Approach 11.3 Partial Molar Properties 11.3.1 Definition 11.3.2 Example 11.1
11.4 Mixture Properties from Partial Molar Properties 11.4.1 The Problem 11.4.2 The Approach 11.4.3 Comments
11.5 Partial Molar Properties from Mixture Properties 11.5.1 The Analytical Expression 11.5.2 Example 11.2
11.6 Interdependency of the Partial Molar Properties 11.6.1 The GibbsDuhem Equation 11.6.2 Example 11.3 11.6.3 Example 11.4
11.7 Evaluation of Mixture Properties II. 7.1 II. 7.2 11.7. 3 11.7 .4 11.7 .5 11.7 .6 11.7.7 11.7. 8
The Problem and Approach The Virial Equation The Pitzer Corresponding States Correlation Cubic Equations of State Example 11.5 Example 11.6 Example 11.7 Comments on the Estimation of Mixture Properties
11.8 Fugacities in Mixtures 11.8.1 11.8.2 11.8.3 11. 8.4
Fugacity of the Components of a Mixture Relationship Betweenfand}; The Case of an Ideal Solution Comments
11.9 Fugacities with Equations of State 11.9.1 Fugacities with the Virial Equation 11.9. 2 Fugacities with Cubic Equations of State 11.9.3 Example 11.8
11.10 Fugacities with the Standard State Method 11.10.1 The Problem and Approach 11.10.2 The Standard State Fugacity for a Condensable Component 11.10.3 Example 11.9 11.10.4 The Standard State Fugacity for a Noncondensahle Component 11.10.5 Example 11.10
340 341 341 342 343 343 343 344 344 344 345 347 347 348 349 349 349 350 353 354 356 357 358 359 360 360 361 362 363 365 365 366 367 368 368 369 370 371 373
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Contents
11.11 An Application of Fugacities: Solubility of Solids and Liquids in Gases 11.11.1 Solids 11.11.2 Comments 11. 11.3 Liquids
11 . 12 Property Changes of Mixing 11.12.1 Definition 11.12. 2 The Case of an Ideal Solution
11. 13 Excess Properties 11.13.1 Definition 11.13.2 The Excess Gibbs Free Energy and the Activity Coefficient
11. 14 Concluding Remarks
12 Equilibrium and Stability
374 374 375
377
378 378 378 380 380 381 382
393
393 394 394 12.3.1 A Closed System of Constant Internal Energy and Volume 395 396 12.3.2 Example 12.1 397 12.3.3 A Closed System of Constant Temperature and Pressure 398 12.3.4 Example 12.2 12.4 The Basis for Phase Equilibrium Calculations at Constant 400 Temperature and Pressure 400 12.4.1 The Equality of Chemical Potentials 401 12.4.2 Example 12.3 401 Equation Clapeyron The 12.4. Example 12.4.3 402 12.4.4 The Equality of Fugacities 402 12.4.5 Comments 12.5 The Basis for Chemical Reaction Equilibrium Calculations 403 at Constant Temperature and Pressure 403 12.5.1 The Problem and Approach 403 12.5.2 Stoichiometric Coefficients and Numbers 404 12.5.3 The Progress of the Reaction Variable 405 12.5.4 Example 12.5 405 12.5.5 Example 12.6 12.1 Introduction 12.2 Objective and Approach 12.3 The Thermodynamic Criteria of Equilibrium
12.5.6 Evaluation of Gi: The Standard State Fugacity and Gibbs Free Energy 12.5. 7 The Expression for the Equilibrium Constant K
406 407
XXV
Contents 12.5.8 Example 12.7 12.5.9 Comments
12.6 The Basis for Mu1tiphase/Multireaction Equilibrium Calculations at Constant Temperature and Pressure 12.6.1 12.6.2 12.6.3 12.6.4
ThreePhase Equilibrium Multiple Reactions in a SinglePhase System Multiple Reactions in a TwoPhase System Comments
12.7 The Complete Specification of an Equilibrium System 12.7 .1 12.7.2 12.7.3 12.7.4
The Phase Rule The Duhem Theorem Example 12.8 A Multiphase, Multireaction System
12.8 Stability in Thermodynamic Systems 12.8.1 Stability Criteria 12.8.2 Comments 12.8.3 Example 12.9
12.9 Phase Transitions 12.9.1 The Unstable Region 12.9.2 The Metastable Region 12.9.3 Comments
12.10 LiquidLiquid Equilibrium in Binary Systems 12.10.1 12.10.2 12.10.3 12.10.4 12.10.5 12.10.6
The Mixing of Two Pure Liquids Evaluation of the Miscibility Limits The Metastable and Unstable Regions The Effect of Temperature on Mutual Solubility Example 12.10 Comments
12.11 Concluding Remarks
13 Low Pressure VaporLiquid Equilibrium 13.1 13.2 13.3 13.4 13.5
Introduction Objective and Approach The VaporLiquid Equilibrium Problem Methodology Activity Coefficients: Determination from VaporLiquid Equilibrium Measurements 13.5.1 Experimental Measurements 13.5.2 Evaluationofthe Activity Coefficient
408 408 409 409 410 411 412 412 413 414 414 415 416 416 417
418 419 420 420 422 424 424 425 426 427 428 429 430
435 435 436 437 438 440 441 441
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13.5.3 Example 13.1 13.5.4 Comments
13.6 Ideal Solutions 13.6.1 13.6.2 13.6.3 13.6.4 13.6.5 13.6.6 13.6.7 13.6.8
Raoult's Law Example 13.2 Example 13.3 Example 13.4 Example 13.5 Example 13.6 Example 13.7 Summary
13.7 Activity Coefficients: Qualitative Remarks 13.7.1 The Effect of Composition 13.7. 2 The Effect of Molecular Nature 13. 7.3 Summary
13.8 Activity Coefficients: The Effect of Pressure and Temperature 13.8.1 Pressure 13. 8. 2 Temperature
13.9 Thermodynamic Consistency of Experimental Data 13.9.1 13.9.2 13.9.3 13.9.4
The Interrelationship Among Activity Coefficients Binary Systems Comments Example 13.8
13.10 Activity Coefficients and Composition: Analytical Expressions 13.10.1 Procedure 13.10.2 Comments
13.11 The Wohl Type Expressions 13. II. I 13 .11. 2 13. II. 3 13.11.4 13.11.5
The General Expression for a Binary System The van Laar Equation The Margules Equation Evaluation of the Parameters Example 13.9
13.12 The Local Composition Expressions 13 .12. I 13.12.2 13.12.3 13.12.4 13.12.5 13.12.6
The Concept of Local Compositions The Wilson Equation The NRTL Equation The UNIQUAC Equation Comments Example 13.10
442 443 445 445 446 448 448 449 450 450 451 452 452 452 457 458 458 458 460 460 460 461 462 463 463 464 465 465 466 466 467 467 469 469 470 471 472 473 474
Contents
13.13 Advantages of the Local Composition Models 13. 13.1 Correlation of Binary Data 13.13.2 Temperature Dependency of the Parameters 13.13.3 Prediction of Multicomponent VLE Behavior
13.14 SingleParameter Expressions 13.15 Evaluation of Vapor Phase Composition from Liquid CompositionTotal Pressure Data 13.15.1 The Approach 13.15.2 Example 13.11
13.16 Bubble and Dew Point Calculations 13. 16.1 Bubble Point 13.16.2 Dew Point
13.17 Prediction of Multi component VLE Behavior from Binary Data 13.17. 1 The Effect of the Method for Evaluating Binary Parameters 13.17. 2 The Effect of the Quality of the Binary Data 13.17. 3 The Effect of Temperature 13.17.4 The Effect of System Type 13.17.5 Example 13.12
13.18 Estimation of VaporLiquid Equilibrium 13.18.1 The GroupContribution Concept 13.18.2 The UNIFAC Model 13.18.3 Comments
13.19 Concluding Remarks
14 High Pressure VaporLiquid Equilibrium 14 .I Introduction 14.2 Objective and Approach 14.3 Qualitative Behavior 14.3.1 Bubble, Dew, and Critical Points of Mixtures 14.3. 2 Comments 14.3. 3 Equilibrium Compositions as Functions of Pressure and Temperature
14.4 Quantitative Description 14.5 Graphical Methods: The KCharts 14.5.1 Charts Using P and T Only 14.5.2 Charts Using the Convergence Pressure
14.6 The ChaoSeader Method
XXVII 475 475 476 478 479 481 481 482 483 484 485 487 487 490 492 492 493 494 495 495 497 498
511 511
511 512 512 514
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518
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XXVIII 14.6.1 The Method 14.6.2 Comments
14.7 The Equation of State Approach 14.7 .I Required Information 14. 7.2 Comments 14.7. 3 Polar Systems
14.8 Bubble Point, Dew Point, and Flash Calculations 14.8.1 Bubble and Dew Points 14.8.2 Example 14.1 14.8.3 Example 14.2 14.8.4 Example 14.3 14.8.5 Flash Calculations 14.8.6 Example 14.4
14.9 Concluding Remarks
15 Chemical Reaction Equilibrium 15.1 Introduction 15.2 Objective and Approach 15.3 The Dependency of K on Temperature 15.3.1 15.3.2 15.3.3 15.3.4 15.3.5 15.3.6 15.3.7 15.3.8 15.3.9
The Problem The Variation of K with the Enthalpy of the Reaction The Temperature Dependency of i!H 0 The Expression for K(T) Comments Example 15.1 Example 15.2 Example 15.3 Example 15.4
15.4 Equilibrium Conversion: GasPhase Reactions 15.4.1 15.4.2 15.4.3 15.4.4 15.4.5
Kin Terms of Composition
Evaluation of the Equilibrium Conversion Example 15.5 Example 15.6 Adiabatic Reactions
15.5 Equilibrium Conversion: LiquidPhase Reactions 15.6 Factors Effecting the Equilibrium Conversion 15.6.1 Temperature 15.6.2 Pressure 15.6.3 Inert Components
52{) 521 522 522 523 528 529 529 531 531 532 533 535 536
545 546 546 548 548 548 550 552 552 553 554 555 556 558 558 558 559 560 561 562 564 564 564 564
XXIX
Contents 15.6.4 Excess Reactant in Feed 15.6.5 Example 15.7. The Ammonia Synthesis 15.6.6 Example 15.8
15.7 The Phase Rule and the Duhem Theorem for Reacting Systems 15.7 .1 The Phase Rule and the Number of Independent Reactions 15.7.2 Example 15.9 15.7.3 Example 15.10 15.7.4 Example 15.11 15.7.5 The Duhem Theorem
15.8 Equilibrium Conversion: Heterogeneous Systems 15.8.1 The Equilibrium Constant in Terms of Composition for a Gas(g)Solid(s) Reaction 15. 8.2 Example 15.12. Solid """Solid + Gas 15.8.3 Example 15.13. Solid + Gas ,... Solid + Gas
15.9 Multiple Reactions 15.9.1 15.9.2 15.9.3 15.9.4
The Problem Example 15.14 Example 15.15 Example 15.16
15.10 Concluding Remarks
16 Elements of Statistical Mechanics 16.1 Introduction 16.2 Objective and Approach 16.3 Description of the Molecular Behavior 16.3.1 16.3.2 a. b. 16.3.3 16.3.4 16.3.5
Classical Mechanics Quantum Mechanics Introductory Remarks The Schrodinger Equation Comments Example 16.1 Example 16.2
16.4 The Translational Wave Function of a Single Particle 16.4.1 The General Solution 16.4.2 Quantum Numbers, Permissible Energy Levels, and the Wave Function 16.4.3 Comments
565 565 567 567 568 569 569 570 570 571 571 572 572 573 573 574 575 575 576 585 585 586 587 588 588 589 590 592 592 593 594 594 595 596
XXX
Contents
16.5 The Averaging Process 16.5.1 The Problem 16.5.2 The Statistical Postulate
16.5.3 Ensembles 16.6 The Canonical Ensemble 16.6.1 The Characteristic Property of the Probability Function 16.6.2 Example 16.3 16.6.3 The Partition Function
16.7 Thermodynamic Properties from the Partition Function 16.8 The Partition Function of an Ideal Gas 16. 8.1 In Terms of the Molecular Partition Function 16.8.2 Factoring the Molecular Partition Function 16.8.3 The Translational Partition Function
16.9 The Description of an Ideal Gas 16.9.1 Equation of State 16.9.2 Thermodynamic Properties
16.10 Concluding Remarks 16.10.I A Rigorous Development of the Second Law 16.10.2 A Philosophical Insight into the Second Law
17 Statistical Mechanics: Application to Real Fluids 17. 1 Introduction 17.2 Objective and Approach 17.3 The Partition Function of a Real Fluid I 7. 3. I The Configuration Integral 17.3.2 Comments 17.3. 3 Example 17 .1. Thermodynamic Properties from the Configuration Integral
17.4 The Radial Distribution Function and Thermodynamic Properties 17 .4. 1 The Radial Distribution Function g(r) 17 .4.2 Thermodynamic Properties from g(r) and f(r) 17.4.3 Comments
17.5 Low Densities: The Virial Equation 17 .5.1 The Vi rial Equation 17.5.2 Comments 17.5.3 Example 17.2. The Second Virial Coefficient of the HardSphere Fluid
596 596 597 598 598 598 599 600 601 602 602 603 604 605 605 606 607 607 608
615 615 615 616 616 617 619 620 620 621 623 624 624 624 625
Contents
XXXI
17.6 High Densities: The Approaches 17.7 Molecular Simulation 17.7.1 17.7.2 17.7. 3 17.7 .4
Monte Carlo (Metropolis et al, 1953) Molecular Dynamics (Alder and Wainwright, 1959b) Comments The HardSphere Fluid
17.8 Perturbation 17.9 Equations of State with the Semiempirical Approach 17.9.1 17.9.2 17.9.3 17.9 .4 17.9.5
The Generalized van der Waals Partition Function The van der Waals Equation of State The CamahanStarlingvdWand RK Equations of State Rotational ami Attractive Contributions Comments
17. I0 Concluding Remarks
626 627 627 628 628 629 631 632 632 633 634 635 636 638
Appendices A B C D E
Conversion Factors Steam Tables Physical Properties for Selected Compounds LeeKesler Tables Computer Programs
643 644
Mollier h,s Diagram
696
Index
699
672 677
695
1 A Glimpse at Thermodynamics
The modern steam engine was designed and first used to facilitate the mining of coal. As mines had to be sunk deeper illto the ground to extract available supplies, it became more difficult to ventilate the mines and to face still another problem. At a certain depth the water table was reached and drainage became a priority. All of these problems required a technological solution. The steam engine was the answer. The first steam pump was patented in 1698 by Thomas Savery. ... The answer to the transportation (of coal) crisis was the invention of the steam locomotive and rail road tracks. Like the steam pump, the steam locomotive was a direct technological response to the needs created by the new coal environment. Together, the steam pump and the steam locomotive laid the technological base for the industrial era thatfollowed. (Emphasis added, Rifkin, 1980, p.74.)
1.1 Introduction Imagine yourself living in the first half of the 19th century. The society you live in is changing rapidly as a result of the Industrial Revolution made possible by the Steam Engine, which replaced the limited human, animal, and waterfall power, by the  what it seemed then  unlimited heat power. Yet, even though the steam engine has been around for over a century it was first introduced by Thomas Savery in 1698  the scientific principles underlying its operation are not established yet. It was not until the year 1850 that Clausius  on the basis of the pioneering work of several individuals, especially Carnot and Joule expressed the foundations of the conversion of heat into work, i.e. the First and Second Laws of Thermodynamics. A new, thus, field of Science and Technology was established: Thermodynamics (from the Greek: therme = heat; dynamis = force, for in those days the terms work and force were used interchangeably).
2
Applied Chemical Engineering Thermodynamics
Finally, in the latter part of the 19th century, J. Willard Gibbs extended the application of thermodynamics to Chemical Reaction and Phase Equilibria, laying the foundations of  what we call today  Chemical Engineering Thermodynamics. As the profession of Chemical Engineering grew into a separate discipline in the early part of the 20th century, thermodynamics played an increasingly important role in its development. Today, it represents together with Chemistry, Mathematics, Physics, Transport Phenomena, Kinetics, and Economics  the cornerstones of our Profession. On the Laws of Thermodynamics
The Laws of1hennodynamics control, in the last resort, the rise and fall of political systems, the freedom or bondage of Nations, the movements of commerce and industry, the origins of wealth and property, and the general welfare of the race. (Frederick Saddy, NobelPrize winning chemist.)
1.2 Objective and Approach Our main objective in this Chapter is to take a quick look  a glimpse at thermodynamics: its Vocabulary, Foundations, and Purpose. We start with a very brief description of the majority of the terms used in this field, the Vocabulary of the field, so that we can begin to learn and understand them and, eventually, communicate through them. We proceed with a discussion  again brief  of the Foundations of thermodynamics: the Laws and the Thermophysical data, so that we become aware of their importance from the very beginning of this endeavor. We continue with a .brief outline of the Purpose of thermodynamics, i.e. the role that thermodynamics plays in the Chemical and Petroleum Industries, and close with a series of Examples that give a brief preview of the type of problems to be addressed in future Chapters.
1.3 The Vocabulary of Thermodynamics There is a variety of terms that are often used in thermodynamics and
A Glimpse at Thermodynamics System and Surroundings
3
Volume, Pressure and Temperature
j
Auxiliary Terms (1)
Energies: PE; KE;
/ 1
I THERMODYNAMICS .
U; H; A; G
.
Work and Heat
/1~
Entropy
(1)
State Properties
Reversibility
Heat capacities, Ideal gas, Residual properties, Ideal solution, Excess properties, Activity coefficients, Chemical Potential and Fugacities
Figure 1.1 The Vocabulary of Thermodynamics.
constitute its vocabulary. Figure 1.1 presents the most common among them in a pictorial fashion, while a brief description of them is presented in Sections 1.4 through 1.13. Our objective here is not, of course, to develop familiarity with  and understanding of these terms. Rather, it is to expose you to them and, hopefully, to refresh your memory, since you have already used several of them in Chemistry and Introductory Chemical Engineering courses. As we proceed through the ensuing Chapters, familiarity and understanding will be developed. After all one learns a language by using it, not by just memorizing words. There is another similarity between learning a language and learning thermodynamics. The same way we can visualize certain objects in a language  for example, chair, or table  we can visualize volume, or temperature in thermodynamics. But when it comes to concepts  such as, love, hate, sorrow, etc. in a language; or, entropy, energy, fugacity, etc., in thermodynamics we only develop a 'feeling' or a 'sense' about them through repeated exposure to and use of them. We cannot visualize them. (It should be kept in mind, however, that the thermodynamic concepts are well defined mathematically.)
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Applied Chemical Engineering Thermodynamics
1.4 Pressure, Temperature and Volume Our understanding of Pressure comes from Physics and it is defined as the ratio of the perpendicular force F exerted on a given area A, divided by this area: p = F
A
(1.4.1)
Consider, next, Temperature. A close examination of this term suggests that, even though we grow up with it from the "warmth of our mother's womb" to medicinal applications and weather forecasting  what we really become familiar with is a 'physiological' sense of 'hotness' and 'coldness' measured with a mercury thermometer in degrees Celsius ec) or Fahrenheit eF), not with a thermodynamic quantity. The latter is a more complex term that will be developed in the next two Chapters, leading to the concept of absolute thermodynamic or just absolute temperature, measured in Kelvin (K) or Rankine (R). The relationship among these scales is: teC)
=
T(K)  273.15
teF) = T(R)  459.67 T(R) = 1.8T(K)
The units for the quantities: F, P, T and V (volume) are given in Table 1.1 for the two main systems: the English System (ES) and the International System (Syst~me Internationale, SI). For conversion from the one set of units to the other see Appendix A. Von Guericke, the Power of the Atmosphere and the Feeble Horses. The credit for making the pressure exerted by the atmosphere entirely explicit belongs to Otto von Guericke who, in 1672, published the famous book in which he described his air pump and the experiments that he had made with it from the mid1650s onwards. A dramatic picture in this book, which was subsequently widely reproduced, showed two teams, each of eight horses, straining ineffectually to pull apart two small hemispheres just over a foot in diameter. These had been evacuated of air by means of a pump and were held together in the form of a sphere solely by the weight of the atmosphere. In this way von Guericke graphically illustrated the power that could be generated by the weight of the atmosphere. It should be added that either
5
A Glimpse at Thermodynamics
Table 1.1 F, P, V, T Units in ES and SI ES Quantity Force Pressure Temperature Volume
SI
lbr
Newton: N
psia: lbr in2 oF, R ft3
Pascal: N m2 K mJ
the sixteen horses were remarkably feeble or von Guericke was exaggerating. The latter is the more probable conclusion. (Cardwell, 1971, p.12)
1.5 Work 1.5.1 Definition Our earliest, probably, experience with work is associated with the lifting of a certain object. In general, we define work (W) through: (1.5.1) dW = Fds where F is the force, and s the displacement, both in the same direction. [For limitations of this definition, and a more rigorous one, see Hatsopoulos and Keenan (1965).] In the ES system work is expressed in ftlbr; in the SI, in J (Joule). In thermodynamics, work involves  in the typical case  a change of the system (Section 1.8.1) volume due to difference in its pressure and that of the surroundings, as demonstrated in the following two Examples.
1.5.2 Example 1.1 A certain amount of air is placed inside a metallic cylinderandpiston arrangement. The piston, assumed to be frictionless, is held in its position with latches so that the air volume is V1 = 5 lit. The system is then placed into a constant temperature (25°C) bath and when thermal equilibrium is reached, the pressure of the air becomes 5 atm. The latches are now removed and the piston is allowed to move against some external pressure P 0 until the air volume becomes V2 •
Applied Chemical Engineering Thermodynamics
6
Develop an expression for the work produced in terms of the air volume. In this case:
(A)
where A is the cross sectional area of the piston. To express the displacement ds in terms of the corresponding volume change of the air dV, we note that:
ds = dVIA
(B)
Introduction of Eqs (A) and (B) into Eq .1.5 .1 yields: dW = P0 dV
and:
1.5.3 Example 1.2 Let us now evaluate the work produced in Example 1.1 for the following three cases: a. there is a vacuum outside the piston; b. the pressure outside the piston is 1 atm; c.the pressure outside the piston is controlled in such a way, that it is always barely (infinitesimally) smaller than that of the air inside it. In all three cases the final volume is twice the initial one.
a.P0
=
0, and no work is produced.
b.P0 = 1 atm, which remains constant throughout the movement of the piston. Hence: W
= P0 (V2  V1) = (1)(10 5) atm lit = 506.6 J
c. In this case, P0 is practically the same as the pressure of the air P throughout the movement of the piston, and, consequently:
W =
Jv PdV 2
(A)
VI
Since P varies with the expansion of the air, we need a relationship between P and V so that the integration can be carried out. Such a relationship is provided by the ideal gas law: PV
= NRT
(B)
The number of moles of air in the cylinder (N) can be calculated from the initial condition: P = 5 atm; V = 5 lit; T = 298.15 K, giving a value of N = 1.022 moles. · Before we introduce Eq.(B) into Eq.(A), we must consider the following question: Does the air temperature Tremain constant during the expansion process?
7
A Glimpse at Thermodynamics
Since the opposing pressures are so close to each other, the process is very slow and it is reasonable to expect that the air temperature will remain constant and equal to that of the bath, i.e. that the system is continuously in a state of internal equilibrium (see Section 1.8.5). Eq.(A) yields, therefore:
W • NRTJn[
~:]· NRT1n2
=
1756.1
Comments: 1. We will see in Section 1.12 that cases (a) and (b), where the driving force  i.e. the difference between external and internal pressures  is large, are referred to as irreversible; and case (c), where the driving force is very small, as reversible. 2. Notice that as we move from the most irreversible case (a) to the reversible case (c), the amount of work produced is increasing. 3. Actually, case (c) produces the maximum possible work. Any attempts to produce more work, by increasing further the external pressure, would lead to compression, i.e. the consumption of work. 4. Only in the reversible case (c), can we replace the external pressure P0 with the air pressure P.
1.6 Energies In its broadest sense, energy reflects the ability of a body, or system, to produce work, for example, to raise an other body to some elevation h above its original position. Notice that we use the term 'sense' for energy, because it is a concept. Several energy terms are encountered in thermodynamics, from the more comprehensible ones, such as potential, kinetic, internal and enthalpy to the more abstract ones, such as the Helmholtz and the Gibbs free energies.
1.6.1 Potential Energy, PE This energy reflects the force of gravity exerted on a body of mass m and it is defined in terms of the elevation of this body h, relative to some reference plane:
Applied Chemical Engineering Thermodynamics
8
PE = mgh
(1.6.1)
where g is the local acceleration of gravity.
1.6.2 Kinetic Energy, KE This is the energy that a body of mass m possesses as a result of its velocity u: (1.6.2)
1.6.3 Internal Energy, U This energy reflects the total energy contained within a body, i.e. independently of its potential and kinetic energies, and is essential in the formulation of the first law of thermodynamics. (We shall return to this subject in Section 1.9.)
1.6.4 Enthalpy, H It is defined through the internal energy:
H = U + PV
(1.6.3)
and its use, among others, simplifies the expression of the first law when applied to a flow process.
1.6.5 The Helmholtz, A, and the Gibbs, G, Free Energies They are defined through the internal energy U and entropy S (Section 1.10): A
=
U TS
= U + P V TS = H  TS
(1.6.4)
(1.6.5) The Helmholtz free energy is of more theoretical importance, used often in statistical mechanics. The Gibbs free energy, on the other hand, is very important from the practical point of view, for it represents the basic quantity in chemical and/or phase equilibrium calculations. G
A Glimpse at Thermodynamics
9
1.7 Heat Here is an other term that we are exposed to from very early in our lives, yet its clear meaning is by no means apparent. (It took, for example, hundreds of years for people to even differentiate between heat and temperature, as we will see in Chapter 4.) Consider a hot body A that comes into contact with a cold body B. We know from experience that, with time, the temperature of the hot body decreases and that of the cold body increases until they become the same. What is responsible for this change? Something, obviously, flows from the one body to the other that causes their temperatures to become the same. We will see in Section 1.9 that this 'something' is just the energy of the molecules of the hot body that is transferred into the molecules of the cold one. It is referred to as heat. Notice, however, that heat exists only as a result of energy flow. It did not exist when the two bodies were not in contact. (For a more rigorous definition, that does not involve the use of temperature, see Hatsopoulos and Keenan.)
s
c Figure 1.2 The heating of benzene with steam: benzene is the system, everything else represents the surroundings. S: Steam; C: Condensate
10
Applied Chemical Engineering Thermodynamics
1.8 System, Surroundings, and Related Terms 1.8.1 System and Surroundings Consider the jacketed vessel of Figure 1.2, where benzene is heated from 20°C to 60°C. The stirrer is used to facilitate the heating process through mixing. We refer to benzene as the system. The steam, vessel, and the atmosphere are the surroundings. In general, the part of the universe that we focus our attention to represents the system. Everything else, the surroundings. Clear identification of these two parts is essential before we proceed with the solution of a problem.
1.8.2 Modes of Interaction Between System and Surroundings If a system can exchange thermal energy with its surroundings, it is considered to be in thermal contact with them and the wall enclosing it, as diathermal. If not, as adiabatic. If a system can exchange mechanical energy with its surroundings, it is considered to be in mechanical contact with them. If a system can exchange mass with its surroundings, it is referred to as an open system; if not, as a closed system. Thus, the aforementioned benzene system is a closed one, in: * thermal  through the vessel walls, and * mechanical  through the stirrer, contact with the surroundings. Finally, a system that cannot exchange with its surroundings thermal energy, mechanical energy, and mass is referred to as isolated.
1.8.3 State It is known from experience that if the temperature T and pressure P of a gas with given composition are known, then its molar volume Vis also known. Actually for a given amount of a homogeneous system with specified composition, knowledge of any two·of these three properties (P, T, V) suffices to determine the condition or state  of the system. (For a more
A Glimpse at Thermodynamics
11
comprehensive definition of the state of a system see Section 1.11 and, especially, Chapters 9, 12, and 15.)
1.8.4 Process A sequence of states that brings a system from its initial state to its final one is referred to as process.
1.8.5 Equilibrium: Internal and External When we refer to a property of a system  say pressure or temperature we assume that the system is uniform and, therefore, this property reflects the whole system. A system within which there are no nonuniformities, that act as driving forces for change, is said to be in a state of internal equilibrium. This is the type of systems that we will be dealing with in this book (see Example 1.2.c). And, if there are no driving forces that can bring about a change between a system and its surroundings, the system is said to be in a state of external equilibrium.
1.9 Internal Energy and Molecular Energy When we say that the internal energy of one mole of steam is U Joules, where is this energy stored? Or when Q Joules of energy flow from system A to system B, where was this energy stored in system A? Where will it be stored in system B? The energy is stored within the molecules of the steam or of the systems in two forms: l.kinetic; and 2.potential. The kinetic energy is contained within the individual molecules and, for each molecule, it is independent of that of the others. For a given species, or mixture of them, it depends on temperature only, and it involves two types of energy: !.translational, and
12
Applied Chemical Engineering Thermodynamics
2.internal (rotational, vibrational, nuclear, and electronic). (We will see in Chapter 17, however, that the rotational and vibrational energies depend on density too but to a much smaller degree than on temperature). The potential energy is contained by all the molecules collectively as a result of the forces exerted among them  referred to as intermolecular forces  and depends on: * the temperature, and * the distance among them, i.e. the system density. The intermolecular forces are of two main types: !.attractive; and 2.repulsive, and are discussed in Chapter 7.
1.10 Entropy It is the most difficult, probably, concept in thermodynamics, introduced in conjunction with the second law. We will use entropy to determine if a process can take place or not; and as the 'yardstick' for measuring how efficiently energy is utilized in a given process. Entropy is also used in Information Theory, Economics, even to describe the development of our civilization (Chapter 6).
1.11 State Properties Assume that we want to determine the value of the volume V of one kg of steam from the Steam Tables presented in Appendix B. To this purpose, we need only specify the state of the steam by specifying (Section 1.8.3), the temperature T and pressure P. The way that the steam was prepared, i.e. the path followed to arrive at this kg of steam, has no bearing on its volume value. Thus, at 400 kPa and 500°C, V = 889.2 m3 /kg, independent of the path followed in making it. And the same applies for the internal energy U (3!29.2 kJ/kg), entropy S, etc. In general, for a homogeneous system of given mass and composition, specification of any two of its properties (P, T, V, U, S, etc.) suffices for
A Glimpse at Thermodynamics
13
the determination of the others, i.e. for the determination of the state of the system, independently of the path followed to arrive at this state. They are referred to, appropriately, as state properties or junctions. (For an extensive discussion of the specification of the state of a system see Chapters 9, 12, and 15.) Let us now contrast state properties to two very important quantities in thermodynamics: work (W) and heat (Q). To this purpose, consider the following question: .how much work and/or heat are required to make the aforementioned kg of steam? i.e. what are the values of W and Q associated with this steam? To answer this question, we need not only specify the starting point, but also  as we will see in the next Chapter  the path followed to arrive at this steam. We will conclude, for example, that if we start with steam at 1 bar and 100°C, and follow two paths: !.constant pressure heating to 500°C, followed by constant temperature compression to 400 kPa; or 2.constant temperature compression to 400 kPa, followed by constant pressure heating to 500°C; we arrive at two different sets of values for Wand Q; they are, in other words, path depending quantities. The characteristics of these two types of thermodynamic quantities, state properties, and work and heat, are contrasted in Table 1.2, which explains why, for example, we use 1U but not 1Q; or why, rigorously speaking, we should use oQ and oW, and not dQ and dW. (In the interest of simplicity, dQ and dW will be used in this book, but it should be kept in mind that they refer to infinitesimal amounts, not changes.)
1.12 Reversibility Processes that proceed so that displacements from equilibrium, i.e. the driving forces, are infinitesimally small, are referred to as reversible processes; if not, as irreversible. We examine, next, the practical implications of reversibility. Consider body A, with a weight of FA = 20 N and at an elevation of 2 meters from the ground, which will be used through a pulley to raise another body B of weight FB from the ground to the same elevation. We assume, for simplicity, that the pulley is frictionless, and proceed to determine the efficiency of the operation:
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Applied Chemical Engineering Thermodynamics
Table 1.2 Characteristics of State Properties and of Work and Heat State Properties (S,T,U, etc.)
Work and Heat
Path independent Instantaneous properties Differential: infinitesimal change Integral: change in property
Path dependent Energy in transit Infinitesimal amount Amount
n
=
Actual Work Produced
~M=ax=im_u_m_p=,od=ucab"""l:e=Wi:::o:r:k
(1.12.1)
as a function of the difference (FA F8 ), i.e. as a function of the driving force of the process. The results are presented in Table 1.3. Comments: l.As the driving force decreases, the efficiency of the process increases, i.e. more work is obtained. 2.At the limit F8 .FA, where the process becomes reversible, the efficiency assumes the maximum value equal to 1.0. 3.We conclude, therefore, that a reversible process provides the maximum possible amount of work, just as we did in Example 1.2. Therefore Eq.1.12.1 becomes:
n
Actual Work Produced Reversible Work
= ::::::::::::
(1.12.2)
4.If work is consumed, it is easily shown that a reversible operation requires the minimum one and the efficiency is defined by:
Reversible Work n = ~~~=~~ Actual Work Consumed
(1.12.3)
S.For the reversible operation, a small increase in F8 would result in a reversal of the process, thus, the name reversible. 6.Actual real life  processes are of course irreversible. The presence of friction, for example, would prevent the movement of the pulley, or of the piston, if the driving force is infinitesimally small; and a heat exchanger operating under such a small temperature difference would require an infinite area for heat transfer. 7 .Reversibility as a concept, hoWever, is of paramount importance in thermodynamic applications, such as in the area of efficient energy utilization, as we will see in later Chapters.
A Glimpse at Thermodynamics
15
Table 1.3 Amount of Work Produced by the Falling Body A with Weight FA = 20 N as a Function of the Driving Force. Work (J) Efficiency F8 (N) FA Fa 20 0 0 0 15 5 10 0.25 10 20 10 0.50 12.5 7.5 25 0.625 30 0.75 15 5 17 34 3 0.85 19 1 38 0.95 19.5 0.975 0.5 39 40 20 0 1.00
1.13 Auxiliary Terms 1.13.1 Heat Capacity The amount of heat required to raise the temperature of one mole of a substance by one degree under conditions of: !.constant volume, or 2.constant pressure, is referred to as heat capacity, Cv and Cp, respectively and is defined by:
cv
=(au) ar v
Cp = (
(1.13.1)
~~t
where U and Hare the molar internal energy and enthalpy respectively.
1.13.2 Ideal Gas This is a hypothetical gas where:
16
Applied Chemical Engineering Thermodynamics
l.there are no intermolecular forces; and 2.the molecular volume is negligible as compared to the volume of the gas. In theory, such a behavior is approached by real gases when their pressure, and thus the density, tends to zero. In practice, however, gases can be assumed to behave ideally at pressures up to one bar (Example 1.5). There are some exceptions, however, mainly the organic acids.
1.13.3 Real Fluid versus Ideal Gas The main difficulty in evaluating thermodynamic properties is the presence of intermolecular forces. (Nature would be very obliging if it could do without them, but could it?) An ideal gas  where such forces are absent  represents, therefore, a useful reference fluid for isolating their effect on the properties of real fluids. Two terms are used for this purpose: * Residual properties: Rea/fluid (T, P, x) property Ideal gas (T, P, x) property. * Departure functions: Rea/fluid (T, P, x) property Ideal gas (T, P 0 , x) property. In the first case, we compare the properties of the real fluid  at the given temperature T, pressure P, and compositionx to those of a hypothetical ideal gas, at the same conditions; in the second, the pressure of the ideal gas is set equal to a specified value P 0 , dictated by convenience (Chapter 9).
1.13.4 Ideal Solution An ideal solution is formed when: l.the forces between like and unlike molecules are the same; and 2.the size and shape of the molecules are the same. As in the case of the ideal gas, an ideal solution is also a hypothetical fluid. Here, however, intermolecular forces are present, but we assume that those between unlike molecules are the same with those between like molecules. If an ideal liquid solution is in equilibrium with the vapor above it that behaves as an ideal gas, the relatio11£hip between the compositions of the two phases is described analytically by Raoult's law: (1.13.2)
17
A Glimpse at Thermodynamics
where, ~ : partial pressure of component i, i.e. the part of the total pressure due to component i; X;,y;: mole fraction of component i in the liquid and vapor phases respectively; P : total pressure over the solution; P/: vapor pressure of pure liquid i, i.e. the pressure that would exist over the liquid if the latter was pure i (at the temperature of the solution). In practice, Raoult's law is valid, approximately, for mixtures of similar compounds only, such as, hexaneheptane or ethanolpropanol.
1.13.5 Real versus Ideal Solution In Section 1.13.3, we used the difference between the properties of a real fluid and an ideal gas to describe the effect of intermolecular forces on the thermodynamic properties of the fluid. Here we use the difference between real and ideal solution behavior of a fluid, at the same T, P and compositionx, to describe the effect of the differences in: * intermolecular forces; and * size and shape; among like and unlike molecules on the properties of the fluid. The comparison is done at the same temperature, pressure, and composition, and the resulting difference is called the excess function (Chapter 11):
Excess Function
=
Real solution property (T, P, x) Ideal solution property (T, P, x)
1.13.6 Activity Coefficient The difference in the behavior of the component of a real solution, as compared to that determined by Raoult's law (Eq.1.13.2), is described by its activity coefficient (Y;):
Y;P
=
Y;X;P:
(1.13.3)
The activity coefficient is of paramount importance in chemical engineering thermodynamics applications, for it represents a key quantity in separation calculations (Example 1.8).
18
Applied Chemical Engineering Thermodynamics
1.13.7 Chemical Potential and Fugacity The presentation of these two very important quantities  in chemical and phase equilibrium calculations  is not possible here. They will be discussed in Chapters 9 and 11.
1.13.8 Intensive and Extensive Properties Properties that are independent of the amount of material present, such as pressure and temperature, and properties that are defined per unit of mass, such as heat capacities, are referred to as intensive properties. On the other hand, properties that depend on the total amount of material present, such as volume, are referred to as extensive properties.
1.14 The Foundations of Thermodynamics Having finished with the Vocabulary of thermodynamics, we turn next to the second objective of this Chapter: a brief discussion of the Foundations of thermodynamics, depicted pictorially in Figure 1.3: 1. the laws of thermodynamics; and 2. the thermophysical data. The Laws provide the framework of the necessary equations for the solution of thermodynamic problems; the Data, the information required to arrive at numerical answers through these equations.
1.14.1 The Laws of Thermodynamics 1. The Zeroth Law: When bodies A and B are in thermal equilibrium with body C, they are in thermal equilibrium with each other. It is used
to establish temperature as a state property. 2.The First Law: Energy is conserved. The basis for energy balances, it relates the state functions: internal energy and enthalpy, to the path dependent ones: work and heat. 3. The Second Law: Any process wh'vse only effect is the transfer of heat from one temperature level to a higher one, is impossible. It introduces
19
A Glimpse at Thermodynamics
~
~ THERMODYNAMICS
.. Carnol .. Joule .. Clausius * Gibbs
I THE
LAWS
I
ARCHITECTS
ITHE
DATA
I
Figure 1.3 The Foundations of Thermodynamics.
the concepts of thermodynamic temperature and of entropy; it also establishes the feasibility of processes and, in combination with the first law, the degree of efficient energy utilization in them. 4. The Third Law: The absolute entropy of a perfect crystal at zero absolute temperature, is zero. It is used to determine absolute values of entropy, that are essential in calculations involving chemical reactions. ~The development of thermodynamics based on these laws follows the 'historical' approach. For other approaches, see the Preface. On the Laws of Thermodynamics
It must be admitted, /think, that the laws of thermodynamics have a different feel from most of the other laws of the physicist. There is something more palpably verbal about them, they smell more of their human origin. The guiding motif is strange to most of physics: namely a capitalizing of the universal failure of human beings to construct perpetual motion machines of either the first or second kind. Why should we expect nature to be interested either positively or negatively in the purposes of human beings, particularly purposes of such an unblushingly economic tinge? (P. W. Bridgman, The Nature of Thermodynamics, Harvard University Press, 1941.)
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Applied Chemical Engineering Thermodynamics
1.14.2 The Thermophysical Data Properties, such as: density, enthalpy, entropy, etc., for given species or mixture of them, are referred to as thermophysical properties and their values, as thermophysical data. A typical example is the Steam Tables. Such data involve two types of quantities: !.Measurable; and 2.Derived. The first type refers to the properties that can be measured in the laboratory: *PressureVolumeTemperature (PVT); * Heat capacities; *Vapor pressures (considered part of the PVT measurements); and, * Enthalpies of vaporization. We will refer to them as physical properties. The second type involves all other properties such as, enthalpy, entropy, etc., that cannot be determined experimentally. There is no such a thing, for example, as an 'entropymeter'. Their values are derived from the measurable quantities through relationships based on the laws of thermodynamics. They are, therefore, 'conceptual' quantities and this is why they are more difficult to comprehend. We will refer to them as thermodynamic properties and their values as thermodynamic data. The availability of thermophysical data is essential to the application of thermodynamics in Chemical Engineering calculations. The first law, for example, provides the necessary relationship for the evaluation of the amount of heat involved in changing the temperature of a given fluid. Unless, however, the required heat capacity data is available, no numerical answer can be obtained. The situation can be simply summarized as follows: "You cannot get much use out of a beautifu1959Porsche without gasoline; You cannot get much information for a given process from the laws of thermodynamics, without the appropriate thermophysical data".
1.15 The Purpose of Chemical Engineering Thermodynamics We turn now to the third objective of this Chapter; a brief outline of the
21
A Glimpse at Thermodynamics
Purpose of chemical engineering thermodynamics: the study of the three broad areas of the field, outlined in Figure 1.4.
1.15.1 Feasibility of Processes and Efficient Energy Utilization The feasibility of processes and the efficient energy utilization in carrying them out represent the original objectives of thermodynamics and the reason for its name. Actually, the second objective has been traditionally of interest to power engineers, mostly mechanical, while chemical engineers paid only lip service to it, when energy cost was very low. The drastic increase in energy cost following the Oil Embargo of 1973, however, and its impact on the product cost of the chemical and petroleum industries, made the subject of efficient energy utilization a major concern of chemical engineers as well. A combination of the first and second laws will be used to establish the criteria for: l.the feasibility of, and 2.the degree of efficient energy utilization in, a process under consideration.
IFeasibility
of Processes and Efficient Energy Utilization
I
IThermophysical
i
Properties
I
i Chemical Engineering Thermodynamics
I Chemical
and Phase Equilibrium
I
Figure 1.4 The Three Broad Areas of Chemical Engineering Thermodynamics.
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Applied Chemical Engineering Thermodynamics
1.15.2 Thermophysical Properties Knowledge of the values of these properties is essential in obtaining numerical results in the other two areas. Our objective here will be twofold: !.evaluation of the thermodynamic  i.e. derived  properties: internal energy, enthalpy, entropy, etc., from data on the measurable ones: pressure, volume, temperature, and heat capacities: and 2.estimation of thermophysical properties. !.Evaluation of Thermodynamic Properties Starting with a combination of the first and second laws we will develop the necessary framework of equations that can be used, in combination with physical property data, in the determination of thermodynamic properties. Because the available in the typical case data cover a limited range of pressure and temperature, the obtained values are not the absolute ones, but relative to some arbitrarily chosen reference state. The entropy and enthalpy values in the Steam Tables, for example, have been developed using these equations, combined with extensive experimental pressurevolumetemperature (PVT) and ideal gas heat capacity data. The given values, however, are relative to zero values for saturated water at 273.16 K. 2.Estimation of Thermophysical Properties On the other hand, considering: * the plethora of compounds involved in the chemical and petroleum processes; * the practically infinite number of mixtures resulting from them; and * the large cost, in time and money, required for the experimental determination of the necessary data for these compounds and mixtures; it becomes apparent that, in the typical case, we must resort to some estimation, i.e. approximate evaluation, of some  or even all  properties, physical and thermodynamic. The importance of thermophysical properties in Chemical Engineering is demonstrated in Table 1.4, using information from a major chemical company. An excellent reference for such estimation techniques is the book of R. Reid, J. Prausnitz, and B. Poling, Properties of Gases and Liquids, 4th Ed., McGrawHill, New York, 1987.
A Glimpse at Thermodynamics
23
Table 1.4 Physical Property Errors Are Costly Property Activity Coefficient Separation: *Very easy *Easy *Difficult * Very Difficult Heat Capacity Heat of Vaporization Density Viscosity Thermal Conductivity Diffusivity
If% Error in Property is
Resulting % Error in Equip. Size Equip. Cost
10
20 15 20 50 20 20 100
3 20 50 100 6 15 16 10 13 6 40
2 13 31 100 6 15 16 10 13 4 23
Data compilations are also available through the Design Institute for Physical Property Data (D/PPR), operating under the auspices of the American Institute of Chemical Engineers (AIChE) and supported by several chemical and petroleum industries; and by several Computerized Data Banks.
1.15.3 Chemical and Phase Equilibrium The importance of these areas becomes apparent the moment one visits a chemical plant or a petroleum refinery: chemical reactors and separation units are the essential blocks of the processes involved in both places, as the typical example shown in Figure 1.5 demonstrates for the manufacture of methanol. [Carbon monoxide and hydrogen react at moderate pressures (50100 atm) and temperatures (250270°C)to form methanol. Following condensation, the unreacted gas is recycled, while methanol is recovered from the liquid mixture by distillation.] Starting with a combination of the first and second laws we will arrive at a framework of equations which, in combination with the appropriate
Applied Chemical Engineering Thermodynamics
24
co
.
r
. COMPRESS:j H2
REACTOR Methanol
~
I z
z
0

CONDENSER
:::;:
>=
::l
::5
___J
0
___J
>= IJl
(.)
0
I Byproducts
Figure 1.5 The manufacture of methanol.
thermophysical data, provide the necessary information for designing these units. Note. The separation of Chemical and Phase Equilibrium from the general subject of Thermophysical Properties is intended to emphasize the importance of the former.
1.16 Examples 1.3 through 1.9 We close this Chapter with a series of Examples that represent a brief preview of the type of problems that chemical engineering thermodynamics deals with: * Examples 1.3 and 1.4 will help develop some familiarity with the SI and ES systems of units. The first is the accepted system in Europe and is becoming slowly, but steadily, the same in the United States, where the second still dominates. * Example 1.5 discusses one method of estimating thermophysical pro
25
A Glimpse at Thermodynamics
perties  gas densities in this case  the ideal gas assumption, and demonstrates the approximate character of, and the errors involved with, this approach. * Examples 1.6 and 1.9 deal with the problem of energy use, the second with efficient energy utilization. *Finally, Examples 1. 7 and 1.8 deal with a subject of major importance to chemical engineering thermodynamics applications: distillation.
1.16.1 Example 1.3 Two systems of units are used today in the U.S.: the traditional English (ES) and the SI (Systeme International d'Unites), towards which the country is moving, albeit slowly. (The SI system was introduced in 1960 by the General Conference of Weights and Measures, an international body that meets every six years in Paris.) a. Why is the adoption of the SI system so slow? b. What are the base units in the SI system? c. What are derived units? a. The slow adoption of the SI system in the U.S. must be attributed to two main reasons: l.The force of habit. The English system has been used in the U.S. for a long time, and we all know that "habit is a second nature". 2. Economic. It has been estimated that the cost of retooling the U.S. industry from the English to the SI system will be well over $10 billions (middle 80's).
b.See Table l.E.3a. Table l.E.3a Base Units in the SI System Quantity
Unit
length
meter
m
mass
kilogram
kg
time
second
s
electric current
ampere
thermodynamic temperature kelvin
SI Symbol
A
K
amount of substance
mole
mol
luminous intensity
candela
cd
c. These are units obtained by combination of the base units. For examples, see Table l.E.3b.
26
Applied Chemical Engineering Thermodynamics Table l.E.3b Examples of Derived Units Quantity density
Unit
SI Symbol
kilogram per cubic meter
force
newton
energy entropy
joule joule per kelvin
pressure
pascal
Formula kg/m3
N J Pa
kg m/s2 Nm J/K N/m2
1.16.2 Example 1.4 What is the weight of 1 kg of water in a place where the local gravity acceleration is g = 9. 70 m s· 2 in the English and the SI systems of units. a.SI Units From Newton's law: F
= mg = (1 kg)(9.70m s·2) = 9.70 N
b.English Units Using the conversion factors of Appendix A: F = (9.70N)(0.2248lbr1N) = 2.18lbr
1.16.3 Example 1.5 Gas densities are important in chemical engineering calculations such as in the sizing of vessels, pipes for gas transport, etc. As a first approximation, such densities can be calculated using the ideal gas law. a. What factors effect the accuracy of the obtained results? b. To test your answer in part (a), calculate the density of steam at: * 3500 kPa, as a function of temperature; and * 300°C, as a function of pressure; and compare the obtained values to those given in the Steam Tables. a. The assumptions behind the ideal gas behavior become more valid as the volume per unit mass increases, i.e. as the density decreases (Section 1.13.2). We should expect, therefore, increasing accuracy: * at a given pressure, as the temperature increases; and * at a given temperature, as the pressure decreases. b. According to the ideal gas law the dt!nsity of steam is given by: d = 18.015/V = 18.015 PI(RT) where R = 8.314 m3 kPa(kmol K)" 1, is the ideal gas constant. The results, pre
A Glimpse at Thermodynamics
27
Table l.E.S Steam Densities (kg/m3) Calculated with the Ideal Gas Law P "'3500kPa t "'300°C Density
Density
t, °C
Calc.
Exp.
%Error
250 275 300 325 350 375 400 450 500 600
14.50 13.83 13.23 12.68 12.17 11.70 11.27 10.49 9.81 8.69
17.04 15.67 14.61 13.75 13.02 12.39 11.84 10.88 10.09 8.84
14.9 11.7 9.5 7.8 6.5  5.6 4.8 3.6 2.8  1.7
P,kPa 100 200 250 350 500 1000 2000 3000 4000 5000
Calc. 0.38 0.76 0.95 1.32 1.89 3.78 7.56 11.34 15.12 18.90
Exp. %Error 0.38 0.76 0.95 1.33 1.91 3.88 7.97 12.32 17.00 22.07
0.2 0.5 0.6 0.8  1.2 2.5  5.1  8.0 11.0 14.4
sented in Table l.E.5 and graphically in Figure l.E.5, support our findings in part (a).
2
2:
·~
v
4
0
"'0
.....v
6
0
"S 8 ~ 0
u
.~ 10
.....
e..... 12
~3500
kPo
!38E3I3EI 300°Celsius
w
I!R 14 160~~~~~~~~~~~~~~~~~~~~~~
5
10
15
Experimental Density (kg/m 3 )
20
25
Figure l.E.S The effect of density on the ideal gas assumption for steam.
28
Applied Chemical Engineering Thermodynamics
Comments: l.As the density approaches zero so does the error, which is consistent with the definition of an ideal gas. 2.Notice the pronounced effect of temperature as compared to pressure on the accuracy of the obtained results. • An increase in temperature by a factor of 1. 7, from 523 K (250°C) to 873 K (600°C), leads to the same reduction in error as a 10folddecrease in pressure from 5000 kPa to 500 kPa. • At densities below about 15 kg m· 3 the errors for t = 300°C are larger (in absolute value) than those for 3500 kPa, because the temperatures involved in the latter case are larger than 300°C. • The trend is reversed at densities above 15 kg m· 3 , because here the temperatures for P = 3500 kPa are lower than 300°C. 3. The explanation for this phenomenon will become apparent when we discuss intermolecular forces. It suffices to say, here, that it has to do with the strong effect that temperature has on the dipoledipole interactions.
1.16.4 Example 1.6 Water at a pressure of 1 atm and a temperature of 100°C enters a boiler where it is vaporized at a rate of 100 kg/min. a.How much heat, in kJ/min and Btu/min, is needed? b. If the heating is done by condensing saturated steam of 1.5 bar, how much steam is needed in kg/min? c. What role does thermodynamics play in solving this problem? a. From the Steam Tables, the heat of vaporization for water at 1 atm and 1oooc is 2256.9 kJ/kg. Hence, heat required: Q = 100x2256.9 = 2.26·105 kJ/min = 2.14·105 Btu/min b.The heat of condensation of saturated steam at 1.5 bar is 2226.3 kJ/kg. Hence: steam = [2.26·105 kJ/min]/[2226.3kJ/kg] = 101.5 kg/min c. The heats of vaporization and condensation are thermophysical data for steam.
In addition, the calculation of the steam requirement is based on the equality: heat absorbed by the water which follows from the first law.
= heat lost by the steam
1.16.5 Example 1.7 One of the most important applications of thermodynamics in our profession is in the design of distillation columns. Such colunms are used to effect the separation of a mixture into its components with desirable purity levels.
29
A Glimpse at Thermodynamics
Consider, now, the separation of a binary mixture of components 1 and 2 into products of, say, 99.9% purity. a. What are the major parts of a distillation unit? b. What are the commonly used types of distillation colunms? c. Describe briefly the operation of the unit. a. A simple schematic diagram of such a unit is shown in Figure l.E. 7 and it consists of three major parts: l.the column and packing or trays 2. the boiler and 3.the condenser.
.~
CONDENSER
...
.. Feed
.....
Water
D
.......... ~ .. ::::: u ·... . .......... (!) .....
~ ~
.. z :·
.·::. < ·: .. .
:. ::.· 0.. ·.. .
v
8 BOILER
Condensate
Steam
Figure I.E. 7 Schematic diagram of a packed distillation column.
30
Applied Chemical Engineering Thermodynamics
b. Distillation columns come into two main types: 1. packed; and 2.tray (mostly sieve). c. The operation of the column is as follows: Vapor (Y), generated in the boiler, rises through the column coming into contact with the liquid descending from the condenser (L0 , called the 'Reflux'). During this contact, the vapor loses some of its less volatile component, which goes into the liquid phase, and the reverse occurs with the liquid. It is apparent that this exchange of constituents between the two streams requires a large contact area, which represents the purpose of the packing or trays. Thus, as the vapor reaches the top of the colunm, it has lost most of its heavy (less volatile) constituent; and, as the liquid reaches the bottom, it has lost most of its light constituent. If components 1 and 2 are the light and heavy components, then: * stream B  known as the 'Bottoms'  can be practically pure 2; and, * stream D known as the 'Distillate'  can be practically pure 1; provided that the height of the column and the reflux are sufficient. The separation, thus, of the feed stream into its components with the specified purity has been accomplished. (Are there any exceptions to this?)
1.16.6 Example 1.8 The essential quantity involved in determining the required height and reflux of a distillation column, such as the one used for the separation of the mixture of components 1 and 2 in the previous Example, is the relative volatility of 1 with respect to 2, a 12 • It is defined as follows: at2
Ytlxt lx
=y
2
2
(A)
where y and x represent the vapor and liquid phase compositions at some point in the column. Assuming that the two phases are at equilibrium with each other, what information is needed to evaluate a 12? From Eq.1.13.3: s
Y;
Y;P;
p
X;
Introduction of Eq.(B) into Eq.(A), yields: • Yt
at2
=
y2
P: P;
(B)
(C)
A Glimpse at Thermodynamics
31
Evaluation, therefore, of the relative volatility requires knowledge of: 1. vapor pressures, and 2.activity coefficients, for the two components. We will see in later Chapters how these quantities are determined.
1.16. 7 Example 1.9 A new chemical plant is expected to have three distillation columns. In the one the boiler operates around 90°C and in the other two around ll0°C. The following discussion reflects the thinking of three members of the design team. Monica: Let us buy a boiler that generates 3 bar steam. Its temperature is somewhat over 130°C, good enough for all three boilers. Besides, being a low pressure one, it should be cheap. George: I agree, but I think it would be better to get a highpressure boiler with a superheater. Expand the steam to 3 bar, so we can generate some electricity, and use the exhaust steam for the boilers. It will obviously be more expensive to buy, but the electricity will come very cheap. Dave: You sound like the Government. You take a simple problem and make it complicated. Monica's boiler will require fuel; George's scheme will just require extra fuel to produce the electricity. Remember, however, that "a Joule, is a Joule, is a Joule", and there is no free lunch. Let us just use electricity. Can you offer anything to this debate? Not yet, but you should, after Chapter 3.
The Social Implications of the Industrial Revolution In the van of the industrial revolution were the twin English towns of Manchester and Salford. Rapid industrial growth meallt equally rapid population increase. In 1801 the combined population of the two towns was about 90,000; by 1831 it had grown to 230,000, and by 1851 it stood at 367,000. This enomwus expansion was accompanied by severe social stress. InAugust 1819the 'PeterlooMassacre' tookplace, when the yeomanry roughly equivalent to the National Guard killed eleven people and injured many more while dispersing a large crowd of demonstrators. (Cardwell, 1983, p.45)
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Applied Chemical Engineering Thermodynamics
References Cardwell, D.S.L., 1971. From Watt to Clausius, the Rise of Thermodynamics in the Early Industrial Age, Cornell University Press, Ithaca, N.Y. Cardwell, D.S.L., in Springs of Scielltific Creativity, Aris, R., Davis, H.T., Stuewer, R.H., Eds, University of Minnesota Press, Minneapolis, Min., 1983. Hatsopoulos,G.H., Keenan, J.H., 1965. Principles of General Thermodynamics, Wiley, New York. Rifkin, J., 1980. Entropy, Viking Press, New York.
Problems 1.1 With reference to system and surroundings, present an example of your own where you: a. identify system and surroundings, and b.demonstrate if the system is in thermal and mechanical contact with the surroundings, and if it is open or closed. 1.2 Express your weight in SI and English units. 1.3 Assuming that inside von Guericke' s hemispheres (insert before Section 1.5) there was an absolute vacuum: a.calculate the total force exerted by the atmosphere; b. which of Cardwell's characterization do you agree with, exaggeration or feeble horses? 1.4 In what areas of thermodynamics are the following properties: H; U; G; and S; important? 1.5 Discuss what you mean when using the terms: a. temperature; and b. heat. 1.6 What are some of the practical applications of Entropy? 1. 7 Discuss the differences between: state properties, and work and heat. 1.8 Is the heat capacity: a. a path dependent, or path independent, quantity? b.an extensive, or intensive, property? 1.9 With reference to gas heat capacities: a.is Cp larger or smaller than Cv? b.can you explain, qualitatively or quantitatively, your answer? 1.10 A certain fluid undergoes the foHowing change: (Pp V1,T1)+ (P2 , V2 ,T2)+ (P3 , V3 ,T3).
A Glimpse at Thermodynamics
33
Is knowledge of the path required for the evaluation of: a. the entropy change of the system? and b. the amount of heat involved? 1.11 What is the difference between: a. the kinetic and potential energy of a body, and b.its internal kinetic and potential energies. 1.12.a. What are residual properties? b. What is their importance? 1.13 What is  in terms of intermolecular forces  the difference between an ideal gas and an ideal solution? 1.14.a.What are excess properties? b. What is their importance? 1.15 In discussing the concept of reversibility we concluded that a reversible operation yields the maximum possible work. Using an example of your own demonstrate that, when work is consumed, a reversible operation requires the minimum one. 1.16 What is meant by the terms: saturated steam and vapor pressure? 1.17 One kg of saturated water at 100°C is placed into a cylinderandpiston arrangement, which is in a constant temperature bath at 100.1 °C. The pressure outside the piston is equal to 1 atm. The water is vaporized slowly, absorbing heat from the bath, pushing the piston until it becomes saturated vapor. a.How much work is produced? b.Could more work be produced, for the same final conditions? c. If yes, how much? 1.18 For what properties is information given at the back of the book Properties of Gases and Liquids? Is this experimental or estimated information? 1.19 Consider the separation of a mixture of nhexane and nheptane, using distillation, into products of 99.99% purity. If the pressure at the top of the column is 800 mmHg, and in the bottom 840 mmHg, estimate the temperature at the top and bottom of the column. 1.20 With reference to Example 1.5: a.why are the calculated with the ideal gas assumption density values smaller than the experimental ones? b. how could you get a better estimate of the steam density? c.what is a dipoledipole interaction? 1.21 In Example 1. 8 it is stated that a key quantity in determining the height of the column is the relative volatility. What else, in your opinion, is also very important to this purpose? 1.22.a. What is a reversible operation? b.Are real life processes reversible? Explain. 1.23.a. What is the origin of the atmospheric pressure? b. What is the weight of the atmosphere that surrounds the earth? 1.24.a. Using the Steam Tables estimate the ideal gas heat capacity of steam at constant pressure as a function of temperature.
34
1.25 1.26 1.27 1.28
Applied Chemical Engineering Thermodynamics
b. Give your estimate of the accuracy of the obtained values.
Why do we define Cv as (8U/8T)v and not as (8Q/8T)v? What are the foundations of chemical engineering thermodynamics? What is the purpose of chemical engineering thermodynamics? For the design of the methanol plant described in Figure 1.5, what is the contribution of: a. the laws of thermodynamics, and b. the thermophysical data. 1.29 What are the perpetual motion machines of the first or second type? (See insert before Section 1.14.2.) 1.30 The heat of vaporization of methanol at 100°C is needed. Could its value be determined from available PVT data including vapor pressure as a function of temperature?
2 The Zeroth and First Laws of Thermodynamics Law: The body of rules governing the affairs of man within a community or among states; common law: the law of nations. Often plural: Principles of conduct perceived to be of natural origin. (Am. Heritage Diet. of English Language)
2.1 Introduction Just as Society is built upon the foundation of its laws, so is a field of Science. There is, however, a difference: Society 1s laws express man 1 s collective wisdom; Science's laws express Nature's 'wisdom'. Thus the Laws of Thermodynamics express Nature's wisdom, as of course perceived by humans, and were developed in two stages: First, observations of a variety of physical phenomena led men to conclude that there is a 'common reason' behind all of them, dictated by Nature and not by the way they are carried out. They arrived, thus, at a verbal statement of this common reason such as, "Energy is Conserved". Second, this verbal statement was expressed analytically so that quantitative results can be obtained. Thus, "Energy is Conserved" becomes: Q = tiU + W. These analytical statements of the laws led eventually to the development of a 'framework' of equations that are used to solve all thermodynamic problems. The whole process  from the collection and interpretation of the observations to the development of the framework of equations was a very lengthy one, as we will see in Chapter 4, where we consider the historical development of the laws of thermodynamics. And this should be kept in mind as we proceed: it is not easy to comprehend and assimilate in a limited amount of time, what took hundreds of years to develop. We will discuss the Zeroth and First Laws in this Chapter, and the Second Law in the following one. The Third Law is of more limited application; it will be stated in Chapter 4 and used in Chapter 15.
36
Applied Chemical Engineering Thermodynamics
2.2 Objective and Approach There are two main objectives in this Chapter: First, to introduce, through the zeroth law, temperature as a state property rather than a physiological sense of 'hotness' and 'coldness'. Second, to develop familiarity with the first law, its formulation and its applications. We will start with the statement of the zeroth law, use it to establish temperature as a state property, and proceed to develop the ideal gas temperature scale. We move then to the first law and discuss its development and analytical expression, especially the form that is convenient for steadystate flow processes that are typical in Chemical Engineering practice. Since applications of the first law involve calculations of work and heat through evaluation of internal energy and enthalpy changes, we proceed to review briefly the fluid for which such changes are the easiest to determine: the ideal gas. We close with some Examples and a few Concluding Remarks.
2.3 The Zeroth Law and the Ideal Gas Temperature 2.3.1 The Statement We know from experience that a piece of wood and one of copper sitting in our back yard for a long time have the same 'hotness' as that of the atmosphere around them. Or, more generally: If bodies A and Bare in thermal equilibrium with a body C, then they are in thermal equilibrium
with each other.
Thermal equilibrium results when two bodies are brought into thermal contact through a diathermal wall, i.e. a wall that allows the flow of energy between the two bodies. The important point to notice here is that bodies A and B are in thermal equilibrium without having been brought into thermal contact with each other.
The Zeroth and First Laws of Thermodynamics
37
This statement, though intuitively obvious, represents a fact of Nature and is referred to as the Zeroth Law of thermodynamics, introduced by Fowler and Guggenheim in 1939. Its practical value is to introduce the concept of temperature as a state property rather than a sense of 'hotness' or 'coldness'.
2.3.2 Is it Really a Law? You may find this statement 'too obvious' to be raised to the level of a Law. Indeed it has been debated in the past, as suggested by the following statement from one of the giants of thermodynamics, James Clerk Maxwell (as quoted by Hatsopoulos and Keenan, 1965): "This law is not a truism, but expresses the fact that if a piece of iron when plunged into a vessel of water is in thermal equilibrium with the water, and if the same piece of iron, without altering its temperature, is transferred to a vessel of oil, and is found to be also in thermal equilibrium with the oil, then if the oil and water were put into the same vessel they would themselves be in thermal equilibrium, and the same would be true of any other three substances. This law, therefore, expresses much more than Euclid's axiom that 'Things which are equal to the same thing are equal to one another'". Redlich (1968), on the other hand, expresses strong objections to this law, while Hatsopoulos and Keenan discuss exceptions to it (Problem 2.51).
2.3.3 Temperature: The Common Property of Systems in Thermal Equilibrium Consider a large bath  large enough that its 'hotness' is not effected by the transfer to, or from it, of a small amount of energy  in which we place a cylinderandpiston arrangement containing one mole of nitrogen. After thermal equilibrium is reached, we record the pressure (P1) and volume (V1) of the gas inside the cylinder. We now move the piston so that the volume changes to V2 , and record the corresponding pressure P2 when thermal equilibrium is reached. We repeat this process with volume values: V3 , V4 , ... and record the corresponding pressure values: P3 , P4 , ... We repeat next the experiment replacing nitrogen with helium, and
38
Applied Chemical Engineering Thermodynamics
record the volume values: V1 ', V2 ', ••. and the corresponding pressure values: P 1 ', P 2 ', ••. Now, according to the zeroth law, all the states of N2 : (P1, V1), (P2, Vv, ... are in thermal equilibrium with: *each other; and * with those of He: (PI', vi I), (P2 ', v2 I), .... All these states of N2 and He, therefore, share a common property which we call temperature. Notice that this property is not some abstract quantity; rather, it has the effect of determining in this case the value of pressure, once the value of the volume is specified. We have, thus, moved  through the zeroth law  from a sense of 'hotness' to the establishment of a state property, the temperature.
2.3.4 The Ideal Gas Temperature Scale The next problem is the assignment of numerical values to this property. Evaluation of the products: pi vl> p2 v2, ... and: PI vi', p2 v2 ', ... indicates that, at low pressures, the values are very close independently of the gases involved. More important, if the products are plotted against the corresponding pressures and then extrapolated to zero pressure, i.e. to the ideal gas state, the lines for the two gases converge to a common value, I
.....
g
e 0.. o
IMH09e ~
I
Gos type 1 Gas type 2
Pressure
Figure 2.1 Extrapolation of the product (PV) toP
=
0.
The Zeroth and First Laws of Thermodynamics
39
which we refer to as (PV) .. as shown in Figure 2.1. If the procedure is repeated using a bath of different hotness, the lines again converge to a common, but different from the previous case, value of (PV)*. We will demonstrate next how these ideal gas values, obtained by extrapolating the product (PV) toP = 0, are used to establish a temperature scale that, in contrast to liquidphase thermometers (mercury, ethanol, etc.), is independent of the fluid used. The procedure, which leads to the ideal gas temperature scale (T ), involves the following two arbitrary steps: l.The product (PV)• is a linear function ofT: (PV)* = RT (2.3.1) 2.The value ofT at the triple point of water is set equal to 273.16 K. Experimental determination of the product (PV)• for several gases at the triple point of water has resulted in an accepted value of 22711.6 bar cm3 mor 1 and, consequently:
R = 22711.6/273.16 = 83.14 bar cm3 mot 1 K
1.
To establish the value of the ideal gas temperature at some other conditions, the value of the product (PV )• is evaluated and the corresponding value ofT is calculated from Eq.2.3.1 (Example 2.1). The ideal gas temperature scale is thus established; and it is the same, as we will see in the next Chapter, with the absolute thermodynamic temperature scale, arrived at through the second law of thermodynamics. Notice, finally, that the ideal gas constant R is a derived quantity from the definition of the ideal gas temperature. Values of R in different units are given in Table 2.1. Table 2.1 Value of the Ideal Gas Constant in Different Units R
= 1.987 cal mol 1 K 1 = 1.987 Btu lbmot 1 R 1 = 8.314 J mol 1 K 1 = 8314 cm3 kPa mol 1 K
1
= 83.14 cm3 bar mol· 1 K 1 = 82.06 atm cm3 mol 1 K 1 .., 10.73 psia ft3 lbmol 1 R 1 0. 7302 atm
rt3 lbmol 1 K 1
= 1.314 atm ft3 lbmot· 1 K 1
Applied Chemical Engineering Thermodynamics
40
2.3.5 Example 2.1 The PV data of Table 2.E.1 were obtained using steam in a pistonandcylinder arrangement, placed in a large bath of constant 'hotness'. What is the corresponding ideal gas temperature of the bath? Figure 2.E.1 presents a plot of (PV) versus P. By extrapolation toP= 0, (PV)• .. 4764.7cm3 MPa mo1" 1• Hence, from Eq.2.3.1: T = 4764.7 cm3 MPa mol. 1/8.314cm3 MPa mo1" 1 K" 1 = 573.1 K = 300°C
Table 2.E.l PressureVolume Data for Example 2.1 0.001 0.010 0.030 0.040 0.050 0.075 0.100
264500 26440 8810.8 6606.5 5283.9 3520.5 2638.7
4765.0 4763.2 4761.8 4760.6 4759.5 4756.6 4753.6
4767r~~,~rr~~,~
......._
0
~4763
...
E 0
0 4761
CL
~ ......... >4759
CL
g4757
"'0
~
CL
4755 4753~~~~~~~~~~~
0.00
F~gure
0.02
0.04
0.06
Pressure (MPo)
0.08
0.10
2.E.l Extrapolation of the product (PV) for steam at a constant
'hotness'.
The Zeroth and First Laws of Thermodynamics
41
2.4 First Law 2.4.1 Background and Statement The first law expresses our empirical evidence that energy is conserved and its origin lies with the experiments that Joule carried out between 1843 and 1850. Joule used several methods, whereby mechanical work was converted into heat absorbed by water placed into a well insulated container. He determined the amount of heat produced by measuring the corresponding increase in water temperature, and concluded that  in all cases: the ratio of heat produced to work consumed was the same. His fascinating work, extremely difficult even with today's capabilities (see Example 2.2), included (Denbigh, 1981, p.15): * paddleandwheel experiments, where the work produced by a falling weight was converted into heat through the friction between paddles and water, as shown schematically in Figure 2.2; * experiments where mechanical work was expended on an electric machine, the resultant electric current being passed through a coil immersed in the water;
w
Figure 2.2 Joule's paddleandwheel experiment.
0=0
42
Applied Chemical Engineering Thermodynamics
* experiments in which mechanical work was expended in the compression of a gas in a cylinder, immersed in the water; * experiments in which the mechanical work was expended on two pieces of iron, which rubbed against each other beneath the surface of the water. Joule, the Brewery Business, Religion and Deflation(!) James, the second surviving child, had an elder and a younger brother and two younger sisters, one of whom died in childhood. He was too delicate for school, so he had private tutors instead. When he was fourteen he and his elder brother, Benjamin, were sent to learn mathematics and science under John Dalton, the doyen of Manchester science. The Joule boys studied for three years under John Dalton at the Manchester Literary and Philosophical Society, after which they assisted their father in the brewery. I doubt if the work was onerous ,for young Joule was able to carry out his early sciellfijic work there before his father built him a laboratory at home. This looks like a rather indulged childhood and adolescence. But considering Joule's social status I doubt if it was untypical. As the family increased in wealth and security, so the sons gave up the dissenting church of their father Dr Roby's Independent Chapel and joined the Church of England, the established church; a transition that was fairly common among those rising in the world. By the time he was thirtysix Joule had given up work at the brewery, which in any case the family were soon to sell. The brothers became gelltlemen of leisure. Benjamin had estates in Ireland and spent his time training church choirs, playing the organ, and composing hymns and chants. John, the youngest, retired to the vacation Isle of Man, where he died comparatively young. James's career was outwardly similar to those of many other nineteenthcentury British scientists who, having made or inherited a comfortable sum of money, went on to devote thetnselves to science. Such was the common career pattern of, for example, W. H. Perkin, David Edward Hughes, Henry Wilde and James Nasmyth. You made your hundred thousand pounds or more, invested it in sound public utilities, railroad stock, or govemment securities to yield, say, three per cent, and retired to a comfortable life of science in the secure knowledge that things were getting better and cheaper all the time. (Cardwell, 1983, pp.4547.)
As a result of Joule's work, the fit:st law was developed and can be stated as follows: Even though energy can be converted from one form to an
other, the total amount is conserved.
43
The Zeroth and First Laws of Thermodynamics
We will discuss Joule's contributions in detail in Chapter 4. It should be mentioned here, however, that by demonstrating the convertibility of work to heat, he tore down the long held axiom that heat, or 'caloric' as it was also called then, is conserved. According to this axiom, caloric consisted of very small particles and was, therefore, neither created nor annihilated. Thus, the steam engine operated through the 'flow' of caloric, not by conversion of caloric to work. This is why there was such an opposition to the conversion of heat to work as demonstrated by Mayer's tragedy. J. R. Mayer, the Establishment, and a Tragedy in Thermodynamics. Actually,just before Joule, J. R. Mayer suggested but did not clearly demonstrate  the convertibility ofheat to work. But so strong was the conviction about the conservation of heat in that period, that Mayer who also had the misfortune of not being part of the then scientific establishment  was ignored, even scorned, leading to his mental collapse. He finally achieved the recognition he deserved, due to Clausius, recovered from his breakdown and was voted into the Royal Society in 1862. (See Chapter 4.)
2.4.2 The Analytical Expression To obtain an analytical expression for the first law, we will accept the following convention: l.Heat added to a system is positive; 2.Work done on a system is negative. Let us now consider a closed system that exchanges an amount of heat Q with the surroundings, but no work. The resulting change in the internal energy of the system U, as it goes from the initial state 1 to the final state 2, is:
11u = u2  u1 Notice that if Q is a positive number i.e. heat is added to a system we would expect that its internal energy will increase; and if negative, that it will decrease. According, therefore, to the first law statement and the accepted convention:
J1U = Q
(2.4.1)
If now an amount of work W is exchanged between system and surroundings, but no heat exchange is involved, then following again the first law
44
Applied Chemical Engineering Thermodynamics
statement and the convention:
&u = w
(2.4.2)
In this case, a positive value for W indicates that the system does work on the surroundings consuming its internal energy which, consequently, decreases; and a negative one, the reverse, both in agreement with Eq. 2.4.2. A combination of Eqs 2.4.1 and 2.4.2 gives the analytical statement of the first law: Q = tJ.U
+
W
(2.4.3)
2.4.3 Comments 1.When small  infinitesimal  amounts of heat and work are involved, Eq.2.4.3 becomes: dQ = dU + dW
(2.4.4)
Remember, however, that dQ and dW refer to small amounts, not small changes, as is the case with the internal energy (Section 1.11). 2.The term W contains all types of work: pressurevolume (PV), i.e. work of compression or expansion; and kinetic KE and potential PE energy changes. 3.1n thermodynamics, however, the term W is reserved for PV work only. Hence Eq.2.4.3 becomes:
Q = tJ.U
+
W + tJ.(KE) + tJ.(PE)
(2.4.5)
where W refers to PV work only. 4.1n case of nuclear reactions, where conversion of mass to energy takes place according to Einstein's expression, then the sum: mass plus energy is conserved. Two simple applications of the first law are presented in Examples 2.2 and 2.3.
2.4.4 Example 2.2 To get some idea about the experimental difficulties, and required accuracy, in the temperature measurements of Joule's work, let us assume that in the paddleandwheel experiment (Figure 2.2): "' the fall of a body with 1 kg mass for 10 meters was used, and "' 1 kg of water was stirred by the paddle.
The Zeroth and First Laws of Thermodynamics
45
Assuming: Cv = 4.184 kJ!k.g K, calculate the corresponding temperature increase. Use g = 9.8067 m sec· 2 • From the first law: ..1 U = .d(PE ). The potential energy change is: .d(PE) = (1 kg)(10m)(9.8067m sec2),. 98.067J
and from the definition of the heat capacity (Eq.1.13.1): ..1 U = Cv.dt. Thus: .dt =0.023°C, a very small temperature change. (See also Problem 2.33.) Indeed, Joule had an excellent instrument maker for whom Cardwell (1971) writes: "John Benjamin Dancer, of Cross Street in Manchester, the inventor of microphotography. I can confirm from personal experience, the superb quality of Dancer's instruments".
2.4.5 Example 2.3 Steam at l25°C and 1 atm, state 1, is placed into a well insulated cylinderandpiston arrangement and heated to 200°C, state 2, with electricity. How much heat is absorbed per kg of steam, if the pressure outside the piston is that of the atmosphere and equal to 1 atm? From the first law:
i.e. the absorbed heat not only increases the internal energy of the steam, but also does work against the atmosphere. We tum now to the Steam Tables to obtain values for U1, U2 , V1, and V2 : u1 =2544.7, u2,. 2658.1, both in kJ/k.g v1 = 1792.7, v2 = 2143.8, both in cm3 /gr Thus: U2  U1 = 2658.1 2544.1 = 113.4 kltkg (V2  V1)(P) = (2143.81792.7)(1)= = 351.1 atm cm3 gr· 1 x 8.314J I 82.06atm cm3 ,. 35.6J/gr + 35.6 ,. 149.0 kJ. 113.4 = and: Q The calculation can be much simplified by noticing that, since the pressure remains constant, we can write Eq.(A) as: Q "' U2  U1 + P2 Y2  P1 vt = H2  H1 Q can be thus calculated by using the enthalpy values only:
Q = H2 HI ,. 2875.3 2726.4 = 148.9 kJ.
46
Applied Chemical Engineering Thermodynamics
2.5 The First Law Expression for a SteadyState Flow Process 2.5.1 The Problem Application of the first law to nonflow, constant mass, systems can be carried out directly by using Eq.2.4.5, as demonstrated in Examples 2.2 and 2.3. Most chemical engineering processes, however, are of the steadystate flow type. Eq.2.4.5 applies to them of course but, as we will see next, it can be recast into a more convenient for this purpose form.
2.5.2 Development of the Expression Consider the general steadystate flow process shown in Figure 2.3 and apply to it the first law, Eq.2.4.5, on the basis of one mole of fluid entering the system in point 1. In evaluating the PV work term we notice that it involves three parts: 1.the work done by the surroundings in pushing the fluid into the system, WI.
2.the work of expansion or compression, W5 , which is called shaft work, because it is communicated through a shaft. 3.the work done by the system in pushing the fluid out of the system into the surroundings, W2 • From the definition of work: WI
I
= PdV = P1.1V
since the pressure at this point is constant and equal to P 1• But what is .::1 V? Simply, the molar volume of the fluid at point 1, since the pressure P 1 must push one mole of the fluid into the system. Hence: WI = P1V1 (Work done on the system) Similarly: w2 = p2 v2 (Work done by the system) Thus, the total work W is:
w = ws  P1 vt
+ P2 Y2
Introduction of Eq.2.5.1 into Eq.2.4.5 yields:
(2.5.1)
47
The Zeroth and First Laws of Thermodynamics Q
1 mol e

1
, 
~
2
1 mole
Figure 2.3 Schematic diagram for the development of the first law expression for a steadystate flow process. System: inside the box; Surroundings: outside the box.
Q
=
(U2
or since: (U2

U 1) + (W6

P 1V1 + P2V2) + Ll(KE) + Ll(PE)
+ P 2 V2)  (U1 + P 1V1) =
Q = LlH
H2
 H 1 = LlH
(2.5.2) The convenience of using Eq.2.5.2, as against Eq.2.4.5, for a flow process is apparent. And since, as mentioned, most processes are of the flow type, enthalpy has become a very commonly used term in thermodynamics. + Ws + Ll(KE) + Ll(PE)
2.5.3 Comments l.We will refer to steadystate flow processes as just flow processes. 2.The potential and kinetic energy changes are usually negligible as compared to the other terms. 3.The importance ofEqs 2.4.5 and 2.5.2lies with the observation that the path depending quantities, Q and W, are related to changes in the state functions U and H. 4.Determination of internal energy and enthalpy changes  from experimental PVI and heat capacity data  is discussed in Chapters 9 and 11.
48
Applied Chemical Engineering Thermodynamics
5.We turn therefore to the simplest case for the calculation of such changes, that of an ideal gas, which is considered next, and then proceed to apply the first law in Examples 2.4 through 2. 7.
2.6 Thermodynamic Relationships of the Ideal Gas 2.6.1 Importance of Ideal Gases As mentioned in Section 1.13.2 an ideal gas is a hypothetical fluid since it represents the behavior of a real gas at zero pressure. We did use this hypothetical fluid, however, to establish the ideal gas temperature through the process of extrapolation to P = 0 of the behavior of real gases. The ideal gas concept is not just useful for the aforementioned purpose. It is a very valuable tool in thermodynamics, because: 1. With very few exceptions ideal gas behavior can be assumed at pressures up to 12 bar and can be thus employed for quick estimates with reasonable accuracy (Example 1.5). (For accurate results, of course, experimental data should be used. If not available, more reliable  than ideal gas behavior  estimation techniques that will be discussed later, should be considered.) 2.1deal gas properties are easier to determine than those of real fluids. By focusing, thus, attention to the difference in the properties of these two types of fluids (Section 1.13.3), we can develop a better understanding of the factors leading to real fluid behavior; and, estimation techniques for the latter (Chapters 8, 9, and 11). In this Chapter, we will take advantage of this easier determination of ideal gas properties by using them in applications of the first law. We can thus enhance our understanding of this law without being concerned with real fluid behavior. We review, therefore, next some relationships of ideal gases.
2.6.2 Equation of State The term equation of state means the functional relationship among pressure, volume, and temperature, which for an ideal gas has the form: PV = RT (2.6.1)
49
The Zeroth and First Laws of Thermodynamics
V is the molar volume, and R is the ideal gas constant, whose value depends on the units used for the other quantities in the equation (Table 2.1).
2.6.3 Internal Energy and Enthalpy Changes Due to Heating or Cooling From the first law for a constant volume process involving one mole: dQ
= dU
and from the definition of heat capacity at constant volume (Eq.l.l3.1): (2.6.2) dU = Cv• dT where the superscript ( • ) indicates ideal gas behavior. Again from the first law, but this time for a constant pressure reversible process involving one mole: dQ = dU + PdV = dU + d(PV) = dH
and from the definition of heat capacity at constant pressure (Eq.l.l3.1): dH =
c; dT
(2.6.3)
2.6.4 The Difference in Heat Capacities From the definition of enthalpy applied to one mole of an ideal gas: dH
=
dU + d(PV)
=
dU + RdT
Introduction of Eqs 2.6.3 and 2.6.2 into this expression yields:
c; c;
=
R
(2.6.4)
2.6.5 The PVT Relationship for an Adiabatic and Reversible Change Consider one mole of an ideal gas that undergoes an adiabatic and reversible change from the initial condition: P 1, VI> and T 1, to a final pressure P 2 . What are the corresponding values of T2 and V2?
50
Applied Chemical Engineering Thermodynamics
From the first law for an adiabatic process: dW = dU. Since the process is reversible, dW =PdV, which for an ideal gas becomes: RT dW = PdV = RdT  VdP = RdT  dP p
dU = C/ dT
From Eq.2.6.2:
C~dT = RdT RTdP
Hence:
C/ dinT= R dinP
or:
p
Integration from T1 and PI> to T2 and P 2 , assuming Cp• constant, yields: T2 R p2 In=  I n TI c•p pl
After rearrangement: (2.6.5)
y1 , where: k = __ y
For the final volume V2 , it can be easily seen that:
P2v{ = Plvr which also suggests that the product (PV r ) remains constant during the process:
pyr where,
= C
(2.6.6)
c = pl vl y = constant.
2.6.6 Comments 1.The ideal gas heat capacity is a function of temperature only, which becomes more pronounced with increasing number of atoms in the molecule as shown in Figure 2.4. 2.The internal energy and enthalpy of an ideal gas are, therefore according to Eqs 2.6.2 and 2.6.3 respectively  also functions of temperature only. 3.Information on C/, expressed as: =a + bT + cT 2 + dT 3 (2.6.7)
C/
51
The Zeroth and First Laws of Thermodynamics
__ ... 100 Water ·   Ethylene
90 ......... 80
.,
~
0
E
., ., .,
., .,
/
70
/ /
........ , ......... 60
/
.
/ / /
IL
/
(.)
50
/ /
/
40 500
700
900
Temperature (K)
1100
1300
Figure 2.4 Ideal gas heat capacities.
is given, among others, by Reid et al (1987) for a variety of gases, and parameter values for some selected ones are presented in Appendix C. C/ information is also given by Daubert and Danner (1985; 1986). 4.Eqs 2.6.5 and 2.6.6 apply to ideal gases of constant heat capacity undergoing reversible and adiabatic changes only.
2.7 Examples 2.4 through 2.7 Some simple applications of the first law are presented in the next Examples, which consider: * Example 2.4, the uncertainties involved in assuming ideal gas behavior; * Example 2.5, an irreversible case; * Example 2.6, the feasibility of a process; and * Example 2. 7, the efficiency of waterwheels.
52
Applied Chemical Engineering Thermodynamics
2.7.1 Example 2.4 Steam is heated from t 1 = 200°C to t2 = 550°C in a constant pressure flow process. Assuming ideal gas behavior, calculate the amount of heat required and compare it to the actual amount as a function of pressure from 100 to 1000 kPa. From the first law for a flow process, neglecting changes in PE and KE, and using Ws = 0, since no shaft work is involved:
where, from Appendix C: CP* =a + bT + cT 2 + dT 3 a= 32.24; b
T: K;
,
= 1.908·10·3 ;
c
C/: J mol"
= 1.057·10· 5 ;
1
K ·I
d = 3.602·10· 9
Thus:
Q =a (T2  T1) + (b/2)(Tl T12 ) + (c/3)(Tl T/) + (d/4)(T24  T14 ) = 129401 mo1" 1 = 718 kJ/kg We calculate, next, the actual Q values as a function of the operating pressure obtaining the H2 and H 1 values from the Steam Tables. The results are presented in Table 2.E.4 and indicate that reasonable accuracy is obtained with the ideal gas assumption up to about 4 bar.
2. 7.2 Example 2.5 One mole of oxygen is compressed in a flow process from state 1 (1 atm and 25°C) to state 2 (6 atm). The compression is adiabatic but not reversible, with an efficiency of 0. 75. Assuming oxygen to behave as an ideal gas under these
Table 2.E.4 Actual Q Values as a Function of Pressure, Example 2.4 p %error HI H2 Q (kPa) (kJ/kg) (kJ/kg) 100 2875.4 3595.6 720.2 0.2 200 2870.5 3594.7 0.8 724.2 2865.5 3593.7 1.3 300 728.2 400 2860.4 3592.8 1.9 732.4 3.1 600 2849.7 3590.9 741.2 800 2838.6 3589.0 750.4 4.3 1000 2826.8 760.3 5.5 3587.1
53
The Zeroth and First Laws of Thermodynamics
conditions, calculate the: a. work requirement; b. discharge temperature, Td; and c.amount of heat that must be removed to cool the gas back to 25°C at the pressure of 6 atm. Comment on the accuracy of the obtained results. 1.41 (asGiven: cP• = 26.625 + 0.0049 T (J mol 1 K" 1); T, in K, and: y sumed constant). a. We calculate first the reversible and adiabatic work W(RA }, and then calculate the real work, W(real ). From the first law: W(RA)
= tlH = (1) J cP• dT = 26.62S(T2 
T1)+(0.0049/2)(T22  T12 )
To evaluate T2 , notice that for a reversible and adiabatic process (Eq.2.6.5): T2 = T1 (P2 /P1)k = 502.0 K Hence: W(RA) = 5827.21 and, since work is consumed, from Eq.1.12.3: W(real) = W(RA )/0.75 = 7169.61 b.From the first law: tlH(real) = W(real ). Thus: 26.625 (Tr T1)+(0.0049/2)(T/ T12) = 7769.61 Solution of this quadratic equation gives Td = 568.4 K. Notice that the extra work, over the W(RA) one, is converted into enthalpy of the gas. This leads to the larger value for the discharge temperature Td, as compared to 502.0 K for the reversible case. c.From the first law:
= H(P2 ,Td) H(P2 ,T1) For an ideal gas, however: H(P2 ,T1) = H(P 1,T1) and consequently all the energy Q = tlH
imparted into the gas through the compression must be removed:
Q
= W(real)
= 7169.61
Comment: Two assumptions are made here: * ideal gas; and * r is temperature independent. To evaluate the impact of these two assumptions exactly, we need actual experimental data for oxygen. In their absence, the following observations can be made: !.Considering the temperature and pressure range involved, the error due to the ideal gas assumption should not be more than 5 %. 2.Evaluation of y at 298 K and 568 K gives values of 1.421 and 1.394 respectively, with an average value of 1.408, close to the one used. The answer, hence, should be within 5% of the correct one.
54
Applied Chemical Engineering Thermodynamics
2.7.3 Example 2.6 Consider the following device proposed by an inventor: Air enters his device at atmospheric conditions, say: T0 = 300 K and P0 = 1 atm, and two streams, 1 and 2, of equal amounts are emerging: stream 1 at T1 = 280 K, and stream 2 at T2 = 320 K, both at 1 atm. No exchange of heat or work with the surroundings is involved. Is this device possible? (Assume that due to the small temperature range involved, the air beat capacity is temperature independent.) To this purpose we examine if the first law, applied to this device, is satisfied. Using as a basis the flow of one mole of air entering the device, and since Q = ws = 0: t!H = (0.5 H2 + 0.5 H 1 )  (1)H0 = 0 ('?) 0.5(H1  H0) + 0.5(H2  H0) = 0 ('?) or: or: 0.5 Cp [T1  T0 ] + 0.5 Cp [T2  T0] = 0 ('?) which is apparently valid and, consequently, the device is possible according to the first law. CNe will see in the next Chapter if it also satisfies the second law.)
2.7.4 Example 2.7. The Efficiency of WaterWheels We have considered, so far, systems where potential and kinetic energy changes have been negligible. A prime example where these are important are the waterwheels. These devices were used extensively as the motive power for the textile industry, one of the early accomplishments of the Industrial Revolution, and were later supplemented by steam engines. To this purpose consider a stream of water that: *is available at an elevation h 1 with a velocity u 1; *enters the waterwheel at an elevation h 1 with a velocity u1 * leaves at an elevation h2 with a velocity u2 and *hits the ground at an elevation h 2 with velocity u2 ; as shown in Figure 2.E.7. Identify the conditions under which the maximum work is obtained. 1
1
1
1
;
From the first law applied to the flow of one mole of water: Q
= Ws
+ t!H + tl(PE) + tl(KE)
Assuming no exchange of heat with the atmosphere, Q = 0:
Ws = t!H  tl(PE)  tl(KE) = C (T1  T2) + g(h 1 1  h2 1 ) + 0.5(u 1 12  u2 12
)
;
55
The Zeroth and First Laws of Thermodynamics
h,h'1
h~
h2Figure 2.E. 7 The waterwheel for Example 2. 7.
where C is the water heat capacity, and T1 and T2 the entrance and exit temperatures respectively. To obtain, therefore, the maximum work: l.(T2  T1) = 0, i.e. no heat must be generated through water turbulence and, consequently, u1 = u1; u2 = u2 • 2.(h 1 ~h2 ') = (h 1  h2), i.e. the whole height of fall should be utilized. Notice that, as expected, these requirements for obtaining the maximum work are those for a reversible operation: u2  u2 = 0. h2  h2 = 0; h 1  h 1 = 0; u 1  u1 = 0; 1
I
1
I
1
1
Comments: 1.To produce the maximum possible amount of work, the exit velocity u2 must be set equal to zero, requiring a wheel moving with an infinitesimally small velocity. This, in tum, requires that u1 = 0 (why?), i.e. that all kinetic energy is converted to potential, before the water enters the wheel. 2. The conditions for obtaining the maximum work from waterwheels were identified by the 17th century (see Cardwell, 1971). It was, thus, possible to determine exactly their efficiency. 3.This, however, was not the case with the efficiency of the steam engine. More than hundred years after its introduction, and even though it had become the dominant source of power, the efficiency of the steam engine was still measured in terms of "work produced per bushel of good coal(!)". They did not know what the maximum obtainable work was.
56
Applied Chemical Engineering Thermodynamics
4.We will see in the next two Chapters that Camot, whose father was a hydraulic engineer, not only was aware of the requirement that for maximum efficiency all the fall of water in a waterwheel must be utilized, but that he used it in establishing the requirement for obtaining the maximum work from a steam engine. 5.We will also see how Clapeyron, using Camot's brilliant approach, arrived at an analytical expression for the determination of the efficiency of a steam engine; and, that this eventually led to the formulation of the second law of thermodynamics.
2.8 Concluding Remarks 1. The zeroth law establishes temperature as a state property rather than a physiological sense of 'hotness' or 'coldness'. 2.Use of the ideal gas behavior exhibited by real gases in the limit of zero pressure leads to the establishment of the ideal gas temperature scale. 3.Application of the first law to a given process provides for the calculation of the important  from the chemical engineering point of view heat and work requirements through changes in the state variables of the system involved: internal energy or enthalpy. 4.If the process involves gases, use of the ideal gas assumption for such calculations allows for reasonable approximations up to moderate pressures (25 bar) in most cases.
References Cardwell, D.S.L., 1971. From Watt to Clausius, the Rise of Thermodynamics in the Early Industrial Age, Cornell University Press, Ithaca, N.Y. Ibid., 1983. In Springs of Scientific Creativity, Aris, R., Davis, H. T., Stuewer, R.H., Eds, University of Minnesota Press, Minneapolis, MN. Daubert, T.E., Danner, R.P, 1985; 1986. Data Compilation Tables of Properties of Pure Compounds, AIChE, New York. Denbigh, K., 1981. The Principles of Chemical Equilibrium, Cambridge University Press, Cambridge. Hatsopoulos, G.N., Keenan, J.H., 1965. Principles of General Thermodynamics, Wiley, New York. Redlich, 0., 1968. Reviews Modern Phys., 40, 556. Reid, R.C., Prausnitz, J.M., Polling, B., 1987. The Properties of Gases and Liquids, 4th Ed., McGrawHill, New York.
The Zeroth and First Laws of Thermodynamics
57
Problems 2.1 Propane at 2 atm is heated from 300 K to 400 K: a.in a constant pressure reversible process; b.at constant volume; and c.in a constant pressure flow process. How much heat is required per mole in each case? 2.2 Why do we write the analytical statement of the first law as: Q = W + .tiU, and not as Q = W .tiU? 2.3 What is the difference between W and Ws? 2.4 Does the expression: Q = .ti U + W + LJ(KE) + LJ(PE) apply to a flow process? Explain your answer. 2.5 Starting with Eq.2.4.5, develop the expression of the first law for a flow process. 2.6.a. What is the enthalpy of an ideal gas a function of? Justify your answer. b. Why is enthalpy important in chemical engineering calculations? 2. 7 It has been suggested that the kitchen in your house could be cooled in the summer by closing it from the rest of the house and opening the door of the electric refrigerator. Show if this is feasible. Make sure you specify system and surroundings, and use data of your choice to demonstrate your conclusion. 2.8 An ideal gas flows through a horizontal tube which has nonconducting walls. No heat is added, or removed, nor is any shaft work involved. The crosssectional area of the tube changes with length and this causes the velocity to change. Develop an expression relating the temperature and velocity of the gas at any point of the tube to their entrance values. 2.9 A gas at a pressure of 5 atm and a temperature of 300 K flows through a well insulated expansion valve with an exit pressure of I atm (expansion through a valve is referred to as 'throttling'). a. Calculate the exit temperature, if the behavior of the gas can be approximated by that of an ideal one. b. Will the same answer hold for a real gas? Explain. Neglect kinetic energy effects. 2.10.a. What are the assumptions behind ideal gas behavior? b. Why are these assumptions failing as the density increases? 2.11 What makes ideal gases important in chemical engineering calculations? 2.12 A certain gas is heated from l00°C to 200°C at a constant pressure of I atm in a flow process involving no shaft work. How much heat is involved per mole of gas? In this temperature range and for this gas: U "' U0 + 28(T 373) J/mol; T,K. 2.13 Methane goes from 273 K and 10 bar to 350 K and 1 bar. a. Devise a nonflow reversible process  any number of steps  to accomplish this change; b. Calculate the enthalpy change per mole of methane.
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2.14 A gas at a pressure of 3 atm is contained in a vessel of 5 1. The vessel is connected through a stop cock with the atmosphere. The stop cock is opened and the gas leaks slowly so that the temperature of the vessel remains constant and equal to that of the atmosphere: a. how much work is produced? b.how much heat is exchanged with the atmosphere? 2.15 Ethylene is compressed from V1 = 20 1, P 1 = 1 bar and T1 = 273 K to V2 = 5 1 and P2 = 6 bar. Calculate any of the following quantities that you have sufficient information for: a.heat requirement; b. work requirement; c.intemal energy change; d. enthalpy change. 2.16 Carbon dioxide at 40 psia and 100°F and at a rate of 400 ft3 /min is compressed to 200 psia and 300°F. The compressor cooling jacket removes heat at a rate of 15 Btu/lbm. What is the compressor's horse power? 2.17 In a steady flow process water at 20°C and saturated steam at 2 bar are mixed to produce 50 kg/min of saturated water at 1.5 bar. In the mixing operation heat is lost at the rate 20 kJ /kg of saturated water produced. What is the flow rate of water entering the mixer? 2.18 One mole of air at 40°C and 10 bar is in a vertical cylinder confined by a piston held with latches. The weight of the piston and the pressure of the atmosphere are equivalent to a downward pressure of2 bar. The latches are suddenly removed and the gas expands to twice its original volume, whereupon the piston hits a stop. The system consisting of the cylinder, piston and gas, is allowed to return to its initial temperature of 40°C. How much heat is transferred to, or from, the surroundings? 2.19 A steel tank whose volume is 3 m3 contains 23 kg of steam and the rest is water at 200°C. The contents of the tank are heated to 250°C. How much heat is added to this purpose? 2.20 One mole of nitrogen undergoes the following series of reversible changes in a cylinder: Isothermal expansion from 10 bar and 90°C to 4 bar, constant volume cooling, and then adiabatic compression to its initial state. Find Q and W for each step and for the entire cycle. 2.21 Consider two well insulated tanks A and B with volumes: VA = 20 1 and V8 =50 1. Tank A contains N2 at 10 bar and 60°C and tank B methane at 3 bar and 20°C, and are connected through a valve. The valve is opened quickly, allowing the two pressures to become equal, and then closed. Determine: a. the common pressure; and b. the temperature and number of moles in each tank. 2.22 One kg of steam at 180°C and having a volume of 0.2 m3 is heated at constant volume to 250°C: a. what is the final pressure of the steam? b.how much heat has been absorbed?
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The Zeroth and First Laws of Thermodynamics
2.23 A centrifugal pump delivers 600 gal/min of water at 80°F from a pressure of 1 atm to a pressure of 3 atm. The diameter of suction inlet to the pump is 6 in and of the outlet 4 in. If the suction and discharge are at the same level, and the pump has an efficiency of 0. 7: a. what is the required work? b.if there is no change in the internal energy of the water, how much heat is exchanged with the surroundings? Hint. Show first that for a reversible pump: W3 =
J
Pz PI
V dP.
2.24 A very large reservoir, containing nitrogen at 12 bar and an unknown temperature T, is connected through a valve with a small tank, whose volume is 0.2 m3 and contains initially 0.2 kg of nitrogen at 27°C. The valve opens and the gas flows into the tank until the pressure in it becomes 3 bar and the temperature 140°C. Calculate the value of Tassuming, for simplicity, ideal gas behavior with Cp• = 29.3 J/mol K. 2.25 A well insulated vessel is divided into two parts. The left contains 1 kg of saturated steam at 1 bar. The right, 2 kg of steam at 6 bar and 300°C. What will be the temperature and enthalpy of the steam if the divider is removed? 2.26 One mole of an ideal gas goes from 1 bar and 20°C to 10 bar and 100 °C. Develop two paths to accomplish this change, and: a.calculate, for each path, the amounts of heat Q and work W involved, and the changes in internal energy and enthalpy, L!U and L!H; b. compare Q, W, Ll U, and L!H for the two paths, and comment on their values. Assume C/ = 7 cal/mol. 2.27 A gas is heated from T1 to T2 at a constant pressure P. If we assume ideal gas behavior to calculate the amount of heat required, what factors will influence the accuracy of the obtained results? 2.28 Starting with the first law, develop Eq.2.6.5. 2.29 One mole of propane at 300 K is compressed from 0.1 MPa to 1 MPa in a batch process. If the compression is adiabatic but irreversible with an efficiency of 0.8: a.how much work is required? b. what is the temperature of the compressed gas? c.how much heat must be removed to cool the gas to its initial temperature? State any assumptions you make and estimate their impact on the accuracy of the obtained results. 2.30 Repeat the previous problem, assuming a heat loss of 3 J/mol during the compression. 2.31 Air enters an apparatus at 1 atm and 30°C and two streams, both at I bar emerge: one containing 40% of the inlet air at 0°C and the other at 45°C. If the apparatus does not exchange any energy with the surroundings, do you see any uncertainty in the data?
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2.32 Assume that in Example 2.5 the compression is done with a twostage compressor with intercooling: the gas leaves the first stage at an intermediate pressure P;. is cooled to the entrance temperature of 25°C and then is compressed to the fmal pressure of 6 atm. Using the same assumptions as in Example 2.5: a.determine the value of P; so that the total work consumed is the minimum; b.explain the difference between this work and that consumed in Example 2.5; c.determine the beat removed in the intercooler. 2.33 In Example 2.2 we assumed for simplicity that the beating of the water takes place at constant volume, while in reality it takes place at constant pressure. What is the actual temperature change assuming that the water is initially at 25°C? 2.34 The following data for propane were measured in a bath of constant 'hotness': Density (kg/m3) 5 10 15 20 25 30 35 Pressure (MPa) 0.2896 0.5571 0.8027 1.0268 1.2295 1.4112 1.5721 What is the corresponding ideal gas temperature? 2.3S.a. What are in your opinion the important items you have learned in this Chapter? b.Give a practical application in each case. 2.51 Discuss exceptions to the zeroth law. 2.S2.a.Demonstrate that the introduction of temperature through the zeroth law can be made without the prior knowledge of heat. b.Proceed then to introduce heat through the first law. (See Denbigh.)
3
The Second Law of Thermodynamics
A good many times I have been present at gatherings ofpeople who by standards ofthe traditional culture are thought highly educated and who have with considerable gusto been expressing their incredulity at the illiteracy of scientists. Once or twice I have been provoked and have asked the company how many of them could describe the Second Law of Thermodynamics. The respond was cold: it was also negative. Yet I was asking something which is the scientific equivalent of· Have you read a work ofShakespeare's?(C.P.Snow: 'The Two Cultures and the Scientific Revolution'.)
3.1 Introduction The origin of the Second Law lies with Sadi Carnot's work: Rejlexions sur Ia Puissance Motrice du Feu, published in 1824, dealing with the establishment of the maximum amount of work that can be obtained from a steam engine. Thus, its efficiency could be rigorously expressed, instead of the method used then: "work (ftlbr) per bushel of good coal(!)". Carnot used in his development the erroneous  but strongly held then  axiom of the conservation of caloric (he did have, however, some doubts about it later, but never articulated them). He died young, of cholera, and his work laid dormant until it was revived a few years later by Clapeyron. In the meantime, the advent of Joule's work in the 1840's showing that caloric was not conserved  led Kelvin, an advocate of the Caloric theory, to declare in 1849 that Carnot and Joule could not be both right. Clausius and Kelvin, who later abandoned the caloric theory, eventually reconciled the work of these two giants of thermodynamics and established the first and second laws in the 1850's and 1860's. (We will discuss the fascinating historical development of these two laws in the next Chapter.)
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3.2 Rationale for a Second Law Let us now return to Carnot's work and consider the following question: Is the maximum work produced by the steam engine sufficient reason for a chemical engineer to get involved with the second law and, especially, with the associated concept of entropy that seems to haunt anyone dealing with thermodynamics? While the development of the second law started with Carnot's work on the efficiency of the steam engine, the law covers a broad spectrum of phenomena. Let us consider some of them.
3.2.1 The Direction of Natural Phenomena We all know from experience that: * if a pipe carrying steam is not insulated, heat flows from the steam to the environment, and not from the latter to the former; * if the valve separating two gases  one at low and the other at high pressure  is opened, the flow occurs from the high pressure to the low one, and not in the reverse direction; * once the wall that separates two pure gases in a vessel is removed, the gases will be soon completely mixed; but if we start with a mixture of these gases, they will not separate by themselves. Nature, thus, seems to dictate a certain direction to processes: heat flow from a hot body to a cold one; gas flow from high to low pressure; mixing rather than unmixing. Yet, the first law does not object to the reversing of any of these processes.
3.2.2 The Feasibility of Processes But these observations, and other similar ones, could still hardly suggest the need for a second law. One could argue that experience suffices to predict these phenomena. Let us, however, consider the following device proposed by an inventor (and discussed in Example 2.6): Air enters his device at atmospheric pressure and temperature, say 300 K, and leaves in two streams that have
The Second Law of Thermodynamics
63
the same flow rate: one at 320 K and the other at 280 K, both being at atmospheric pressure. The device requires no external energy, thermal or mechanical, and it would obviously be of immense practical use in providing cooling during the summer and heating during the winter. Is this device possible? We concluded, in Example 2.6, that the device does not violate the first law: the enthalpy lost by the cold stream can be found in the hot one. Notice, also, that the device does not involve flow of heat from a cold to a hot body; or gas flow from a low pressure to a high one; or any unmixing of gases. It should, therefore, be feasible. Yet, our intuition suggests that this may not be the case.
3.2.3 The Big Question: Where is the Energy Gone? Let us next raise the following very important question: If energy is conserved, according to the first law, how come the world is running out of it? Why can't we use all the energy surrounding us in the air and sea? Is it that we have not developed the proper devices yet, or is it really impossible?
3.2.4 Summary All these observations and questions  and of course a variety of others can be summed up as follows: 1. The first law is a necessary but not sufficient condition for the occurrence of a process. It makes no provision with respect to the direction in which the process will proceed. 2.Natural (spontaneous)processes, such as the flow of heat or the mixing of gases, etc., cannot be reversed on their own. 3.The direction of a process seems to be determined by the initial and final states of the substances involved: their pressure, temperature, and composition. It is possible, therefore, that some state function is involved, whose characteristics determine the direction of the process. 4.Experience and intuition may not always suffice for the establishment of the feasibility of a given process. We need a more definite and quantitative criterion.
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3.3 Objective and Approach The overall objective of this Chapter is to present the second law and to introduce, through it, the concept of entropy as the quantitative criterion  the 'yardstick'  for: l.the establishment of the feasibility of a given process, and 2. the efficient utilization of thermal energy. More specifically, to understand: 1. The concept of a reversible heat engine, as it is described by the Carnot cycle, and that it establishes the upper limit for the amount of mechanical energy that can be produced from a given amount of heat flowing from a given temperature level to a lower one. 2.The concept of absolute thermodynamic temperature and its relationship to that obtained through an ideal gas fluid. 3.The approach and methods used in the calculation of entropy changes. 4. The application of the second law in achieving the overall objective of this Chapter, i.e. in: * establishing the feasibility of a given process, and * obtaining the maximum possible amount of mechanical energy from a given amount of thermal one. The latter is, by far, the most important one, for in the final analysis  thermal energy is the motive power of our civilization. We start with the development of the second law, which will be carried out in the following five steps: !.Presentation of two  but equivalent statements of the law that represent a verbal description of our experiences with respect to the way certain processes occur (or, do not). 2.Discussion of the conversion of heat to work with a reversible heat engine the Carnot cycle essential for the development of an analytical statement of the second law; (and, to bring this 'ideal' engine closer to reality, the very 'real' power plant cycle  for the production of electricity from fuels  is described briefly). 3.Presentation of four Propositions, that will be proved through a combination of the Carnot cycle and the verbal statements of the second law. 4. Use of the first three Propositions to introduce the concept of thermodynamic temperature and the concept of entropy. 5.Use of the latter in the fourth Proposition to develop the analytical statement of the second law, i.e. the desirable quantitative criterion for the establishment of the feasibility of a given process and for the efficient utilization of thermal energy.
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Application of this criterion requires the determination of entropy changes, which are considered next, followed by a series of Examples that demonstrate how these entropy changes are used in applications of the second law, developing thus our ability to use it. (fo provide for some understanding of the concept of entropy, however, a molecular interpretation of it is presented before the Examples). We return, then, to the discussion of steam power plants, a major application of the second law, to enhance our familiarity with using this law, and close with some Concluding Remarks.
3.4 Second Law: The Verbal Statement 3.4.1 Statennents We will use the following two  equivalent statements of the second law: Statement A: It is impossible to make a transfer of heat from a heat
bath, at a uniform temperature, and obtain an equivalent amount of work, without causing a change in the thermodynamic state of an other body (Denbigh, 1981, p.26). Statement B: Any process whose only effect is the transfer of heat from
one temperature level to a higher one, is impossible.
3.4.2 Example 3.1 Consider the following two observations: a. An ideal gas expands isothermally in a constant temperature bath absorbing an amount of heat Q and producing an amount of work W. From the first law, noting that the isothermal change of the internal energy of an ideal gas is equal to zero, we conclude that: W= Q
which indicates that all heat absorbed is converted into work. b. Aircondition units transfer heat from the colder room to the hotter atmosphere. Do you see any disagreement between these observations and the second law statements?
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a.No, because one of the thermodynamic properties of the ideal gas, its pressure, bas changed (Statement A). b.Again, no, for this is not the only effect in the process. Electricity is used (Statement B).
3.5 The Carnot Cycle In the annals of science Carnot's work is almost unique,for it had no discernible predecessors and was built up from the assemblage of unordered opinions and problems, concepts, theories and measurements that were available at the time. Carnot, in short, was not standing on the shoulders ofgiants; he saw further than his contemporaries because he had a much clearer vision (Cardwell, 1971, p.193). Indeed we cannot (explain Carnot's thought process), for he was one of the most truly original thinkers in the whole history of science. It might not be wholly inappropriate to say that in the independence ofhis spirit he was English, and in the rigor and clarity of his mind he was French,· but in the depth and scope ofhis insight he transcends all such classifications (Ibid, p.211).
We will start with a description of the cycle, develop an expression for its thermal efficiency, and then demonstrate that the cycle provides the maximum possible efficiency.
3.5.1 Description Assume that we have: * a heat source, large enough that removal of an amount of heat Q1 does not change its temperature T1; it is referred to as a heat reservoir. And, * a heat receiver at a temperature T2, lower than T1, large enough that addition of an amount of heat Q2 does not change its temperature; it is referred to as a heat sink. We consider now a hypothetical cycle whose net effect is to: * absorb an amount of heat Q1 from the heat reservoir; * produce an amount of work W; aqd * reject an amount of heat Q2 to the heat sink. To this purpose, the cycle uses one mole of an ideal gas, which undergoes the following four changes shown in Figure 3.1:
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The Second Law of Thermodynamics
p
3
v Figure 3.1 The PVbehavior of the gas in the Camot Cycle.
1.It expands isothermally (1+ 2), at T1' = T1  L1 T1, absorbing an amount of heat Q1 from the heat reservoir and producing an amount of work W12 • 2.It expands adiabatically (2+ 3) to temperature T2 ' = T2 + LlT2 , producing an amount of work W23 • 3.It contracts isothermally (3+ 4) at T2 ' rejecting an amount of heat Q2 to the heat sink and consuming an amount of work W34 . 4.lt is compressed adiabatically to its original state (4+ 1), consuming an amount of work W41 • All PV changes, expansions and compressions, are reversible.
3.5.2 The Expression for the Thermal Efficiency The thermal efficiency of the cycle is defined by:
n
=
net work produced heat absorbed
W =
Q1
(3.5.1)
According to the first law applied to the whole cycle, and considering the fact that the gas returns to its original state, i.e. L1 U = 0: Q = W, where: Q
= Ql
+ Q2
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Hence:
(3.5.2) To express Eq.3.5.2 in terms of temperatures we notice that for the two isothermal steps:
I
v: p v2 PdV = RT/ln~ = RT/ln~ v, v. pl I p4 = RT2 ln
Q2
and:
p3
while from the two adiabatic steps (Eq.2.6.5):
c;
where: k = yl
y =
y
c~
Thus: and Eq.3.5.2 becomes:
n
=
T2 1ln(P4 1P 3 ) 1+  :     T/ln(P2/ P 1 )
It is apparent that in order for the cycle to produce the maximum work .dT1 and .dT2 must approach zero, i.e. the heat transfer must be also reversible. Thus:
n
=
T2 1Tl
(3.5.3)
and the cycle efficiency is determined by the temperature of the heat reservoir and that of the heat sink only. This reversible cycle was introduced by Carnot in 1824, and was described  in terms of the volumetric behavior of the gas (Figure 3.1) by Clapeyron several years later.
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Bath at t 1
Bath at t 2 Figure 3.2 The two engines of Section 3.5.3.
3.5.3 The Maximum Thermal Efficiency We conclude, thus, that as the cycle approaches reversible operation, its thermal efficiency approaches its maximum value. We will demonstrate next that, of all heat engines operating between two specified temperatures, the reversible one has the maximum thermal efficiency. Consider the two heat engines shown in Figure 3.2. They operate between the two temperatures t 1 and t 2 , absorbing heat at t 1 and rejecting heat at t2 (t 1 > t2), and produce the same amount of work. Engine A is a reversible one; engine B is irreversible. We will demonstrate next that if B were more efficient than A, the second law would be violated. If B is more efficient, and both engines produce the same amount of work, then from the definition of thermal efficiency (Eq.3.5.1):
lqtl > IQtl
where the absolute values are used to avoid concern with signs. Also: lqzl = lqtl W, IQzl = IQtl W, hence lq2l > IQzl· Let us now reverse engine A and drive it using the work produced by B. Because A is reversible it will now absorb an amount of heat Iq2 1 and reject an amount of heat Iq1 l.
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We have, therefore, a process  the combination of the two cycles whereby the only effect is the transfer of an amount of heat q = I q2 1 I Q2 1 , from a lower to a higher temperature. This, of course, violates statement B of the second law and, consequently, the irreversible engine cannot be more efficient than the reversible one.
3.5.4 Comments l.It is apparent that: a. The efficiency of the Carnot Cycle is independent of the nature of the working fluid (the ideal gas was used for convenience; the same result is obtained using a real gas, Problem 8.56). b.For maximum efficiency, there should be no temperature difference between the working fluid and the: * heat source, and *heat sink. 2.Notice the similarity between the cycle's efficiency and that of a waterwheel, discussed in Example 2. 7. Indeed, Carnot was aware of this his father was a military hydraulic engineer  and it influenced his thinking in working with the efficiency of the steam engine. Carnot himself states: "In accordance with the principles we have now established, we can reasonably compare the motive power of heat with that of a head of water: for both of them there is a maximum which cannot be exceeded, whatever the type of hydraulic machine and whatever the type of heat engine employed. The motive power of a head of water depends upon its height and the quantity of water; the motive power of heat depends also on the quantity of caloric and on what may be called  on what we shall call  the height of its fall, that is on the temperature difference of the bodies between which the caloric flows." (Cardwell, p.196) 3.Notice that Carnot does not talk about conversion of caloric to work, for it was indestructible according to the caloric theory, but about flow of it. 4. Carnot did, however, recognize the difference between the two efficiencies: for the waterwheel it does not depend on the absolute values of the two heights, only their difference; for the thermal efficiency, it does depend on the absolute values of the two temperatures. The recognition of this difference reflects the genius qf Carnot. 5.The cycle is, of course, an imaginary one: it represents the limit of a real cycle as the driving forces become infinitesimally small. This is
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71
why we call it an 'ideal heat engine' but, as with the case of the ideal gas, this makes it no less useful. 6.Eq.3.5.3leads to two very practical conclusions with respect to the amount of work produced from a given amount of thermal energy in a cyclic process: a.it is always less than that of the thermal energy; and b.it depends on the temperature: *at which the thermal energy is available (the higher the better); and * of the heat sink (the lower the better). 7 .Eq.3.5.3 provides also the answer to the 'big' question posed in Section 3.2, as to 'where the energy is gone': use of the thermal energy contained by the atmosphere, or the oceans, would require a large heat sink at a lower temperature, which is not available. 8.The importance of the relationship between thermal energy and the temperature it is available at is demonstrated in Examples 3.2 and 3.3; and it is considered in detail in Chapter 5. 9.The maximum efficiency of the reversible heat engine is consistent with our observations in Chapter 1, where it was demonstrated that a reversible operation produces the maximum amount of work, such as in the expansion of a gas, when P(gas) is just larger than P(opposing); or, in raising weight B with weight A, when WA is just larger than W8 . lO.Some authors refer to Eq.3.5.3 as defining the efficiency of the Carnot cycle without, in other words, including the word 'thermal'. This, however, can be misleading for: *since efficiency is defined, Eq.1.12.1, as:
net work produced maximum producible work * and the Carnot cycle provides the maximum producible work, its efficiency is always equal to one. Its thermal efficiency, on the other hand, is always less than one. When, hence, we refer to the efficiency of a heat engine, we imply its thermal one. ll.ComparisonofEqs 3.5.2 and 3.5.3 yields: Ql
Tl
(3.5.4)
i.e. the ratio of heat absorbed to heat rejected depends on the ideal gas temperatures of the two baths only, and is equal to their ratio. We will use Eq.3.5.4 in Section 3.9 to demonstrate that the ideal gas temperature (T) represents also the thermodynamic temperature.
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To relate the somewhat 'abstract' Carnot cycle to the real world, we consider after the Examples 3.2 and 3.3 the very 'real' power plant cycle, used for the production of electricity from thermal energy.
3.5.5 Example 3.2 It bas been suggested that the temperature difference between the surface and deep water of the oceans can be used for the production of electricity. Assuming that the two temperatures are 25°C and 10°C respectively: a. calculate the maximum amount of electricity that can be produced per Joule of absorbed heat; b.explain why the real amount will be less.
a.From Eq.3.5.3:
l
w = Q,[ [ ~: = 0.050 J In other words, the maximum amount of produced work is only 5% of the absorbed beat. b. In obtaining this answer, we assumed: * complete reversibility  through the Camot cycle  in the compression and expansion of the working fluid; and in the heat transfer. * complete convertibility of mechanical power to electricity, which is theoretically feasible but  in reality  the efficiency is about 90%. In practice, thus, the work produced will be still less. In spite of this low thermal efficiencies, the available amounts of energy in this source are so large, that it was estimated in the middle 1970's that it could provide from 1 to 5% of the U.S. energy needs by the year 2000 (Seltzer, 1976). This, of course, was before the low oil prices of the 1980's severely curtailed work on alternative energy sources.
3.5.6 Example 3.3 One kg of water at 80°C is available. Calculate the maximum work that can be obtained from it. The atmosphere at 20°C will be the heat sink. Assume a constant heat capacity for water of C = 1 kcal/kg K.
If the maximum work is to be produced, a Camot cycle operating between the water and the atmosphere must be used: Notice, h~wever, that in this case the temperature of the water does not remain constant but it varies from the initial value, T1 = 353.15 K, to the fmal one, T2 = 293.15 K. Let us assume that at a given moment:
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73
*the temperature of the water is T; and * a small amount of heat dq is absorbed from it. The amount of work produced from this amount of heat will be: dW
T. T
T. T
= dq(I2) = (l)(CdT)(I2)
and:
Since the amount of heat absorbed in this case is 60 kcal, the overall thermal efficiency of the cycle is 5.41160 = 0.09. Note. The small amount of work produced per unit of heat absorbed in both Examples supports the characterization of thermal energy as a 'second class' form of energy.
3.6 A Real Heat Engine: A Simple Power Plant Most, by far, electric energy is produced by the combustion of fossil fuels. The resulting thermal energy is converted in part into mechanical energy which is then converted into electrical one while the remaining is discarded or, on occasion, used for process heating purposes. The process is carried out in what is referred to as a power plant cycle.
3.6.1 Description For small units, a 'simple' cycle is often used, consisting of the following steps (Figure 3.3): 1. water at high pressure is vaporized in a boiler and the steam is then superheated (1+ 2); 2.the superheated steam expands through a turbine, where part of its energy is converted into mechanical (2+ 3), which in turn is converted into electric; 3.the exiting steam enters a condenser where it is condensed (3+ 4), using cooling water; and 4.the condensate is pumped back into the boiler (4+ 1).
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3
4
Figure 3.3 A simple power plant cycle.
3.6.2 The Expression for the Thermal Efficiency From the definition, Eq.3.5.1, the efficiency of the cycle is given by: W+W (3.6.1) n= I p Ql where, W,: turbine work WP: pump work Q1: heat absorbed in the boiler/superheater.
3.6.3 Comments l.The turbine is usually adiabatic. Hence from the first law for a steadystate flow process: W, =  (H3  H 2). 2.In the same fashion: WP =  (h 1  h4), where the symbol h is used for the liquid phase and H for the vapor. 3.Again from the first law: Q1 = H 2  h 1. 4.We are not ready yet to carry out calculations involving power plants. We will do it after we discuss entropy change calculations. We proceed now with the development of the analytical statement of the second law.
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75
3. 7 Second Law: The Analytical Statement Once we understand the concept of a reversible heat engine, we are ready to develop the analytical statement of the second law. This is carried out
with the use of the ensuing four 'Propositions' discussed briefly here (for an in depth presentation see Denbigh p.26): I.If a heat engine performs a reversible cycle by: *absorbing an amount of heat Q1 at a temperature t 1, and * rejecting an amount of heat Q2 at a temperature t 2 , then the ratio (Q 11Qv is a function of t 1 and t2 only, i.e.:
IQll IQzl
=
f(tl ,tz)
where the absolute values are used to eliminate concern with signs. II. This ratio is given by:
IQll IQzl
/U1) f(t2)
where f (t) is the thermodynamic temperature, which can be set equal to the 'ideal gas' temperature, i.e.:
IQII IQzl
Tl
T2
III. The entropy S, defined by:
dS
=
dQr T
where the subscript r stands for reversible, is a state function. IV.For any natural (spontaneous) process, the total entropy change (.dS10r) equal to: * the entropy change of the system (L1Ssys), plus *that of the surroundings (L1Ssw.), is larger than zero. It is equal to zero, only when the process is reversible, and it can never be negative: L1Sror
= L1Ssys
+
L1Ssur ~ 0
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This expression represents the 'analytical' statement of the second law. We proceed, next, to discuss these four Propositions. [Undergraduate students, however, can move to Section 10 (Proposition III), and accept that: the ideal gas temperature represents also the thermodynamic temperature, called such because its establishment is based on the second law of thermodynamics.]
3.8 Proposition I Consider two reversible heat engines A and B operating between the same two baths with temperatures t 1 and t 2 (t1 > t 2 ). Cycle A absorbs an amount of heat q 1 at t 1 and rejects an amount of heat q2 at t 2 . Cycle B operates in the opposite direction: it absorbs an amount of heat Q2 at t 2 and rejects an amount of heat Q1 at t 1• We operate engine A n times and engine B N times so that n Iq2 l = Nl Q2 1, i.e. the total amount of heat rejected to the cold bath by engine A equals the amount of heat absorbed from the same bath by engine B. We will demonstrate next that, since engines A and B have returned to their initial states (n and N are integers), then: n Iq1l = Nl Q11. Assume first that n Iq 1 I > Nl Q1 1. The net result of the operation of the two cycles will be that an amount of heat equal to: (n Iq 1 I  Nl Q1 1) is absorbed from the bath at t 1• Now, according to the first law applied to the whole process, this heat has been converted completely into work, which is impossible according to statement A of the second law. If now n I q 1 I < Nl Q1 1 , we can reverse the two cycles  remember they are reversible  and end up with the first case. We conclude, thus, that n Iq1l = Nl Q11, which combined with the starting equation, n Iq2l = Nl Q21, yields:
lq1l lq2l
=
IQ11 IQ21
The ratio is, thus, the same for both cycles, independent of the operating pressures and volumes of the gas involved in each cycle. It must, therefore, be a function of the temperatures of the two baths, t 1 and t2 , only:
(3.8.1)
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The Second Law of Thermodynamics
3.9 Proposition II 3.9.1 Approach We will show first that the functionf(t 1h) of Proposition I is given by: f(tl) f(tl,t2) = f(t2) and proceed to introduce the concept of thermodynamic temperature, as contrasted to the ideal gas one. We will then demonstrate that the two temperatures can be set equal to each other.
Consider three reversible heat engines that operate among three different temperatures t 1, t 2 , t3 (t1 > t 2 > t3 ), as shown in Figure 3.4: * the first between t 1 and t 2 ; *the second between t 2 and t3 ; and
p A
F
v Figure 3.4 The three cycles used in the proof of Proposition II.
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Applied Chemical Engineering Thermodynamics
* the third between t 1 and t 3 •
As demonstrated in Proposition I, Eq. 3. 8.1:
IQtl
Cycle ABCD:
IQ2 l
Cycle DCEF:
IQ3 l
Cycle ABEF:
IQ3 l
IQ21
IQtl
=
f(tpt 2)
= f(t2 ,t3 )
=
f(tpt 3)
Elimination of IQ3 1 between the last two equations leads to:
IQtl
f
IQ2I
JU2>
I Q21 IQ31
J J
IQ31
JU3>
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79
3.9.3 The Thermodynamic Temperature We have arrived, thus, at a new temperature, established by the second law and not arbitrarily through, for example, the Celsius scale or even through an ideal gas. Rather, it is defined in terms of the ratio of the heats absorbed and rejected in a reversible heat engine:
IQ.I _ f(t•) IQ21  /Ct2)
(3.9.3)
We have also seen, Eq.3.5.4, that:
IQ.l
Tl
IQ21
T2
where Tis the ideal gas temperature. It follows then that this temperature is proportional to T: ft..t) = aT To avoid the existence of two scales, we set a = 1, and we call T the absolute thermodynamic temperature, or just the absolute temperature. Finally, since Q1 and Q2 have opposite signs, we recover from Eq. 3.9.3, Eq.3.5.4: (3.5.4)
3.10 Proposition III Let us now take a closer look at the ratio (QIT). We notice that no matter what the values of the heat absorbed Q; and rejected Qi are, when divided by the corresponding temperatures 1'; and Tj, they must satisfy Eq.3.5.4: Qi

1';
=
Q.
...!..
1j
There may be, thus, more to this ratio (QIT) than meets the eye. We define, therefore, the quantity S such that:
dS
=
dQr T
(3.10.1)
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Applied Chemical Engineering Thermodynamics
where the subscript r depicts the reversible character of the operation, and call it Entropy. (As we will see in Chapter 4 this term was introduced by Clausius in 1865.) We will demonstrate next that entropy is a state junction, i.e. that there is indeed more to the ratio (Q!T) than meets the eye. To this purpose consider a gas that undergoes a Carnot cycle by absorbing an amount of heat Q1 at T1 and rejecting an amount of heat Q2 at T2 • To calculate the entropy changes of the gas going through this cycle, we notice that they occur only in the heat absorption and heat rejection steps since  according to Eq.3.10.1  there is no entropy change in the two adiabatic ones:
and:
QJ T1
Q2 (3.10.2) T2 Thus  according to Eq.3.5.4 the total entropy change of the gas going through the whole cycle is zero: ..1S = ..1S1 + ..1S2 = 
fas
+
(3.10.3)
= ..1s = o
Eq.3.10.3 applies to any reversible cycle, i.e. other than those involving two isothermal and two adiabatic steps (for proof see, among others, Denbigh), and demonstrates that entropy is a state function. Consider, for example, a body going from state A to state B following path 'a', or path 'b', reversibly in both cases. Application of Eq.3.10.3 to the cycle: A to B through path 'a' and then back to A through path 'b' gives:
J: J:
dS(a) +
dS(a)
J~
=  J~
dS(b) = 0; dS(b)
=
J:
or, dS(b)
The entropy change of the body is thus independent of the path followed  provided it is a reversible one and, consequently, entropy is a state function.
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The Second Law of Thermodynamics
3.11 Proposition IV 3.11.1 Objective In discussing the concept of reversibility in Chapter 1 we noted that all real life  spontaneous  processes are irreversible. We will demonstrate next that when a system undergoes such a spontaneous (irreversible) process, the total entropy change  that of the system plus that of the surroundings is positive. This represents the formal (analytical) statement of the second law.
3.11.2 Entropy Change of a System Undergoing a Spontaneous Process Consider a system which undergoes a spontaneous (irreversible) change from state A to state B within an adiabatic enclosure. Since the process is irreversible (spontaneous), the entropy change (S8  SA) cannot be evaluated through Eq.3.10.1. We bring now the system back from state B to state A using a reversible engine. This, of course, cannot be accomplished adiabatically, for then the process A to B would be also reversible. So, let Q8A be the amount of heat involved in the process B to A, and WAB and W 8A, the amounts of work associated with the forward and backward directions of the process. From the first law applied to the whole cycle: QBA
= WAB
+ WBA
Clearly, QBA must be negative, for: * If it were zero, the whole process would be reversible; and * If it were positive  i.e. heat was absorbed by the system  it would correspond to the conversion of this heat completely into work in a cyclic process  since the system returns to its initial state  and it would violate statement A of the second law. We conclude, therefore, that returning the system reversibly from B to A involves the rejection of an amount of heat QBA· Let, now, T be the system temperature at which the heat was rejected. Then:
S
A
s
B
=
JA
B
dQr
T
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Applied Chemical Engineering Thermodynamics
In other words, S8 is larger than SA, which indicates that the original spontaneous process leads to an increase in the system entropy.
3.11.3 The Total Entropy Change If we are now to remove this entropy from the system i.e. if we want S8 to equal SA  then during the spontaneous process an appropriate amount of heat must be transferred to the surroundings. This, of course, will result in an increase of the entropy of the surroundings. On the other hand, it is also apparent from our discussion that when the process A to B is reversible, there will be no entropy change for the system or the surroundings. We conclude, therefore, that the total entropy change LiS101  that of the system (LiSsys) plus that of the surroundings (LiSsur)  must be positive for any natural (spontaneous) process:
LiS101
=
LiSsys + LiSsur
~
0
(3.11.1)
Eq. 3 .11.1 represents the formal statement of the second law, the equal sign referring to the case of a reversible process, and provides the desired criterion that establishes: *the direction and feasibility of a given process, and *the efficiency in the use of energy in this process.
3.12 Entropy Change Calculations Application of Eq.3.11.1 to a given process requires evaluation of the entropy changes involved, which is considered next.
3.12.1 The Problem and Approach Consider a system undergoing a spontaneous (irreversible) change from state 1 to state 2. How will the entropy change of the system be evaluated? Since entropy is a state function, the change: LiS = S2  S1 is independent of the path followed by the system in going from state 1 to state 2.
We devise, therefore, a hypothetical reversible path from state 1 to state
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83
2 and use it to evaluate the entropy change with Eq.3.10.1. We consider
next some applications.
3.12.2 lsothennal Removal or Addition of Heat This occurs in the case of a bath large enough that the addition or removal of an amount of heat Q does not effect its temperature T, such as a bath consisting of a system undergoing an isothermal phase change. Then from Eq.3.10.1: .tiS
=
I dQrT
=
Q
T
(3.12.1)
It takes, for example, 2256.9 kJ to vaporize one kg of saturated water at 373.15 K. Hence, according to Eq.3.12.1, the entropy of vaporization is 6.0485 kJ/kg K.
3.12.3 Heating or Cooling of a Body from T1 to T2 To this purpose, we combine Eq.3.10.1 with the definition of heat capacity, Eq.1.13.1, which yields on a molar basis: a. Constant volume (isochoric) process: dQr = dU = Cv dT, dQ c .tiS = JT2 _r = T2 ~ dT (3.12.2) Tl T Tl T
I
b.Constant pressure (isobaric) process: dQr = dU + PdV = dH = CpdT, dQ .tiS = J T2 _ r = J T2 _!_dT (3.12.3) Tl T Tl T
c
3.12.4 Entropy Changes of an Ideal Gas Consider one mole of an ideal gas that goes from state 1 (T1, V1, P 1) to state 2 (T2 , V2 , P 2). Since our hypothetical path is reversible, dW = PdV, and Eq.3.10.1 yields:
c·
dQ dS = _r = dU+PdV = ....!dT+~ dV T T T V
and:
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Applied Chemical Engineering Thermodynamics
(3.12.4) To obtain the final answer in terms of P 1 and P2 we introduce:
T2Pl V2 T1P2 Vl leading, after some algebra, to:
~s
an
J
=
* CpR
c*
P
T
p2
.!ar + Rln.!.
72 Tl
=
* Cv
d
(3.12.5)
Notice that while the enthalpy and internal energy changes of an ideal gas are functions of temperature only (Section 2.6.3), the entropy change depends on both, temperature and pressure (or volume), changes.
3.12.5 Entropy of Mixing Two pure ideal gases 1 and 2, both at the same temperature T and pressure P, are mixed isothermally with final compositiony 1 and y2 • What is the entropy of mixing per mole of mixture formed? We start by noticing that as a result of the mixing process the pressure of component 1 goes from P to its partial pressure p1 , i.e. its contribution to the mixture pressure; and the same applies to component 2. Thus, according to Eq.3.12.5: ..
(j
c
;g 0.29 w Q)
+'
c 0
0:: 0.28 Steam Temperature: 475°C Condenser Pressure: 1.0 bar
1...
Q)
3: 0
0...
0.27
0. 2 6
''JI....J...........'L..J.....L...L..L''JI....J.'''L..J.....L...L..L''JI....J.''''.J.....L...J....C_~I....J.~
25
30
35
40
45
50
55
Boiler Pressure (bar)
60
Figure 3. 7 Effect of boiler pressure on power plant efficiency.
65
70
The Second Law of Thermodynamics
99
predicted by the Carnot cycle, higher boiler operating pressures, and thus vaporization temperatures, lead to increased efficiencies. Since the degree of superheat effects also the average temperature at which heat is absorbed, in the results presented in Figure 3.8 we keep the boiler pressure constant at 35 bar, and vary the temperature of the superheated steam. The condenser operates as before at 1 bar. Again, as predicted by the Carnot cycle, higher operating temperatures lead to increased efficiencies. Turning to the heat rejection step, we notice that the temperature at which the condenser operates is determined, as with case of the boiler, by the operating pressure. The effect on the cycle efficiency is presented in Figure 3.9. Here we keep the boiler pressure and steam temperature constant (35 bar and 475°C). Again as predicted by the Carnot cycle, low operating pressures and, thus, temperatures, lead to higher efficiencies. ~Calculations of efficiencies are based on a reversible and adiabatic turbine.
3.15.2 Comments l.We conclude that, as predicted by the Carnot cycle, for increased thermal efficiency: a.high boiler pressure and degree of superheat, and b.low condenser pressure, are desirable. 2.Modern power plants, thus, operate at high pressures and temperatures, within the limitations imposed by the materials of construction, but rarely much above 100 bar and 600°C (Smith and Van Ness, 1987). 3.They also operate at low condenser pressures, determined by the temperature of the cooling water, with typical values of around 0.1 bar. 4.More complex cycles are also used involving multistage turbines with part of the exhaust from each stage steam used to increase the feed water temperature (why?). See, for example, Problem 3.57. 5.Under such conditions, overall thermal efficiencies of about 0.45 are realized. 6.A new approach, actually an old one that is revived, provides improved utilization of the fuel heat by combining the generation of mechanical power with the production of process steam. This is referred to as Cogeneration and is discussed in detail in Chapter 5, where the efficient utilization of thermal energy is discussed in detail. A simple case, is demonstrated in the next Example.
100
~
~ '()
Applied Chemical Engineering Thermodynamics
Boiler Pressure: 35 bor Condenser Pressure: 1 bor 0.275
:::: l.J
....c: 0 a: ~ 31:
0.260
0
a..
0.2 45 .__,__,__,__L.....JL.....JL.....J'J'J'lII1LLLLLLL.......LL.......L.......Ll 300 350 400 450 500 550
Steam Temperature (°C)
Figure 3.8 Effect of steam temperature on power plant efficiency.
0.34
>
0
~ 0.32
Boiler Pressure: 35 bor Steam Temperature: 475"C
·u
:::l.J
.... 0.30 c 0
a:
~
31: 0 0..
0.28
0.26
0.2
0.4
0.6
0.8
1.0
1.2
1.4
Exhaust Pressure (bar)
1.6
1.8
Figure 3.9 Effect of condenser pressure on power plant efficiency.
2.0
The Second Law of Thermodynamics
101
3.15.3 Example 3.13. A Look at Example 1.9 Let us consider again the debate among George, Monica, and David and see if we can enter into it. We start with David's argument: "A Joule is a Joule". Even modem plants will give no more that 0.45 Joule of electricity per Joule of fuel heat . Here "A Joule is not always a Joule". Monica's position is more valid. Her boiler should produce in theory at least one Joule of steam enthalpy per Joule of fuel heat. About 0.9 in practice. George's position is much better, however. Using the same ratio of 0.9 Joule of steam enthalpy per Joule of fuel heat, as in Monica's case, he will be getting about 0.9 Joule of electricity per extra over Monica's Joule of fuel heat. This is about twice what modem power plants get. But George read about this approach somewhere. It is called Cogeneration. To better understand George's argument, and the concept of Cogeneration, let us assume that the cycle he proposes operates as follows: l.Steam at 100 bar and 500°C leaves the boiler/superheater; 2. The turbine is adiabatic, with an efficiency of 0.9 and an exhaust pressure of 3 bar, i.e. the desired pressure for the process steam; and 3.0ne GJ/hr of process heat is required. Following Figure 3.3, with the condenser replaced by the distillation boilers where the exhaust steam will be used, we arrive at the following operating conditions: *Point 2: P 2 = 100 bar; t 2 = 500°C; H2 = 3374.6 kJ/kg * Point 3: P3 = 3 bar; H3(RA) = 2570 kJ/kg. Hence: H 3 = 3374.6 0.9(3374.6 2570) = 2650.5 kJ/kg; t 3 = 133.5°C * Point 4: Assume, for simplicity, that only the heat of condensation of the process steam is used. Hence: P 4 = 3 bar; t4 = 133.SOC; h4 = 561.4 kJ/kg *Point 1: WP = 1.073·10·3 (P1  P 4) = 10.4 kJ/kg; Hence: hl = 571.8 kJ/kg Summarizing the cycle: Ql = 3374.6 571.8 = 2802.8 kJ/kg Q2 = 2650.5 561.4 = 2089.1 kJ/kg (process heat) ~ = 3374.6 2650.5 = 724.1 kJ/kg n = (724.110.4)/2802.8 = 0.255 To produce, therefore, the required 1 GJ/hr of process steam, the amount of fuel heat  using a boiler/superheater efficiency of 0. 9  is:
Q = _1 2802.8 = 1. 49 GJ/hr
0.9 2089.1 The amount of mechanical energy produced is: W = 1.49x0.9x0.255 = 0.342 GJ/hr Hence, George's approach uses (1.491/0.9) = 0.38 GJ/hr more fuel heat than Monica's approach. But 90% of it is converted into mechanical energy. This is the advantage of Cogeneration.
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Note. The water used in power plant cycles must be very pure to avoid corrosion of the turbine blades. In practice, therefore, the exhaust steam is used to generate the process steam through a heat exchanger, not directly (see Chapter 5).
3.16 Concluding Remarks 1. The feasibility of a process should be tested against both, the first and the second law. To this purpose the latter introduces the state function entropy, defined by Eq.3.10.1, and requires that Eq.3.1l.l be satisfied. Note that the entropy change of both, system and surroundings, should be considered. 2. The second law establishes also the criterion for the efficient utilization of energy in a given process: the less total entropy system plus surroundings it generates, the more efficient it is. (This subject is considered in more detail in Chapter 5.) 3.Conversionofthermal energy our main source of energy into mechanical takes place through cyclic processes. The thermal efficiency (Eq.3.5.1) in these processes depends on the temperature: * at which the thermal energy is available, and * of the heat sink, where the thermal energy not converted into mechanical is rejected to. 4. The maximum thermal efficiency is established by the Carnot cycle, Eq.3.5.3, which indicates that complete conversion of thermal energy into mechanical is not possible. 5.Conversionofthermal energy into mechanical is carried out in power plants with maximum efficiencies of about 45%. 6.1mproved efficiency in the utilization of thermal energy for the production of mechanical one can be accomplished by using Cogeneration.
References Cardwell, D.S.L., 1971. From Watt to Clausius, the Rise of Thermodynamics in the Early Industrial Age, Cornell University Press, Ithaca, N.Y. Denbigh, K., 1981. The Principles of Chemical Equilibrium, Cambridge University Press, Cambridge.
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Seltzer, R.I., Feb. 9,1976. Chemical & Engineering News, 19. Smith, J.M., Van Ness, H.C., 1987. Imroduction to Chemical Engineering Thermodynamics, McGrawHill, New York. Van Ness, H.C., 1969. Understanding Thermodynamics, McGrawHill, Paperbacks, New York.
Problems 3.1 With reference to Example 3.1: a. how much work is produced per mole, if the expansion takes place at 350 K, from 3 atm to 1 atm? b.how much work is required to bring the gas back to 1 atm at 350 K? c. would Q = W be valid, if the gas were not ideal? Explain. Assume reversible operation. 3.2 Discuss the similarities and dissimilarities between the efficiencies of the Carnot cycle and a waterwheel (see Example 2. 7). 3.3 Develop the expression for the efficiency of a Camot cycle. 3.4 A Camot cycle operates between the constant temperatures of 100°C and 50°C. For each Joule of work produced how much heat is rejected? 3.5 With reference to Example 3.3, what is the minimum amount of work required to heat the 1 kg of water from 20°C back to 80°C, using the atmosphere as a heat source? 3.6 What is the maximum amount of work that can be produced from 1 kg of saturated steam at 1 atm? The atmospheric temperature is 25°C. 3. 7 Discuss the practical usefulness of the Camot cycle. 3.8 Assume that in Section 3.13 we used two crystals containing six atoms each. a.How many arrangements are possible in each state? b. Which one of them will the system be in? 3.9.a. What conclusion do you reach about entropy when considering the molecular nature of matter? b.Compare it to the classical thermodynamics view of entropy. 3.10 In Example 3.5, assume that the flow rate of steam is such that the condensate leaves at 60°C. a. Will the entropy change be the same as it was with the case of saturated liquid? Explain your answer. b.lf the answer is no, what is the entropy change? 3.11 One mole of gas A is heated from 50°C to 100°C using gas B, whose temperature changes from 150°C to l20°C. a.Calculate the entropy changes: gas A, gas B, and total, for cocurrent operation.
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Applied Chemical Engineering Thermodynamics
b. Would the values be the same, if the operation were countercurrent? Assume ideal gas behavior with: Cp* = 7 cal/mol K for both gases. 3.12 One hundred kg of water are heated from 5°C to 90°C at P = 1 atm using the condensation of saturated steam at 1 bar. a. Calculate the entropy change of the water and of the steam. b.Does the process satisfy the second law? c.If 15 kg of the steam are used for the same purpose, what will be the entropy change of the heating medium? d. Which of the two heating processes is less irreversible? 3.13 One kg of saturated steam at 1 bar, and one at 1 bar and 200°C, are mixed adiabatically. a. What is the entropy change of each? b. What is the entropy of the mixture? c.Is the process reversible? 3.14 One mole of an ideal gas is compressed from [T1, Vd to [T2 , V2 ]. a. Starting with Eq.3.10.I, develop an expression for the entropy change of the gas. b. What is the total entropy change? 3.15 One mole of methane at 300 K and 1 bar is mixed with two moles of ethane at 400 K and 2 bar. What is the entropy change? 3.16 Superheated steam at 20 bar and 550°C enters a turbine and leaves at O.I5 bar. a. If the turbine is reversible and adiabatic, what is the temperature of the exiting steam? b.If there is a heat loss of 7 kJ/kg of steam, and the turbine has an efficiency of90% (as compared to the reversible and adiabatic operation), what is the enthalpy of the exiting steam? 3.17 One mole of propane is compressed in a flow process from I bar and 60°C to 10 bar. The compressor is adiabatic and reversible. a. What is the entropy change of the gas? b. What is the temperature of the exiting gas? c.How much work is required? Assume that in this range propane behaves approximately as an ideal gas. 3.18 With reference to the previous Problem, what would the temperature of the compressed gas be, if there were a loss of0.2J/mol and the compressor efficiency is 0.8 (based on the reversible and adiabatic operation)? 3.19 Starting with: ..:18101 > 0, show that it is impossible to develop a process whose only effect is the transfer of one Joule of heat from 273 K to 373 K. 3.20 Ethane expands from 10 bar and 100°C to I bar and 20°C. a. What is the entropy change of the gas? b. Can you calculate the work produced? 3.21 Show whether the following process is possible: Air at 7 bar and 40°C enters an apparatus that can exchange heat with the surroundings but no mechanical work. Half of the air leaves at 70°C and the other half at 20°C, and
The Second Law of Thermodynamics
105
both streams have the same composition with the inlet stream. Assume ideal gas behavior with Cp• = 7 cal/mol K. 3.22 Assuming steam to be an ideal gas find the entropy change per kg when the pressure and temperature change from 1 bar and l25°C to 10 bar and 450°C. What is the error resulting from the ideal gas assumption? 3.23 Steam in steady flow is expanded reversibly and adiabatically as it flows through a tube of variable crosssection. The steam enters the tube at 18 bar and 375°C with a negligible velocity, and leaves at 5 bar. What is the velocity of the steam leaving the tube? 3.24 An inventor claims that his apparatus, which exchanges no energy with the surroundings, can separate an equimolar mixture of nitrogen and oxygen at 2 bar and 300 K into practically pure products at the same temperature and a pressure of 1 bar. Is his claim valid? 3.25 Ten Ibm of ice (Cp=0.5 Btu/Ibm R) at 10°F are placed in an insulated tank containing 100 Ibm of water at 90°F. Calculate the resulting entropy change when the system reaches equilibrium. Heat of fusion of ice: 144 Btu/Ibm. 3.26 Which operating pressure in a power plant condenser will lead to higher cycle efficiency: a.O.l bar; b.1 bar. The boiler/superheater operation is the same in both cases. Justify your answer. 3.27 A simple power plant operates as follows: l.stearn at 600°C and 50 bar leaves the superheater; 2. the turbine is adiabatic but irreversible with an efficiency of 0. 90; 3.saturated water leaves the condenser which operates at 0.1 bar. Calculate: a. the cycle efficiency, and b.the condenser duty. 3.28.a.Describe the operation of the power plant cycle of Problem 3.27 on a TS diagram by plotting the temperature of the water as a function of its entropy as it goes through the cycle. b. Show on this diagram the effect of irreversibility in the turbine operation. 3.29 Process heat at a rate of 500 MJ lhr, as saturated steam at 5 bar, is needed. a. How much mechanical energy can be produced, if a cogeneration scheme is used? b. What is the thermal efficiency of the cycle, based on the heat used for the production of the mechanical energy only? c. If a conventional cycle were used to produce this mechanical energy, with a turbine exhaust pressure of 0.4 bar, what would its efficiency be? Use any of the information given in Example 3.13 that you find appropriate. 3.30 Show that entropy is a state function. 3.31 Discuss the importance of the second law in the Chemical Industry. 3.32 Explain why in the HilschRanque tube the air enters tangentially and close to the orifice of the tube.
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Applied Chemical Engineering Thermodynamics
3.33 What is the major source of irreversibility in a steam power plant using a fossil fuel as source of thermal energy?
3.34 An inventor claims that he can avoid the condenser in a power plant cycle and the resulting loss of energy there  by compressing the exhaust from the turbine vapor directly into the boiler. Identify the problems you see with this approach. 3.35 With reference to this Chapter: a. What are the most important items that you have learned? b.Give an example in each case.
3.51 Show, using the verbal statement of the second law, that if two reversible heat engines operate between the same temperatures, they have equal efficiencies.
3.52.a.What is the difference between the thermodynamic and the ideal gas temperature? b.Do they have to be the same? 3.53 Two moles of a given liquid are available. One is at 400 K and the other at 300 K. The heat capacity is 20 cal/mol K and is temperature independent. a. What is the final common temperature of the system that yields the maximum amount of work that can be obtained from it? b. What is this work? c. What is the common temperature that leads to the minimum work produced? 3.54 Using the mathematical statement ofthe second law, prove the verbal ones. 3.55 Discuss the purpose of the four Propositions. 3.56 Show that for a real process the total entropy change is positive. 3.57 Steam at 30 bar and 600°C is expanded in the first stage of a turbine to 3 bar. At this point a portion of the steam is extracted for feedwater heating and the rest is expanded in the second stage of the turbine to 0.1 bar, the condenser pressure. The saturated liquid condensate from the condenser is pumped into the feedwater heater where it mixes with the extracted steam, the mixture leaving as saturated water at 3 bar. a. Find the amount of steam extracted per kg of steam generated. b. What advantage do you see in using the extracted steam in the feedwater heater? Assume reversible and adiabatic expansion in the turbine and adiabatic operation of the feedwater heater. 3.58 With reference to the previous problem, calculate the thermal efficiency of the cycle for: a. the proposed operation; and b.one where all the steam expands through the second turbine. Explain the difference.
4 Thermodynamics: A Historical Perspective
4.1 Introduction In the last two Chapters we concluded that the first and second laws represent the basis for the development of the framework of equations necessary for the: a. calculation of energy balances; b.determination of the feasibility of processes; c.establishment of guidelines for the efficient utilization of energy. Furthermore, as we will see in later Chapters, these two laws provide also the framework of equations used for the solution of two other important chemical engineering problems: a.the evaluation of thermodynamic properties: enthalpy, entropy, etc., from the measurable properties: pressure, volume, temperature, and heat capacities; and b. the calculation of chemical and/or phase equilibrium. Even though these two laws were developed over a somewhat short period of time, about 50 years, their evolution scans a period of several centuries. The real impetus for the birth of thermodynamics, and especially the first and second laws, was the Steam Engine. Actually, it was exactly the conversion of thermal energy into work that gave the name to this new field of science and technology. (From the Greek: therme = heat; dynamis = force = work  for in that era, force and work were not differentiated.) This unlimited  as it seemed then  source of energy replaced the limited human and animal ones, and provided literally the motive power for the establishment of the Industrial Revolution.
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The development of the steam engine, however, while accelerated by the urgent needs of the Industrial Revolution for power, was a natural consequence of early very crude by today's standards, of course experiments that demonstrated the power of heat. From the theoretical point of view, the socalled Science of Heat was in such an embryonic state that contributed rather little to the development of the steam engine. (Actually, some important contributions in this field took place during  and to some degree, as a result of  the work on the steam engine.) It does represent, however, the theoretical precursor of thermodynamics.
4.2 Objective and Approach Our objective, as we take this trip in the past, from the very far to the more recent, is to: 1. Trace the development of the two laws enhancing, in the process, our understanding of them and of the quantities associated with them: heat, work, internal energy, and entropy; and of the concept of reversibility. 2.Appreciate the broader importance of thermodynamics, by recognizing the role it played in the establishment of the Industrial Age, in which we still live today the suggestion that we are already in the 'PostIndustrial' Age not withstanding. Our underlying objective, however, is a more fundamental one: To provide a sense of the history ofthermodynamics,for afield of Science and Technology  just like a Society  cannot function and grow without it. We start with a brief discussion of the early recognition of the power of heat, followed by a historical account of the development of the steam engine and of the evolution of the science of heat through the 17th century. We proceed, then, with the establishment of the first and second laws through the contributions of Carnot, Clapeyron, Joule, Kelvin, and Clausius, relying on the excellent bookofD.S.L. Cardwell (1911)From Watt to Clausius, the Rise of Thermodynamics in the Early Industrial Age. We close with a brief account of the developments associated with the evaluation of the thermodynamic properties of matter, a field that was founded with the pioneering work of J. W. Gibbs.
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4.3 The Power of Heat: Early Recognition Heat is, in fact, the grand movingagent of nature. Inertia, together with central forces, accountfor the local motion ofplanets and satellites; on the earth heat is the source of motion; without it the accidental factors of air resistance and friction would soon destroy all motion. There would be no rain, no winds, no variations in climate, no crops and indeed no life at all. (Cardwell, p.187)
Fire has been of paramount importance to humans from the beginning of their existence, its power so overwhelming that the early people worshipped it. Harnessing this power has been the aspiration of philosophers, dreamers, scientists, and engineers, through the years. Even today, the use of the 'nuclear fire' of the sun represents the biggest, perhaps, dream of the human race. Table 4.1 presents a brief list of events that demonstrate the recognition of the fire  and the resulting heat  power, going back to the preChrist period. For more details on the fascinating devices associated with these events the interested reader should see: The Great Engineers by J.B. Hart (1928).
The First James Bond was a Thermodynamicist!!!
The Marquis of Worcester was a man ofmany accomplishments. In addition to a bent for scientific investigation, he was also a man of affairs. He had a somewhat chequered career. Becoming involved in the Civil War between Charles I and the Parliament, he took up the cause of his King with a vigour which, unfortunately for himself, lost him his fortune and brought him to imprisonment in Ireland. He succeeded, however, in escaping to France, and there joined the Court of the exiled Charles. Returning to England in the role of secret agent to the Royal Family, he was detected and thrown into the Tower of London. Here he had the time and leisure to exercise his bent for scientific reflection. The story is told of him, ... that whilst in the Tower his attention was attracted to the continual lifting of the lid of a saucepan in which his dinner was cooking. His reflections led him along the path of invention with which his name has since become associated. (Hart)
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Table 4.1 Major Early Events in the Recognition of the Power of Heat 1st Century B.C.
Heron designs (builds?) the Aeolopilel, a toy operating with steam.
11th Century
Gunpowder is probably used in China and India in cannons.
15th Century
Leonardo da Vinci proposes the Architronito a steam using cannon.
16th Century
Blanco de Garay uses steam power to move a ship out of port(?).
1601
Giovanni Battista della Porta uses steam power to transfer water out of a vessel.
1629
Giovanni Branca describes a steampowered machine that can be considered the forerunner of the steam turbine.
16281647
Marquis of Worcester erects (?) two machines that use steam power to raise water.
16781680
Jean Hautefeuille and Christiaan Huygens design cylinderandpiston arrangements operating with gunpowder, which can be considered the forerunner of the internal combustion engine.
I
I
I,
4.4 The Steam Engine On Steam Power and Good Horses(!)
Water being evaporated by fire, the vapours require a greater space (about two thousand times) than that occupied by the water, and rather than submit to imprisonment, it will burst a piece of ordnance. But, being controlled according to the laws of Statics and, by science, reduced to the measure of weight and balance, it bears its burden peaceably (like good horses), and thus may be of great use to mankind, especially for the raising of water. { Sir Samuel Morland, first Master Mechanic of the Royal Household, 1682 (Hart, p.66)}.
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Following the events outlined in Table 4.1 we arrive to the last part of the 17th century. In England , it is the dawn of the Industrial Revolution that was destined to transform the world. Industry, however, needed power. The flooding of mines, for example, becomes a serious problem as the available horses are not sufficient to remove the water. Attention to the whole problem of steam and its practical use is becoming more rigorous, and King Charles II creates the post of 'Master Mechanic' to the Royal Household. It is in this atmosphere that the development of the steam engine, outlined iri Table 4.2, took place. The Savery engine represents the first, probably, commercial steam engine to come to existence using high pressure steam to remove the water from the mines, thus its name. It had a rather limited success however. "Exactly how many of these engines were made is unknown; whether any of them worked satisfactorily is uncertain." (Cardwell, p.13) A major improvement came soon by Thomas New co men's steam engine. He introduced in 1712 a pistonandcylinder arrangement operating by the weight of the atmosphere, thus avoiding the problems with high pressure steam. "The Newcomen engine became, therefore, the most important technological innovation of the time. Such engines were set up all over England and Scotland, as well as abroad: Sweden, Schemnitz (Slovakia)  which was at the center of one of the leading mining areas of Europe France, Spain, Russia and the Americas ... And the systematic work of John Smeaton in England led to doubling efficiency of the Newcomen engine from 4. 7 to about 9.0 million ftlbr per bushel of coal." (Cardwell, p.32) Notice the way efficiency was measured at that time!
Table 4.2 Major Events in the Development of the Steam Engine 1698
Savery presents a model of his steam engine (the Miner's Friend) to the Royal Society of England.
1712
Newcomen presents a pistonandcylinder steam engine.
1769
Watt patents a new type of steam engine that employs a separate condenser.
1774
The firm of 'Boulton and Watt' presents the first really successful singleacting steam engine.
1781
Watt receives a patent for the doubleacting steam engine.
Early 1800's
Trevithick introduces the high pressure steam engine.
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4.4.1 James Watt The consequences of James Watt's doubleacting engine were far reaching. Reasonably efficient in construction, commercially successful in practice, it created a new industrial situation. By linking up the crankshaft of the engine with suitable shafting in the workshop, the largescale use of machinery in industry became possible. The whole outlook of industrial life was changed, and parallel inventions and improvements in the machinery of the textile and other industries were rapidly effected. Thus was brought about what is known as the Industrial Revolution. A new era for industry was at hand. (Hart)
The next major development comes from James Watt (17361819) who was working at Glasgow University, a place that was becoming a center for the study of heat and the technology of power. Watt was an instrument maker and a scientist, not an engineer. His first experience with a Newcomen engine was in the winter of 1763 while repairing a small replica of it, used in a natural philosophy class. He soon became aware of the tremendous loss of steam involved in heating the cylinder after the condensation step  which was more pronounced in the small engine due to 'scaling' effect. This was one of the major problems of the Savery engine which, of course, remained with the Newcomen one as well. Watt spent the following two years trying to resolve this problem. (For a detailed account of this effort during which Watt stumbled into the phenomenon of latent heat and also determined the vapor pressure of water with remarkable accuracy the reader is referred to Cardwell.) His elaborate analysis of the problem led him to the following dilemma: "For maximum economy the cylinder must be kept hot all the time, but for maximum power it must be cooled down once every cycle. These desiderata are quite incompatible, for a cylinder cannot be both hot and cold at one and the same time. Or rather they can be compatible if we have two cylinders in place of one; the first to be kept hot all the time, the second cold." (Cardwell) The introduction of the separate condenser was a major breakthrough in the development of the steam engine. Watt obtained his patent in 1769 and turned to the construction of the first commercial engine, not an easy task. Hart (p.84) describes the commercialization of the Watt engine : "In order to produce an engine model on these lines, Watt hired a room in an old pottery, and here, through the good offices of Dr. Black, he became acquainted with Dr .Roebuck, who had just started the famous
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Carron Iron Works .... The expenses to which Watt was put as a result of his experiments considerably impoverished him, but now Dr. Roebuck agreed to finance him, and the two went into partnership. The first engine made by Watt in 1769 was not a great success. The condenser was of the surface type and was not sufficiently water and steamtight, and the piston leaked very seriously. Watt became very despondent, and in his anxiety to relieve Dr. Roebuck of the financial losses that threatened he looked around for other help. This luckily came to hand in the person of Matthew Boulton, the son of a Birmingham engineer, whom Watt had met during a journey to London for the purpose of securing his patent. Thus was set up the famous partnership of Boulton and Watt that so greatly enriched the profession of engineering." Nevertheless, it was not until 1774 that the first really successful singleacting engine was effected. The orders started to come in to the firm and Watt, with the financial anxiety removed, continued to improve the engine, receiving in 1781 a patent for a doubleacting steam engine. Steam Engine and the Urbanization of Society ... the textile revolution ... demanded that every inch offall on every reasonably big river or stream should be harnessedfor production; and so it came about that textile mills were spread out, like beads on a thread, along the rivers of the textile counties. During this phase the industrial revolution was essentially rural in its location; concentration into great industrial cities and conurbations came later, as a consequence of the wider use of the steam engine. (Cardwell, p.68)
4.4.2 Richard Trevithick and the Railway Locomotive The final major improvement of the steam engine is due to Richard Trevithick who pioneered the use of highpressure steam, much to the elderly Watt's disapproval. Cardwell comments: "The advantages of highpressure steam were obvious enough: the higher the pressure the smaller the cylinder needed to generate the same power. Moreover, the higher the pressure the less necessary the condenser became; to do without one and exhaust the steam directly into the air meant, for an engine working at an excess pressure of five atmospheres, losing only onesixth or about 17 per cent of the total available pressure difference, while gaining enormously in simplicity, cheapness and mobility. Without the cumbersome condenser there was no real obstacle to put
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ting a small and powerful highpressure engine on wheels. Hence the railway locomotive was born from the supersession of Watt's lowpressure condensing engine by Trevithick's highpressure engine. Further and incidental advantages were that the highpressure engine, being small, might be expected to lose less heat by convection, conduction and radiation than the big lowpressure ones; and having fewer moving parts for example no airpump was required  it would lose less power through friction. It was therefore reasonable to expect that it would be markedly more efficient than the lowpressure engine." (Cardwell, p.84)
4.5 The Science of Heat While the study of the steam engine provided the main impetus for the development of thermodynamics, its theoretical background even though a very limited one  was provided by the Science of Heat. We present therefore briefly, the development of this field through the 18th century, following the major events outlined in Table 4.3. Cardwell suggests that Fire, Air, Water, and Earth were the four Elements of the Empedoclean (5th century B.C.) classification. Other sources, however, indicate that 'Aether', and not Fire, was one of the Empedoclean Elements. Anyway, this is as much of a scientific description of
Table 4.3 Major Events in the Development of the Science of Heat 5th Century B.C. Empedocles suggests that Fire, Air, Water, and Earth are the four Elements of the World. 17th Century
The Thermometer is introduced; The first theories of heat are suggested by Galileo and Descartes.
1727
Boerhaave proposes the axiom that 'Fire is Conserved'.
17601799
Black replaces Fire with Heat in Boerhaave' s axiom; he also introduces the concepts of 'Specific' and of 'Latent' heat.
1779
Cleghorn presents the postulates of the Caloric theory of heat.
1798
Count Rumford shakes the foundation of the Caloric theory.
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fire as the ancient Greek aphorism 'Fire, Women, and Sea'(!). It represents however, along with the conjugated qualities of 'hot' and 'cold', the extent of knowledge on the subject of heat up to the 16th century.
4.5.1 The Thermometer The first important development towards the establishment of the science of heat was the invention of the thermometer in the beginning of the 17th century by Galileo. A friend of his recorded the following in 1638 (Hatsopoulos and Keenan, 1965): "Galileo took a glass vessel about the size of a hen's egg, fitted it to a tube the width of a straw and about two spans long; he heated the glass bulb in his hands and turned the glass upside down so that the tube dipped in the water contained in another vessel; as soon as the ball cooled down, the water rose in the tube to the height of a span above the level in the vessel; this instrument he used to investigate the degrees of heat and cold". (p.xv) The first liquid thermometer was built in 1631 by Jean Rey a French physician and was used to take the temperature of patients. The stem was, however, open and the water evaporation effected the accuracy of reading. The first thermometer with a scaled stem was built in 1641 by the Grand Duke Ferdinard II of Tuscany, using alcohol instead of water. The calibration of the thermometer involving the fixing of two points, the lower and upper and the divisions in the between, took quite a while. Fixed points involving cold winter and hot summerdays; the melting of butter; and the somewhat constant temperature of cellars; among others, were proposed. Finally, in 1694 Carlo Renaldini of Padua proposed the freezing and the boiling water as fixed points. When it was realized that atmospheric pressure had an effect on these points, pressure of one atmosphere was specified. (The wellknown scales of Fahrenheit and Celsius were not established until later, in 1709 and 1750 respectively.) The introduction of the thermometer early in the 17th century, even in its crude forms, helped focus attention into two very important, but also very hazy at that time, concepts: 'Heat' and 'Hotness'. Cardwell, for example, comments (p.2): "Unfortunately there was for a long time doubt about what it (the thermometer) actually measured. As a clinical instrument it certainly reduced the subjective element in the diagnosis of fevers, but beyond this the exact significance of its reading was uncertain and interpretation of them depended on the observer's philosophy of nature."
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And Hatsopoulos and Keenan write on the same subject: "Inventors of the thermometer were trying to devise an instrument which would measure a quantity as yet undefined. The physiological sensations of hotness and coldness were involved, but they were inadequate. The thermometer would indicate the same temperature for a block of wood and a block of copper exposed to the same 'severe winter cold' even though these two blocks seemed to be of quite different coldness by the sense of touch. An equilibrium concept was doubtless involved  namely, that all bodies exposed to the same cold atmosphere would ultimately attain a uniform degree of coldness, despite contrary evidence of the senses." (This remark should remind us of the arguments in support of the Zeroth law.) "A second concept played its part, namely, the idea that temperature was a driving potential which caused some influence to pass from one body to another of unequal hotness or coldness. Thus, when a body was taken from a warm room to the cold outdoors, its length began immediately to change. The difference in coldness between the body and its environment caused the environment to affect the body." (p.xv)
4.5.2
The Nature of Heat
Two, thus, important questions had to be answered: 1. What was responsible for the sense of 'hotness' and 'coldness'? 2. What was responsible for the observable effects? What, for example, made a copper wire increase in length when heated? Two main schools of thought, that of Galileo and that of Descartes, were advanced for this purpose providing a very hazy answer however. Cardwell writes: "Galileo argued, in 'II saggiatore' and again in the 'Discorsi', that the sensation of 'heat' is caused by the very rapid motion of certain specific atoms. 'Heat' is thus an illusion of the senses, a product of the mental alchemy which transmutes the eternal scurrying of inert atoms of the 'real' world into our sensual world of colour and warmth, sweetness and bitterness, comfort and discomfort. There is therefore no such thing as 'heat' if there is no one present to experience it: in exactly the same way there is no such thing as a real pain at the tip of a dentist's drill, or a real tickle at the end of a feather. (The examples are from the writings of Sir Arthur Eddington and Galileo respectively.) Yet the sensation we call 'heat' is correlated with definite physical changes: bodies do expand as they seem to get 'hotter' and if they are 'hot' enough they change state, melting or vaporizing as the case may be.
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Clearly, while we cannot talk about a quantity of 'hotness' (how can one quantify a subjective sensation?), we must try to account for expansion and change of state. This the Galilean atomic model can do, at least qualitatively, and if we have a measuring instrument, or thermometer, we can talk reasonably about a 'quantity of heat'. For the expression now indicates the total quantity of motion of all the specific atoms; the assumed cause of such (objective) phenomena as expansion and change of state. The views of Rene Descartes (15961650)are more explicit than those of Galileo ... To account for the world as we know it Descartes requires three types of particles, which are distinguished from one another by their geometrical properties only  that is, by 'extension'. These are the small 'fire particles' some of them are infinitely small  the intermediate 'boules' and the 'gross material particles' which in aggregate constitute all material bodies. The first two sorts of particles form the allpervading aether which circulates in a gigantic but invisible whirlpool or vortex about that large hot body, the sun. It is the motion of this vortex that carries the planets round in their circular orbits, and it is the pressure due to the aether flow round the earth  which has its own relatively small vortex that accounts for the phenomenon we call 'weight'. Light is a sensation caused by pressures transmitted through the aether to the retina of the eye, while the sensation of hotness is due to the vibrations of (probably) the material particles stimulating the nervous system. On the other hand the phenomena of expansion and change of state are caused by fire particles and boules penetrating between the material particles and forcing them farther and farther apart. It may even be that vibrations of fire particles and boules agitate material particles and are thus the first cause of the sensation of hotness. Dr G.R. Talbot has pointed out that later Cartesians who did not share their master's anxiety about the void, or completely empty space, saw no reason for both fire particles and boules and so lumped them together as a 'subtle fluid': a rather hazy concept which developed in the course of the eighteenth century into the doctrine of material 'caloric'." (Cardwell, p.2)
4.5.3 The Conservation of Fire We arrive, thus, in the first half of the 18th century. In this period, in spite of some relevant to the science of heat work by Isaak Newton and Jean Bernoulli's son Daniel, the major contribution comes from Herman Boerhaave (16681738), professor of Medicine, Chemistry and Botany at
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the University of Leyden. He , one could argue, sowed the seeds for the first law. Boerhaave was an excellent lecturer, that students from all over the civilized world came to listen to. His work was first published  in a pirated(!) edition  in 1727, and is described by Cardwell as follows (Cardwell, p.27): "Boerhaave's lucidly expounded ideas set the pattern for his contemporaries and successors. He was basically a Cartesian and his concern was therefore with 'Fire', which he held to be distributed throughout all bodies and throughout 'space'. He maintained that Fire is the great universal agent of change, and is abundantly present even in the coldest things under the coldest conditions; for it had been shown that one could strike sparks from a flint in the depths of an arctic winter. An ultimate limit of cold can be imagined, he says, when a body would be wholly deprived of Fire, but such a state cannot be reached either by nature or by technics: it is therefore not worth bothering about it. In fact, Boerhaave's attitude throughout was one of common sense, so even when he was wrong, by later standards, he was never unreasonably or dogmatically so. He sets out as an essential axiom the proposition that Fire is always conserved; he argues that it cannot be created de novo, as claimed by certain English philosophers, and that in all changes the total quantity remains unaltered. Now this statement was perhaps crucial, for it is very difficult to see how a science of heat could have developed without a
basic conservation principle: such a principle was plainly essential for the formulation of the concepts of quantity of heat, of specific heat and of latent heat." (emphasis added.)
4.5.4 The Differentiation of Heat and Temperature In addition to the fundamental axiom of 'the conservation of fire' a second, but somewhat diffuse development, took place in the first half of the eighteenth century: the notion of 'quantity of heat' as a concept distinct from that of 'intensity or degree of heat' was introduced. Cardwell comments (p.31): "Curiously, no one claimed to be the inventor of this concept. It seems that a number of scientists started using it independently, more or less simultaneously, and without any fuss. It was a concept that led inevidently to the postulation of an absolute zero of temperature and, a little later on, to the discovery of specific and latent heats."
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4.5.5 The Conservation of Heat The third quarter of the 18th century is dominated by Joseph Black ( 172899) who was a professor of medicine and chemistry first at Glasgow and from 1764 in Edinburgh. He was an excellent teacher, but published little during his career. (The edict  aphorism?  of modern academia: 'publish or perish' had not , I guess, appeared yet!). What we know is based on notes from his lectures. Black was influenced by Boerhaave's ideas but his philosophy was different. Cardwell contrasts the two philosophies and presents Black's contribution as follows: "The latter (Boerhaave) had postulated Fire as the general principle of change but had been unable to bridge the gulf between familiar thermal phenomena and their postulated metaphysical cause. In contrast, Black's attitude smacked of the pragmatic positivism that sometimes characterized Newton: we do not know the cause of gravity but we can devise quantitative laws that account for the behaviour of bodies when gravity acts upon them. Accordingly, Black confessed himself agnostic on the nature of heat, and with a sound grasp of the realities of the situation ignored the elusive 'Fire' to concentrate on the operationally intelligible measures of temperature and 'quantity of heat'. He formulated the latter quite naturally, and was almost certainly unaware of the great advance that it represented. The conservation principle is maintained but is tacitly transferred from Fire to heat." (Cardwell, p.35)
The next step towards the establishment of the first law was, thus, taken.
Having formulated the concept of 'quantity of heat', Black introduced the ideas of 'specific heat' and of 'latent heat'. Black's discussion of the second item is indicative of the level of understanding of thermal phenomena at that period: "Black gives us two examples from common experience which show how important this phenomenon of latent heat really is. A kettle of water at a temperature marginally or infinitesimally below the boilingpoint would, on the commonsense view, require the addition of only a trivial amount of heat to raise its temperature above the boilingpoint. But if this happened the water would be all converted into steam, technically into superheated steam, with explosively disastrous consequences! This does not occur because a vast amount of latent heat is required if water is to be converted into steam. Again, on the commonsense view only a trivial amount of heat would serve to convert a vast snowfield just below free
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zingpoint into (disastrous) flood water. This does not happen; snow is notoriously slow to melt and on the northern slopes of the highest Scottish mountains it lingers on through summer. Once more the doctrine of latent heat provides the explanation. In this way an advance in the science of heat gave new insights into a range of meteorological and other 'transformation' processes. "(Cardwell, p.38) At this point we should remember that James Watt had stumbled upon the phenomenon of latent heat, which was an important factor in the development of his engine. And since he was working at the same place with Black, Glasgow University, a longlived legend developed according to which Black's discovery of the latent heat was instrumental in the development of Watt's engine. For example, as late as 1961 Charles Singer in A Shon History of Scientific Ideas (Oxford University Press, p.300) stated: "Watt (1765) applied Black's discovery (of latent heat) to his contrivance of the separate condenser." A fascinating account of the origin and growth of this legend is presented by Cardwell (p.41) who demonstrates rather clearly that it is a false one.
4.5.6 The Caloric Theory and its Problems We come, hence, to the last quarter of the 18th century. The 'caloric' theory of heat, whose origin lies  as we have seen  with Descartes' 'subtle fluid', is fairly well accepted but there is some strong opposition to it. Hatsopoulos and Keenan describe the situation as follows: "According to the caloric theory, caloric or 'matter of heat' was an allpervading subtle fluid which could be neither created nor destroyed. This idea was in tune with its era, which was an age of subtle fluids like electricity and phlogiston. Probably no authoritative statement of the caloric theory ever existed. A fairly complete statement is contained, however, in the following postulates of William Cleghorn (1779): l.Caloric is an elastic fluid, the particles of which repel one another strongly. 2.Particles of caloric are attracted by particles of ordinary matter in different degrees for different substances and different states of aggregation. 3.Caloric is indestructible and cannot be created. 4.It is either sensible or latent. Latent caloric is caloric which has combined chemically with particles of so1id to form liquid or of liquid to form vapor. 5. Caloric has weight.
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These postulates were ingeniously contrived to account for certain observed phenomena. Postulate: (1) accounts for expansion and contraction upon heating and cooling; (2) for variations of specific heat capacity; (3) for calorimetry; (4) for latent heat; and (5) for gain in weight of certain metals when heated (calcined) in the presence of air. The plausibility of such postulates prompted Joseph Black, a dedicated experimentalist and a sceptic regarding all theories, to say that 'such an idea of the nature of heat is the most probable of any I know'. The gap in the armor of the caloric theory was the heating effect of friction, a phenomenon so common that it seems incredible that any theory of heat which ignored it should ever have been seriously entertained. Thus, John Locke (1772) observed: 'The axle trees of carts and coaches are often made hot, and sometimes to a degree that it sets them on fire, by the rubbing of the naves of the wheels upon them.' In an attempt to close this gap some argued that friction reduced the attraction between caloric and material particles and so freed the caloric to flow to the surroundings. Count Rumford conducted a relentless campaign against the caloric theory. He showed, first, that caloric is weightless; second, that metal which has been subjected to friction has the same heat capacity as the native metal; third, that the ability of brass to produce heat by rubbing is inexhaustible. This last point, which he demonstrated by his famous cannonboring experiments in his arsenal at Munich, proved to be the crucial one. In the words of Rumford, 'The source of the heat generated by friction in these experiments, appeared evidently to be inexhaustible. It is hardly necessary to add that anything which any insulated body, or system of bodies, can continue to furnish without limitation cannot possibly be a material substance'." (pp.xvixvii)
4.6 The Establishment of Thermodynamics We enter thus the 19th century that saw the establishment of thermodynamics through the development of the first and second laws. But first let us summarize the state of the Science of Heat and of the Steam Engine in the early part of the 19th century. Heat is made of fire particles (caloric), even though some disagree
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Table 4.4 Major Developments Leading to the Establishment of Thermodynamics 1824
Carnot publishes his book: Rejlexions sur Ia Puissance
Motrice du Feu.
1834
Clapeyron presents the Carnot cycle in terms of a PV diagram and develops an expression for its efficiency.
1842
Mayer publishes his theory about the conversion of heat to work.
18431850
Joule, through a series of experiments, establishes the convertibility of work to heat.
1849
Kelvin expresses doubts that both Carnot and Joule can be right; also introduces the term Thermodynamics.
1850
Clausius reconciles the findings of Carnot and of Joule and presents the analytical statement of the first law and the verbal one of the second.
1851
Kelvin accepts Joule's findings; introduces the 'Absolute Temperature' and the analytical statement of the second law for reversible processes.
1865
Clausius introduces the concept of Entropy and the analytical statement for the second law for irreversible (and reversible) processes.
strongly and hold that it is the 'vis viva (living force)' of the atoms. It is also conserved, i.e. the total amount of heat is constant and it can not be converted into anything else. The steam engine has been perfected: from the embryonic Savery type to the Watt type with a separate condenser and to the high pressure type of Trevithick. Yet, while the efficiency of waterwheels, the main source of power up to this period, can be properly determined work produced as compared to the maximum obtainable one (Example 2. 7)  the performance of the steam engine is still measured in terms of work (ftlbr) produced per bu
shel of good(!) coal.
It is also a revolutionary period: Cardwell comments (p.190): "In France political revolution had let loose an abundance of creative talent while at the same time providing scientists with the organization, the in
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centives and the moral stimulus to attack the widest range of problems. In England revolution was revealed in the rise of new forms of industry wherein manufacturing processes were analyzed into their major components so that each stage could be mechanized." In addition, the continuing economic growth demanded more energy and more efficient use of it. Energy consumption, actually, became  and still is today the barometer of economic growth. Thus Leibig stated:
Civilization is the Economy of Power.
This was the picture scientific, technological, and social in the early part of the 19th century that the two fathers of Thermodynamics: Sadi Carnot and James Prescott Joule, entered in. A new field of science and technology was established in about half a century, following the major developments presented in Table 4.4. (The 'christening' was done by Lord Kelvin in 1849).
4. 7 Sadi Carnot Sadi Carnot was born in 1796 when his father was just 43 years old. The young Sadi entered the Ecole Polytechnique in 1812 with the intention of becoming, like his father before him, an officer in the engineers, the 'Genie'. He saw action at Vincennes in 1814, but the long war was coming to an end and the years of peace that followed Waterloo afforded less opportunityfor distinction and promotion to an army officer; especially to one whose father was a political exile. In 1820 Sadi Carnot went on halfpay and, Professor Mendoza tells us, devoted his energies to study, making the best use of the educational and research facilities of Paris. He studied principally physics and economics and, according to Professor Mendoza, spent a good deal of time visiting factories and studying the organisation and economics of various industries, becoming an expert on the commerce and industries of different European countries. The fruit of this devoted labour was a short book, some 118 pages long, which he published in 1824 under the title, 'Rejlexions sur lapuissance motrice dufeu'. (Cardwell, p.191)
We will start with a discussion of Carnot's motivation for his work and proceed with his arguments and conclusions. We will close with Cardwell's commentary on the fate of his work (and see what happens when one is born before his time).
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4. 7.1 Motivation Carnot's main motivation came from his admiration for the steam engine and the lack of theory about its operation. Cardwell writes: "He is impressed by the widespread utilisation of steampower in England, and he is unstinting in his praise for the engineers  most of them British  who have developed the steamengine up to its latest, highpressure form. With profound insight he remarks that the steamengine is now more important for England's existence than is her Navy evidently his readings in economics and his comparative studies of industry and technology had not been a waste of time. Nevertheless, in spite of all this scientific knowledge and technological development there is no general theory of the heatengine, applicable to all conceivable forms of heatengine, whatever their working substances or their mechanical principles. He observes: Machines which are not driven by heat, those which are driven by the power of men or of animals, by a fall of water, by the movement of air, etc. , can be analyzed down to their last details by mechanical theory. Every event is predictable, all possible movements are in accordance with established general principles which are applicable in all circumstances. This, of course, is characteristic of a complete theory. A similar theory is obviously required for heat engines." (Cardwell, p.192)
4. 7.2 Arguments We discussed Carnot's cycle in Chapter 3 and demonstrated its two basic points, the need for: * a heat reservoir and a heat sink; and * a reversible engine. Here now is how Cardwell presents Carnot's arguments: "He begins by grasping and emphasizing two essential points. The first one is that if you want to use heat to generate power you must have a cold body as well as a hot one. That is to say, a 'fall of caloric'  or a temperature difference is absolutely essential. The waterpower analogy is very close at this stage of the argument: a given quantity of water will only generate power if there is a lower level to which it can flow; a tailrace as well as a penstock. For Carnot, of course, the actual motive agent in a heatengine, the analogue to the water in a hydraulic machine, is the highly elastic fluid, caloric.
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The second essential point is that for the generation of the maximum amount of power there must at no point and at no time be a useless flow of heat. In other words, heat which could cause the working substance to expand and do work must not flow uselessly from the hot body to the cold one. This means that ideally there must be no contact between bodies at different temperatures during the operation of a heatengine; a condition which, the logician will object, would make a perfect or ideal engine quite impossible. But this objection could, with equal force, be raised against Lazare Carnot's conditions for the maximum efficiency of a hydraulic engine; that the water enter the machine without shock of impact of any sort and leave it without relative velocity (Example 2. 7). Again the analogy between the working of a hydraulic engine and a heatengine is a close one; the head or, since the one determines the other, the velocity of the water being the analogue of the temperature difference. In the case of the heatengine Sadi Carnot answers the logician's objection by asserting that vanishingly small temperature differences are, for his purposes, sufficient to cause heat to flow without violating the prime condition that in a perfect heatengine bodies in contact must be at the same temperature." (Cardwell, p.193) We should notice here that Carnot had accepted the axiom of the 'conservation of heat'. Hence, when he talks about a 'fall of caloric' he means that all caloric that left the heat reservoir finds itself in the heat sink, just as all the water that falls through a hydraulic engine arrives at the bottom. Work is produced because of the 'fall' of caloric, not through conversion of it.
4. 7.3 Conclusions By using Carnot's cycle in Chapter 3 we concluded that: a.A reversible engine has the highest efficiency which is independent of the working fluid; b. This efficiency depends on the temperature difference (t 1  t 2), where t 1 and t 2 refer to the heat reservoir and heat sink respectively; and c.It also depends, for a given (t 1  t 2 ) value, on the temperature level: it is higher when t 1 = 200°C and t 2 = 100°C, than when t 1 = 300°C and t 2 = 200°C. On the first Cardwell writes: " ... Carnot has no difficulty in showing that a heatengine working according to the cycle he prescribed, and of course mechanically perfect, is the most efficient possible. For let us suppose that there is an even more efficient engine; then if it is used to
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drive Carnot's engine backwards so that it acts as a heatpump, all the heat 'let down' through the more efficient engine will be restored to the source. But the latter, by definition, will deliver more power than is required to drive the heatpump. The consequence will therefore be the generation of a net amount of power without any net 'fall' of heat and this would, in fact, mean the production of perpetual motion. Thus no heatengine can be more efficient than one which can be reversed in the way in which Carnot prescribes. It is obvious that the nature of the working substance cannot affect this conclusion; all such engines, no matter what the working substance air, steam, alcoholvapour or anything else are necessarily of exactly equal efficiency." (Cardwell, p.199) On the second, Carnot himself states: "In accordance with the principles we have now established, we can reasonably compare the motive power of heat with that of a head of water: for both of them there is a maximum which cannot be exceeded, whatever the type of hydraulic machine and whatever the type of heat engine employed. The motive power of a head of water depends upon its height and the quantity of water; the motive power of heat depends also on the quantity of caloric and on what may be called  on what we shall call  the height of its fall, that is on the temperature difference of the bodies between which the caloric flows." The situation with respect to the third item is more interesting for the analogy to the hydraulic engine would suggest to Carnot that the fall of caloric from 100 to 90°C and from 10°C to ooc should produce the same effect. (Remember that the expression: n = 1  (T2/T1) was not developed until later). Yet he was more cautious and proceeded to demonstrate that for a given (t 1  t 2), the lower the temperature of the heat sink t 2 , the higher the efficiency. (See Cardwell, p.204.) But how could Carnot using the false axiom of the conservation of caloric arrive at the correct conclusions? For the first two, the proof is independent of the conversion of heat to work. As for the third, this is one case where two wrongs make a right. Again from Cardwell, p.205: "The conclusion was drawn from an argument based on the false assumption of the conservation of heat and from misleading experimental evidence that the specific heat of gases increases with volume. Nevertheless the conclusion happens to be correct."
4. 7.4 The Acceptance of his Work "Rejlexions received a long and favourable notice in the 'Revue Encyclo
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pedique' and was mentioned in the 'Bulletin des Sciences Technologiques', but for the rest it was simply ignored in France, Britain and elsewhere. When, many years later, the young and enthusiastic student William Thomson (Kelvin) tried to buy a copy of the book he found that the Paris booksellers had never heard of it, or its author; the only Carnot they knew of was the father, the 'Organiser of Victory', Lazare Carnot. It is not difficult to understand why Rejlexions was a failure. To the engineer, the postulation of the ideal engine and the reversible cycle must have seemed closer to the more imaginative passages in the 'Timaeus' than to the needs of powerengineering. Christian, Poncelet and others had advised French engineers to keep practical requirements always in mind and to avoid being too abstract and theoretical. Always there was the example of the Cornish engines, whose extremely high standards of performance had been achieved by careful attention to details and whose designers seemed to have made little use of abstract theory. In fact, Carnot had very little to teach the engineers of his time: the superiority of the highpressure condensing steamengine had been established, so had the advantages of expansive operation, and by 1824 engineers knew well enough that, all things considered, there was no better workingsubstance than steam. We must conclude that Carnot' s message was essentially for engineers of a later generation with the resources of a more advanced technology. On the other hand Carnot's ideas were general or cosmological in their scope and therefore, in principle, relevant to physics, chemistry and meteorology. In these fields however his mode of reasoning would, one infers, be unfamiliar and thus perhaps unacceptable to those who had had little or no experience of powerengineering. In sum, the 'Rejlexions' fell  inevitably, when one recalls its scope between a number of specialist stools: the recognition of its technological importance lay in the future, while for the physical sciences the clarification and demonstration of its relevance called for the further, and highly original, work of Clapeyron and Kelvin (in physics), Gibbs (in chemistry) and Napier Shaw (in meteorology)." (Cardwell, pp.211212)
4.8 Emile Clapeyron The next major step in the development of thermodynamics belongs to Emile Clapeyron.
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Emile Clapeyron (1799 1864) was an engineer, a graduate of the Ecole Polytechnique, whose interests seem to have been in the theory of structures and in civil engineering generally. His posthumous distinction however rests on one paper, published in 1834, in which with superb timing (as it transpired) he rescued Carnot's ideas from the oblivion into which they had fallen and restated them with such conviction and clarity that they aroused the interest of the perceptive minority that exists in every generation. More, he made some highly original contributions to the science  thermodynamics  which Carnot had founded, for all that he, like Carnot, subscribed to the axiom of the conservation of heat. (Cardwell, p.220)
4.8.1 Contributions Clapeyron made three main contributions: First, he represented the Carnot cycle on an 'indicator diagram', i.e. a pressurevolume diagram (Figure 3.1). Carnot's abstract ideas are now presented in a graphical form. (How Clapeyron came to know about this diagram, that was so closely guarded by Boulton and Watt, is still a mystery.) Second, he developed an expression for the maximum amount of work (W ) produced by the fall of an amount of caloric Q from temperature (t+dt) to t, i.e. the efficiency of the Carnot cycle: dW
Q
=
_!_ dt
c
(4.8.1)
where (1/C) is the maximum amount of work that a unit of heat can perform when falling through 1°C at a temperature t; in other words, C is a function of temperature. It was later called the 'Carnot factor'. Third, using the Carnot cycle, he developed as shown in Example 4.1  the 'Clapeyron equation' for the determination of the heat of vaporization L of a pure fluid at some temperature t: (4.8.2) where: V1 and V2 are the saturated molar volumes, liquid and vapor respective! y; and dPs ldt,is the slope of the vapor pressure curve; all evaluated at t. The development of Eq.4.8.2 was of extreme importance for Clapeyron used it, along with available at that time vapor pressure and heat of
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vaporization data, to determine experimental values for C (see Example 4.2 for water). He then demonstrated using this expression that, just as Carnot anticipated, (1/C) decreased with increasing temperature. (Undergraduate Students can now move to Section 4.9.)
4.8.2 The Efficiency of the Carnot Cycle Once Clapeyron described the Carnot cycle on a PV diagram he proceeded to determine its efficiency. In those days the ideal gas equation had the form: PV
=
R(t
+ 267)
with t in °C. He could thus calculate the work involved in the two isothermal steps (Figure 4.1). There was no relationship, however, available between P and V for the two adiabatic steps. To get around this point, Clapeyron very shrewdly considered that the two temperatures differed by a very small amount dt. As a result, the two isotherms in Figure 4.1 are parallel, the segments 14 and 23 are straight lines, and the amounts of work involved in them cancel each other. Hence the work produced in the cycle is: dW
I:
=
I~
=
R(t + dt + 267)ln(P 1 I P2 ) + R(t + 267)ln(P3 I P4 )
p dV +
p dV
=
Clapeyron had also developed (Cardwell, p.222) the following expression for the total heat content of the gas: Q = R(B ClnP)
where B and C depend on temperature only. Thus, the heat absorbed by the gas is: Q = R(B ClnP2) R(B ClnP1) = RCln(P11P2)
Since lines 12 and 43 are parallel, (P 11P4) = (P21P3), or (P21P1) (P31P4 ), and we conclude after some algebra that: dW
Q
=
__!_ dt
c
(4.8.1)
Thus, (liC) becomes the measure of the work produced by a unit of heat falling by 1°C at a temperature t, and is a function of temperature only.
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3
Volume
Figure 4.1 P V diagram for the development of the Carnot cycle efficiency by Clapeyron.
4.8.3 Example 4.1. The Clapeyron Equation Using one mole of a saturated liquid undergoing the following cycle (Figure 4.E.l): 1. Vaporization at constant pressure P s from the saturation liquid volume V1 to the saturation vapor volume V2 ; 2.Heating from ps tops + dPs at constant volume V2 ; 3. Cooling at constant pressure to V1 ; 4.Cooling at constant volume from ps + dPs tops; develop Eq. 4.8.2. We note that: *work produced by the cycle is: dW = (V2  V1)dPs; * heat absorbed: the heat of vaporization, L. Hence: dW
Q
(v2  v) I dPs L
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v
3 ~
Q) L
p•
   '                '
dP"
a...
Volume
Figure 4.E.l PV diagram for the development of the Clapeyron equation. Clapeyron then argued in his 1834 paper that the maximum amount of work produced by a unit of heat cannot, by Carnot's axiom, exceed the work that it could do if it were applied to any substance. It must therefore be given by Eq. 4. 8.1. Thus:
After rearrangement we obtain the Clapeyron equation.
4.8.4 Example 4.2. Evaluation of the Carnot Factor C Use data from the Steam Tables, say in the interval 2°C to 80°C, to obtain numerical values for C: L
C=
at different temperatures.
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All pertinent information and C values are presented in the top of the next page:
c
(1/C)104
2496.8
277.6
36.0
0.145
2454.3
293.7
34.1
12.04
0.610
2382.9
324.5
30.8
3.408
1.920
2308.8
352.8
28.3
t CO C)
v2 vi
(m3 /kg)
drldt (kPa/0 C)
(kJ/kg)
2
179.90
0.050
20
57.84
50
80
L
Comments 1. ( 11 C) increases with decreasing temperature just as Carnot had anticipated. But Clapeyron proved it using experimental data available at that time (water, ethanol, ether, and turpentine). 2. If we compare Clapeyron's efficiency of the Carnot cycle, Eq .4. 8.1, with the one we developed in Chapter 3, Eq.3.5.3, it follows that C = T, i.e. the absolute temperature. (This, of course, was not done until later.) And this is why the calculated values of C are, within the accuracy of evaluating the slope (dP' ldt), equal to the absolute temperature in Kelvin.
Carnot's somewhat abstract ideas about the heat engine and Clapeyron's quantitative interpretation of them laid the foundations for the establishment of the second law. This, however, was not done until the 1850's and 1860's by Kelvin and Clausius. In the between, the foundations for the first law were laid by Joule.
4.9 The Conversion of Work to Heat This idea, as expected, generated tremendous opposition because it was contrary to the doctrine of the 'conservation of heat' that was deeply rooted as the basis of the science of heat at that time. Its establishment belongs to Julius Robert Mayer and, especially, to James Prescott Joule.
4.9.1 J.R. Mayer (18141878) Mayer was trained as a medical doctor and was a shipsurgeon on a vessel trading to the East Indies. His first ideas about the conversion of work to heat came from his observation of the difference in venous blood color between the tropics and higher latitudes. He concluded that in the warm
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tropical climate, the inhabitants needed less oxygen to maintain their body's temperature, hence the bright red color of their venous blood. Mayer went on to establish a balance between energy intake, as measured by the oxygen used, and the energy consumption, determined by the heat lost by the body and by manual work. To determine the equivalence between heat and work Mayer relied on the difference in gas heat capacities at constant pressure and at constant volume. In the case of the constant pressure heat capacity CP• the amount of added heat not only raises the temperature of the gas, but also must account for the work of expansion of the gas against the external pressure. This work is not, of course, involved in the constant volume heat capacity Cy, and thus Cp is larger than Cy. Arguing that all the extra heat was converted to work, he arrived at the number of 356 kg.m per kilocalorie (664.8 ftlbf/Btu), about 15% less than the accepted value today. Mayer published his findings in 1842, but his ideas were not accepted: First, he did not prove that all the extra heat (for Cp) was used for work against the external pressure. Second, the idea of a loss of caloric (heat) by conversion to work was against the aforementioned 'conservation of heat' axiom. Finally, as Cardwell suggests (p.231), "he was outside the tradition of research and scholarship in the science and technology of heat; he stood apart from the then Establishment." This rejection of Mayer's pioneering ideas, along with some personal problems, led to his mental collapse. He did achieve, however, the recognition he deserved due to the efforts of Clausius, recovered from his breakdown and was voted into the Royal Society in 1862.
4.9.2 J.P. Joule (18181889) Joule faced also a lot of opposition. But his deep scientific insight, outstanding experimental work, and the fact that he was a member of a scientific establishment  in Manchester  helped him overcome it. Joule carried out, as we mentioned in Chapter 2, a series of very careful experiments whereby he demonstrated that work was converted to heat; and that there was an exact relationship between work consumed and heat produced. Caloric was thus generated, leading eventually to the collapse of the caloric theory. Here is how Cardwell describes Joule's work, his resolution of the heat capacities problem that Mayer faced, and the reaction of the then scientific community.
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4.9.3 Early Experimental Work "Joule was led to consider the problems of energy conversion through an early interest in magnetoelectric machinery. His first papers were of a practical, engineering nature: he looked forward to the day when electric motors would replace steamengines for driving machinery. From electric motors he graduated to a study of the generation and the heating effects of electric currents. In 1841 he discovered the important law that the heating effect of an electric current is proportional to the resistance of the circuit and the square of the current flowing. Two years later came the very important paper: 'On the Caloric Effects of MagnetoElectricity, and on the Mechanical Value of Heat'. In this paper he demonstrated that the heat caused by the passage of an electric current is not transferred from another part of the circuit, which is correspondingly cooled, ... but is actually generated. In his own words, he showed that: ... the mechanical power exerted in turning a magnetoelectric machine is converted into the heat evolved by the passage of the currents of induction through its coils; and, on the other hand, that the motive power of the electromagnetic engine is obtained at the expense of the heat due to the chemical reactions of the battery by which it is worked.
The mechanical power was provided  and measured  by the descent of a pair of weights placed, one each, in two scale pans; the latter were connected by means of fine strings passing over two pulleys to a vertical axle, round which a great length of each string was wound. As the weights descended a measured distance so the axle was rotated and drove the magnetoelectric machine (dynamo). As a result of a number of experiments Joule concluded that the mechanical value of a unit of heat was 838 ft.lbs of work expended to raise one pound of water by one degree (°F). These experiments, together with the conclusions drawn, were announced to the Chemical Section of the British Association meeting at Cork in 1843. They were received in silence; partly, no doubt, because some of his audience were unable to grasp the full import of what was at stake, partly because others were unwilling to throw overboard the grand principle of the conservation of heat on which the entire science had been erected, and partly because still others felt that too much was being inferred from just one set of experiments,
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But Joule had already, in a postscript to this paper, transferred his attention from the electrical generation of heat to the purely frictional. By forcing water through tiny holes in a perforated cylinder, he had been able to evolve a detectable amount of heat, which compared with the work expended yielded a mechanical value for a unit of heat of 770 ft.lbs. This was in tolerable if not good agreement with the value obtained by means of the electrical experiments." Joule's Some Other Experiments (!)
Another of his early and little known experiments was to examine the effects of electricity on human beings. His subject was a servant girl who reported her sensations as the voltage was stepped up  he used a larger battery  to the moment when she became unconscious, whereat he felt it advisable to stop the experiment. Unfortunately the young lady's opinions were not recorded. (Cardwell, 1983)
4.9.4 The Expansion/Compression of Gases and Later Work "But if expending mechanical power on water in this way can generate heat, what of the famous phenomenon of the 'adiabatic' heating of gases? Joule's next experiment was to compress a gas in a strong copper vessel, which was placed in a large calorimeter. The walls of the calorimeter were made as impermeable to heat as was possible, and it was filled with water. By comparing the heat imparted to the water with the work performed in compressing the gas Joule found a new value for the unit of heat. This was 823 ft.lbs. At this point Joule faced the unresolved difficulty that had made Mayer's determination of the mechanical value of heat uncertain. What proof was there that all the work performed on the gas had been converted into heat; alternatively, that all the heat produced was due to the work performed? Might it not be that in compressing the gas  in reducing its volume some of the heat released may have been due to 'latent' heat being made 'sensible'? These objections did not apply to the experiments made using water and a perforated piston, for water is virtually incompressible. But until they could be answered in the case of gases, determinations based on 'adiabatic' compression or expansion would be entirely uncertain. Joule's answer to this dilemma was another and very famous experiment. . .. Two large copper vessels are connected by means of a pipe containing a stopcock and are placed in a large calorimeter full of water; the walls of the calorimeter are as impermeable to heat as is possible.
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One of the copper vessels is evacuated and the other is filled with dry air at a pressure of 22 atmospheres. The stopcock is opened so that the air rushes into the empty vessel, but no detectable change in the temperature of the water in the calorimeter is noticed. In other words, when air expands without doing work no heat is lost. Joule now proceeded to analyze this experiment. He provided each of the two copper vessels with its own separate calorimeter and repeated the experiment. The results indicated that the calorimeter containing the evacuated vessel gained exactly the same amount of heat as the other one lost. The sum of the two quantities of heat imparted to and abstracted from the two identical calorimeters was, therefore, zero. The implication was plain: merely changing the volume of a gas does not produce or consume heat; this happens only when work is done on or by the gas. The 'latent' heat of expansion (or compression) is therefore an illusion: it is no more than the heat value of the work done by or on the gas, against or by an external force. Joule's calculation of the mechanical value of a unit of heat by measuring the amount produced when the gas is compressed to a given extent was therefore justified in principle; and so, incidentally, was that made by Mayer. This splendid paper was rejected by the Royal Society. But it was printed in the 'Philosophical Magazine', and in this way the scientific world learned not only of Joule's fundamental insights and confirmatory experiments but also of his penetrating criticisms of Clapeyron and Carnot. For the first time the fundamental ideas of thermodynamics were subject to the (constructive) criticism of a firstclass scientific intellect: I conceive that this theory ... is opposed to the recognized principles of philosophy because it leads to the conclusion that vis viva (i.e. energy) may be destroyed by an improper disposition of the apparatus: thus Mr Clapeyron draws the inference that 'the temperature of the fire being from I()()(J>C to 2()()(J>C higher than that of the boiler there is an enormous loss of vis viva in the passage of the heat from the furnace to the boiler'. Believing that the power to destroy belongs to the Creator alone I affirm ... that any theory which, when carried out demands the annihilation offorce, is necessarily erroneous.
In the following year (1845) Joule read, to the British Association meeting at Cambridge, an account of a new method of determining the mechanical value of heat. A paddlewheel, driven by the same mechanism of falling weights that he had used tD provide power for his dynamo in 1843, rotated horizontally in a calorimeter full of water. A comparison of the heat generated with the work performed yielded a mechanical value
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of 819 ft.lbs per heat unit. But no one seemed very interested in Joule's ideas or experiments. No one, that is, in Britain; abroad he was not without honour, for when in 1847 Helmholtz published his important paper 'On the Conservation of Energy' he cited Joule's experiments in support of his arguments. In that year Joule described his improved paddlewheel apparatus for the measurement of mechanical value of heat and this time, at the British Association meeting at Oxford, his ideas aroused much more interest. This was largely due to the keen appreciation of the young William Thomson, later Lord Kelvin. The Oxford meeting of the British Association marked an historic stage in the history of thermodynamics: the real beginning of the acceptance of the first law (emphasis added). It is, of course, true that the meausurements on which Joule had had to rely for his determinations were small fractions of a degree, but he was a masterful experimentalist and he was supported by a most able instrumentmaker (see Example 2.2). The implication of Joule's ideas that one could heat up water merely by shaking it, seemed in the context of scientific  or common sense  thought in the 1840's quite extraordinary, and the confirmation that this was indeed the case was taken as strong confirmation of his theory. The way ahead was now, at last, clear for Joule. He had soldiered on and had won through to recognition and acceptance. In 1850 the Royal Society published his account of new and very accurate paddlewheel experiments which gave the figure of 772 ft.lbs for the mechanical value of a unit of heat. From this time onwards Joule could extend and consolidate his theory, devise even more accurate methods for determining his constant, and find further instances of the great principle of the convertibility of energy in nature." (Cardwell, pp.232236)
4.10 Kelvin, Clausius and the Establishment of the First, and Second Laws 4.10.1 Lord Kelvin (1824;.1907) William Thomson was knighted in 1866for his services in connection with the great Atlantic cable and was made Lord Kelvin of Largs in 1892.
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Joule won the battle but the war was not over, by far. No other than the very sympathetic to Joule as we have seen  Kelvin, in his 1849 paper on 'Account of Carnot's Theory of the Motive Power of Heat; with Numerical Results derived from Regnault's Experiments on Steam' still has very strong doubts. Cardwell writes: "Kelvin mentions Joule's ideas only to reject them: ' ... the conversion of heat, or caloric, into mechanical effect is probably impossible, certainly undiscovered'. But he then goes on to raise an important question; what happens when heat flows by conduction from a hot to a cold body? When thermal agency is thus spent in conducting heat through a solid what becomes of the mechanical effect which it might produce? Nothing can be lost in the operations of nature  no energy can be destroyed. What effect then is produced in place of the mechanical effect which is lost? A perfect theory of heat imperatively demands an answer to this question,· yet no answer can be given in the present state of science.
This was, at one and the same time, an unoriginal and a very penetrating question. It was unoriginal in that the key phrase 'Nothing can be lost in the operations of nature  no energy can be destroyed' is hardly more than a rephrasing of Joule's remark: 'Believing that the power to destroy belongs to the Creator alone I affirm ... that any theory which, when carried out, demands the annihilation afforce, is necessarily erroneous' (see Section 4.9 .4). Therefore Kelvin had picked up the gauntlet thrown down by Joule, even if he did not know what to do with it. On the other hand he had posed the question in the framework of thought established by Carnot: that relating to the efficiency of heatengines. We can regard the mere conduction of heat from a hot to a cold body as the operation of a heatengine of zero efficiency; in which case we must ask what happens to the work which a more efficient engine, in exactly the same situation, would produce? In this lay the key to the further development of thermodynamics." (Cardwell, p.241) What is fascinating about Kelvin's doubts and demonstrates how difficult it was for the 'conservation of heat' doctrine to be abandoned  is that by the time he published this 1849 paper, he had in his own laboratory duplicated Joule's findings. In a letter of his, dated December 5, 1847, he writes: "Perhaps before then (Friday 7) I may have succeeded in boiling water by friction. This is not very probab•e, however, as the machine I have made being not strong enough. In the first experiment made with it the temperature rose from 45, 46 to 57 at about the rate 1 every five minutes
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during a continued turning of the instrument (a very flat disc with narrow vanes turning in a thin tin box about 8 inches in diameter of which the bottom and lid were furnished with thin fixed vanes. The bearings of the paddle were entirely within the box which was full of water). In a second experiment we began with water at 98 & the temperature rose to about 99 1/2 when part of the machine gave way before the experiment could be considered quite decisive. My assistant was preparing to make an experiment today with water at 80, 90 & to go on grinding (along with another for relief) for about 4 hours, but I have not heard yet whether it went on." (Cardwell, 1983, p.56) The debate,,thus, for and against the conservation of heat went on. Cardwell comments on Kelvin: "The arguments on both sides seemed very strong, and it is hardly surprising that in the words of Kelvin's biographer, Silvanus Thompson, 'The apparent conflict took possession of his (Kelvin's) mind and dominated his thoughts.' There can be little doubt that at the root of Kelvin's objections to Joule's theory lay his conviction that the whole science of heat rested on what since 1783 had been the basic axiom of conservation. If that were rejected what would happen to the impressive structure of experimental knowledge and theoretical development that had been built up by the labours of men like Delaroche and Berard, Fourier, Dulong and Petit, Poisson, Victor Regnault and many others? Thus he quoted Carnot as saying, of the axiom of conservation, that: 'To deny it would be to overturn the whole theory of heat, in which it is the fundamental principle'. And he himself adds the comment that if the axiom of conservation is rejected: ' ... we meet with innumerable other difficulties  insuperable without further experimental investigation, and an entire reconstruction of the theory of heat from its foundation.'" (Cardwell, p.244) It is also noteworthy that in this 1849 paper, Kelvin coined the word "Thermodynamics", becoming the 'godfather' of the new field.
4.10.2 R.J.E. Clausius (18221888) Cardwell presents the next phase in the development of the laws of thermodynamics as follows: "While Kelvin was mulling over these problems he was being overtaken by two other men, R.J.E. Clausius (18221888)and W.J.M. Rankine (18201872). In some ways Clausius could almost be described as a disci
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pie, at one remove, of Kelvin. He had not read the original Carnot volume and he knew of the theory only through the writings of Clapeyron and Kelvin. Possibly this was an advantage, for he could consider the theory dispassionately, without being unduly influenced by the persuasiveness of the whole work, and in the light of Kelvin's important observations about Joule's ideas. In any case, after comparing Kelvin's work very favorably with that done by Holtzmann, ... he takes (in his 1850 paper) as his starting point the basic problem to which Kelvin had drawn attention: what happens to the mechanical effect which is lost when heat flows from a hot to a cold body not by way of a heat engine but by simple conduction? He sees that, in fact, the position is more open, there are rather more possibilities, than Kelvin had supposed. 'On a nearer view' and how much is contained in that modest phrase! the new (Joule) theory is opposed not to Carnot's theory but to the assertion that no heat is expended or lost in a cyclic operation. For it is quite possible that in the production of work both processes take place at the same time: '... a certain portion of heat may be consumed and a further portion transmitted from a hot body to a cold one; and both portions may stand in a definite relation to the quantity of work produced.' These words deserve to be pondered very carefully. In Clausius' view, Joule had established the consumption of heat in a cyclic process, Carnot the transmission of heat, and both phenomena are necessarily related to the work produced." (pp.244245) Clausius proceeds then to develop the analytical expression of the first law in the form: dQ = dU +~(a+ t) dV
v
J where J is the Joule factor for converting heat to work, and a is a constant that converts t to in today's terms absolute temperature (Clapeyron used a = 267, in developing the expression for the efficiency of the Carnot cycle, Section 4.8.2). Hence, according to Clausius, an amount of heat Q entering a gas can be broken down into two parts: * the first accounts for changes in sensible heat (i.e. temperature) and internal work (i.e. to overcome the intermolecular forces); and * the second accounts for externaf work. Total energy is therefore conserved, not just heat, as the caloric theory had it.
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Clausius then turns to Carnot S cycle and examines the issue of the maximum efficiency of the reversible engine in the light of the new theory. This leads him to the statement of the second law. Cardwell writes: "Clausius asserts that if there were some working substance, or engine, to provide more work for the fall of a given amount of heat than another engine; or, what amounts to the same thing, were it to provide the same work for the fall of a smaller amount of heat, then if the first engine drives the second one in reverse we would have an entirely anomalous result: no net work would be done and heat would, in effect, flow from a cold to a hot body. This, he postulates, is quite impossible; and thus he proves Carnot's fundamental principle. This, in fact, constitutes the first historical statement ofthe second law of thermodynamics (emphasis added). Carnot, we remember, went on to argue that a more efficient engine than a completely reversible one is impossible, since it would logically imply the feasibility of perpetual motion; and so indeed it would if the caloric theory, or the axiom of conservation, were true. The situation would be exactly the same as if a perfect columnofwater engine were to be driven in reverse, as a pump, by an even more perfect engine! But we are no longer dealing with fluids, subtle or gravitating; we are dealing with energy, and the extension of the argument merely leads us to the conclusion that the flow of heat from a cold to a hot body would enable us to perform work by applying the heat to a third engine, which would thus be deriving its energy from that stored in the cold body. Now there is, a priori, nothing in such an arrangement to contradict the energy principle as enunciated by Joule and Mayer and, more recently, in a very general form, by Helmholtz. All that would be happening would be that the thermal energy in the cold body would be being converted into the equivalent amount of mechanical energy and there would be nothing absurd or selfcontradictory about that. It certainly does not violate the great new principle of the conservation of energy. Clausius therefore rests his proof on the flat assertion that it is impossible for heat to flow of its own accord from a cold to a hot body. It is, of course, common experience that hot bodies always tend to cool down, that you cannot boil a kettle of water by putting it on a block of ice, and so on. Now the common experience of countless generations from time immemorial is to be elevated into a scientific axiom. We shall discuss the admissibility of this procedure later". (Cardwell, pp.247248) Indeed Cardwell presents a fascinating discussion on the acceptance of Clausius assertion. 1
I
I
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4.11 The Absolute Temperature and the Analytical Expression of the Second Law 4.11.1 The Absolute Temperature Kelvin formally accepted the dynamical theory of heat in his paper of March 1851 (Cardwell, p.254). He went on to work on the relationship between heat absorbed and work produced and between heat absorbed and heat rejected. In his 1854 paper, he accepts Joule's suggestion that the Carnot function C is simply the value of the temperature measured from 273°C, and demonstrates that: (4.11.1) where Tis the Absolute Temperature defined through the Carnot cycle. As Kelvin put it " ... the absolute values of two temperatures are to one another in the proportion of the heat taken in to the heat rejected in a perfect thermodynamic engine working with a source and refrigerator at the higher and lower of the temperatures respectively. "(Cardwell, p.258)
4.11.2 The Analytical Statement of the Second Law: Reversible Processes Cardwell continues: "And then, in a moment of penetrating insight, Kelvin saw that this simple relationship (Eq.4.11.1) amounted to a quantitative, or mathematical statement of the second law of thermodynamics. If we consider the heat absorbed as positive and that transmitted as negative then the relationship becomes: Ql + Q2 TI
=
0
(4.11.2)
T2
He is able to generalize this to include complex reversible cycles  we have come a long way from the simple expansive operation of steamengines  and to lay down that in ~uch complex cycles the quantities of heat are related to the temperatures (absolute) at which they are absorbed or transmitted by the linear equation:
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Ql(_!_)+Q2(_!_)+Q3(_!_)+ ... +Qn(_!_) TI T2 T3 Tn
=
0
(4.11.3)
This, Kelvin asserts, can be taken as ' ... the mathematical expression of the second fundamental law of the dynamical theory of heat'. The corresponding expression for the first law he writes as:
W +J(Ql + Q2 + "' + Qn)
=
0
Thus the first law of thermodynamics expresses the equivalence between heat energy and mechanical energy, while the second law imposes a definite restriction on the ways in which the conversion from one form of energy to the other can take place. In the case of completely reversible changes the second law can be expressed mathematically by Kelvin's linear equation." (Cardwell, pp.258259)
4.11.3 Real Life Processes and Entropy: Irreversible Processes The two laws of thermodynamics have been thus developed, even though Kelvin's analytical statement of the second law, Eq.4.11.3, is limited to reversible processes only. What happens with real life, irreversible processes? The analytical statement of the second law for such processes came from Clausius a few years later, in 1865. To this purpose he introduced the concept of Entropy which he defined as follows: "We might call 'S' the 'transformational content' of the body, just as we termed the magnitude' U' the 'thermal and ergonal content'. But as I hold it better to borrow terms for important magnitudes from the ancient languages so that they may be adopted unchanged in the modern languages, I propose to call the magnitude 'S' the 'Entropy' of the body from the Greek word 7p01r1l, transformation." He proceeded to demonstrate that "the entropy of the universe tends to a maximum", i.e. that for a real life, irreversible process, the total entropy change is positive: ..1S > 0. It took fifteen difficult years for Clausius  from 1850 when he presented, as we saw, the verbal statement of the second law to 1865 to arrive at the concept of entropy. It is no wonder, therefore, that we have such a hard time comprehending this term. Cardwell presents a detailed account of Clausius' development of Entropy. It represents a fascinating reading for any one interested in this concept. Clausius' starting, probably, point is described by Cardwell as follows (p.260):
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"Clausius saw clearly that the Carnot cycle represents a limiting, or ideal case. It is one end of a wide spectrum of thermal transformations, which go on in nature and indeed under man's control as part of his developed technology. In this respect the Carnot cycle is similar in kind to Galileo's intuition of the laws of falling bodies and of (circular) inertia: eliminate all imperfections  air resistance, friction, etc.  and under these ideal conditions all bodies fall with equal acceleration, or continue in uniform motion for ever." In other words, the same way that these imperfections prevent the continuous motion of a body or the conversion of its kinetic energy completely into work, the generation of entropy resulting also from such physical imperfections  prevents the realization of the Carnot cycle efficiency. It should be emphasized, however, that in the first case we are referring to complete conversion of kinetic energy to work; in the second, to
the conversion of the part of the flowing heat into work that is prescribed by the Carnot cycle. Complete conversion is not, of course, possible in a cyclic process.
4.12 Thermodynamic Properties of Matter With the development of the two laws  and the publication of books such as "Die Mechanische Warmetheorie" by Clausius in 1876 the field of thermodynamics was established. The rigorous analysis ofthe conversion of thermal energy to work, of the 'motive power of heat', that started with the pioneering work of Sadi Carnot, was achieved. Our discussion of the laws of thermodynamics in the 2nd and 3rd Chapters indicates that their application, combined with knowledge of thermophysical properties, provides for the evaluation of the feasibility of processes and, as we will see in the next Chapter, for the efficient energy utilization, i.e. the first objective of chemical engineering thermodynamics. The theoretical framework for the accomplishment of the second and third objectives, i.e. the evaluation of thermophysical properties and of phase and chemical reaction equilibria based of course again on the laws ofthermodynamics was provided by J.W. Gibbs, who is thus considered the father of this filed. The application of this framework to the solution of actual problems, however, has involved  and continues to do so  several investigators.
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4.12.1 J.W. Gibbs (18391903) Gibbs grew up in his native New Haven, attended Yale College and graduated in 1858, having won a number of prizes for his work in Latin and, especially, mathematics. He continued his studies in the new graduate school at Yale, and was one of the first students to be awarded a Ph.D. by an American university. Since Gibbs is nowfamous for the abstract and theoretical character of his scientific work, it is a little startling to learn that his doctorate was earned in engineering and that his thesis has the title, "On the Form of the Teeth of Wheels in Spur Gearing" ... After receiving his doctorate Gibbs was appointed a tutor in Yale College where he gave elementary instruction in Latin and physics for three years. During this period he continued to work on engineering problems, and in 1866he obtained a patent on his design for an improved railway car brake. That same year Gibbs left New Haven for three years of study in Europe, the only time he would ever be away from his native city for an extended period... In 1871 Gibbs was appointed to the newly created professorship of mathematical physics at Yale, but the official notice ofthis appointment stated flatly that it was "without salary". The Yale administration was presumably aware that Gibbs' financial independence would make it possible for him to accept this unendowed professorship. In any case his teaching duties would be very light: he had no undergraduate courses, and very few of the students in the small graduate department were ready to tackle the courses Gibbs offered. He received no salary for the first nine years of his appointment, and seems not even to have been tempted by an offer from Bowdoin College that would have given him $1800 a year and his choice of the chairs of mathematics and physics. Only in 1880, when Gibbs was on the verge of leaving Yale to join the faculty of the appealing new Johns Hopkins University, a university devoted primarily to graduate study and research, did his home institution offer him a salary. Even though Yale offered only two thirds of what Hopkins would have paid, the advantages of remaining in his familiar environment, of being at home, convinced Gibbs to stay. His decision was surely influenced by the warm way in which his colleagues reacted to the threat of losing him. "Johns Hopkins can get on vastly better without you than we can. We cannot. " These were the words of James Dwight Dana, the distinguished Yale geologist; Gibbs found it impossible to leave. " (Klein, 1982)
A very interesting account of the man and his contributions is given by Klein (1982; 1983). Very briefly: l.In the first paper (1873): 'Graphical Methods in the Thermodynamics
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of Fluids', he introduced the fundamental equation, which for a closed homogeneous system has the form (Chapter 9): U=f(S,V)
Gibbs places the emphasis on the properties of the material system rather than the 'motive power of heat'. Thus the state functions U, S, and V, take precedence over the quantities that depend on the process carried out by the system, the work and heat it exchanges with the surroundings. The latter was the approach used until then, such as in the aforementioned book by Clausius. Thermodynamics, thus, expands its domain beyond
work and heat, into the description of material properties.
In the same paper he argued in favor of representing the second law in terms of a S T diagram rather than a PV diagram, the one that  as we have seen  Clapeyron used about forty years earlier and was first introduced by Watt (see Problem 4.51). 2. In the second paper published three months later 'A Method of Geometrical Representation of the Thermodynamic Properties of Substances by Means of Surfaces', he arrived at an answer to a very important question at that time: the determination of the pressure at which a liquid and a vapor coexist, e.g. the pressure at which a liquid boils at a given temperature. [Maxwell arrived at the 'equality of areas' approach in 1875, and Clausius at the same, but independently, five years later (Problem 9 .57)]. 3.In the third, his masterpiece, paper 'On the Equilibrium of Heterogeneous Substances' (1878), he further extended the world of thermodynamics by treating chemical  and other  phenomena. He introduced the chemical potential and its use in describing phase and/or chemical equilibria. To appreciate the latter, consider the following simple problem: what are the equilibrium compositions of a methanolwater mixture at a specified temperature and pressure? The solution of this problem is carried out in the threestep procedure depicted in Figure 4.2: Step I. Translation of the real problem into the abstract level of pure thermodynamics; Step II. Solution of the problem in the abstract level; Step III. Translation of the abstract solution into the real world. The purpose of Step I is to define the appropriate function to be used in Step II. Gibbs introduced this function, the chemical potential JJ.;, and provided a very simple answer for Step II: the chemical potential of each component of the equilibrium mixture in the one phase equals that in the other. He provided, thus, the solution to the theoretical part of the problem.
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Abstract Level of Pure Thermodynamics
Step I
Solution
,u,'
= JJ~
Projection ·Of the Real Problem into the Abstract Level
Translation of the Abstract Solution into the Real World
Step Ill
Real World
Problem
Answer
Figure 4.2 The threestep procedure for the application of thermodynamics to the solution of the phase equilibrium problem. (Prausnitz et al, 1986, p.4, reprinted by permission of PrenticeHall, Englewood Cliffs, New Jersey.)
The really difficult step in practice is Step III. The translation of theory into the real world, i.e. the expression of the theoretical function through the measurable quantities: pressure, volume, temperature, composition, and heat capacities. While this translation was greatly facilitated by G .N. Lewis, who introduced early in this century the concepts of partial property, fugacity and ideal solution, it has been  and still is  the task of many people, who have been using two main tools to this purpose: a.Equations of State, and b.Activity coefficients. In addition, the need of absolute entropy values in chemical reaction equilibrium calculations relies on Nernst's third law of thermodynamics.
4.12.2 Equations of State The pioneer of the filed is J.D. van der Waals (18371923) who introduced in 1873 the first equation of state (EoS) applicable to real fluids:
P=
RT
Vb
a y2
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where V is the molar volume and a and b are positive constants, specific for each fluid. Here is how Otto Redlich (1976) presents it: "The famous equation of van der Waals has the advantages of a simple theoretical basis and of elucidating the whole field of the gaseous and liquid states. The two reasons for deviations from the perfect gas equation at moderate pressures had been discussed even before van der Waals. The volume available to a molecule for its motion is the total volume V reduced by the volume b occupied by all other molecules present in the same volume V. The true concentration in the freely available space is therefore NA/(Vb) rather than NA/V. (NA: Avogadro number). Therefore the term RT!V of the perfect gas is replaced by RTI(Vb). The second reason is the intermolecular attraction (which leads to condensation at high pressures); it increases with decreasing distances between the molecules. Its effect is to pull back the molecules from the wall to the bulk. Thus it results in a reduction of the pressure; for each molecule near the wall of the vessel the effect is proportional to the concentration of the retracting molecules or to 1/V. Since the concentration of the molecules hitting the wall during a second is also proportional to the concentration, the total effect on the pressure is proportional to its square, or to 1/V2 . Van der Waals not only derived these two effects but also discussed the consequences of his equation for the equilibrium between gas and liquid, for the critical state, and for the correspondence in the behavior of different substances. Qualitatively the equation was tremendously successful. The quantitative representation of experimental data was unsatisfactory." (Redlich,p.47) The van der Waals (vdW) EoS, and the 'corresponding states' principle that follows from it (Section 8.10.3), represents the basis on which the famous compressibility charts of Hougen, Watson, and Ragatz were developed around the middle of this century. It also provided the information necessary for Dewar and Onnes to liquify many of the socalled permanent gases. Vander Waals received the Nobel prize in Physics in 1910. The next major improvement came from the modification of the vdW EoS by Redlich and Kwong (RK, 1949):
p
=
RT _ Vb
a {iV(V+b)
for which Redlich writes: "There is no really good theoreti'cal justification for the two changes in the denominator of the aterm but the agreement with observed data is considerably improved so that the equation is sufficient for a number of practical applications."
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The RK EoS dominated the area of prediction of the volumetric behavior of fluids for about two decades. Within the same period, 1942 to 1951, the famous BenedictWebbRubin (BWR) EoS was developed to meet the need for the description of the volumetric and phase equilibria behavior of hydrocarbons for the Petroleum Industry. It is a noncubic EoS containing eight parameters per species, whose evaluation is very cumbersome (a whole floor of the M.W. Kellogg Co., where the EoS was developed, full of handoperated calculators was used; one can imagine the noise!) The BWR EoS has been used for years, even today, for calculations involving mainly hydrocarbon systems. The major step in the application of EoS in phase equilibrium calculations came in 1972 when G. Soave, following the pioneering work of G. Wilson (1964) and J. Joffe and D. Zudkevitch (1966), presented his famous modification of the RK EoS, that was successful in the correlation and prediction of phase equilibria in nonpolar systems. After the complicated BWR EoS, the field of EoS applications returned to the simplicity and elegance of the van der Waals form. In the same period J. Prausnitz (1969); C. Tsonopoulos (1974, 1975); and J. O'Connell (1975), following the pioneering work of K. Pitzer in the middle 50's, presented generalized correlations for the second virial coefficient. The theoretically based virial equation introduced by K. Onnes in 1901, could be now used in the prediction of the volumetric behavior of nonpolar and often polar gases and vapors, but only at low pressures. Question and ... Question
* What did money contribute to thermodynamics? Or, * What would Joule and Gibbs have contributed without it? 4.12.3 Activity Coefficients Around the same period with K. Onnes, FrancoisMarie Raoult introduced his famous relationship: 
s
P; = X;P;
known as Raoult's law: the partial pressure of component i over its equilibrium solution is proportional to its concentration; the proportionality constant is its vapor pressure Pt.
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Raoult's experimental work was unsurpassed, agreeing in many instances with modern measurements to within 0. 0001 °C. His thermometer was considered by many antediluvian (i.e. belonging to an era before the Flood), but as Van't Hoff expressed it: "with this antediluvian thermometer the world was conquered." Raoult's law is, of course, valid for mixtures of very similar compounds, i.e. 'ideal' solutions. For 'nonideal' solutions: 
s
P; = Y;X;P;
where the activity coefficient Y; can be viewed as the 'correction factor' to Raoult's law and is a function of concentration and temperature, for low to moderate pressures. The first theory for nonideal solutions was proposed in 1910 by J.J. van Laar who was one of van der Waals' students and, later, collaborators. His development was based on the vdW EoS and it is very elegantly described by Prausnitz et al (1986), leading to an expression for the activity coefficient involving two temperaturedependent parameters. It took a little over half a century for the next major development to occur. G. Wilson in 1964 introduced the idea of 'local compositions' which led to an expression for Y; bearing his name and later to the NRTL and UNIQUAC expressions by J. Prausnitz and his coworkers (1967 and 1975). The expressions have two main advantages over the van Laar model; first, their parameters have a builtin temperature dependency making them very useful in distillation applications (why?); second, they provide successful prediction of multicomponent activity coefficients from binary ones. Finally, based on the UNIQUAC model, Fredenslund and his coworkers (1977) have developed a groupcontribution model, UNIF AC, that can be used for the estimation of activity coefficients for a large variety of systems.
4.12.4 The Third Law of Thermodynamics Application of the first and second laws requires, as we have seen, the evaluation of differences of the thermodynamic properties: internal energy, enthalpy, and entropy. No absolute values of these properties are thus required. Evaluation, on the other hand, of equilibrium conversions in chemical reactions requires absolute values of the entropy. This evaluation utilizes the Third Law (Section 1.14.1) introduced in 1906 by Walther Nernst
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(18641941). For an interesting account of the development of this law, see Denbigh (1981) and Hatsopoulos and Keenan.
4.12.5 Other Contributors The application of chemical engineering thermodynamics in the industrial world is the result of the experimental and theoretical contributions of a large number of individuals in addition to those already mentioned. Among them: K. Denbigh, R.L. Dodge, P. Flory, E. Guggenheim, J. Hildebrand, 0. Hougen, D. Katz, W. Kay, W. Lacey, G.N. Lewis, B. Linnhoff, K. Pitzer, I. Prigogine, R. Reid, F. Rossini, J. Rowlinson, B. Sage, T. Sherwood, and H. Van Ness, stand out. Finally, as judged by the relevant literature and international conferences, over 500 individuals are working today in the field, especially in thermophysical data and in phase equilibria evaluation.
References Cardwell, D.S.L., 1971. From Watt to Clausius, the Rise of Thermodynamics in the Early Industrial Age, Cornell University Press, Ithaca, N.Y. Cardwell, D.S.L., 1983. In Springs of Scientific Creativity, Aris, R., Davis, H.T., Stuewer, R.H., Eds, University of Minnesota Press, Minneapolis, MN. (Copyright 1983 by the University of Minnesota, reprinted with permission.) Cleghorn, W., 1719./nauguralDissertation at the University of Edinburgh, (as quoted by Hatsopoulos and Keenan, op. cit.). Denbigh, K. 1981. The Principles of Chemical Equilibrium, Cambridge University Press, Cambridge. Hart, I. B., 1928. The Great Engineers, Books for Libraries Press, Freeport, N.Y. Hatsopoulos, G.N., Keenan, J.H., 1965. Principles of General Thermodynamics, Wiley, N.Y. Klein, M.J., 1982. Foote Prints, 45, 2, 15. Klein, M.J., 1983. In Springs of Scientific Creativity, op. cit. Locke, John, 1772. Elements of Natural Philosophy, quoted in The Early Development of the Concepts of Temperature and Heat, by Duane Rotter, Harvard University Press, Cambridge, Mass., 1950. Prausnitz, J.M., Lichtenthaler, R.N., de Azevedo, E.G., 1986. Molecular Thermodynamics of FluidPhase Equilibria, 2nd Ed., PrenticeHall, Englewood Cliffs, N.J.
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Redlich, 0., 1976. Thermodynamics: Fundamentals, Applications, Elsevier,
N.Y.
Problems 4.1 Describe the operation of the following devices: a.Heron's Aeolopile; b.della Porta's; c.Branca's. 4.2 For the Savery engine: a.describe the method of operation; b. what is the maximum depth that the engine could raise water from? c. what should the boiler pressure and temperature be to raise water 150ft from its level in the mine? d. why was the presence of air detrimental to the engine's operation? 4.3 The same as the previous Problem parts (a) through (c) but for the Newcomen engine. 4.4 What is the origin of the atmospheric and of the steam pressure? 4.5 Explain Watt's dilemma that: "for maximum economy the cylinder must be kept hot all the time, but for maximum power it must be cooled down once every cycle. • 4.6 What was the main improvement of Watt's engine over that of Newcomen? 4.7 The same as the previous Problem, but for Trevithick's one over that of Watt. 4.8 Why would Galileo' s thermometer give different readings from day to day, even when the actual temperature was the same? 4.9 With reference to Galileo's and Descartes' theories about the nature of heat: a. what questions were they trying to answer? b.discuss and compare the two theories. 4.10 What makes the "Principle of Conservation of Fire  and later of Heat", important in the development of thermodynamics? 4.11 Who discovered the 'latent' heat? 4.12 What do you consider as the first most important step in the development in the science of thermodynamics? 4.13 With reference to the 'Caloric' theory: a.state it; b. what is its origin? c.how did it explain phase changes? d. give the main argument against ii.
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4.14 In comparing the efficiency of a waterwheel to that of the Camot engine, discuss the: a.similarities, and b. differences. 4.15 Discuss the most important  in your opinion contribution that Clapeyron made. 4.16 With reference to the development of Eq .4. 8.1 by Clapeyron, explain why the amounts of work involved in steps 41 and 23 cancel each other. 4.17 Using data for a fluid of your choice, other than steam, calculate values for the Camot factor using the Clapeyron equation. 4.18 What are the main reasons  according to Cardwell  that caused Mayer's failure to convince his contemporaries about the convertibility of work to heat? 4.19 Discuss the importance of Joule's work. 4.20.a.Why did Joule have to prove that the difference between Cp and Cv for a gas is a consequence of the work involved in the change of the gas volume? b.Describe the experiments that he used for this purpose. c. Where else could the heat absorbed by the water come from? d.Did he prove the same for liquids? Explain. 4.21 With reference to Joule's experiments involving the expansion of air from one container to an other, what is responsible for the observed temperature changes of the two baths? 4.22 Describe one of Joule's experiments for the conversion of work to heat, and using what you consider reasonable data calculate the obtained temperature increase. Any comments on the required accuracy? 4.23 Explain Joule's criticism of Camot's and of Clapeyron's work. 4.24 How did Clausius reconcile the findings of Camot and of Joule? 4.25 With reference to the verbal statement of the second law: a. who made it? b.how did he prove it? 4.26 Who presented the analytical statement of the second law, for: a.reversible processes; and b.real life, irreversible, processes? 4.27 Compare Camot's and Clausius' proofs that a reversible heat engine produces the maximum amount of work for given temperatures of the heat reservoir and heat sink. 4.28 How did Clausius define entropy? 4.29 Present the major steps in the development of the first and second laws. 4.30 Discuss, briefly, the contributions of: a. Gibbs; b. van der Waals; c.Raoult, in the development of chemical engineering thermodynamics. 4.31 Discuss the origin of the name 'thermodynamics'. 4.32 Present a flow diagram that identifies the important steps in the development of chemical engineering thermodynamics.
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4.51.a.Develop an expression for the thermal efficiency of a Camot cycle using: (i) a P V, and (ii) a ST diagram. b. Use the approach you consider easier to prove that for a real engine: LlSwtal > 0. 4.52 Discuss briefly the importance of Gibbs' contributions in determining the thermodynamic properties of fluids. 4.53 An equimolar liquid mixture of water with ethanol at 80°C is in equilibrium with the vapor above it, whose composition is sought. a. Can it be determined without resorting to an experimental measurement? b.If yes, how and what problems do you expect? 4.54 Develop the van der Waals EoS following Redlich's arguments. 4.55 Explain the practical advantages of Wilson's equation over that of van Laar.
5 Efficient Energy Utilization: Energy Conservation
As for energy conservation progress thus far, I believe only the suiface has been scratched in the process industries. At present, inefficient units have been shut down. The conservation measures implemented on those units still operating are largely in the realm of howsekeeping and simple additional heat recovery. Few fundamelltal improvements in process or equipment design have been implemented. These will be necessary for dealing with the economics of shortage in the future. (Kenney, 1984, p.ix)
5.1 Introduction Until the early 1970's chemical engineers did not pay much attention to efficient energy utilization. Energy was very cheap, at least in the United States, and its impact on plant operating cost relatively small. The oil embargo of 1973, however, and the resulting drastic increase in oil prices, changed the picture substantially. This is demonstrated in Figure 5.1, which shows the energy cost profile from 1972 to 1980 of a typical Gulf Coast petrochemical plant. Notice that the energy cost jumped, from about 20% of the total in 1972 to about 60% in 1980. And this in spite of the significant effects of energy conservation that yielded, as shown in Figure 5.1, savings in 1980 equal to the total operating cost in 1972. Efficient energy utilization, i.e. energy conservation, became consequently a factor of paramount importance in the Chemical and Petroleum industries. Thus the U.S. Chemical industry committed to, and achieved, a 113 reduction in energy consumption by the early 1980's. But cost  in spite of the dramatic impact demonstrated in Figure 5.1  was not the only reason. It became also apparent  somewhat all of a sudden that the availability of energy was not unlimited.
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350
...
300
~Nonenergy IHHHHI Energy Cost
Cost
Energy Conservation +++++ Total Cost ( excl. conservation) ~
c: 250
,g
.E 2oo .s u;150 0
u
100 50
1973
1975
Year
1977
1979
1981
Figure 5.1 A comparison of energyrelated and nonenergyrelated costs (excluding feedstock costs) for a typical U.S. petrochemical plant, including the effect of energy conservation. (Information from Kenney, p.3.)
The conservation effort, however, is really in its beginning as Table 5.1 demonstrates: * first, appreciable energy savings can be achieved using existing technology, but * second, even with this improved operation, we are a long way from the minimum energy consumption suggested by thermodynamics. In the petroleum refining, for example, use of 1973 technology will reduce the energy consumption from 11 times to only 8 times the minimum amount required.
5.2 Objective and Approach Efficient energy utilization  energy conservation  is a very broad and complex subject to be covered in the limited amount of space allocated here. Our objective, therefore, is to focus attention to the problem and
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Table 5.1 Comparison of the Specific Fuel Consumption of Known Processes with the Theoretical Minimum for Selected U.S. Industries(a). (From Kenney, p.3.) Potential Process Specific fuel Theoretical consumption specitic fuel minimum 1968 consumption specific fuel (MBtu/ton) using 1973 consumption (MBtu/ton) (b) technology (MBtu/ton) Iron and steel manufacturing Petroleum refining Paper manufacturing
26.5
17.2
6.0
4.4
3.3
0.4
39.0 (c)
23.8
> 0.2 (d) < 0.1
Primary aluminum production (e)
190.0
152.0
25.2
Cement production
7.9
4.7
0.8
(a) Gyftopoulos et at (1974) (b) Based on thermodynamic availability analysis (c) Includes 14.5 MBtu/ton of paper produced from waste products consumed as fuel by the paper industry (d) A negative value means that no fuel is required (e) Does not include the effect of scrap recycling
provide the necessary background for dealing with it. More specifically, it is to: I .Introduce the concepts that are necessary to evaluate the efficiency of a given process and, thus, determine the available room for energy conservation. 2.Demonstrate the effect that the temperature and pressure of steam  the major agent for the utilization of our main source of energy, thermal have on its ability to produce useful work. 3.Emphasize the importance of the temperature, that an amount of thermal energy is available at, on the value of this energy. 4.Apply these concepts to a major approach towards energy conservation, Cogeneration.
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We will start with establishing the 'standard of excellence' for efficient energy utilization: the concept of ideal work. When this is reached, no further energy conservation is possible. We continue with a discussion of the 'usefulness' of the various forms of energy and the method for determining it through the use of a relatively new concept, exergy. We proceed with a presentation of the quantitative criteria used to determine the degree of efficient energy utilization in a given process:.first and second law efficiencies, followed by a brief description of one of the methods used to obtain increased such efficiencies in the use of thermal energy, Cogeneration. We demonstrate the energy savings achieved and discuss the limitations involved and the ways around them. We continue with Kenney's commentary on the favorable economics of energy conservation, but also on the Institutional barriers that limit its wider applicability, and close the Chapter with some concluding re
marks.
5.3 Ideal Work 5.3.1 Definition For a given process, involving a specified change, there is an amount of work associated with it, referred to as Ideal Work W(id), which is: * the minimum one, if the process is work consuming; *the maximum one, if the process is work producing. To evaluate the ideal work, we set up a process that brings about the same change as the real one, whereby: l.all operations are carried out reversibly; 2.all heat transfer  to, or from, the system  is carried out using the surroundings reversibly, of course as a 'heat reservoir' or as a 'heat sink' respective! y. Thus, if heat is transferred to the system then, no matter what its origin in reality is, it must come from the surroundings in calculating the ideal work. The same applies when heat is rejected by the system. A temperature of T0 = 298 K is typically accepted for the surroundings. Such a process is, of course, a hypothetical one. If an expansion is involved, for example, it would require a 'perfect', i.e. a completely reversible turbine; if heat is to be transferred from the system at temperature T to the surroundings (at T0), a Carnot cycle must be employed.
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Efficient Energy Utilization: Energy Conservation
The practical usefulness of such a hypothetical reversible process lies with the observation that, since it effects the same change as the real process, it can be used as a basis for evaluating the efficiency of energy use in the latter. In the following Example we demonstrate the concept of ideal work and then proceed to develop the general formulation for its determination.
5.3.2 Example 5.1 One kg of steam at P 1 = 60 bar and t 1 = 540°C flows through a turbine with an exhaust pressure of P2 = 1 bar. Assuming that the turbine is adiabatic but irreversible, with an efficiency of 0. 9 calculate the ideal work Wiid ). To calculate the ideal work, Wpd ), we must first establish the state of the steam leaving the turbine, since we only know the exit pressure. To this purpose, we calculate first the reversible and adiabatic work, WiRA). * From the first law: Ws(Q 1) = 3241.0(1 T0 /T) = 2775.1 kJ Hence the second law efficiency is: n11 = 713.7/(2775.1+ 3.55) = 0.257
5. 7.4 Connnnents l.Notice that our definitions of first and second law efficiencies, Eqs
5. 7.1 and 5. 7.2 respectively, are pragmatic for they consider the useful
output streams only. If we consider all exit streams, then: a. The first law efficiency will be equal to 1, since energy is conserved, which can be misleading. b.The second law efficiency will be higher, which can be also misleading. It is recommended, however, that the enthalpies and exergies of the nonuseful streams be evaluated to identify possible sources of substantial energy losses.
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2.Thus, in the power plant discussed in Example 5.8, the exergy of the flue gases and of the cooling water leaving the condenser should be considered. Evaluation of the first requires knowledge of the stoichiometry of the combustion reaction and of the excess air used (see Sussman for details). Evaluation of the second, is straightforward (see Problem 5.17). In general, use of the flue gases for preheating purposes leads to low exit temperature and, consequently, low exergy losses, typically about 2% of the fuel exergy value. 3.In calculating the fuel exergy so far, we have used the heat of combustion at the temperature of the flue gases (1800°C, for example). The true fuel exergy, however, i.e. the maximum possible useful energy that can be obtained from it, is evaluated by assuming that the fuel reacts reversibly with oxygen in a so called 'fuel cell'. We will discuss the method of evaluating this exergy which is referred to as chemical exergy  when we develop the necessary background in Chapter 15 (Example 15.4). We will see that in this case it is the Gibbs free energy of the combustion reaction that is used and, as a result, the obtained chemical exergy is larger than that based on the heat of combustion. We discuss next how improved first and second law efficiencies in the use of thermal energy can be realized through a very interesting approach, Cogeneration.
5.8 Cogeneration 5.8.1 Rationale How can the low second law efficiency of the boiler in Example 5.7 and the low first and second law efficiencies of the power plant of Example 5.8 be improved? How, in other words, can we improve the utilization of the fuel heating value, thus effecting energy conservation? The answer is suggested by the results of Example 5.6: High temperature and pressure steam has high exergy value and should be used for the production of mechanical work and electricity; low pressure steam has low exergy, but still high heating value, and should be used for heating purposes. This combination is called Cogeneration. The following Example demonstrates the advantages of cogeneration.
Efficient Energy Utilization: Energy Conservation
5.8.2
~aunple
175
5.9
A chemical plant needs 1000 kW (1 MW) of electric power and 17282 MJ!hr (4.8 MW) of process heat as 10 bar saturated steam. Two schemes, shown in Figures 5.E.9a and 5.E.9b, are considered for this purpose. In the 'conventional' scheme, two separate boilers are used, that produce: ... the one, high pressure steam (110 bar and 550°C), which is expanded to a pressure of 0.1 bar for power generation; • the other, the 10 bar saturated steam. Both boilers are efficient in the first law sense, with an efficiency of 0. 90. In the 'cogeneration' scheme, a single boiler is used generating steam at 110 bar and 550°C which flows through a turbine with an exhaust pressure of 15 bar. This steam is used then in a heat exchanger to generate the I 0 bar steam for process heating. This boiler also has a first Jaw efficiency of 0.9. The following information is given: • turbines are adiabatic but irreversible, with an efficiency of 0. 8; • the process steam returning to the boiler can be assumed to be saturated liquid at 10 bar; • similarly, the water from the condenser is saturated liquid at 0.1 bar, and from the heat exchanger, at 15 bar; • pump duties can be neglected. The pertinent data are presented in Figures 5.E.9a and 5.E.9b. Determine the: a.Fuel heat requirement for the two schemes; b.First law efficiencies; and c. Second law efficiencies; assuming the fuel to be methane. Given: Enthalpy of combustion: 212.8 kcal/mol = 49.5 MJ/kg Chemical Exergy of CH4 : 195.5 kcal/mol = 45.5 MJ/kg (Sussman) (Undergraduate students can proceed to the Comments at the end of the Example.)
a.Fuel Heat Requirement 1. Conventional Scheme Ws = 0.80(3490 2120) = 1096 kJ/kg = 1.096 MJ/kg Steam generated: (1000 kW)(3600 kJ/kWh)/(1096 kJ/kg) = 3285 kglhr Heat absorbed in power plant boiler: (3285 kg/hr)(3490 192) kJ/kg = 10834 MJ/hr Heat absorbed in process steam boiler: 17282 MJ lhr Total fuel heat: (10834 + 17282)·106 /0.9 = 31240MJ/hr
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176
P = 0.1 bar t = 46•c H = 2394 kJ/kg S = 7.56 kJ/kg K
110 bar 55o•c = 3490 kJ/kg 6.7 kJ/kg K
P t = H S =
12038 MJ/hr
Boiler
Condenser
P
t
= 0.1 bar = 46•c
h = 192 kJ/kg
P
t
= 10
bor
= 1ao•c
H = 2776 kJ/kg
19202 MJ/hr
Process
Steam
Q
17282 MJ/hr
Boiler
P
t
= 10
bor
= 1ao•c
h = 763 kJ/kg
Figure 5.E.9a Conventional Scheme for Example 5.9.
\0
Ul
"0 (P
3
tT1 >< ::»
0' ....
3~
~
::r'
()
(/)
::l
~
ac;·
g
(/Q
() 0
0"
"'
tr1
01
;;!
;tc:
23202 MJ/hr
Boiler
P = 110 bar t = 550° c H = 3490 kJ/kg S = 6.7 kJ/kg K
ws =
c__.,,
1000 kW
c
P = 15 bar t = 19ft c h = 845 kJ/kg
Heat Exchanger
=
P = 15 bar t 298° H = 3034 kJ/kg S = 6.91 kJ/kg K
Q
=
10 bar 18Cf c 2776 kJ/kg
P = 10 bar t = 18Cf c h = 763 kJ/kg
17282 MJ/hr
=
= H =
P t
::l
...] ...]
::l
~
:1 !e. c;·
"'
g
()
~ .... ~
tT1
P.
c;·
~·
~
'
.
.....e 0.
4
c:
w
2
12.0
8.0
4.0
Pressure (MPa)
Figure 9.E.3 Variation of (S 0

16.0
20.0
S) with pressure foributane.
9.7.4 Evaluation of Departure Functions ExaminationofEqs 9.3.2, 9.3.10, 9.3.15, and 9.3.19 indicates that evaluation of isothermal changes respectively for U, H, A, and G in terms of PVT data  directly or through an equation of state  can be accomplished readily for the last two only. Thus:
I
v•
A 0 A = v PdV
(9.7.3) (9.7.4)
Since most equations of state (EoS) are in the form P = f(V,T), rather than V = f(P,T), we choose Eq.9.7.3 and, to eliminate the problem of integrating between an ideal gas and a real gas state, we break the integral into two parts: A 0 A = J 00 PdV Jv•PdV = J 00 PdV Jv•RTdV ooV V oa V
since the second integral covers the ideal gas region.
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302
Finally, to avoid the difficulties involved with the infinity limit, we add to and subtract from the righthand side the integral:
Iv v
ooRTdV
Then, since also (VIV 0 ) = z: A 0 A =
I ;(p
(9.7.5)
R:)dv +RTlnz
The other departure functions can be now easily developed:
Ivoo[(ap) arv
l!_]dVRlnz
(9.7.6)
H 0 H = (A 0 A)+T(S 0 S)RT(z1)
(9.7.7)
U 0 U = (A 0 A)+T(S 0 S)
(9.7.8)
G 0 G = (A 0 A)RT(z1)
(9.7.9)
sos = [a(AoA)]
ar
v
=
v
Evaluation of departure functions from experimental PVT data can be carried out graphically using these expressions. This approach, however, involves evaluation of derivatives from plots of experimental data which introduces substantial uncertainties. In the typical case, therefore, the PVT data are fitted to an accurate EoS to facilitate the calculations. Or, in the absence of such data, some generalized EoS can be used.
9.7.5 Evaluation of Enthalpy and Entropy Differences Our objective of evaluating enthalpy and entropy differences for a homogeneous phase undergoing a change from state 1(P1,T1) to state 2(P2,T2), through the measurable quantities: * PVT, and *C* p has been, thus, accomplished:
(9.7.10) (9.7.11)
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Thermodynamic Properties of Pure Fluids
9.7.6 Development of Enthalpy and Entropy Tables and Charts Since we are interested in enthalpy and entropy differences, the values of these properties presented in Tables and Charts are not the absolute ones but relative to an arbitrarily chosen set of values at some reference point. In the Steam Tables, for example, the H and S values are relative to a zero value arbitrarily assigned to saturated water at 273.16 K. Once the reference point is set, enthalpy and entropy values are determined by using the following information: !.PressureVolumeTemperature, PVT, data to evaluate the parameters in an accurate equation of state, through which the departure function values will be obtained; 2.Ideal gas heat capacities; and 3. Vapor pressure as a function of temperature to establish: * the phase boundary, and * in combination with saturated vapor and liquid volumes, the enthalpies and entropies of vaporization, if they are not available from direct measurements. For details see, among others, Goodwin and Haynes (1982) who used this approach to develop extensive tables of thermodynamic data for ibutane. (From which the values presented in Table 9.E.2 were obtained). In the typical case, the extensive PVT data needed for the evaluation of the departure function values through an accurate EoS are not available. We resort, therefore, to the estimation techniques discussed next.
9.8 Estimation of Departure Functions 9.8.1 The Methods Estimation of departure function values is carried out by using either: * Pitzer's corresponding states approach, or *equations of state (EoS). Using the corresponding states principle, Pitzer proposed that:
[] [ro> (A)[cso;s>r> =
+
(9.8.1)
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304
(9.8.2) For best results the Lee and Kesler (1975) Tables presented in Appendix D should be used. Use of the SRK EoS (Table 8.3) for the Helmholtz free energy departure, Eq.9.7.5, yields (see Example 9.4): Ao A = RTln Yb +~In Y+b yo b Y
(9.8.3)
IntroductionofEq.9.8.3 into Eq.9.7.6yields:
S 0 S
=
Yb
Rlnyo
mac [1 +m(l Tro.l]
+
b
T To5 c r
Y+b Iny
(9.8.4)
l
Introduction of Eqs 9.8.3 and 9.8.4 into Eq.9.7.7 yields:
a + macbTro.s [1+m(lTr.J o~] Y+b H 0 H= [ b lnV RT(z1) (9.8.5) Similar expressions are obtained with other EoS. Development, finally, of the departure expressions for the internal energy (U 0  U ), and the Gibbs free energy (G 0  G), is straightforward through Eqs 9.7.8 and 9.7.9.
9.8.2 Example 9.4 Develop the Helmholtz free energy departure function using the SRK EoS. Introduction of the SRK EoS into Eq. 9. 7.5 yields:
Ao A =
J ""[v
RT a  RT]dY + Yb Y(Y+b) Y
= RTln Yb Y
+~In Y+b + RTln(~) b
Y
= RTln Y b + ~In Y + b
yo
i.e. Eq.9.8.3.
b
Y
yo
RTln(~) yo
Thermodynamic Properties of Pure Fluids
305
9.8.3 Example 9.5 In calculating enthalpy and entropy departures foributane in Examples 9.2 and 9.3, we assumed ideal gas behavior at 0.01 MPa (0.1 bar). Strictly speaking, of course, ideal gas behavior is approached only as P ""* 0. Evaluate the corrections to the (H 0  H) and (S 0  S) values at 400 K required as a result of this assumption. Given: z(0.01 MPa) "' 0.9979. a. Enthalpy
Assuming that the SRK EoS provides an accurate description of the nonideal behavior of iC4 from 0 to 0.01 MPa, and using Tc "'408.1K; Pc "' 3.648 MPa; and Ct.l "'0.176, we have: ac "' 1.349·106 MPa cm6 /(moli b "' 80.58 cm3 /mol m "' 0. 7533, a "'1.007 From Eq.9.8.5: [H 0  H(P = 0.01 MPa)] = 14.01/mol. b.Entropy Similarly, using Eq.9.8.4: [S 0

S(P
= 0.01 MPa)] .. 0.027 J/mol K.
The (H 0  H) values in Figure 9.E.2 must be increased, therefore, by 14 J/mol; and the (S 0  S) values of Figure 9.E.3 by 0.027 J/mol K, both very small quantities, as it should be expected.
9.8.4 Example 9.6 Estimate the entropy and enthalpy departures foributane at 400K and 2. 80 MPa, using the Pitzer correlation and the SRK EoS, and compare them to the experimental values. Using the Pitzer correlation with T, "' 0. 980, and P, = 0. 768: [(S 0  S)IR] "' 0.908 [(S 0  S)IR] O> "' 1.183 and: (S 0  S) "' 9.28 J/mol K. [(H 0  H)IRTc] (O) = 1.196 [(H 0  H)/RTc ] (I) "' 1.235 and: (H 0 H) = 4796 J/mol. From Examples 9.2, 9.3, and 9.5, the experimental values for the two departure functions are: (S 0  S) "' 8.87 + 0.03 .. 8.90 J/mol K (H 0  H) = 4659 + 14 .. 4673 J/mol;
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306
which indicate errors of 4.3% and 2.6% in the predicted values respectively. For the SRK EoS, we must first calculate the volume at 400 K and 2.8 MPa obtaining: V = 733.5 cm3 /mol; and the ideal gas volume at the same conditions: V 0 = 1187.7 cm3 /mol. Use of these values yields: (S 0  S) = 8.25 J/mol K; and (H 0  H) = 4353 J/mol with errors of 7.3% and 6.8% respectively.
9.8.5 Comments on the Estimation of Entropy and Enthalpy Departures l. The Pitzer corresponding states approach, with the LeeKesler Tables of Appendix D, and the cubic equations of state of Table 8.3 give reasonable accuracy for nonpolar/weakly polar compounds. 2.Typical results with the Pitzer approach and the SRK EoS are presented in Table 9.1 and similar ones are obtained with the PR and vdW711 EoS. For the latter t = t 0 should be used to avoid the pronounced and distorting effect that (dt/dT) has on enthalpy and entropy departure values close to the critical point (Androulakis et al, 1989). 3.At very high pressures the Pitzer approach should be preferred over the aforementioned, and other, cubic EoS. 4.For polar systems, the virial EoS truncated after B gives reasonable estimates of departure values within the range of applicability of the equation, provided that the required temperature dependency of B is available from experimental data. Use of an empirical correlation forB could be made for nonpolar/weakly polar compounds only. 5.For polar systems at higher pressures, the Pitzer correlation should be used, but the uncertainty increases as the polarity of the system increases. Table 9.1 Estimation of Entropy and Enthalpy Departure Values foriButane p (Ho H) • %Error (So S) • %Error T J/mol
Pit. SRK
4673 7.3 4.3 8.90 5046 1.5 10.5 440 8.69 4.7 5.7 5606 700 6.63 (*)Calculated from the data of Goodwin and Haynes.
2.6 6.8 7.1 0.2 1.4 4.4
K
400
MPa 2.8 4.0 20.0
J/mol K
Pit.
SRK
Thermodynamic Properties of Pure Fluids
307
6.For the estimation of departure functions for pure liquids, see Reid et al. 7. The departure functions are also used for mixtures, provided that L and L 0 refer to the same composition, and the above comments apply to them as well. 8.0ur discussion so far has dealt with the estimation of the nonideal gas behavior contribution to (L,. L 1), i.e. (L,.0  L,.) and (L 1° L 1). For (L,.0  L 1°) the required ideal gas heat capacity depends on the internal molecular structure (e.g. bond lengths, vibration frequencies, and configuration of the constituent atoms) and, unfortunately, cannot be estimated through generalized correlations, such as those used for (L 0  L), which reflect the intermolecular forces. 9.C/ values are available for a large number of compounds, however, mostly from spectroscopic measurements (see Section 8.15). Estimation, based on group contribution techniques, is also possible but for a limited number of compounds (Reid et al, 1987). We proceed now with a discussion of two other very important thermodynamic properties: chemical potential and fugacity.
9.9 The Chemical Potential The chemical potential is the key property in the most important application of chemical engineering thermodynamics, chemical and phase equilibrium. The ease of separation, for example, of methanol from its mixture with water by distillation is determined by the relationship between the concentration of methanol in the liquid and vapor phases, which  as Gibbs showed (Section 4.12)  is dictated by the chemical potential of methanol in the two phases. We saw in Section 9.4, Eq.9.4.16, that the chemical potential is defined in terms of all four energy functions. Since, however, we are interested in chemical and in phase equilibrium reached at constant temperature and pressure, we use the one through the Gibbs free energy. For a pure compound of interest in this Chapter Eq.9.4.15 becomes:
~  = ( ~~L.P
(9.9.1)
Thus, the chemical potential is the Gibbs free energy per mole. It follows then from Eq.9.3.19 that:
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(9.9.2)
dp. = SdT+ VdP
where S and V represent the molar entropy and volume respectively. To evaluate the chemical potential at some temperature T and pressure P, p.(T,P), we integrate Eq.9.9.2: p.(T,P)p.(T1 ,P1) =
J~,SdT+
J:,VdP
(9.9.3)
where p.(T',P') is the chemical potential at some reference pressure P' and temperature T'. Hence p.(T,P) can only be evaluated relatively to its value at some other state. This uncertainty is made worse by the following observation. Assume, for simplicity, that we: a. choose as a reference temperature that of the system, i.e. T' = T ; and b.assign some finite value, say 10 J/mol, to the chemical potential at T and an arbitrarily chosen reference pressure, say P' = 1 atm; and proceed to evaluate the chemical potential at the temperature T = T' as a function of the pressure P, for P < 1 atm. Since in this range we can assume ideal gas behavior, Eq.9.9.3 gives: p.(T,P)
= 10
+
RTln(P/1)
We note that asP becomes progressively smaller, the chemical potential assumes large negative values, approaching negative infinity as P approaches zero. Conversely, if we assign a value of zero to the chemical potential at T as P + 0, then very large values will be obtained for it at higher pressures.
9.10 Fugacity 9.10.1 Definition To avoid the aforementioned uncertainties associated with the evaluation of the chemical potential G. N. Lewis introduced the concept of fugacity in 1901. Consider first the change in chemical potential of a pure ideal gas from pressure P 1 to pressure P 2 at a constanttemperature T. From Eq. 9.9.2:
Jl2 p. 1
= JL (P2 , T)
 p.(P1 , T) = RTin(P2 / P 1 )
(9.10.1)
309
Thermodynamic Properties of Pure Fluids
This expression indicates that, for an ideal gas, the isothermal change in chemical potential can be determined through its initial and final pressure. Turning to real systems, Lewis introduced a new function called jUgacity (/ ), such that the isothermal change in the chemical potential for any compound in any system (gas, liquid, or solid) is given by an expression similar to Eq.9.10.1:
ILl p. 1
=
p.(P2 , 1) p.(Pp 1)
=
RTln(h If.)
(9.10.2)
Obviously, p.1 and/; refer to the same state. Consider now the ratio of the fugacity of a pure compound at a pressure P divided by P. This ratio is called the .fUgacity coefficient:
q,
=
.L p
(9.10.3)
Since ideal gas behavior is approached asP approaches zero:
limq,
PO
=
lim
PO
!_ p
=
1
(9.10.4)
Eq.9.10.4 along with the constant temperature differential form of Eq. 9.10.2: (9.10.5) dp. = RTdlnf provide the complete definition of fugacity for a pure species. Usually we calculate the fugacity through the fugacity coefficient. Before, however, we discuss how the fugacity coefficient is calculated, let us consider a physical interpretation of fugacity.
9.10.2 A Physical Interpretation Fugacity has units of pressure as Eq.9.10.4 indicates, and a comparison of Eqs 9.10.1 and 9.10.2 suggests. Beyond this, however, its physical meaning becomes elusive. Webster's Dictionary defines 'fugacity' as the "quality of being fugacious" and 'fugacious' as "fleeing, or apt to flee; passing quickly away; ephemeral". (Fugacity is indeed ephemeral; one forgets it very fast!). Consider, now, the case of liquid water at 100°C and 1 atm, in equilibrium with its vapor, i.e. saturated water. According to Webster's definition of fugacity we would expect that, due to the equilibrium between the two phases, the molecules of the liquid are as "apt to flee" their
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310
phase as are the molecules of the vapor phase. (And this is physically the case: vaporliquid equilibrium is not static, but dynamic, with molecules equal in number fleeing continuously from one phase to the other.) The fugacity of the liquid should be, thus, equal to that of the vapor. We will demonstrate in Chapter 12, through a combination of the first and second laws, that this also represents a thermodynamic requirement: (9.10.6) where the superscripts and V indicate saturated liquid and vapor phases respective! y. Hence, at least in the case of phase equilibrium, fugacity may be viewed as the 1escaping tendency 1 of the molecules in each phase. 1
/
1
1
1
9.11 Evaluation of Fugacities from Experimental Data 9.11.1 The Mathematical Formulation CombinationofEqs 9.10.5 and 9.9.2 yields, at a constant temperature T: (9.11.1)
RTdlnf = VdP For an ideal gas, at the same temperature, Eq.9.11.1 becomes:
RTdlnP = RT dP
(9.11.2)
p
Subtraction of Eq.9.11.2 from Eq.9.11.1 yields:
RTdtn( ~) = RTdln = ( Vand:
In
= 
1
RT
~)dP
JoP(v RT)dP P
(9.11.3)
(9.11.4)
Eq.9.11.4 can be used along with experimental PVT data for the evaluation of fugacity coefficients through graphical integration. As with the evaluation of enthalpy and entropy departures, however, the integration is carried out by first fitting the PVT'data to an accurate equation of state. And since such EoS express P =f(V,T), rather than V =f(P,T), direct use of Eq. 9 .11.4 is not possible.
311
Thermodynamic Properties of Pure Fluids
To circumvent this problem, let us develop an expression for the difference in the chemical potential of a pure substance between its: * real state at T and P, and * ideal gas state at the same conditions, through the corresponding difference in fugacities. From Eq.9.10.5: p.(T,P) p. 0 (T,P) = RTln([_) = RTinq, p
We notice that, since according to Eq.9.9.1 the chemical potential equals the molar Gibbs free energy, the lefthand side of the above equation equals the negative of the departure function for the latter. Thus: G0

(9.11.5)
G = RTinq,
From Eqs 9.7.9 and 9.7.5: In = (z  1) lnz  _ 1_ RT
Jv
oo
(p
RT) dV V
(9.11.6)
Evaluation of fugacities from experimental PVT data can be thus accomplished with two approaches: a. with Eq.9 .11.6, using the data directly, i.e. graphically, or better, after they have been fitted with an accurate EoS; and b. with Eq.9.11.5, using enthalpy and entropy values generated from the fitted PVT data, since: RTln = (GG 0 ) = (HH 0 )T(SS 0 )
(9.11.7)
9.11.2 Evaluation of Fugacities with an Equation of State We will demonstrate this approach with the virial EoS. Use of its volume series form truncated after C into Eq.9.11.6 yields: In
=
2B 3C lnz +  + V 2V2
(9.11.8)
while use of the pressure series, directly in this case into Eq.9 .11.4, yields: ln =BP  + RT
CB2 (P) RT 2

2
(9.11.9)
As with the case of volume calculations, Eq.9.11.8 performs better than Eq.9.11.9. But when truncated after B, the expression:
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Applied Chemical Engineering Thermodynamics
In
= BP
(9.11.10) RT performs better than the corresponding expression obtained from Eq. 9.11.8 (Van Ness). Evaluation of fugacity coefficients from experimental data is demonstrated in the following Examples.
9.11.3 Example 9.7 Using the Steam Tables, determine the fugacity of steam at 300°C and 70 bar. Two approaches are possible: a. Use of the enthalpy and entropy values; and b.Graphically, using the PVI data directly. The first one will be used here, since it is the easiest and should provide good accuracy, at least comparable to the second. From the Steam Tables at 300°C and 70 bar (by interpolation): H ,. 2839.3 kJ/kg; S ,. 5.9327 kJ/kg K; and at P ,. 0.1 bar and 300°C: H ,. 3076.6 kJ/kg; S ,. 9.2820 kJ/kg K. Since we can reasonably expect ideal gas behavior at the latter conditions: H 0 ,. 3076.6kJ/kg so ,. 9.2820 + (R/18) ln(0.1/70),. 6.2561 kJ/kg K (see Example 9.3) Hence, from Eq.9.11.7: ln(f/P),. 18 [(2839.3 3076.6) 573.15(5.9327 6.2561)]/[(8.314)(573.15)] = 0.196 and f,. 57.6 bar.
9.11.4 Example 9.8 Goodwin and Haynes give the following value for the second vi rial coefficient of ibutane (iC4) at 410 K: B = 324.6 cm3 /mol. Calculate the fugacity coefficient of iC4 as a function of pressure at 410 K and compare it to the recommended values of Goodwin and Haynes presented in Table 9 .E. 8, and arrived at by using enthalpy and entropy values. From Eq.9.11.10: ln = BPIRT,. 0.0952P(MPa). The results are presented in Table 9.E.8 and indicate that: a. Within the range of applicability of the virial EoS truncated after B: P'!i!.(T/2)(Pc I Tc ), i.e. below about 1.8 MPa, good results are obtained. For still better accuracy, the third virial coefficient must be used. b.At higher pressures the results become progressively poorer, as expected.
313
Thermodynamic Properties of Pure Fluids
Table 9.E.8 Performance of the Virial Equation in the Evaluation of Fuga
city Coefficients foriButane at 410 K as a Function of Pressure. (Experimental Values from Goodwin and Haynes, GH). (/)
P(MPa) 0.101325 0.5 1.0 1.4 1.8 2.2 2.6 3.0 3.4 3.8 4.2 4.6 5.0 6.0 7.0 8.0 10.0 12.0 14.0 16.0 20.0
GH
Virial
%Error
0.983 0.948 0.892 0.858 0.824 0.789 0.155 0.719 0.683 0.641 0.594 0.553 0.519 0.453 0.405 0.370 0.322 0.290 0.269 0.254 0.235
0.990 0.954 0.909 0.875 0.843 0.811 0.781 0.752 0.723 0.696 0.670 0.645 0.621 0.565 0.514 0.467 0.386 0.319 0.264 0.218 0.149
0.7 0.6 1.9 2.0 2.3 2.8 3.4 4.5 5.9 8.6 12.8 16.6 19.7 24.7 26.7 26.2 19.9 10.0 1.9 14.2 36.6
9.11.5 Liquid Phase Fugacities Our discussion about fugacities has focused, so far, on vapors and gases. In addition, since the fugacity of a saturated liquid is equal to that of the saturated vapor in equilibrium with it (Eq.9.10.6), our discussion applies also to saturated liquids as well. We will consider next how the fugacity of a subcooled liquid, i.e. one where the pressure on the liquid is higher than its vapor pressure at the given temperature (P > ps ), is determined. We start again with Eq.9.1l.4 and break the integral into two parts: In = _1 RT
_1 JP ( ~ RT)dP  RT)dP+ JP'(v RT P' l p p v 0
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Applied Chemical Engineering Thermodynamics
where the superscript's' stands for saturation, and the subscripts 'v' and 'I' refer to vapor and liquid phases respectively. Notice that the first integral represents the fugacity coefficient of the saturated vapor and, consequently: p V, ps In = ln 8 + J dP +In() p• RT P
(9.11.11)
Eq. 9.11.11 gives, after some algebra, the expression for the liquid phase fugacity,fi: (9.11.12) where:
Pe
=
exp
v, JpP' dP RT
(9.11.13)
is known as the Poynting effect, and represents the effect of pressure on the fugacity of a pure liquid, from saturation to system pressure. Evaluation hence of the liquid phase fugacity at a given temperature T and pressure P requires: a. the value of the fugacity coefficient of the saturated vapor at the temperature T, and b.evaluation of the integral in Eq.9.11.13, either graphically or through an equation of state applicable to the liquid region. As we have seen, however, liquid volumes are fairly insensitive to pressure  except close to the critical point and V1 can be replaced by its average value V1(avg) in the range P 8 toP. Under this assumption, Eq. 9.11.12, becomes: F
' 1
= s psexp
I
V,(avg)(P P 8 )]
RT
(9.11.14)
Evaluation of liquid phase fugacity is considered in the next Example.
9.11.6 Example 9.9 Determine the fugacity of ibutane at 310.9 K and 1OMPa. Given (Goodwin and Haynes): ps = 0.5 MPa; VI (0.5 MPa) = 108.8 cm3 /mol; VI (10 MPa) = 104.8 cm3 /mol; cps= 0.8725.
315
Thermodynamic Properties of Pure Fluids
From Eq.9.11.14, using 1j (avg) = 106.8 cm3 /mol, we obtain.fi = 0.646 MPa, which is in agreement with the value given by Goodwin and Haynes of 0.647 MPa. Notice the insensitivity of the liquid molar volume to pressure: a twentyfold increase in pressure has an effect of less than 4% on the volume; but the significant effect of pressure on the liquid phase fugacity as indicated by the value of Pe = 1.48.
9.12 Estimation of Fugacities 9.12.1 The Methods Estimation of fugacity coefficients  and hence fugacities  can be accomplished with two approaches: a. The Pitzer corresponding states principle; and b.Equations of state. According to Pitzer's method:
(/)
log10 P
=
log10 (1)~ P +
(I)
(1)ru
log10 P
(9.12.1)
and best results are obtained using the Lee and Kesler Tables presented in Appendix D. Estimation of fugacities with the virial EoS truncated after Buses Eq. 9 .11.1 0, with second virial coefficient values obtained from the empirical correlations of Tsonopou1os or of Hayden and O'Connell (Chapter 8). Use of the SRK EoS into Eq.9.11.6leads to: ln
6 MPa (60 bar), the errors are not that high. Similar results should be expected with the PR and vdW711 EoS. 2.For nonpolar and polar compounds the virial equation truncated after B, with the latter obtained from empirical correlations, gives reasonable accuracy within the range of its applicability. 3.At high pressures the Pitzer correlation should be used. The results become poorer, however, with increasing polarity. 4.The importance of fugacity in solving chemical engineering problems will become apparent when we discuss phase and chemical reaction equilibrium. We consider, for example, ·in the next Chapter a simple such application, the use of fugacities in calculating vapor pressures through cubic EoS.
Thermodynamic Properties of Pure Fluids
317
9.13 Summary We can summarize our discussion on the evaluation of thermophysical properties of pure fluids, presented in the last two Chapters, as follows: l.In the typical case, only limited experimental information is available mostly the critical properties and acentric factor, and the ideal gas heat capacity as a function of temperature. We must, consequently, resort to some estimation (predictive) technique. 2.At low pressures, the virial equation truncated after B, with the second virial coefficient values predicted by the Tsonopoulos or the HaydenO'Connellcorrelations, gives the best results. Cubic EoS are also successful, but for nonpolar/weakly polar compounds only. 3.At higher pressures and for nonpolar compounds, cubic EoS can be used except close to the critical temperature. More reliable results in this region can be obtained with Pitzer's corresponding states approach, combined with the LeeKesler Tables, which should be also used at high pressures. [While the Tables cannot be easily adapted for computer use, the generalized BWR EoS, from which these Tables (Appendix D) were developed, can be used for this purpose.] 4. The PitzerLeeKesler approach can be also used for polar compounds at higher pressures. The accuracy, however, decreases with increasing polarity and the presence of hydrogen bonding. S.Finally, before an estimation approach is used, all available sources of data should be examined.
References Androulakis,J., Kalospiros, N., Tassios, D., 1989.FluidPhaseEquilibria,45, 135. Goodwin,R.D., Haynes, W. M., 1982. ThermophysicalProperties oflsobutane from 114 K to 700 Kat Pressures to 70 MPa, Nat. Bur. Standards (U.S.), Washington, D.C. Lee, B.I., Kesler, M.G., 1975.AIChEJ., 21,510. Modell, M., Reid, R.C., 1974. Thermodynamics and its Applications, PrenticeHall, Englewood Cliffs, N.J. Reid, R.C., Prausnitz, J.M., Poling, B.E., 1987. The Properties of Gases and
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Liquids, 4th Ed., McGrawHill, New York. Smith, J .M., Van Ness, H. C., 1987. Introduction to Chemical Engineering Thermodynamics, 3rd Ed., McGrawHill, New York. Van Ness, H.C., 1964. ClassicalThermodynamicsofNonElectrolyteSolutions, McMillan, New York.
Problems 9.1 What is the importance of the fundamental equations? 9.2 Is the expression: I = U + TS, where I is some energy, a fundamental equation? Justify your answer.
9.3 Why is the chemical potential called so? 9.4 With reference to the fundamental postulate: a.state it; b. why is it called a postulate? c. what is its importance? 9.5 With reference to the Maxwell relations: a. what is their mathematical origin? b. give an application of them; c.develop Eq.9.6.6. 9.6.a. What is the purpose of the departure functions? b. Show that evaluation of the departure functions for HandS suffices for the determination of those for U, A, and G. 9. 7 With reference to the Steam Tables: a.are the enthalpy and entropy values absolute or relative (to what)? b. does this create any problems in chemical engineering calculations? 9.8 With reference to the estimation of the enthalpy difference between two gaseous states of a pure compound: a. what is the minimum information needed? b. what method(s) do you recommend? c. what accuracy do you expect as a function of pressure? Consider: (i) nonpolar; and (ii) polar compounds. 9.9.a.Show that use of the virial equation in evaluating departure functions gives:
Thermodynamic Properties of Pure Fluids
319
b. Calculate the enthalpy and entropy departure values foributane at 700 K and 2.80 MPa and compare them with those obtained from the data in Table 9.E.2. 9.10 At 0.01 MPa and 350 K the enthalpy of ibutane is 470441/mol. Calculate the enthalpy at 2 MPa and 400 K (exp. value: 50238 J/mol). 9.11 Saturated steam at 1 bar is compressed to 30 bar in a flow process. If the compressor is adiabatic and irreversible with an efficiency of 0. 75: a.how much work is required per kg of steam? b. what is the temperature of the steam leaving the compressor? c. what is the second law efficiency of the process? 9.12 Propane at 1 bar and 20"C is compressed to 10 bar in a flow process. If the compressor is reversible and adiabatic, how much work is required per mole of propane? Compare your answer to that obtained in Example 8.12. 9.13 With reference to the previous Problem, if the compressor is adiabatic but irreversible with an efficiency of 0. 75: a.how much work is required? b. what is the temperature of the gas leaving the compressor? c. what is the lost work? 9.14 Methane is compressed isothermally at 300 K from 1 bar to 10 bar in a nonflow process. If the compressor is irreversible with an efficiency of 0. 75, calculate on a per mole basis: a. the work required; b. the amount of heat exchanged with the surroundings. Given: B=161 cm3/mol. 9.15 The enthalpy and entropy of saturated liquid water at 273.16 K are set arbitrarily equal to zero. a. Calculate the enthalpy and entropy of saturated vapor and liquid at 10 bar and compare them with the values given in the Steam Tables. b. How would you use the extensive PVI and vapor pressure data available for water in carrying out the calculations in part (a)? 9.16 Wet steam at 11 bar and with a quality of95.0% is throttled until it becomes dry and saturated. What are the fmal pressure and temperature of the steam? 9.17 Estimate the enthalpy difference between saturated steam at 10 bar and at 100 bar. Compare your value with the experimental one. 9.18 Between 0 and 80°C the cubical expansion of liquid water is given by: V = V0 [15.3255·106t + 7.61153 ·1071 2  4.37217·1091 3 + 1.64322·10111 4] where V0 is the volume at 0°C (cm3 /mol) and t in °C. Calculate the entropy change in J/mol K, when liquid water is compressed from 0 to 50 atm at 50°C. Neglect the pressure effect on volume. 9.19 For a certain fluid the functional relationship between enthalpy and temperature and pressure is given by: H =aT +bP where a and b are constants. Show that for this fluid the JouleThomson coefficient: (i!J'/CP)8 is a constant, and express it in terms of a and b.
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9.20.a. What is the minimum information needed for the estimation of the fugacity of a pure vapor or liquid? b. What method(s) do you recommend? c. What accuracy would you expect as a function of pressure? Consider: (i) nonpolar; and (ii) polar compounds. 9.21 Estimate the fugacity coefficient of nbutane at 340 K and 0. 7 MPa using the virial, SRK, and Pitzer methods and compare them with the recommended value of0.862. 9.22 The vapor pressure of nbutane at 357.0 K is 1.1 MPa. Estimate the fugacity of: a.saturated liquid at this temperature. b. liquid at this temperature and 2MPa. 9.23 Estimate the fugacity of methanol at 380 K and 0.25 MPa. What accuracy do you expect? 9.24 Explain the difference in errors in the prediction of q, and z of Example 9.10 for iC4 • 9.25 If the compressibility factor of a gas is equal to 1 at some temperature and pressure, will its fugacity coefficient be also equal to 1 at the same conditions? Explain. 9.26 With reference to this Chapter: a. what is the main objective? b. write a short essay describing the important items that you have learned, including some applications. 9.51 Starting with the fact that for a constant composition gas: S = f(T,P), show that for the isentropic change of an ideal gas: pyr =constant (y = C/!Cv*). 9.52 Show that the departure values for H and U do not depend on the value of the reference pressure P 0 , but those for S, A, and G, do. 9.53 Examine the variation of enthalpy with pressure, at constant temperature, for a pure gas. 9.54 Show that for a gas that obeys the Berthelot EoS: p =
RT ~ Vb TV 2
(~~L. :;, ;(:VL. V~b (7~ 2 ~;), +
; [
2a (TV)2
9.55 Show that use of the van der Waals EoS leads to the following expression for the fugacity coefficient:
Inq,
= In
RT
P(Vb)
+ _b_ 
Vb
2a
VRT
Thermodynamic Properties of Pure Fluids
321
p P,
v Figure 9.P.l A pressurevolume isotherm for a pure fluid at a temperature T1 generated from an equation of state.
9.56 Starting with Eq.9.10.6present a flow diagram that demonstrates how you would calculate the vapor pressure of a pure fluid at some temperature T using a cubic EoS. 9.51 The isotherm (T = T1) in Figure 9.P.l is generated by some equation of state for a pure fluid. Show that if: Area A = Area B, then P1 represents the vapor pressure of this fluid at T1• 9.58 We have seen that corresponding states methods give progressive} y poorer results in the prediction of thermophysical properties as the polarity of the fluid increases. How do you explain this in terms of intermolecular forces? 9.59.a.Show that use of an EoS combined with Cp • data can provide a complete description of the properties of a pure fluid. b.Is this in conflict with Section 9.5.1? 9.60 Develop Eq.9.4.16.
10 Cubic Equations of State
10.1 Introduction Fluidphase equilibria: vaporliquid, gasliquid, and liquidliquid, represent a most important application of chemical engineering thermodynamics in the petroleum and chemical industries. Quantitative description and/or prediction of such equilibria is essential in the design of distillation, extraction, and gas absorption units. Cubic Equations of State (EoS) are progressively becoming the main tool for phase equilibria calculations and, even though they are  so far successful for nonpolar/weakly polar systems only, it will not be long before they can handle polar systems as well. The SoaveRedlichKwong (SRK, Soave, 1972) and the PR (Peng and Robinson, 1976) EoS modifications of the first EoS proposed, that of van der Waals (vdW) are the most commonly used among them. For the successful description of phase equilibria EoS must meet the following requirements: 1.Provide reliable prediction of vapor pressures of pure compounds over a wide temperature range; 2.Reflect a threeparameter corresponding states principle, i.e. predict successfully vapor pressures for nonpolar/weaklypolar compounds where Tc, Pc, and (o) values are available; 3.Provide reliable values of the saturated vapor and liquid volumes, especially the of latter; 4.Provide successful correlation and prediction of phase equilibria. Application of cubic EoS in phase equilibria calculations of mixtures will be discussed in Chapter 14. Here, we will concentrate on pure fluids only. This Chapter is intended for Graduate Students. Undergraduates should read through and, if time allows, should go into the details and do the problems.
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10.2 Objective and Approach The main objective of this Chapter is to: !.Demonstrate the methodology used in developing cubic EoS for phase equilibria calculations; and 2.Explore their capabilities and limitations with respect to the aforementioned requirements (Section 10.1) with emphasis on pure compounds. To develop the appropriate background we will begin by addressing the following question: Given an EoS, how is the vapor pressure of a pure fluid evaluated at a specified temperature? The results will suggest that the traditional cubic EoS  such as the vdW and RedlichKwong (RK)  fail to provide satisfactory values for vapor pressures. We proceed, thus, to discuss how this problem is dealt with  using the vdW EoS as a starting point  presenting the approach that is typical for this type of cubic EoS, such as the SRK and PR ones. We turn next to the second requirement  mentioned in the Introduction  by outlining the procedure whereby this modified vdW EoS can be generalized to provide prediction of vapor pressures for nonpolar compounds, just as the SRK and PR are, from knowledge of the critical properties and acentric factor values. We proceed then with the third requirement and demonstrate how this vdW EoS is modified further, to provide successful predictions of saturated liquid molar volumes. This will lead to the vdW711 EoS, which is typical of cubic EoS as far as the first two requirements are concerned, and somewhat novel with respect to the third one. We follow with some comments on the capabilities and limitations of these cubic EoS and close with the concluding remarks.
10.3 Evaluation of Vapor Pressures with an Equation of State 10.3.1 The Problem The problem can be stated as follows: Given an EoS, evaluate the vapor pressure of a pure fluid at a specified temperature, i.e. the pressure
325
Cubic Equations of State
where a vapor and a liquid phase coexist in equilibrium with each other. In the previous Chapter we stated  and will prove it in Chapter 12 that the thermodynamic requirement for equilibrium between two phases is that their fugacities be equal. Thus, for vaporliquid equilibrium of pure compounds: (9.10.6) where the superscripts I and v stand for saturated liquid and vapor phase respectively. The problem, therefore, becomes: find the pressure where  at the specified temperature the two fugacities, calculated through the given EoS, are equal. We demonstrate the method for accomplishing this task in the next Example using the RK EoS.
10.3.2 Example 10.1 Calculate the vapor pressure of benzene at 373.15 K using the RK EoS (experimental value: 1.801 bar). Given: Tc = 562.16 K; Pc = 48.98 bar. The fugacity coefficient of a pure compound according to the RK EoS is given by: In£ P
= ln
0.5; for "' S 0.5, see Table 8.3 (3) Magoulas (1990)
333
Cubic Equations of State
Table 10.3 Prediction of Vapor Pressures with the vdW711, PR, tvdW and tPR EoS (from 0.1 mm Hg to the Critical) Max. Abs. Error %
Aver. Abs. Error %
PR
tvdW
vdW
No
vdW
c. c2 c3
nC3)
1.7 1.9 1.5 1.3 1.3 1.2 1.3 1.3 1.4 1.7 1.7 1.8 2.1 2.1 2.3 2.7 3.0 3.3 3.1 3.8
0.8 2.9 3.0 3.1 3.0 2.9 3.3 3.6 3.7 3.9 2.7 2.6 2.2 2.3 1.9 1.7 1.4 1.3 1.5 1.6
1.8 1.9 1.4 1.3 1.3 1.2 1.3 1.3 1.4 1.6 1.7 1.7 2.1 2.1 2.3 2.7 3.0 3.5 3.0 3.3
1.0 1.5 1.7 1.9 2.0 2.0 2.1 2.1 1.9 1.8 1.7 1.7 1.7 1.6 1.8 1.7 2.0 2.4 1.9 2.0
2.9 4.9 2.4 2.1 2.4 2.3 2.3 2.5 2.6 2.9 3.1 3.3 3.6 3.9 4.2 4.6 4.9 5.4 5.5 6.1
Ovcnill
2.1
2.5
2.0
1.8
3.6
ne.
nCs nC6 nC7
nCs nC9 nCto nCu nCt2 nCt3 nC14 nCts nCt6 nCt7 nCts nCJ9
tPR
PR
tvdW
tPR
1.6 19.8 20.4 21.1 22.8 21.4 24.3 25.2 24.3 23.9 18.6 17.2 14.5 14.5 11.9 9.7 6.8 4.5 7.0 6.0
2.6 4.5 2.2 2.1 2.4 2.3 2.4 2.8 2.6 3.0 3.2 3.4 3.6 3.9 4.1 4.6 4.8 5.3 5.3 5.7
2.8 9.4 10.9 11.7 13.4 12.1 13.1 12.8 16.0 8.5 7.6 6.8 4.7 5.0 3.2 3.4 4.2 4.8 3.5 3.5
15.8
3.5
7.6
711
711
to much lower pressures. Thus they predict the very accurate data for propane of Goodwin and Haynes (1982) down to the triple point of 1.6·109 bar with remarkable accuracy. 5.The poorer predictions at very low pressures are due in addition to uncertainties of the experimental data in that region  to the fact that, as suggested by Table 10.1, the m value is not a true constant in all these EoS. This, however, is important only at low Tr values as Eq.10.4.2 indicates and Figure 10.5 demonstrates. Here the error in predicted vapor pressures for nbutane with the PR EoS is plotted versus Tr for specified errors in m. Notice, for example, that an error in m of 2% has a negligible effect on the predicted ps value at Tr = 0.8, but a considerable one at Tr = 0.4. 6.Prediction results for saturated liquid volumes with the tPR and tvdW EoS are shown in Figure 10.1. Notice the significant improvement of the tPR over the PR EoS, while for the tvdW EoS the improvement over the vdW711 EoS occurs only at high cu values.
Applied Chemical Engineering Thermodynamics
334 120
(I)
,',
::J
'
*****vdW711 and tvdW
'
PR
(I)
~HHHtE~tPR
(l.
0
a. 0
>
.s
e'

45
'
w c
(I)
~
(I)
(l.
10 ...
10 _,
10...
10
~
Vapor Pressure (bar)
10
~
10
Figure 10.3 Prediction of vapor pressure for propane. (Experimental data from Goodwin and Haynes, 1982.)
oeeeePR ~HMHH~tPR
***** vdW711 and tvdW
'e ~20
c
(I)
~
(I)
(l.
10...
10 ...
Vapor Pressure (bar)
10
2
Figure 10.4 Prediction of vapor pressure for toluene. (Experimental data from Daubert and Danner.)
335
Cubic Equations of State
7 .In applying cubic EoS for the prediction of vapor pressures attention should be paid to the accuracy of the critical properties and "' values. Small errors in these values can have a significant impact on the accuracy of the obtained P s predictions, as demonstrated with the typical results shown in Figure 10.6 for the critical temperature, where the effect is most pronounced. Notice, for example, that an uncertainty of 3 K (0.4%) in the ~ value for nC 16 leads to a tripling of the error in P' obtained with the SRK EoS. And such an uncertainty is not unusual for large compounds. The following, for example, Tc values for nC 16 are given in the literature: 717 K, 720.6 K, and 724.3 K. 8. Vapor pressures and saturated liquid and vapor volumes with the SRK, PR, vdW711, tPR and tvdW can be calculated with the Program EOS presented in Appendix E. 9 .For polar compounds the simple dependency of alpha on Tr indicated by Eq.10.4.2 is not, as should be expected (why?), sufficient; rather a temperature dependency that involves nongeneralized parameters is needed (Mathias, 1983; Stryjek and Vera, 1986; Gupte et al, 1986; Androulakis et al, 1989).
Q)
20.00
I..
:I
en en
2li5
Q)
~
I..
Q.
I..
10.00
0
Q_
0
> "0 Q)
+'
~
0.00
"0
Q)
I..
Q.
c
· 10.00 ~
0.40
0.50
0.60
0.70
0.80
Reduced Temperature (T,)
Figure 10.5 The effect of the uncertainty in the m value on the accuracy of vapor pressure predictions for nbutane with the PR EoS.
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Applied Chemical Engineering Thermodynamics
nee nc,e .s ~
15
~
10
~
Q)
a..
5 0
~~~~~~~~~~~~~~~~~~~~~~
2
1
0
Percent Error in Tc
2
Figure 10.6 Effect of the uncertainty in Tc on the accuracy of predicted vapor pressures using the SRK EoS, holding the Pc and "' values constant.
10.7 Concluding Remarks 1.For nonpolar/weaklypolar compounds very good prediction of vapor pressures is obtained with the cubic EoS discussed here. At very low ps values, however, only the tvdW and vdW711 EoS seem to give reliable results, down to about 105 bar, as suggested for example by the cases of propane and toluene presented in Figures 10.3 and 10.4 respectively. 2.Use of the correct Tc, Pc, and "'values is essential for the reliable prediction of vapor pressures. 3.Both the tPR and tvdW EoS give excellent prediction of saturation liquid volumes. 4.Prediction of vapor and gaseous volumes, enthalpies of vaporization and enthalpy and entropy departures with cubic EoS are discussed in Sections 8.10.8, 8.14.1 and 9.8.5 respectively. 5.Application of cubic EoS in tlie correlation and prediction of vaporliquid equilibria of mixtures is considered in Chapter 14.
Cubic Equations of State
337
References Androulakis, 1., Kalospiros, N., Tassios, D. P., 1989. Fluid Phase Equilibria, 45, 135. Daubert, T.E., Danner, R.P., 1985; 1986. Data Compilation Tables of Properties of Pure Compounds, DIPPR, AIChE, N.Y. (They provide correlations of available experimental data. Use, thus, of these correlations outside the range of the data involves extrapolation.) Dymond, J.H., 1985. The Second International/UPAC Workshop, Paris. Goodwin, R.D., Haynes, W.M., 1982. Thermophysical Properties of Propane from 85 to 700 Kat Pressures to 70 MPa, Monograph 170, National Bureau of Standards, Washington, D.C. Gupte, P.A., Rasmussen, P., Fredenslund, Aa., 1986. Ind. Eng. Chem. Fundam., 25, 236. Magoulas, K., Tassios, D., 1990. Fluid Phase Equilibria, 56, 119. Mathias, P.M., 1983. Ind. Eng. Chem. Process Des. Dev., 22, 385. Peneloux, E., Rauzy, E., Frez.e, R., 1982. Fluid Phase Equilibria, 8,7. Peng, D.Y., Robinson, D.B., 1976. Ind. Eng. Chem. Fundam., 15, 59. Robinson, D.B., Peng, D.Y., 1978. The Characterization of the Heptanes and Heavier Fractions for the GPA PengRobinsonPrograms. GPA Res. Rep. 28. Salerno, S., Cascella, M., May, D., Watson, P., Tassios, D., 1986.FluidPhase Equilibria, 21, 15. Soave, G., 1972. Chem. Eng. Sci., 27, 1197. Soave, G., 1979. Chem. Eng. Sci., 34, 225. Stryjek, R., Vera, J.H., 1986. Can. J. Chem. Eng., 64, 323. Van Ness, H. C., Abbott, M.M., 1982. Classical Thermodynamics of Non Electrolyte Solutions, McGrawHill, New York. Watson, P., Cascella, M., May, D., Salerno, S., Tassios, D., 1986.FluidPhase Equilibria, 21, 35.
Problems 10.1 What requirements, with respect to pure compounds, must be met by EoS used in phase equilibria calculations? 10.2 Develop Eq.(A) of Example 10.1. 10.3 Using the original vdW EoS calculate the vapor pressure ofnoctane at 370.76 K and compare it with the value given in Table 10.P.l. 10.4 Using the experimental value of the vapor pressure for nC 8 at 370.76 K of
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Table 10.P.1 Vapor Pressure Data for nOctane 404.98 423.15 T(K) 370.76 ps (bar) 1.905 1.200 0.433
473.15 5.472
Table 10.P.1,calculate the value of a in the vdW EoS that reproduces this vapor pressure. Repeat the calculations, using the second and third vapor pressure data points given in Table lO.P .1, and in each case determine the value of a = alae. What correlation of alpha to Tr do you find best for the obtained values? 10.5 Using the correlation of alpha to Tr developed in the previous Problem, determine the vapor pressure of nC8 at 473.15 K. 10.6.a.Show that volume translation, i.e. the replacement of V by V' = (V + t) in any EoS, does not effect the calculated vapor pressure. b.Demonstrate it numerically by redoing Problem 10.5 using an arbitrary translation. 10.7 Calculate the vapor pressure ofnC6 at 286.2 K using the vdW711, PR, and SRK EoS and the computer Program EOS of Appendix E (expt. value: 0.117 bar). 10.8 Compare the saturated liquid volume obtained in the previous Problem with that given by the correlation of Hankinson and Thomson (Chapter 8). 10.9.a.Develop expressions for the second virial coefficient from the vdW, vdW711, SRK, and tPR EoS. b.Calculate B values from these expressions for noctane and plot them, along with the recommended values (Dymond, 1985) given in Table 10.P.2, versus Tr. Table 10.P.l Second Virial Coefficients (cm3/mol) for nOctane T(K) 380 400 420 460 480 500 525 550 575 B 1990 1690 1490 1170 1025 920 815 725 650 10.10 How do you explain: a. the difference between the B values obtained from the vdW711 EoS and the recommended ones? b. the fact that, in spite of this difference, the EoS gives good predictions for saturated vapor volumes (except close to the critical point)? 10.11 Examine the effect of the uncertainty in the Tc, Pc and(&) values on the predicted vapor pressures and saturated liquid volumes for a compound of your choice using the SRK, tPR, and tvdW EoS. 10.12.a.Discuss the problems associated with the sharp change of the translation with Tr close to Tr = 1. (See Androulakis et al.) b. Answer the question posed in comment 9 of Section 10.6. 10.13 Develop the following recursive formula to be used in calculating vapor pressures through an equation of state {Van Ness and Abbott, 1982):
P]+t = Pj[(zvz 1)iln(fvli))l(zvz 1 )i
11 Properties of Mixtures
11.1 Introduction Evaluation of thermophysical properties represents the second main objective of Chemical Engineering Thermodynamics. Pure compounds were considered in the previous three Chapters, a general discussion of mixtures in this one, and phase and chemical reaction equilibria in the ensuing Chapters 12 through 15. It is important that we state clearly from the beginning the nomenclature to be used in this Chapter. to eliminate potential confusion in the representation of pure compound and mixture properties. We define: M: Mixture molar property (V, H, etc.) N: Number of moles in a mixture NM: Total mixture property (NV, NS, etc.) M;: Pure compound molar property (V;. H;. etc.) Consider now a homogeneous mixture containing N1 moles of compound 1, N2 of 2, ... and Nk of k, where k is the number of components in the mixture, for a total of N moles; and the following question: How is the total property of this mixture, for example the total volume NV, to be determined at some specified temperature T and pressure P ? The simplest answer to this question would be provided by the equation: (11.1.1) with all M;'s evaluated at the same T and P. Unfortunately, nature is not very obliging and this equation is not valid. Instead the correct expression is: NM =
or. dividing by N:
M
EN;M;
~x.M. = L..J I I
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where M; is called the partial molar property of component i. Thus, the molar volume V of a binary mixture is not given by:
v = xl v. +x2 v2,
but by:
v
= xl

 +x2 v2 v.
In other words, a given property of compound i has a different value when
i is the component of a mixture from that in its pure state, both calculated
at the same temperature and pressure. This difference results from differences in the: * inter(unlike) and intra(like)molecular forces, and * molecular size and shape, of the components of the mixture. If such differences are not present, a hypothetical case of course, the mixture is referred to as an ideal one and, as we will see in Section 11.12, Eq.11.1.1 holds except when entropy is involved, where the entropy of mixing must be accounted for (Section 11.12.2).
11.2 Objective and Approach The main objective in this Chapter is to develop familiarity with mixture properties and the methods used for their evaluation. More specifically to: 1. Understand the concept of partial molar property, which describes the behavior of the component of a mixture, and contrast it to its behavior in the pure state. 2.Discuss the adjustments that must be made in the methods that we used in the prediction of pure compound properties: * the Corresponding States Principle; and * Equations of State, so that they can be used for mixtures. 3.Extend the concept of pure compound fugacity into that of the component of a mixture, and establish the methodology for determining it. 4.Develop expressions that isolate the mixing effect by comparing mixture properties to their values: a.in the absence of mixing, and b.in a hypothetical ideal solution. We start with the definition and a brief description of partial molar properties (pmp) and then examine: 1.how mixture properties are obtained from pmp; 2.the reverse case, i.e. how pmp are evaluated from mixture properties;
341
Properties of Mixtures
3.the interdependency among the pmp of a mixture. We proceed with a discussion of the socalled mixing and combining rules, which allow us to extend pure component and binary system information into the determination of mixture properties. We turn then to the very important concept of the fugacity of the component of a mixture, and proceed to outline the two main approaches used in its determination: 1.equations of state; and 2. the standard state fugacity. To help us appreciate the importance of fugacities, we apply them to the determination of the solubility of solids and liquids in gases. (Extensive application of fugacities will be made in the ensuing four Chapters where phase and chemical equilibria are discussed.) We proceed with a discussion of two auxiliary mixture quantities, which help us isolate the mixing effects from: 1. pure component behavior, through the socalled properties of mixing; 2.ideal solution behavior, through the socalled excess properties. We close with some concluding remarks.
11.3 Partial Molar Properties 11.3.1 Definition Consider a homogeneous mixture made up of N1, N2 , ••. , Nk moles of components 1, 2, ... , k, respectively. To this mixture we add a small amount of component i, dN;, while maintaining the temperature T, pressure P, and number of moles of the other components~ constant. We measure the resulting change in the total property, d(NM ), and define the ratio d(NM) over dN;:
M;
=(a~~>] 1
(11.3.1) T,P,~
as the partial molar property M; of component i of this mixture at the specified composition, temperature, and pressure. (The subscriptj, here and throughout the Chapter, assumes all values from 1 through k, except i.)
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Notice that the change must be determined at constant temperature, pressure, and number of moles of the other components. If, for example, the addition of the dN; moles takes place at constant pressure  say the open atmosphere  we must make sure that any change in temperature, resulting from the mixing, is accommodated for by the addition or removal of the necessary amount of heat so that it remains also constant. It follows from Eq.11.3.1 that, for a mixture of specified components, the partial molar properties are functions of its composition, temperature, and pressure: (11.3.2) We demonstrate the physical meaning of partial molar properties with the simple case of the volume of a binary mixture in the next Example.
11.3.2 Example 11.1 Consider a mixture containing 3 moles of methanol(m) and 7 moles of water(w) at 25°C and 1 atm. To this mixture we add a small amount of water, say 0.1 mole (see Problem 11.2.a), while maintaining the temperature, pressure, and number of moles of methanol constant. The volume of the mixture increases by 1.78 cm3 • a. What is the partial molar volume of water? b. What is the effect of mixing on the molar volume of water? c. What would the partial molar volume of water be equal to, if the solution were ideal? Given: Vw (25°C, 1 atm) = 18.1 cm3/mol. a. From the definition:
vw  (a(NV)l a(Nw)
T,P,Nm
..
(.1(NV)l .d(Nw)
T,P,Nm
1.78  0.10 
17 8 cm3 . mol
b. The molar volume of water in the mixture, i.e. its partial molar volume, is smaller by 1. 7% than its value in the pure state at the same T and P. c.Inthiscase, Yw = Vw = 18.1cm3 /mol. Comments 1. The mixing effect on volume is usually small, when the mixture is well below its critical point, but it can be very pronounced as this point is approached. Thus, Prausnitz et al (1986) report that for the saturated liquid mixture C02(1)  nbutane(2),at x 1 = 0.6 and 71°C: V2 = 20 cm3 /mol, while V2(saturated liquid at 71 °C) = 115 cm3 /mol. Actually at the critical point, which at this temperature
343
Properties of Mixtures
occurs at x 1 = 0.713 and P = 81.7 bar (Olds et al, 1949), y2 = 155 cm3 /mol. 2. The mixing effect can be also be very pronounced for other properties, especially fugacity, even well away from the mixture critical point.
11.4 Mixture Properties from Partial Molar Properties 11.4.1 The Problem We stated in the Introduction that, unfortunately, the simple Eq .11.1.1 cannot describe mixture properties. If, on the other hand, the fundamental equation for this mixture were available, all properties of the system could be determined as we saw in Section 9.5 .1. Thus, if the fundamental equation for the Gibbs free energy of the mixture which, using the nomenclature of this chapter becomes: (9.4.13) were available, then its volume  as an example  could be determined:
NV=
(o(NG)) oP
T,N,
Unfortunately, the analytical relationship for Eq. 9 .4.13 is not available as we also saw in that Section.
11.4.2 The Approach We resort therefore, as we did in Section 9.5.2, to the fundamental postulate according to which the total property NM of a homogeneous mixture  at a specified temperature T and pressure P  is a function of the amounts of its components only: (11.4.1) Differentiation yields: d(NM) =
[o(NM)l oN l
dN + .•. + •
T,P,~
[o(NM)l aN k
dN, T,P,~
k
(11.4.2)
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Notice that the derivatives are, according to the definition, Eq.11.3.1, the partial molar properties of NM. Thus Eq.11.4.2 becomes: (11.4.3) Integration of Eq.11.4.3 from zero to the actual values of the number of moles, at constant temperature, pressure, and composition a path along which all M1 values remain constant according to Eq.11.3.2  yields: NM =
E~Mi
(11.4.4a)
or, dividing by the total number of moles N:
M =
Ex;M;
(11.4.4b)
i.e. the equations presented in the Introduction.
11.4.3 Comments l.Example 11.1 and Eq.11.4.4 indicate that the partial molar property of a given component i represents the value of this property when i is surrounded by other compounds and itself, while the pure compound value when i is in the pure state. 2.The partial molar property can be thus visualized as an 'effective' value of this property in the given mixture. For example, the total volume of a mixture is not determined by using the values of the molar volumes of the pure components but rather the 'effective' values of the volumes of its components in the solution, the partial molar volumes (Eq.11.4.4). 3.If the partial molar properties for the components of a mixture are known at a given T, P, and overall composition, the corresponding property of the mixture at the same conditions can be evaluated.
11.5 Partial Molar Properties from Mixture Properties 11.5.1 The Analytical Expression In the previous Section we developed an expression for the evaluation of a mixture property M in terms of the corresponding partial molar properties. In this Section we consider the reverse case: assuming that we
345
Properties of Mixtures
have an expression for M in terms of composition, at a given pressure P and temperature T(Eq.11.4.1): M = f(x 1, x2,
••• ,
x.J
(11.5.1)
we will develop an expression for the evaluation of the mixture partial properties. Starting with the definition of M;. Eq.11.3.1, we obtain:
ii.
= o(NM) = M +NoM
(11.5.2) oN; oN; where the partial derivatives are taken at constant T, P, and number of moles of the other components. To evaluate the derivative (oM/oN;) we note that in terms of independent variables, since ~; = 1, Eq .11.5 .1 should be written as: I
M =f(x 1, x2 , ••• ,
X;_ 1,
xi+ 1, ••• , x.J
(11.5.1a)
Using chainrule differentiation:
M. = M+NE[oMoxil = M+NE[oMo(~IN)l = MEx.oM I
0Xj 0N;
OXj
0N;
J OXj (11.5.3)
where the summation is over all components except i and (oM/i!Jxi) is taken at constant T, P, and x1 (l~i andj). Eq.11.5.3 is more convenient to use than Eq.11.3.1 because, in the typical case, we have an expression of M as a function of composition rather than of NM as a function of the number of moles of the mixture components.
11.5.2 Example 11.2 The molar vapor volume of a mixture of acetonitrile(1) and acetaldehyde(2) at 100°C and 1. 013 bar is given by the following expression: (A)
where: a = 1935; b = 530; c function of composition.
= 2440, all in cm3/mol. Calculate V 1 as a
From Eq.11.5.3: (B)
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~
0
E
~
E 28000 0
''
....
§ 27000 ::::!:
:e 0
0
a..
v,· 2sooo~~~~~~~~L~~~~~~~L~~~~
0.00
0.20
0.40
0.60
0.80
1.00
Figure ll.E.2 Variation of the partial molar volume of component (1) with y 1 for the system acetonitrile( 1)  acetaldehyde(2) at 100°C and 1. 013 bar.
where the partial derivative is taken at constant T and P, but not y 1:
av

ay2
"'2ay1 (1)+2by2 +2cy1 +2cy2(1)
(C)
Introduction of Eqs (A) and (C) into (B) yields, after some algebra: vl "'
RT
p
2
2
b
cm 3
+a(yl +2yly2) +y2 (2c ) ' mol
The results are presented in Figure ll.E.2, where also shown are: the molar volume of pure 1, V1, and the value of Y1 at the limit y 1 = 0, referred to as "the partial molar volume of component 1 at infinite dilution in 2"'
vl "".
Comments !.Notice that as the composition of acetonitrile becomes smaller, its deviation from its pure compound behavior becomes more pronounced: it is only 0.3% at y1 "'0.8, but 8.4% in the limit y 1 "'0.0. 2.We will see that this phenomenon is typical with other thermodynamic properties as well. It seems that the molecules of componenti in this case those of acetonitrile  'feel' the molecules of t4,e other components  in this case those of acetaldehyde as 'hostile'. And this 'feeling' becomes stronger, as the number of the other molecules increases.
Properties of Mixtures
347
3. This 'hostility' increases also with increasing dissimilarity. Thus water mixes completely with methanol in all proportions but not with pentanol, where the dissimilarity is increased (why?). 4.For ideal solution behavior, V; = v;, as it was the case in the methanolwater system of Example 11.1 (Part c.). 5.Notice that as y 1 approaches 1, V1 tends to V1, i.e. ideal solution behavior is approached with respect to this component. 6.It should be emphasized that ideal solution behavior of a gas mixture does not necessarily mean ideal gas behavior. For, if this mixture were to behave as an ideal gas, then:

y1
RT
.
= V1(zdeal gas) = p = 30625
cm 3 mol
7. This is why an ideal solution of gases is also referred to as an "ideal solution or mixture of real gases". Thus an ideal gas is an ideal solution, but the reverse is not necessarily the case.
11.6 Interdependency of the Partial Molar Properties 11.6.1 The GibbsDuhem Equation In the general case the total mixture property NM is a function of the temperature T and pressure P in addition to the number of moles of its componentsN1, N2 , •.• , Nk and Eq.ll.4.1 becomes: (11.6.1) Consequently: d(NM)
=
(a(NM)) ar =
P,~
(aM)
NaT
dT+(a(NM)) ap
(aM)
dT+Nap 'i
Px
From Eq.11.4.4a: d(NM)
T,N,
=
dP+E[a(NM)l aN.
T.x
I
dN. T.P N '
=
I
' j

dP+EM1dN;
'i
E N;dM; + E M;dN;
Comparison with Eq .11.6.2, followed by division by N, yields:
(11.6.2)
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348
Ex;dM; (aM) dT(aMJ dP oT
oP
P,x,
T,x,
=
o
(11.6.3)
This expression represents the GibbsDuhem equation and indicates that the intensive properties of the mixture: temperature, pressure and partial molar properties, cannot vary independently. Restricted to constant T and P, Eq.11.6.3 becomes: (11.6.4) which is referred to as the restricted form of the GibbsDuhem equation. We will present two applications of the restricted form of the GibbsDuhem equation in Examples 11.3 and 11.4, while Eq.11.6.3 will be used in Chapter 13 to evaluate the thermodynamic consistency of vaporliquid equilibrium data.
11.6.2 Example 11.3 The following expressions are proposed for the composition dependency of the partial molar volumes of a binary mixture at a given temperature and pressure: V1
= V1 (1 +ax2 )
(A)
V2
=
(B)
V2 (1+bx 1 )
V1 and V2 are the molar volumes of pure 1 and 2 respectively, at the same temperature T and pressure P as the mixture; and a and b are constants, functions of T and P only. Is there any restriction on the values of a and b?
Introduction of Eqs (A) and (B) into the restricted form of the GibbsDuhem equation yields: x 1(aV1)dt2 +x2 (bV2 )dt1
Since, dt1 =  dt2:
(x 1V1a x 2 V2b)dt2
=0
=0
(C)
In order that Eq.(C) is valid for all values of dt2 : x 1V1a x2 V2b = 0 or:
x2 v2
a= h  xl vl
The only way that a and b are constants, independent of composition, is that: a= b = 0.
We conclude, therefore, that the interdependency between the two partial molar volumes, required by the GibbsDuhem equation, restricts the values of a and b to zero. Under these conditions, Eqs (A) and (B) refer to an ideal solution.
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Properties of Mixtures
11.6.3 Example 11.4 The partial molar volume of water(w) in its mixture with methanol(m) at 25°C and 1 atm can be approximated by:
Vw = 18.1 +ax,.2
;
a
= 3.2 cm 3/mol
(A)
Develop an expression for the partial molar volume of methanol at the same conditions. Given: V,. (1 atm and 25°C) = 40.7 cm3 /mol From the restricted form of the GibbsDuhem equation, since P and T remain constant:

 =0
xwdVw +xmdVm Solution for
dVm , and evaluation of dVw
from Eq.(A), yields:
Integration yields: where c is the integration constant. At xw = 0, c = Vm, and:
Vm
= 40 • 7 +axw2
(B)
We conclude, therefore, that if the partial molar property M; for the one component of a binary mixture is known as a function of composition at a given temperature T and pressure P from experimental measurements, the partial molar property Mj of the other component can be determined at the same temperature and pressure without the need for further experimental work. The value for the property in the pure state ~· at the same T and P as the mixture, must also be known however.
11.7 Evaluation of Mixture Properties 11.7.1 The Problem and Approach Eq.11.4.4 indicates that determination of mixture properties requires values of the corresponding partial molar properties. But according to Eq.11.5.3, evaluation of the partial molar properties requires knowledge of the corresponding mixture property.
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Mixture properties are consequently obtained from experimental measurements or, in their absence, through estimation methods. Experimental data are available for several binary systems, but for very few multicomponentones that represent the common industrial case. Use of estimation techniques represents the typical methodology for mixtures, therefore, just as it did with pure compounds. In discussing physical and thermodynamic properties of pure fluids in Chapters 8 and 9, we concluded that the main estimation techniques are: l.Pitzer's formulation of the Corresponding States Principle (CSP), and 2.Equations of State (EoS), especially the virial and cubic ones. Furthermore, the success of these methods in the estimation of pure compound properties suggests their application to mixtures as well. This, in turn, leads to the following question: what adjustments must be made to these techniques, so that they can be used for mixtures? If, for example, we are to estimate the molar volume of a mixture of nitrogen and nbutane  at a specified temperature, pressure, and composition using the Pitzer correlation, what values will be used for Tc and Pc to determine the reduced temperature and pressure values needed for the corresponding Tables (Appendix D)? Or, if we were to use the RedlichKwong  or any other  cubic EoS for the same purpose, how will the parameter values for the EoS be determined? It should be apparent that, whatever the method used, it should rely on the information more likely to be available: l.pure component properties; and, 2.on occasion, binary data. Otherwise, if extensive mixture data are needed, the predictive character of the technique is lost. Evaluation of mixture parameters from pure component and, if available, binary parameters is carried out by using the socalled mixing and combining rules. We will discuss first the virial equation, because of its theoretical basis, and then consider the empirical approaches: Pitzer's CSP and cubic EoS.
11.7.2 The Virial Equation a.Mixture Coefficients Because of its theoretical basis, i.e. its development from Statistical Mechanics, the virial equation (Section 8.9) represents the only EoS where rigorous mixing rules for the mixture coefficients are available. Thus:
Properties of Mixtures
351 (11.7.1)
Bii is the second virial coefficient of pure i, Bjj of purej, and Bij is the second virial coefficient for the pair ij, referred to as cross virial coefficient (not to be confused with the mixture second virial coefficient given by Eq.11.7.1). Obviously Bii = Bi; and, as with the case of a pure compound, Bij is a function of temperature only. The summations, finally, are over all mixture components. Thus, for a binary mixture: B = Y1 2Bu +y/B22 + 2y1y2B12 Similar expressions exist for the third and higher coefficients. We will restrict, however, our discussion to the virial equation truncated after the second term since higher coefficients, as we have seen, are rarely available. Values for the cross coefficient Bii are determined from experimental data, such as PVT ones, for the binary system ij. But such experimental Bu values are not often available and, consequently, we must resort to some estimation technique, just as we did with B for pure compounds. Use of the Tsonopoulosand the Abbott correlations, Examples 8.5 and 8.6, for the pair ij requires values for the: * critical temperature, Tc , and pressure Pc ; and * acent nc . 1ac &: tor, c.Jii. u u The following rules are recommended for this purpose: (11.7.2) p
Cy
zc RTC u
=
vCy
u
(.,); + (J)j
2
(11.7.3)
(11.7.4)
where: (11.7.5)
v
Cy
=
[
v
1/3+ c1
2
v
CJ
1/3]3
(11. 7.6)
A single subscript refers to pure component values and a double one to binary parameters.
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Eqs 11.7 .2 through 11.7 .6 provide recipes for the determination of these binary parameters by combining pure component properties. They are, hence, referred to as combining rules. Notice that: l.The equations are purely empirical. Their only justification is that, when used with the Tsonopoulos correlation, they lead to reasonable estimates of Bu values. 2. These binary parameters have no particular physical meaning. They are just constructs, useful for the estimation of mixture properties. 3.Eq.11. 7.2 expresses what is referred to as the geometric mean combining rule, whose origin and limitations are discussed in Section 7 .6.4. 4. The performance of these combining rules becomes poorer, as we should expect, with increasing dissimilarity between species i and j. It would indeed be naive to expect that pure component properties alone would suffice for the prediction of mixture properties, considering the difference between intra and inter (like and unlike species) molecular forces. And this difference becomes, of course, more pronounced as the dissimilarity between the species increases. For improved thus results, Eq.11.7.2 is written as: Tc = (1/IJ.. )(Tc Tc ) 112 (}
I
(11.7.7)
j
The parameter lu is referred to as the interaction coefficient and it is assumed to be independent of temperature and pressure. Values for lu are obtained from experimental Bu values. b.Comments on the Interaction Coefficient
l.lu can be set equal to zero, only for mixtures of compounds that are very similar. 2.For methane i with some hydrocarbon j, Tsonopoulos (1979) suggests that the following empirical expression: lu
=
0.0279 (In Nci ) 2
(11. 7 .8)
correlates successfully the available lu values obtained by fitting experimental data  to the number of carbon atoms in the hydrocarbon j, Nci. Notice that, as expected, increasing dissimilarity  difference in the molar volumes in this case, as expressed through increasing Nc. values leads to larger lu values. Thus, luis equal to 0.013 for the C 1 2 binary, but to 0.148 for the CcC 10 one. 3.For ethane or ethylene i with some other hydrocarbonj, Tsonopoulos suggests:
C
(11. 7 .8a)
Properties of Mixtures
353
This expression can be used for other hydrocarbon pairs, with less confidence however. 4.The same reference gives lij values for several gases (CH 4, C2H4, C02, N2 , etc.) with nonpolar and some polar compounds. S.In general lij values for nonpolar/weakly polar systems are small, typically in the range 0.0 to 0.2, and can be often estimated with reasonable accuracy. The reliability, thus, of the obtained Bij values  and consequently of the calculated mixture properties  for such systems is rather high. 6.For systems containing polar components, however, lij values can be larger; and in the few cases where there is strong solvation, i.e. hydrogen bonding between the molecules of the different species, lij can assume negative values that are functions of temperature. Thus for the acetonitrileacetaldehyde system, ld40°C) = 0.32; ld100°C) = 0.23 (see Problem 11.41). 7 .Even though some guidelines for estimating lij values for polar systems are given by Tsonopoulos (1974; 1975), the reliability of the obtained Bij values is rather low (unless, of course, lij for the system under consideration is available from experimental data). 8.For the estimation of Bij values in the Hayden and O'Connell (1975) correlation, see the original article and also Prausnitz et al (1980). 9.When truncated after B, the virial equation should be used in the pressure range given by the following empirical expression (Prausnitz et al, 1980):
P
~
rLYtPc1
=
2 LY;Tcl For a pure compound, this equation reduces to Eq.8.9.5.
(11.7.9)
11.7.3 The Pitzer Corresponding States Correlation In using the Pitzer correlation (Section 8. 7 .2) we need values of the mixture critical properties and acentric factor, which are calculated from empirical mixing rules. The obtained values are not, consequently, the true critical properties of the mixture. They are referred to, appropriately, as pseudocritical properties. When using Pitzer's correlation with the Tables of Lee and Kesler, discussed in Chapters 8 and 9, the following mixing rules are recommended by the authors (the subscript m denotes mixture):
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Applied Chemical Engineering Thermodynamics
Tc = _1_"t"'"t"'y,.yJ.(Vc113+Vcl/3i(Tc Tc)l/2 "' 8V LJLJ I J I J
(11.7.10)
c,.
RTC Pc = (0.2905+0.085c.Jm) V "' "'
c
(11.7.11)
"'
where: (11.7.13)
vc
RTC
= (0.2905+0.085c.Jj)1 PC
I
(11.7.14)
I
For improved results, an interaction coefficient in the product (TcI TcJ ) 112 is required in Eq.11.7.10(see Reid et al, 1987), but it is rarely used in practice.
11.7.4 Cubic Equations of State a.Mixture Coefficients The following general mixing rules are used with cubic (Chapters 8 and 10) equations of state: =
E:Ex1xiaii
(11.7.15)
b =
E E X;Xjbij
(11.7.16)
a
where:
a11
=
ac1a1
b;; = b;
(11.7.17) (11.7.18)
Typically, a geometric mean combining rule is used for the interaction parameter au:
aii
=
('a;;ajj)tn
and an arithmetic mean one, for the parameter bu:
(11.7.19)
355
Properties of Mixtures
b .. +b ..
b = ..!!..___!!. 00
I)
2
(11.7.20)
In the latter case, Eq .11. 7.16 becomes: b = :£x;b;
(11.7.21)
Finally for translated EoS, the following mixing rule is required for the translation parameter of the mixture t (Peneloux et al, 1982): (11.7.22) As with the case of Bij, for improved results an interaction coefficient kij is needed in Eq.11.7.19: aij = (lkij)(a 11 aii) 112
(11.7.23)
Notice finally the similarity of Eqs.11. 7.15 and 11.7.16 with Eq.11.7.1. The first two, however, are purely empirical; the latter is developed from Statistical Mechanics. b. Comments on the Interaction Coefficient 1. Values for the interaction coefficient are obtained from experimental data involving a mixture of compounds i andj. 2.Again as with the case of the virial equation, they are small, typically between 0.0 and 0.2 for nonpolar or weakly polar systems, and can be larger  or even negative  for polar ones. 3.The values vary from one EoS to the other. 4.For hydrocarbons that are not very different in size, zero values for the interaction coefficient can be used. 5.For systems containing C02 and nparaffins, kij values for the PengRobinson (PR) EoS are given by Kato et al (1981). 6.For systems containing C02 , H 2S, N2 , and CO with hydrocarbons, fGi values for the SoaveRedlichKwong(SRK) EoS are given by Graboski and Daubert (1978). 7.For H2hydrocarbonsystems, see Lin (1980), and Tsonopoulos and Heidman (1986). 8.For the volumetric properties of gaseous water with some nonpolar gases, see De Santis et al (1974). We examine the prediction of the volumetric behavior of binary mixtures in the next two Examples 11.5 and 11.6; and the evaluation of interaction coefficients, in Example 11.7. We proceed, then, with some general comments on the estimation of mixture properties, followed by the discussion of fugacities in mixtures.
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356
11.7.5 Example 11.5 Estimate the compressibility factor of a methane(1) nbutane(2)vapor mixture at 400 psia, 100°F, and y 1 = 0. 8942. Use the virial equation, Pitzer correlation, and SRK. EoS. Experimental value: z = 0.9135 (Sage et al, 1940). Critical property and Ca> values are presented below:
vc
6)
T,
P,
0.288
99.1
0.011
1.632
0.599
0.274
255.1
0.193
0.732
0.726
(K)
(bar)
PC
Zc
c.
190.58
46.04
nC4
425.18
37.97
Tc
(cm3 /mol)
a. The Virial Equation From Eq.11.7.9,P 12 = 0.102; PCt2 = 38.1 bar. Thus: B 12 = 145.7 =Y12Bu + YlB22 + 2Y1Y2B12 = 65.6 cm3/mol,
which gives: z = 1 + BPIRT = 0.930, in very good agreement with the experimental value (error 1.8%). To examine the effect of 112 , we calculate B 12 using 112 = 0. 0, obtaining a value of B 12(112 = 0.0) = 164,6cm3 /mol. This leads to B = 69.1 cm3 /mol and z = 0.926, which is also in very good agreement with the experimental value (error: 1.4%). (The slightly better result for 112 = 0.0 may be due to the uncertainty in the experimental data.) b. The Pitzer Correlation
= 100.3; Vc2 = 285. 7; Vc,. = 116.7, all = 224.0 K; 6>,. = 0.030; p c, = 46.7 bar. = 1.389; P, = 0.563; and: zl0) = 0.932; z< 1> = 0.046; and z =
From Eqs 11.7 .tO through 11.7.14: Vc
1
in cm3 /mol; Tc, Therefore: T, 0.933, which is also in very good agreement with the experimental value (error: 2.1 %) c. The SRK. EoS Using the expressions from Table 8.3: = 2.331·106 bar cm6/mol2 ; a 11 = i.727·106 bar cm6/mol2 a ell
b1
= 29.82 cm3 /mol
357
Properties of Mixtures
a = 1.407·107 bar cm6 /mol2 "22 b2 = 80.67 cm3 /mol
;
~2 = 1.742·107 bar cm6/mol2
From Eq.11.7.23,using k12 = 0.0, valid for this mixture and the SRK EoS:
a 12 = 5.485·106 bar cm6/mol2 Thus, from Eqs 11.7.15 and 11.7.21 respectively:
a = 2.6136·106 bar cm6 /mot2 b = 35.21 cm3 /mol To solve for the compressibility factor, it is more convenient to write the SRK EoS in the following form:
r r + (ABW)zAB =0
where:
aP (RD 2
bP = 0.0375 RT
A =   = 0.1078 ; B = 
Solution of the cubic equation for the largest z value gives: z = 0. 931, which is also in very good agreement with the experimental value (error 1.9%). The results are summarized in Table 11.E.5. Table ll.E.S Tabulation of Results for Example 11.5
z
%Error
Virial (112 =0.054)
0.930
1.8
Virial (/12 =0.0)
0.926
1.4
Pitzer
0.933
2.1
SRK
0.931
1.9
Method
11.7.6 Example 11.6 Estimate the compressibility factors of the methane  nbutane mixture of Example 11.5  at the same temperature and composition, but at increased pressures  and compare them with the experimental values (Sage et al) given in Table 11.E.6. Use the SRK EoS. Following the same approach as in Example 11.5, and using the computer Program PVT of Appendix E, we obtain the results presented in Table 11.E.6, which can be considered satisfactory up to 1500 psia ( 100 bar). Notice, however, that: both compounds are nonpolar, and * the mixture is about 90% methane, whose reduced temperature is high ( 1.632).
*
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Applied Chemical Engineering Thermodynamics Table ll.E.6 Prediction of z Values for a Methane nButane Mixture as a Function of Pressure with the SRK EoS. y(CH4) = 0.8942; T = 100°F P (psia)
z (exp)
z (SRK)
%Error
400
0.9135
0.931
1.9
600
0.8769
0.897
2.3
800
0.8444
0.866
2.5
1000
0.8154
0.837
2.6
1500
0.7587
0.781
2.9
2000
0.7274
0.759
4.3
2500
0.7212
0.768
6.5
3000
0.7470
0.796
6.6
11.7.7 Example 11.7 Using the saturated vapor compressibility factor data for the system nbutanecarbon dioxide (Olds et al) presented in Table 1l.E. 7: a.evaluate the performance of the SRK, PR, and vdW711 EoS using a zero value for the interaction coefficients; b. determine the interaction coefficient values for the three EoS. a. Compressibility factor values were calculated using the Program PVT of Appendix E and are presented in Table 1l.E.7. The larger errors at 618 psia should be attributed, in addition to the saturation condition and the dissimilarity of the components of the mixture, to the higher pressure. b. Using the same Program, the interaction coefficient values presented in Table
Table ll.E. 7 Compressibility Factors and Interaction Coefficients for nButane  Carbon Dioxide % Error (k12 =0.0)
z
7{°F) P(psia) y(nC4)
Inter. Coefficient
PR vdW711
SRK
PR
vdW711
3.0
2.9
0.077
0.188
0.195
7.3 11.6 15.5
0.081
0.144
0.213
SRK
0.822
100
372
0.1694 1.2
0.495
280
618
0.8273
Properties of Mixtures
359
ll.E. 7 were obtained by trialanderror. (They could of course be obtained directly by regression.) Notice that the obtained values vary between the two sets of conditions. Two reasons account for this. First, kij is not a constant, strictly speaking. Second, and the most important in this case, the obtained values reflect the uncertainty in the experimental data. In practice, therefore, kiJ values are obtained through regression of a set of data, typically vaporliquid equilibrium ones.
11.7.8 Comments on the Estimation of Mixture Properties 1. The predictive techniques for pure compounds can be applied to mixtures as well. The quality of the obtained results, however, depends on the need for, and the availability of, interaction coefficient values. 2.For the volumetric behavior of vapor and gaseous mixtures of hydrocarbons up to moderate pressures, the cubic equations of state discussed in Chapter 10 give reasonable results even with kij = 0. At very high pressures, the Pitzer correlation (with the LeeKesler Tables) should be used. 3.In the presence of gases such as C02 , CO, H2S, N2 , etc. in the hydrocarbon mixture, however, interaction coefficients should be used for accurate results, especially at high pressures. They are also needed, but to a lesser degree, in the presence of CH 4 • 4.For the volumetric behavior of polar  and, of course, nonpolar mixtures at low pressures, the virial equation gives reasonable accuracy, provided that Iii values are available. 5.For polar systems at high pressures the cubic equations of state modified for polar compounds (Section 14.7.3), and for still higher ones the PitzerLeeKesler approach, give the most reliable results. The accuracy is poorer, however, than that of the previous cases. 6.For volumes of nonpolar/weaklypolar saturated liquid mixtures, the translated EoS give very good results as we will see in Chapter 14. The correlation of Hankinson and Thomson, Chapter 8, gives also good results. 7 .For enthalpy and entropy departure values of nonpolar/weakly polar mixtures, the cubic EoS discussed here give reasonable accuracy up to moderate pressures. At higher pressures, the PitzerLeeKesler approach should be used. 8.For more details on the estimation of the volumetric behavior and of enthalpy and entropy departure values for gas and liquid mixtures, see Reid et al.
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360
11.8 Fugacities in Mixtures Fugacities of the components of a mixture are extremely important in Chemical Engineering applications for they are essential in chemical and in phase equilibrium calculations. We will start our discussion of this subject by presenting in this Section: l.the definition of the fugacity of component i in a mixture and how it can be evaluated in principle; 2.the relationship between the fugacity of a mixture and those of its constituents, which will demonstrate that the component fugacities are not partial molar properties of the mixture fugacity; and 3.the relationship between the fugacity of a component i in a mixture and that of pure i, for the special case of an ideal solution. We proceed then with a presentation in Sections 11.9 and 11.10 of the two approaches used in the evaluation of fugacities of mixture components: a. equations of state, and b.the standard state fugacity, and apply them  in Section 11.11  in the determination of the solubility of solids and liquids in gases. Extensive application of fugacities is made, of course, in the Chapters where phase and chemical equilibria are discussed (Chapters 12, 13, 14 and 15).
11.8.1 Fugacity of the Components of a Mixture The fugacity of the component i, };, in a homogeneous mixture is defined in a fashion similar to that of a pure compound:
dp.; = dG; = RTdln}; = ii;dP, T=const.
(11.8.1)
with the proviso that:
. }; IIm
p ... o X;P
=
1
For the case of an ideal gas Eq.11.8.1 yields:
RT RTdln(x;P) = V;(id)dP = pdP
(11.8.1a)
Properties of Mixtures
361
Subtraction from Eq.11.8.1 yields after rearrangement and integration:
1 lnq,. =In = A
];
x.P I
1
RT
I
p (RTJ dP V.o 1 P
(11.8.2)
where 4> i• is the fugacity coefficient of component i of the mixture. Determination, hence, of the fugacity of the component i of a mixture is carried out  as in the case of a pure compound  through its fugacity coefficient. Evaluation of the latter, however, requires partial molar volume data for component i in this mixture  from zero pressure to that of the system  all at the temperature and composition of the system. Such information is of course very rare, and fugacities are typically calculated using one of the following two approaches: 1.through equations of state, and 2.through the concept of standard state fugacity. The first approach is discussed in Section 11.9, and can be successfully applied to vapor mixtures; and to liquid mixtures of nonpolar or weakly polar compounds (i.e. hydrocarbons, CO, C02 , H2S, H2 , etc.). For polar liquid mixtures, the available equations of state do not provide yet reliable results, especially in multicomponent cases. Fugacities in such mixtures, therefore, are calculated with the second approach, discussed in Section 11.10. For a physical interpretation of the fugacity of the component i of a mixture, we will demonstrate in the next Chapter that when two phases for example liquid (/) and vapor (v)  are in equilibrium with each other at a given temperature and pressure, then: };1
= };v
(11.8.3)
We can interpret, therefore, fugacity in this case as the escaping tendency of component i in each phase, just as we did for pure compounds.
11.8.2 Relationship Between I and j i Following Eq .11. 8.1 the fugacity of a mixture,/, is related to its chemical potential, p,, through the expression: dp,
= RTdlnj, T = const.
(11.8.4)
Integration at constant T and composition, from a low pressure p•, where the mixture behaves as an ideal gas, to pressure P, yields: p,[T,P,xpx2 , ••• ]p,*[T,P*,xi'x2 , ••• ] = RTln(l..) (11.8.5)
p•
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362
l
Integration of Eq .11. 8.1, under the same conditions yields:
P.;[T,P,x1 ,~,
••• ]
p./[T,P*,x.,x2 , ... ] = RTln[
J;
x.P* I
(11.8.6)
Multiplication by x1 of both sides of Eq.l1.8.6 yields:
~
x1p.. x.p.1* = x.RTin[ p* ] I I I
(11.8.7)
X;
where the terms in the brackets are dropped for simplicity. Summation over all the components of the mixture leads to:
E[x,,,x,,.,·J ",,· "RTEx,[l•(~)lnP"l
(11.8.8)
Comparison with Eq.l1.8.5 gives:
In/" Ex1 ln(~)
(11.8.9)
Subtracting lnP from both sides yields: ln4> = Ex;ln~;
(11.8.10)
Eqs 11.8.9 and 11.8.10 indicate that ln(J/x;) and In~; are partial molar properties;]; and ~; are not. This is why we use the symbol (''') and not to differentiate the fugacity and fugacity coefficient of a mixture component i from that of pure i.
n
11.8.3 The Case of an Ideal Solution Subtraction of Eq.9.11.4 from Eq.11.8.2 yields:
J;
In = 
X;!;
1
RT
 ~)dP Ip (V.0
(11.8.11)
I
where V; is the partial molar volume of i, and V; the molar volume of pure i, both at the same P and T. In the case of an ideal solution, V; = V;, and consequently: (11.8.12)
Properties of Mixtures
363
0.97
....c
cu :Qo.94 ::::cu 0
(.)
~0.91 0 0
0>
~
0.88
0.20
0.40
0.60
Vapor Mole Fraction of Acetonitrile
0.80
1.00
Figure 11.1 Fugacity coefficients for the system acetonitrile(!) acetaldehyde(2) at 1 bar and l00°C with the virial equation and experimental Bii and Bii.
This equation is also known as the LewisRandall rule. It can be used for the determination of the fugacity of a mixture component through its fugacity in the pure state  at the same P and T as the mixture and in the same phase. Eq.11.8.12is applicable to ideal mixtures only, i.e. mixtures of similar compounds. It does provide, however, a good approximation for ]; in nonideal systems at high mole fractions of component i, where  as we saw in Example 11.2  ideal solution behavior with respect to component i is approached. The accuracy declines, of course, with decreasing mole fraction of component i; and with increasing nonideality of the solution.
11.8.4 Comments l.At low pressures, say below 5 bar, vapor phase fugacity coefficient values are not very different from unity as Figure 11.1 demonstrates. 2.Systems containing organic acids, where dimerization due to hydrogen bonding occurs, demonstrate large deviations from ideality, even at subatmospheric pressures. Thus, Prausnitz et al (1980) report for the system acetaldehydeacetic acid at 90°C and 0.25 atm fugacity coefficient values in the range 0.4 to 1.5. 3.At high pressures, fugacity coefficients deviate substantially from unity as demonstrated in Figures 11.2 and 11.3.
Applied Chemical Engineering Thermodynamics
364 1.25
...
1.20
...
... ... ...
0
u
...
...
.... 1.15 0
..... c
...
...
...

·~ 1.10
:0: 4)
...
0
u 1.05 ~
·u
...
0
01
~
... ... ...
...
...
1.00
0.95
0
150
75
300
225
Pressure (bar)
...
375
Figure 11.2 Fugacity coefficient of C02 (1) in its mixture with nC4 at 478 K and y 1 = 0.17. (Calculated from the volumetric data of Olds et al.)
Q)
c
Q)
0
:E 10 a.

l
0
c 0
+'
c
Q)
·u
**
:0:
'Qi10J 0 0
0 0
* 0
50
100
* 0
150
Pressure (bar)
*
~
A
200
250
Figure 11.3 Fugacity coefficients of napthalene in ethane at saturation. (Calculated from the solubility data of Johnston et al, 1982.)
365
Properties of Mixtures
11.9 Fugacities with Equations of State To express the fugacity of component i of a mixture through an EoS, we note that  in the typical case  the EoS is in the form: P =f(V, T, camp.)
(11.9.1)
rather than in the form:
=f(P,
V
T, camp.)
(11.9.2)
that is needed to evaluate V; in Eq.11.8.2. For this reason, the fugacity coefficient is determined using the following thermodynamic transformation of Eq.11. 8.2, where the independent variable is V rather than P (Smith and Van Ness, 1987):
( aNz] aN.
.. , = z1lnz JvQO ln. J I b.From the equality of fugacities of C02 in the vapor and liquid phases:
Yt4>tP = xt Yt• Ht From the ideal vapor assumption: 4> 1 = 1.0; and, since in this case x 1 is close enough to zero, we can safely assume y 1* = 1.0. Hence: H 1 = 1/x1 = 1053 bar. c.At p1 = 2 bar: x 1 = 2/H1 = 0.0019, which is again close enough to zero for Yt*= 1.0.
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Applied Chemical Engineering Thermodynamics
Note: At low pressures, the solubility of practically all gases in a variety of liquids is very small and Henry's law, i.e. 1.0, can be safely assumed. This is not the case, however, at high pressures as we will see when considering high pressure vaporliquid equilibrium in Chapter 14.
rt "'
11.11 An Application of Fugacities: Solubility of Solids and Liquids in Gases 11.11.1 Solids Consider the equilibrium between a solid, component 2, and some gas, component 1, at some temperature T and pressure P. Since the solubility of the gas in the solid is negligible, the equality of fugacities for component 2 in the two phases yields: Y2~2P = p2sq,2s(Peh
where: (Pe)2 = exp (
(11.11.1)
s)l
V: (PP 2 RT 2
P28 , V2 are the solid vapor pressure and molar volume respectively at T. The solubility of the solid in the vapor phase is, therefore: ps
y2
=
2 PE
(11.11.2)
where: E = q,2s(Peh
~2
(11.11.3)
At very low pressures, the two fugacity coefficients and the Poynting effect are equal to unity, and: Y2
= p2sfp
is referred to as the ideal gas law solubility of the solid in the gas. Eq. 11.11.2 becomes, therefore: Y2 = y2(ideal gas law) E As the pressure increases, however, E becomes progressively more important. Actually, E is always larger than one, leading to increased 
375
Properties of Mixtures
over the ideal gas law  solid solubility. It is referred to, appropriately, as the enhancement factor.
11.11.2 Comments !.Evaluation of y2 involves a trialanderror calculation. 2.The enhancement factor can assume very large values at high pressures as demonstrated with the case of fluorene in ethylene (Figure 11.5). 3.Notice that at a given pressure the solubility enhancement is more pronounced at temperatures just above the critical of the gaseous solvent (Tc of ethylene equals 282.4 K). 4.Theenhancementfactorcontainsthreeterms: ¢ 2s, (Pe) 2 , and ¢ 2 . The first is practically equal to one, since the solid vapor pressure is low. The second represents the effect of pressure on the fugacity of the solid, and it increases with increasing pressure. It rarely, however, assumes values larger than 2 to 3. The third, which reflects mainly the difference in the 11, 22, and 12 intermolecular forces, is thus the most important one for it can assume values much smaller than 1 (Figure 11.3). 5.These large solubilities of solids in supercritical gases represent the basis for the increasingly popular field of Supercritica/ Extraction. Consider, for example, a solid mixture containing a valuable component A. If a gas can be found that through strong interaction with A  but not with the other components of the mixture  can act as a good gas solvent, then recovery of A can be effected. For once the resulting, from the treatment of the solid, gas mixture is decompressed, E becomes equal to 1 and the solid precipitates. · 6.For best results, temperatures just over the gas Tc and pressures substantially larger than its Pc should be used as Figure 11.5 suggests. Under these conditions liquidlike solubilities are approached combined I
I
with the large gas diffusivities.
7. The main advantage of supercritical extraction, over the conventional one or distillation, is that solvent free products  of essential importance in the food and pharmaceutical industries can be recovered. Substantial energy savings are also possible. 8.For a review of the status of industrial applications of supercritical extraction, see Basta (1985); McHugh and Krukonis (1986); and Johnston and Penninger (1989) . .9.The thermodynamic treatment of supercritical extraction is rather difficult for three main reasons: a. the system is highly asymmetric, both in terms of molecular size and
Applied Chemical Engineering Thermodynamics
376
....c: cu
E cu 0
c:
_g 10 c:
3
w
100
200 300 Pressure (bar)
400
500
Figure U.S Variation of the enhancement factor with pressure as a function of temperature for the system fluoreneethylene( from the solubility data of Johnston et al, 1982). Lines for identification only.
of intermolecular forces, as demonstrated by the difference in the critical properties of the gas and the solute; b.the system is close to the gas critical temperature; and c. pure component property values: Tc, Pc, cu, and vapor pressures are often unavailable, or unreliable, for the soluteS. 10.Quantitative calculations are carried out through equations of state [see, for example: Prausnitz, et al (1986); and Johnston et al, 1982]; or empirical approaches (Ziger and Eckert, 1983). For a review of the subject, see Williams (1981); Starling et al (1984); and especially McHugh and Krukonis; and Johnston and Penninger. We discuss, next, the calculation of solubilities of liquids in gases. Superficial(!) Extraction
•Theron Brayman of Orrville, Ohio, passed along a report of a new process that is said to remove 95% of the cholesterol from various dairy products without effecting their appearance, consistency, or taste. The process is reported to use pure carbon dioxide and to be similar to a technique known as supeificial(!)fluid extraction. " (Chern. & Eng. News, Dec. 22, 1986, p.44.)
377
Properties of Mixtures
11.11.3 Liquids The approach for determining the solubility of liquids in gases is similar to that of solids, except for the complication introduced by the fact that, here, the gas is soluble in the liquid. At high pressures, the solubility of the gas in the liquid phase can be substantial and the subject will be discussed under high pressure vaporliquid equilibrium. Here we will concentrate in the simpler case of low pressures, where the gas solubility is typically low enough for Henry's law to apply. Component 1 is again the gas, and 2 the liquid. From the equality of the fugacities of component 2 in the two phases:
Y2=
x2P2s 2 does not deviate much from unity and the solubility of the liquid does not depend on the nature of the gas. 2.At increased pressures, however, very small values of 4> 2 are realized that lead to large solubilities, just as it was the case with solids, and the nature of the gas becomes very important. 3.Thus, Prausnitz et al (1986) report that the solubility of ndecane in carbon dioxide, at 350 K and 100 atm, is about: * ten times larger than that in nitrogen; and, * forty times larger than that in hydrogen. 4.The solubility of liquids (and solids) in gases must be taken into account in gascooling and gastransportation processes, especially of natural gas, to avoid plugging by precipitating liquids (and solids) due to drop in temperature and pressure. 5.The concept of supercritical extraction can be of course applied to mixtures of highboiling liquids as well.
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Applied Chemical Engineering Thermodynamics
11.12 Property Changes of Mixing 11.12.1 Definition Consider the mixing of x 1 moles of compound 1, x2 moles of 2, ... , and moles of compound k, for a total of one mole. Pure compounds and mixture are at the same temperature T and pressure P so that the change, from pure compounds to mixture, reflects only mixing effects. The property change due to mixing or just the property of mixing, as it is referred to  is defined by:
xk
LlM = M 'r' x.M. (11.12.1) L..J I I and represents the measure of this change. LlH, for example, is the amount of heat that must be added or removed so that the mixture attains the same temperature as that of the pure compounds, both at the same pressure. Replacement of M in Eq.11.12.1 through Eq.11.4.4b yields: LlM
= LX;(M;M;)
(11.12.2) (11.12.3)

where:
LlM;
=

M;M;
(11.12.4)
represents the effect of mixing on component i. Rearrangement of Eq .11.12 .1 yields: M
= L..J 'r' x.M.+ LlM I I
(11.12.1a)
indicating that knowledge of LlM  along with the corresponding properties of the pure compounds, all at the same T and P  suffices for the determination of the mixture property M. For example, knowledge of LlH and of the pure compound enthalpies, all at the same P and T, determines the enthalpy of the mixture. We consider next the property changes associated with the formation of the simplest possible mixture, an ideal one.
11.12.2 The Case of an Ideal Solution An ideal solution (also referred to as an ideal mixture) meets the following requirements:
379
Properties of Mixtures
l.the forces between unlike and like molecules are the same, and 2.the molecular sizes and shapes are also the same. Using (id) to designate such behavior, we proceed to develop expressions for .dM(id), where M = V, H, G and S. From Eq.11.12.1: .1V(id)
= V(id)Ex1V1 = Ex;[V1(id)Y;]
.dH(id)
= H(id)Ex1H 1 = Ex1[H1(id)H1] (11.12.6)
.tiG(id)
= G(id)Ex1G1 = :Ex;(G1(id)G1] (11.12.7)
.1S(id) = S(id)Ex1S1 = Ex;[S;(id)S1]
(11.12.5)
(11.12.8)
In the first two cases, the ideal solution assumption dictates that: Y;Cid) =
v1
H 1(id)
H1
=
and, consequently:
0
(11.12.5a)
.1H(id) = 0
(11.12.6a)
.1 V(id)
=
For the case of .d G(id) and .1S(id ), however, the answer is not so apparent. Rather, to develop the appropriate expressions, we integrate Eq.11.8.1 from the fugacity of pure i, .f;, to that of i in an ideal solution at the same T and P which, according to Eq.11.8.12, is equal to (x1.f; ): x.f (11.12.9) G.(id)G. = RTln!;1 1 = RTlnx.I I I Hence: .1G(id)
Finally, since G
=
= RT'Ex11nx1
(11.12.7a)
H TS, .tiG = .dH T.tiS: .dS(id) = R:Ex1 1nx1
(11.12.8a)
Comments l.An ideal mixture is, of course, a hypothetical one. Since, however, the second requirement for such behavior becomes important only at large differences in size, ideal mixture behavior is approximated by molecules of similar chemical structure.
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2.Thus a mixture of pentane and hexane, or of ethanol and propanol, would approximate an ideal mixture, especially in the gaseous phase, where the larger distances among molecules tend to reduce the effect of the difference in inter and intramolecular forces. 3.According also to Section 11.8.3, the larger the composition of a component of a mixture, the closer its behavior approaches that in an ideal mixture. 4.Notice that tl.M(id) = 0 for volume, internal energy and enthalpy; but it is different from zero, as it should be expected, for entropy and the associated properties, the Gibbs and Helmholtz free energies. 5.It is apparent, finally, that since from Eq.11.12.1a: M(id) = 'L,x;M;+tl.M(id)
the properties of an ideal mixture are known provided that the corresponding values of the components in their pure state, at the same P and T as the mixture, are known. M(id) represents, therefore, a convenient reference for comparing real mixture behavior to, as discussed next.
11.13 Excess Properties 11.13.1 Definition The difference between the value of the property M of a mixture at a given pressure, temperature and composition, and its value M(id) if the mixture were to behave as an ideal one at the same conditions, represents the excess property ME: ME= MM(id) (11.13.1) Similarly, for the component i of a mixture: E
M;
=


M; M;(id)
(11.13.2) (11.13.3)
From the theoretical point of view, excess properties provide a quantitative measure of the effect of the differences: * between inter and intramolecular forces; and * in the size and shape of the molecules; on the mixture behavior.
381
Properties of Mixtures
From the practical point of view, the most important excess property is that of the Gibbs free energy, because of its relationship to the activity coefficients of the components of the mixture, as it will be demonstrated next.
11.13.2 The Excess Gibbs Free Energy and the Activity Coefficient Integration of Eq.11.8.1:
dG;
=
RTdln}i
between the ideal and real solution states at constant temperature, pressure, and composition, yields: E G. I
=
G.G.(id) I I
=
1 1 1  X. RT In [x.y.f/1 F_O
=
IJi
RTlny.I
(11.13.4)
and, according to Eq.11.13.3: E
E
G =ExiGi
=
RTEx;lnyi
(11.13.5)
which indicate that RTln Y; is a partial molar property of the excess Gibbs free energy. It follows then, that: lnr;
=
[a(N~;:RT)l 1
(11.13.6) T,P,~
Comments 1.Notice that, since for an ideal mixture the excess Gibbs free energy is equal to zero, all activity coefficients in such a mixture are equal to one. 2.Eq.ll.13.6 provides the basis for the development of analytical expressions for the activity coefficient, such as those of van Laar, Margules, Wilson, etc. To this purpose, qualitative and quantitative arguments about the deviation of a given solution from an ideal one at the same conditions, are used to develop analytical expressions for G E as functions of the composition, temperature, and pressure. Appropriate differentiation, using Eq.11.13.6, yields then the analytical expression for the activity coefficient of each solution component. We will discuss this approach in Chapter 13.
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Applied Chemical Engineering Thermodynamics
11.14 Concluding Remarks !.Differences in: a.inter and intramolecular forces, and b.the size and shape of the molecules, lead to nonideal mixture behavior. 2.As a result, mixture properties cannot be determined from pure compound properties alone. 3.Experimental data are available for a variety of binary mixtures, but rarely for multicomponent ones that are needed in the typical industrial application. 4.We resort therefore, as with pure compounds, to predictive techniques, mainly the corresponding states principle and equations of state. S.Vapor phase mixture properties at low pressures are calculated with the virial equation. Interaction coefficients are useful for nonpolar systems, important for polar ones, and essential where solvation occurs. 6.For vapor and gaseous nonpolar/weaklypolar systems at higher pressures cubic EoS are used. Reasonable accuracy is obtained for hydrocarbon systems without interaction coefficients, which are essential however in the presence of gases such as CO, C02 , H 2 , H 2S, etc., and to a lesser degree of CH 4 • 7 .No reliable method exists for vapor and gaseous polar systems at high pressures; best estimates are obtained with cubic EoS, modified for polar compounds (Section 14.7 .3), and for very high pressures with the PitzerLeeKesler approach. 8.Component fugacities in nonpolar/weakly polar liquid mixtures are calculated with EoS, typically cubic ones, and use of kij is essential for accurate results, except for mixtures of similar hydrocarbons, such as of alkanes (that do not contain CH 4 , however). 9 .For polar liquid mixtures, the standard state fugacity approach is used for this purpose. For a subcritical component, it is expressed through its vapor pressure; for a supercritical one, through its Henry's constant in the solution, requiring thus mixture data (see also Section 14. 7.3). 1O.ln gassolid and gasliquid systems, large deviations from ideal solution behavior in the vapor phase at high pressures can lead to high solubilities of solids and of liquids in compressed gases. 11. We have just touched on the subject of phase equilibria which, along with chemical reaction equilibria, are of paramount importance in our profession representing, as we have seen, the third main objective of
Properties of Mixtures
383
Chemical Engineering Thermodynamics. Our discussion of phase equilibria in mixtures will continue in Chapters 13 and 14, while chemical reactions in them will be considered in Chapter 15. The thermodynamic basis for such equilibria will be developed in the next Chapter.
References Basta, N., 1985. Chern. Eng., Feb. 4, 14. De Santis, R., Breedveld, G.J.F., Prausnitz, J.M., 1974. Ind. Eng. Chern. Process Des. Dev., 13, 574. Graboski, M.S., Daubert, T.E., 1978. Ind. Eng. Chern. Process Des. Dev., 17, 448. Hayden, J.G., O'Connell, J.P., 1975./nd. Eng. Chern. Process Des. Dev., 14, 221. Hermsen, R. W., Prausnitz, J .M., 1963. Chern. Eng. Sci., 18, 485. Johnston, K.P., Penninger, J.M.L., Editors, 1989. Supercritical Fluid Science and Technology, ACS Symposium Series No 406, ACS, Washington, D.C. Johnston, K.P., Ziger, D.H.,Eckert, C.A., 1982./nd. Eng. Chern. Fundam. ,21, 191. Kato, K., Nagahama, K., Hirata, M., 1981. Fluid Phase Equilibria, 1, 219. Lin, H., 1980./nd. Eng. Chern. Process Des. Dev., 19, 501. Lin, C., Daubert, T.E., 1980. Ind. Eng. Chern. Process Des. Dev., 19, 51. McHugh, M.A., Krukonis, V.J., 1986. Supercritical Fluid Extraction, Butterworths, Boston. Olds, R.H., Reamer, H.H., Sage, B.H., Lacey, W.N., 1949./nd. Eng. Chern., 41, 475. Peneloux, A., Rauzy, E., Freze, R., 1982. Fluid Phase Equilibria, 8, 7. Prausnitz, J.M., Lichtenthaler, R.N., de Azevedo, E.G., 1986. Molecular Thermodynamics of FluidPhase Equilibria, PrenticeHall, Englewood Cliffs, N.J. Prausnitz,J.M., Anderson, T.F., Grens, E.A., Eckert, C.A.,Hsieh, R., O'Connell, J.P., 1980. ComputerCalculationsfor MulticomponentVaporLiquidand LiquidLiquidEquilibria, PrenticeHall, Englewood Cliffs, N.J. Prausnitz, J.M., Carter, W.B., 1960. A/ChE J., 6, 611. Reamer, H.H., Sage, B.H., Lacey, W.N., 1951. Ind. Eng. Chern., 43, 976. Reid, R.C., Prausnitz, J.M., Poling, B.E., 1987. The Properties of Gases and Liquids, McGrawHill, New York. Sandler, S., 1977. Chemical Engineering Thermodynamics, Wiley, New York. Sage, B.H., Budenholzer,R.A., Lacey, W.N., 1940.Ind. Eng. Chem.,32, 1262. Smith, J.M., Van Ness, H.C., 1987. Introduction to Chemical Engineering Thermodynamics, McGrawHill, New York.
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Starling, K.E., Khan, M.A., Watanasiri, S., 1984.AIChE Meeting, San Francisco. Tsonopoulos, C., 1974;1975. AIChEJ., 20, 263; 21, 827. Tsonopoulos, C., 1979. In Equations of State in Engineering and Research, Chao, K. C., Robinson, R. L., Editors. Advances in Chemistry Series, Vol. 182, Am. Chern. Soc., Washington, D.C. Tsonopoulos, C., Heidman, J.L., 1986. Fluid Phase Equilibria, 29, 391. Williams, D.F., 198l.Chem. Eng. Sci., 36, 1769. Ziger, D.H., Eckert, C.A., 1983. Ind. Eng. Chern. Process Des. Dev., 22,582.
Problems 11.1 What is responsible for the difference in values between partial molar and pure component properties? 11.2 With reference to Example 11.1: a. why is a only small amount of water added? b. can you determine the partial molar volume of methanol from the given data? 11.3 Under what conditions will a mixture behave as an ideal one? 11.4 What factors effect the values of the partial molar volume of a component of a binary mixture? 11.5 The enthalpy of a binary liquid mixture at a given temperature and pressure is given by: where: a = 15000; b = 20000; and c = 2000; all in J/mol. a. Calculate the enthalpies of pure 1 and 2; b.Plot the partial molar enthalpies of 1 and 2 versus x 1 ; c.Compare H1 at x 1 = 0.2 and at x 1 = 0.8, to H 1; d.Discuss the effect of composition on ii1 and ii2 • 11.6 With reference to the GibbsDuhem equation: a.give two applications; b.develop a specific example where you apply it. 11.7 The partial molar enthalpy of component 1 in a binary mixture, at a specified temperature T and pressure P, is given by:
H1 = H1 +ax/+bx23 where: H 1 = 100001/mol, is the enthalpy of pure 1 at Tand P, and a = 500and b = 1000, both in J/mol, are functions ofT and P only. Develop the expression for ii2 as a function of x 1 (any problems?).
385
Properties of Mixtures
11.8 With reference to the mixing rules: a. what is their purpose? b. what is the combining rule? c. what is the purpose of the interaction coefficients? 11.9 Smith and Van Ness (1987) suggest that the second virial coefficients of Prausnitz and Carter (1960) for the system acetonitrile(1) acetaldehyde(2)can be represented approximately as follows in the range 325 K to 375 K:
Bu=8.55(1~r50; B22=21.5(10~r2S; B.2=1.74(1~r3S where: Tin K, and B in cm3/mol. a. Calculate the molar volume of an equimolar mixture at 370 K and 1.5 bar. b. Give your justified estimate of the accuracy of the obtained value. 11.10.a.In Problem 11.9, you used experimental values for the second virial coefficients. Explain if use of the term 'experimental' is justified. b.Repeat the Problem, using values obtained through the empirical correlation of your choice. 11.11 Repeat Problem 11.9, using the PR equation of state. 11.12 Tabulate your results on the acetonitrileacetaldehyde system (Problems 11.9 through 11.11), compare them to the correct value, and explain the observed differences. 11.13 With reference to Problem 11.9: a. plot the partial molar volume of component 2 versus y 2, and comment on the effect of y2; b.repeat the plot, assuming the mixture to behave as: (1) an ideal mixture of real gases, and (2) an ideal gas, and discuss the accuracy of the obtained values. 11.14.a.What is the minimum information needed to estimate the volume of a nonpolar vapor or gaseous mixture? b. The same for a polar one. c. What method would you use in each case? d. What accuracy would you expect? Consider both, low and high pressures, and complete the following Table: System
Pressure
Nonpolar
Low
Nonpolar
High
Polar
Low
Polar
High
Proposed method
Information needed
Expected accuracy
11.1S.a.With reference to the interaction coefficient 112 in the Tsonopoulos correlation for B12 :
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Applied Chemical Engineering Thermodynamics
1. what is 112 a function of ? 2.can it be estimated from pure component properties? If yes, when? 3.discuss the type of experimental measurement that you would need to evaluate it. b. The same, but for k12 (SRK). 11.16 For an equimolar vapor mixture of methane and hydrogen sulphide at 160 °F and 1000 psia: a. what method would you use to estimate the molar volume V of this mixture? b. present the reasons for your decision; c.calculate V using this method; d.could you get a better answer? Experimental value: 5.52 ft3 /lbmol (Reamer et al, 1951) 11.17 A gas mixture consisting of 80% (mole) methane and the rest propane exerts a pressure of 6 MPa when stored in a cylinder at 280 K. What is the maximum temperature that the cylinder may be exposed to if the pressure is not to exceed 12 MPa? 11.18 A gas mixture consisting of 75% (mole) methane and the rest propane enters a pipeline of constant crosssection at 310 K and 70 bar with a linear velocity of 10 mls, and leaves at 280 K and 20 bar. If the flow is steady, what is the linear velocity of the gas leaving the pipe? 11.19 One thousand moles of an equimolar mixture of nitrogen and carbon dioxide at 1 bar and 325 K are compressed isothermally to 200 bar. What is the corresponding enthalpy and entropy changes? 11.20.a.Show that if the volumetric behavior of a gas mixture is described by:
z
BP
= 1+
RT
then :
lncf> = BP RT
b. Show that if this mixture is a binary one, then:
p 2 lncf>.I  (B RT II.. +y.J oI).. ) A

11.21 Why do we use],. and not j.? I I 11.22 Starting with: RTdlni; = ~dP, develop the LewisRandall rule. 11.23 Express the partial pressure of a component of a vapor mixture in terms of its fugacity. 11.24 For the acetonitrile(!) acetaldehyde(2) system at: y 1 = 0.1, T = 373 K and P=1 bar, calculate the fugacity of component 1 using: a.the virial equation with the Bii and Bii values given in Problem 11.9; b. the same equation, with Bil and Bij estimated with the method of your choice; c.the PR EoS. Tabulate your results, and explain the obtained % errors. 11.25 The fugacity coefficient of nbutane(l) in a vapor mixture with nitrogen(2) at y 1 =0.5024, 370°C, and 100 bar is needed.
387
Properties of Mixtures
a. What method would you recommend? b. What accuracy would you expect? c. Calculate the fugacity coefficient with the recommended method. d. What information is needed to determine this fugacity coefficient ex· perimentally? 11.26 If component 1 of a binary equimolar vapor mixture is nonpolar, complete the following Table: Component2
Pressure
Range of 4>1 values Comp. 1
Nonpolar
Low
Nonpolar
High
Polar
Low
Hbonded
Low
Comp. 2
11.27.a.Calculate the fugacity of steam at 300°C and 10 bar. b. Using this answer, estimate the fugacity of steam(l) in a mixture with nbutane(2) at the same conditions, and at two mole fractions, y 1 = 0.2 and y 1 = 0.8, using the LewisRandall rule. c. Which of the two values is more accurate? Explain. d.How could you get a better estimate? 11.28 In Figure 11.4, the curves for the two activity coefficients are terminated beforex1=1.0. Why? 11.29 At 346 K and 1 atm, a liquid mixture of acetone( I) and benzene(2) with x 1 = 0.20 is in equilibrium with a vapor phase where y 1 = 0.40. Assuming ideal gas behavior for the vapor phase, calculate: a. the fugacity of acetone in the liquid phase; b. the standard state fugacity for acetone at these conditions; c. the activity coefficient values. Estimate the error involved in each case as a result of the ideal gas assumption. 11.30 What experimental measurements would you carry out to determine the Henry's constant of methane in ethanol at 25°C and 10 bar? 11.31 The solubility of methane in nheptaneat 220°F and a partial pressure of 1 atm is: x = 2.83·104 . a. What is the solubility of methane at the same temperature and a partial pressure of 5 atm? b. State any approximations you make and estimate their impact on your answer.' 11.32 Do Problem 11.61, part c. 11.33 The fugacity of component 1 in a binary liquid mixture at some temperature T and pressure P is given by:
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Applied Chemical Engineering Thermodynamics
a. What error would result from evaluating } 1 by using the LewisRandall rule at x 1 = 0.5 and the same T and P? b.Calculate the Henry's constant of component 1 in this mixture at the given Tand P. 11.34 With reference to supercritical extraction: a. what is it? b. what are its advantages over other separation techniques? c. what operating conditions are desirable? d. what are the difficulties in its thermodynamic modelling? e. give a possible application of your choice. 11.35 The vapor pressure of solid ndodecane at 10°C is 0.6 Pa. Calculate its solubility in methane at 10°C and 100 bar. 11.36 For an ideal mixture: a. which property changes of mixing are equal to zero and which are not? b.show that all activity coefficients are equal to one. 11.37 Discuss the importance of excess properties. 11.38 Redlich and Kister have proposed the following polynomial expression to represent GE of a binary mixture: GE
= x 1x2 [A +B(x1 x2 ) +C(x1 x2 i+ ... ]
For moderately nonideal systems, the two first terms are usually sufficient. Show that for such systems:
RTln y 1
= (A +3B)x/4Bx23
RTlny2 = (A3B)x 12 +4Bx 13 11.39 A binary solution has the following properties: a.its entropy is the same as that of a hypothetical ideal solution of the same composition, and b. its enthalpy  per mole of mixture  exceeds that of the corresponding ideal solution by an amount (bx 1x2 ), where b is a function of pressure only. Show that the activity coefficients are given by:
RTlny1 = bx/ RTlny2 = bx 12 11.40 What advantages, if any, does the virial equation have over the other methods in the estimation of mixture properties? 11.41 Explain positive and negative values for lij in Eq.11. 7. 7 in terms of intermolecular forces. 11.42 With reference to this Chapter: a. what is the main objective?· b. describe briefly three important items that you have learned including an application in each case.
389
Properties of Mixtures
11.51 Demonstrate, with whatever method you want, that k12 (or 112) is more important for the accurate determination of the partial molar volume of a component of a mixture, than for that of the mixture volume. 11.52 For the system acetonitrileacetaldehyde: a. using the B12 data of Prausnitz and Carter, Problem 11.9, calculate k12 in the Tsonopoulos correlation as a function of the temperature T. b.how do you explain this variation with T ? 11.53 Dave and George want to estimate the molar volume of a methane nbutane mixture at: y 1=0.8, 150°C, and 100 atm. George: Let us use the SRK EoS; Dave:Whataboutk12? George: They are both alkanes, we can use k12 = 0.0; Dave: Yes, but with methane present we would get better accuracy if we could find a value for k12 • Perhaps we can use the virial equation, where we have 112 from the Tsonopoulos correlation; George: Ya, but the pressure is too high; Dave: Well, even if we cannot use the virial, there must be a way of using this 112 value. a. What advice would you give to them? b. Calculate the molar volume following your advice; c. What accuracy do you expect? Hint: Develop first an expression for the second virial coefficient through the SRK EoS. 11.54.a.Show that use of the SRK EoS for the entropy departure of a binary system gives:
where:
Q
= [ Y1 2 +y1y2(1k12)
(a
l
l
2 a11 a22 da11 + [ Y2 +y1y2(1k12) (a a )112 a )112 dT
11 22
11 22
da22 dT
b.Do Problem 11.19, using the SRK EoS with k 12 = 0.0. c.Compare the answer with the one you obtained in Problem 11.19 and comment on their accuracy. 11.55 Show that:
v."' [alnHi.s] aP r I,S
where H;,s and ~.s"' are the Henry's constant and infinite dilution partial molar volume respectively of component i in solvents.
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Applied Chemical Engineering Thermodynamics
11.56 In Example 11.10 we neglected the effect of pressure on the Henry's constant. What error does this lead to? 11.57 The solubility of methane in water, at 25°C and a partial pressure of 1 atm, is x = 0.00025. a. Calculate the solubility of methane at a total pressure of 5 atm making any simplifying assumptions you want. b. Comment on the impact of these assumptions on the accuracy of your final answer. c. Is there a better method? If yes, which one? 11.58 Consider the solubility of a solid(2) in a gas(1) as a function of pressure Pat some temperature T. Assuming that the vapor phase nonideality is described by the virial equation truncated after B, show that there is a minimum solubility of 2 in 1 given by:
y (minimum) 2
=
P(at minimum) =
cB ps 12 2
RT
RT
=:==
Bu +2Bl2
where cis a constant depending only on the units used. Neglect x 2 , .•. , xk_ 1 = k  1 *vapor phase compositions: y 1, Y2· ... , Yk 1 = k 1 * pressure (P) and temperature (T) = 2 *total = 2k
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Applied Chemical Engineering Thermodynamics
p
T
y,, y2, ... , yk
x,, x2, ... ,xk Figure 13.1 A vaporliquid equilibrium system.
2.Accordingto the Phase Rule (Section 12.7.1): F = k + 22 = k i.e. specification of k of these 2k variables suffices for the complete description of the intensive state of the system. The remaining k variables would thus have to be determined experimentally. This approach is timeconsuming and costly, however, and our
objective is the evaluation of these k variables using the minimum amount of experimental information possible. This represents the general statement of the vaporliquid equilibrium problem. The typical cases encountered in practice, are presented in Table 13.1.
13.4 Methodology We have seen in Chapter 12 that when a system containing two phases reaches equilibrium  at constant temperature and pressure  the total Gibbs free energy assumes its minimum value; and that this, in turn, leads to the equality of fugacities for any component i in the two phases (Section 12.4.4). In the case of vaporliquid equilibrium, thus:
i/ = i/
(13.4.1)
where the superscripts (v) and (I) refer to vapor and liquid phase respectively. We examine next the methodology used to obtain numerical answers from the equality of these two abstract quantities. To this purpose, we
439
Low Pressure VaporLiquid Equilibrium
Table 13.1 Typical VLE Problems Specified Variables
Variables to be determined
Method
x's and T x's and P
y's and P Bubble Point P y's and T Bubble Point T y's and T x's and P Dew PointP y's and P x's and T Dew Point T An other type of VLE calculation, the flash calculation, will be discussed in Chapter 14. consider  for simplicity  a binary system where, according to Section 13.3, two variables must be specified in order that the intensive state of the system be completely determined. Let us, then, assume that T and x 1 are specified, and examine how P and y1 are evaluated. The two equations to be used for this purpose are: A
v
A
I
ft = ft Av
AI
/2 = /2
Following Section 11.8:
IP(v.v RT)dP p In¢~ = In!!_ = RT IoP(v. 1 RT)dP Ay
In¢~ = In~ Y;P I
= 1
RT
0
I
AI
1
x.P I
1
1
P
If an equation of state that can describe the volumetric behavior of both phases is av~ilable, Athen } 1v and ] 2v can be expressed in terms of P, T andy 1; and// and/21 in terms of P, Tandx1• We have, thus, a system of two equations that can be solved for the two unknowns: y 1 and P. The same methodology would be used for a multicomponent system. In practice vaporliquid equilibrium calculations using an equation of state, that is applicable to both phases, can be carried out with sufficient accuracy only for systems that are up to moderately nonideal, typically hydrocarbon mixtures alone or with such gases as: H2S, CO, C02, etc. Furthermore, since separation of such systems by distillation because of their typically low boiling point temperatures at atmospheric pressure is carried out at high pressures, this methodology represents what is traditionally referred to as the High Pressure or Equation of State approach to
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Applied Chemical Engineering Thermodynamics
vaporliquid equilibrium calculations. For nonideal systems, however, this approach does not provide yet sufficient accuracy, especially for multi component cases, because of limitations in the available equations of state. Such systems, in addition, have typically higher boiling points at atmospheric pressure and their separation by distillation is carried out at low pressures. These systems are described using the following methodology: l.The vapor phase fugacity is expressed through: Ay
A
(11.8.2)
/ 1 = y1 ~ 1 P
and the vapor phase fugacity coefficient 4>1( phi) is evaluated with sufficient accuracy, as we have seen in Section 11.9, through an equation of state. 2. The liquid phase fugacity is expressed through the standard state fugacity Ut> approach (Section 11.1 0): A/
s s
ft = x1 Y;./; = x1 y1P1 ~~ exp 0
[Y;(PPt>] RT
(11.10.3)
where: * y 1 is the activity coefficient (gamma); * Pt is the vapor pressure of pure i at the system temperature T; * ~t is the fugacity coefficient of saturated pure liquid i at T, that can be again evaluated from an equation of state; * V1 is the average molar volume of pure liquid i at T from P/ to the system pressure P; and *the exponential term expresses the Poynting effect (Pe) 1• This methodology represents what is traditionally referred to as the Low Pressure or GammaPhi approach to vaporliquid equilibrium calculations, and it is discussed in this Chapter; and the high pressure one, in the next.
13.5 Activity Coefficients: Determination from VaporLiquid Equilibrium Measurements To this purpose, we present first a typical method for such measurements, and then discuss how the activity coefficients are evaluated from them.
Low Pressure VaporLiquid Equilibrium
441
13.5.1 Experimental Measurements Let us consider the following experiment: 25 ml of a mixture of ethanol and water is charged into a small (100 ml) flask, which is connected to a mercury manometer and a vacuum source, and is placed into a constant temperature bath. In the typical case, liquids contain dissolved gases  mostly air  which would effect the equilibrium conditions and must be removed. To this purpose the bath temperature is lowered until the mixture solidifies and the vacuum is activated. The dissolved gases are removed, while only a small amount of the solvent mixture is lost due to its very low vapor pressure at this temperature. The flask is then disconnected from the vacuum source and the bath temperature is raised to, say, 60°C. We notice that the pressure in the flask is increasing with time, even after the bath has reached the temperature of 60°C. As a matter of fact, the increase continues for several hours, probably over 20, until no discernible variation in pressure can be observed. At this point we say that the liquid and vapor phases of the ethanolwater system are at equilibrium. (In practice, a magnetic stirrer is placed in the flask which helps the attainment of equilibrium, typically in one to two hours. It should be realized, of course, that 'true' equilibrium is reached only after an infinite amount of time.) Let us now assume that, for all practical purposes, equilibrium has been reached and proceed to determine the compositions of the two phases by some analytical technique, such as chromatography. We know, thus, the following quantities for the equilibrium system: * pressure: P; * temperature: T; * composition of the liquid phase: x1 and x2 ; * composition of the vapor phase: y 1 and y2 •
13.5.2 Evaluation of the Activity Coefficient We proceed now to demonstrate how the activity coefficient for the component i of the mixture is evaluated from these data. To develop the appropriate thermodynamic expression for Y; we combine Eqs 11.10.3 and 11.8.2 (Section 13.4):
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Applied Chemical Engineering Thermodynamics
(13.5.1)
Y;
The vapor pressure of pure liquid i, P/, can be calculated from some appropriate relationship, such as the Antoine equation. (In practice, however, it is recommended that for improved accuracy of the calculated activity coefficients it is better to measure P/ directly, by running an experimental VLE measurement with pure i, see Example 13.1.) Determination, thus, of the activity coefficient value requires the evaluation of the two fugacity coefficients and of the Poynting effect. To this purpose, we recall from Chapters 9 and 11 that at the low pressures of interest here: !.Fugacity coefficients of pure saturated liquids, and of the components of a vapor phase, can be successfully evaluated through the virial equation truncated after B:
Incp: In¢;
(9.11.10) p RT
= (2~y.B .. B)J
u
(11.9.5)
where B;;, Bij, and B are the second virial coefficients of the pairs ii and ij and the mixture respectively. 2.In the absence of experimental values for the second virial coefficients, predicted ones through the Tsonopoulos (1974; 1975) or the HaydenO'Connell (1975) correlations, provide reliable estimates of the fugacity coefficients. 3. The Poynting effect can be determined using Eq. 9. 11.14. Determination, thus, of the fugacity coefficients through the virial equation and of the Poynting effect through Eq. 9.11.14 allows the evaluation of activity coefficients from the experimental data. An appropriate computer program (GAMMA) is presented in Appendix E.
13.5.3 Example 13.1 The vaporliquid equilibrium data of Table 13.E.l are reported (Gmehling et al, 1979) for the system: acetone(l)cyclohexane(2) at 25°C. Calculate the activity coefficients assuming, because of the low pressures involved and for simplicity in the calculations, ideal vapor behavior.
443
Low Pressure VaporLiquid Equilibrium
Table 13.E.l VaporLiquid Equilibrium Data and Results for the System Acetone(1)Cyclohexane(2) at 25°C P(mmHg) y, Xt 97.45 118.05 124.95 137.90 145.00 172.90 207.70 227.70 237.85 253.90 259.40 261.10 262.00 258.70 252.00 243.80 230.40
0.0 0.0115 0.0160 0.0250 0.0300 0.0575 0.1125 0.1775 0.2330 0.4235 0.5760 0.6605 0.7390 0.8605 0.9250 0.9625 1.0000
From Eq.13.5.1:
0.0 0.1810 0.2250 0.3040 0.3450 0.4580 0.5670 0.6110 0.6325 0.6800 0.7050 0.7170 0.7390 0.8030 0.8580 0.9160 1.0000
y 1P
r,
Y2
ln(y,ly2)
8.064 7.626 7.278 7.237 5.977 4.543 3.402 2.802 1.769 1.378 1.230 1.137 1.048 1.015 1.007
1.004 1.010 1.010 1.005 1.020 1.040 1.105 1.169 1.446 1.852 2.233 2.689 3.749 4.896 5.604
2.084 2.022 1.975 1.975 1.768 1.475 1.124 0.874 0.202 0.296 0.596 0.860 1.275 1.574 1.716
y2 P
Yt =  ; Y2 = x2 ~ x1
P:
To calculate P 11 and P 2" we can use the Antoine constants given by Gmehling et al, obtaining: P 11 = 229.9 mm Hg and P2s = 97.6 mm Hg, which are very close to the experimental values of 230.40 and 97.45 mm Hg respectively given in the Table. The latter values, however, should be used  as already discussed  which give the activity coefficient values shown also in Table 13.E.l. Notice the pronounced effect of composition: as X; decreases, the deviation of Y; from one increases. We will discuss this further in Section 13.7.
13.5.4 Comments l.It is apparent that vaporliquid equilibrium measurements do not provide all the information needed for the evaluation of the activity coefficients: the vapor phase nonideality must be estimated from an equation of state. Because, however, equations of state  especially the virial  provide reliable estimates of vapor phase fugacities at low pressures, we refer to the activity coefficients calculated through Eq.13.5.1 as
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tli.li nHeptone
~pDioxone
3.00
COWO Butyl omine OQQQO 1Proponol
1
Figure 13.2 Experimentally obtained activity coefficients (lines generated from the Margules equation using infinite dilution y's). *Positive deviations from Raoult's law: nheptane(l)/pdioxane(2) at 80°C. * Negative deviations from Raoult' s law: butyl amine( 1)/1propanol(2} at 45°C.
experimental ones.
2. Typical results for two binary systems are shown in Figure 13 .2. Infinite dilution activity coefficients Y; oo, obtained through extrapolation to X; = 0 of the experimentally determined gammas, are also shown. 3.Eq.13.5.1 can be written in the following 'compact' form: y.P (13.5.2) Y; =  ' 8 F; X;P;
where: F;
¢;
= 
: (Pe );
(13.5.3)
F; represents thus the contribution to Y; of the nonideality in the vapor
phase, ¢; and /, and of the effect of pressure on the liquid phase fugacity (Pe);. The latter is practically equal to 1.0 at the low pressures of interest here. 4.At low pressures, say up to 23 bar, F; is typically in the range of 0.9 to 1.1. (This is not the case for systems containing organic acids
Low Pressure VaporLiquid Equilibrium
445
where, as discussed in Section 11.8.4, large deviations from ideality are encountered, even at subatmospheric pressures.) 5.Activity coefficient values, on the other hand, deviate much more from one, as demonstrated in Figure 13.2. Thus, at low pressures typi
cally below 10 atm  the most impor,tant, by far, quantity involved in vaporliquid equilibrium calculations is the activity coefficient.
6.Ifthe pressure is low, typically 1 atm or below, and the components of the equilibrium mixture similar in chemical nature, we can safely assume ideal vapor behavior. Under these conditions the resulting error is very small, since F1 is very close to 1, and Eq.13.5.2 becomes: Y;P (13.5.4) Y; =  5 x1P1 7. The experimental system mentioned in Section 13.5 .1 is referred to as a 'static' one. In practice, determination of the vapor phase composition in such a system is difficult and can lead to substantial errors. 8. To avoid this problem and also the determination of the liquid phase composition as well  specitied amounts of predegassed liquids are charged into a small flask, called the 'equilibrium cell'. Since the vapor space above the liquid phase is small, the change in the liquid composition as the system reaches equilibrium is negligible and, thus, all x1's are known. Vapor phase compositions are then estimated with the approach that will be discussed in Section 13.15. 9.Complete degassing of a liquid requires more than one freezing and thawing cycles. lO.Ifvapor phase compositions must also be determined experimentally, a different technique referred to as 'dynamic'  is used. For details on experimental methods, see Hala (1967). Before we discuss further the activity coefficient, we consider next the simple case of an ideal solution, where all gammas are equal to one.
13.6 Ideal Solutions 13.6.1 Raoult's Law We saw in Section 11.13.2 that in an ideal solution all activity coefficients are equal to 1.0. Under these conditions Eq.13.5.1 gives: (13.6.1)
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This represents the wellknown Raoult's law, and indicates that the partial pressure, ~. of component i is simply proportional to its mole fraction in the liquid phase; and the proportionality constant is the vapor pressure of pure i at the system temperature, independently of what the other components of the mixture are. In other words, the tendency of component i to leave the liquid state, its liquid phase fugacity, is not effected by the other components of the mixture; just by its own vapor pressure in the pure state and its mole fraction in the liquid mixture. In terms of molecular characteristics, this would require (Section 11.12.2) that: * the forces between like and unlike molecules be the same, and * the size and shape of the molecules be also the same. An ideal solution represents, therefore, a hypothetical case. Because size and shape differences, however, are important only in extreme cases such as polymersolvent systems we would expect that Raoult's law will be a reasonable approximation for mixtures of similar compounds, such as methanolethanol, hexaneheptane, etc. Indeed, the activity coefficients for the methanolethanol system at atmospheric pressure, for example, are to within 10% from one. In the next six Examples we use Raoult's law in some typical VLE problems including applications to distillation column design.
13.6.2 Example 13.2 In Section 8.4.2 we recognized the vapor pressure of a pure liquid  at a specified temperature  as the pressure read by a manometer when this liquid is in equilibrium with the vapor above it. Thus, if in the vaporliquid equilibrium experiment of Section 13.5.1 we introduced a pure liquid, the pressure read by the manometer represents the vapor pressure of this liquid at the bath temperature. Similarly, the value read in the case of a mixture represents the vapor pressure of this liquid, at the corresponding composition and bath temperature. It is referred to as the total pressure because it represents the contribution to pressure by all components of the mixture. Let us now assume that the mixture is one of nhexane(l) and nheptane(2), and that after equilibrium is reached at a bath temperature of 60°C, analysis of the liquid phase gives: x 1 = x2 = 0.5. Since experimental measurements are timeconsuming and costly, they are avoided whenever possible. To this purpose, let us then consider the following two questions: 1. What information would be needed to calculate the pressure that would be read by the manometer  i.e. the vapor pressure of this liquid  without resorting to an experimental measurement?
447
Low Pressure VaporLiquid Equilibrium 2. The same, but for the vapor phase composition.
1. The pressure over the solution results from the collision of the vapor molecules with the walls of the flask and the surface of the mercury in the manometer. Since the contribution of eac~ component is given by its partial pressure Pi , the total pressure P is:
How are we now to evaluate P
p = Pi+P2

and P 2 ? The answer is provided by Eq.13.5.2: x.y.P.s P; = y.P =  '  '' (A) I F;
Thus:
p
1
S xi r1P1
= Fi
pS
x2r2 2 ~'
F2
(B)
Since x 1 and x2 are given, we need the values for the: "' two activity coefficients, "' vapor phase fugacity coefficients, and "'vapor pressures of pure liquids (1) and (2). For the activity coefficients we notice that becau~ of the similarity between hexane and heptane we can assume ideal solution behavior. It follows then (why?) that the vapor phase can be assumed to behave as an ideal gas, and: (C) The solution pressure can be thus determined if the vapor pressures of the pure components are available. To this purpose we use the Antoine equation, with the parameter values from Appendix C, obtaining: P1s = 572.8 mm Hg; and P2s = 210.6 mm Hg, and, from Eq.(C): P = (0.5)(572.8)+ (0.5)(210.6) = 391.7mm Hg 2.From Eq.(A): (D)
Comments l.The calculation we have just carried out determines also the pressure at which this liquid mixture  at the specified temperature  will start boiling, i.e. where the first 'bubble' will appear. (What is the bubble's composition?) It is referred to, appropriately, as bubble point (B.P.) pressure calculation; and the liquid, as was the case with a pure compound, as saturated. 2.In the next Example, we consider a bubble point temperature calculation.
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13.6.3 Example 13.3 Let us now consider the following case with reference to the system nhexane( 1) nheptane(2)of Example 13.2: the liquid phase composition of the equilibrium system is again: x 1 =x2 =0.5, but the pressure is equal to 300 mm Hg. What will the equilibrium temperature of the system and the vapor phase composition be, again without resorting to an experimental measurement? Using Eq.(C) of Example 13.2 but expressing the two vapor pressures through the Antoine equation we obtain by trialanderror: t = 52.5°C. Evaluation of the vapor phase composition is then straightforward: y1
= (0.5)(443.0)/300 = 0.738;
y2
= 0.262
Comments 1.In this case, our calculation is referred to as a bubble point temperature one. 2.0ur two bubble point calculations required only knowledge of the pure component vapor pressures. No mixture data were needed. 3. This, however, resulted from the assumption of ideal solution, valid here because of the similar nature of the two components, which led to activity coefficient values equal to one. We will see in Section 13.16 how bubble point calculations are carried out in real solutions. 4.In these two Examples, the liquid phase composition was specified, and: a. for a given T, we calculated P andy's; or b. for a given P, we calculated Tandy's. In the next Example, we consider the reverse case.
13.6.4 Example 13.4 Let us now assume that in the hexaneheptane system of Example 13.2 the temperature of the equilibrium system is 60°C again, but this time we know the composition of the vapor phase: y 1 = 0. 731; y2 = 0.269: a. what is the system pressure? and b. what is the liquid composition? a.From Eq.(A), Example 13.2: yiP x 1 =  and x 2
P(
But since: x 1
(A)
+ x2 = 1.0, it foiiows that: Y1
Yz
P(
Pf
+
P
(B)
Low Pressure VaporLiquid Equilibrium Hence: P
449
= 1/[0.731/572.8+ 0.269/210.6] = 391.6 mm Hg.
b.For the liquid phase composition: x 1 = y 1 PIP13 = 0.5; x2 = 0.5. Comments 1. The fact that the values for the pressure and the liquid phase composition are the same with those of Example 13.2, should not be surprising. The one case is the reverse of the other. 2.In this Example, we have also calculated the pressure at which this vapor mixture at the specified temperature will start condensing, i.e. where the first liquid 'droplet' or 'dew' will appear. (What will the droplet's composition be?) It is referred to, appropriately, as dew point (D.P.) pressure calculation; and the vapor at these conditions, again as with the case of a pure compound, as saturated. 3.Assume now that the pressure and vapor phase composition are specified, say: P = 300 mm Hg; y 1 = 0.738 and y 2 = 0.262, and we want to calculate: a.the temperature at which the first droplet will appear, i.e. the dew point temperature, and b. the composition of this droplet. In this case, we express the vapor pressures in Eq.(B) through the Antoine equation and solve  by trialanderror  for the temperature, obtaining t = 52.5°C. And: x 1 = y 1 PIP 13 = (0.738)(300)/443 = 0.5. · 4.The fact that the values fort and x are the same with those of Example 13.3 should not be, again, surprising. The one case is the reverse of the other. 5.It should be apparent from our discussion so far that, if a liquid and a vapor phases are in equilibrium with each other, then: a.each phase is saturated, b. the liquid phase is at its bubble point, and c. the vapor phase is at its dew point. 6.Calculation of dew points for real solutions will be discussed in Section 13.16. We proceed now to use bubble and dew point calculations in the following Examples that represent typical applications of vaporliquid equilibrium to distillation column design.
13.6.5 Example 13.5 In distillation operations it is desirable, for economic reasons, that the condenser operate with water as cooling medium. Assuming, thus, that during the summer months the water from the cooling tower enters the condenser at 45°C  and using a 10°C minimum •approach. temperature  we arrive at a condensate temperature of 55°C. Consider now an equimolar mixture of nhexane(l)and nheptane(2), that is to be separated through distillation into products of 90% (mole) purity. What is the minimum pressure that the condenser should operate at?
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In order that all the vapor entering the condenser be converted into liquid, the operating pressure must be at least equal to the saturation pressure of the distillate at 55°C, i.e. its bubble point value: P
=x1 P1 + x2 P/ = (0.9)(483.4) + (0.1)(173.5) = 452.4 mm Hg 8
If the pressure is lower, we will not have complete condensation; if it is higher, there is no problem for we will end up with a subcooled liquid. The latter is actually desirable in practice, to avoid problems in pumping a saturated liquid back into the column as reflux, and the remaining to storage.
13.6.6 Example 13.6 From the economic point of view it is also desirable to operate distillation systems at atmospheric pressure. (Vacuum costs money and low operating pres~ sure in the column results in large vapor volume and, consequently, large diameter). Assuming: "' that the condenser of Example 13.5 operates at 1 atm, and "' that due to pressure drop between the top of the column and the condenser the pressure there is 800 mm Hg, consider the following question: If the vapor leaving the column is in equilibrium with the liquid at the top tray, what is the temperature there and the liquid phase composition? The vapor has, of course, a composition of y 1 = 0.9 and y 2 = 0.1 and, since it is assumed to be in equilibrium with the liquid, it must be at its dew point temperature. Thus: Yi fpis
+ Y2 fp2s = 1/800
and by trialanderror, t = 75.1 °C. Also: x 1 = (800)(0.9)/(924.1) = 0.779 and x2 = 0.221.
13.6.7 Example 13.7 Let us now assume that due to pressure drop through the column the boiler operates at 900 mm Hg. a. What is the temperature in the boiler? b. What is the composition of the vapor leaving the boiler?
a. Assuming that the liquid and vapor phases in the boiler are in equilibrium, the liquid is at its bubble point temperature:
Low Pressure VaporLiquid Equilibrium x 1 P1s
where: x 1 = 0.10 and x 2 t = l00°C.
451
+ x2 P/ = 900 mm Hg
= 0.90, from Example 13.5 (why?). By trialanderror,
b.It can be easily seen that: y 1
= (0.1)(1845.2)/900 = 0.205 and y2 = 0. 795.
13.6.8 Summary Let us now summarize our work with the distillation column calculations for the separation of the hexaneheptane mixture, Examples 13.5, 13.6 and 13.7. 1. The economic requirement that water be used as a cooling medium in the condenser set the temperature of the condensate. 2.A bubble point pressure calculation on the distillate determined the minimum operating pressure in the condenser, which was below the atmospheric one. 3.Economic, again, considerations led to atmospheric pressure operation for the condenser. 4.Using a reasonable pressure drop between the condenser and the top of the column, we estimated the pressure there. A dew point temperature calculation then on the vapor leaving the top of the column determined: * the temperature of the vapor and liquid of the top tray, and * the composition of the liquid leaving it. 5. Using again a reasonable pressure drop along the column, we estimated the pressure in the boiler. A bubble point temperature calculation then on the bottoms stream determined: *the temperature in the boiler, and * the composition of the vapor stream leaving it. 6.We know, thus: * the temperature and pressure at the two ends of the column, and * the compositions of the two phases at these ends. This information is sufficient to start the socalled 'traytotray' calculation, that provides the information needed for the design of the column. 7. Unfortunately, the typical mixtures encountered in practice are not ideal, and calculation of bubble and dew points requires  as we have seen  knowledge of the activity coefficients of the components involved. We proceed, therefore, with a qualitative and quantitative consideration of activity coefficients and later, Section 13.16, we will see how they are used in carrying out bubble and dew point calculations for real solutions.
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13.7 Activity Coefficients: Qualitative Remarks Considering the importance of activity coefficients in VLE calculations and the high cost of the experimental measurements required for their evaluation  a qualitative understanding of the factors that effect their values: a. composition, b.molecular nature of the components in the mixture, and c.pressure and temperature, can be very helpful in avoiding unnecessary work. We consider the first two here and the third one, in the next Section.
13.7.1 The Effect of Composition Consider a binary system with x1 = 0.95 and x2 = 0.05. The molecules of component 1 are surrounded mostly by their own, being oblivious of the existence of the molecules of component 2. We would expect, hence, its activity coefficient to be practically equal to one, i.e. to follow Raoult's law. This is indeed the case as Figure 13.2 demonstrates. On the other hand, the molecules of component2 are surrounded mainly by molecules of 1, i.e. they find themselves in a 'strange' environment. We would expect, therefore, that its activity coefficient would be very different from one; and, using the same physical argument, that this difference would decrease with increasing values of x2 • This is indeed the case, as shown again in Figure 13.2. [There are, however, a few rare cases where due to a combination of physical and chemical interactions  the activity coefficient's dependency on concentration passes through a maximum or minimum (see, for example, the system acetonemethanol in Severns et a!, 1955).] These qualitative arguments about deviations from ideality (Raoult's law) should remind us of the LewisRandall rule (Section 11.8.3). Indeed Raoult's law, even though it was proposed before the LewisRandall rule, represents a special case of the latter.
13.7.2 The Effect of Molecular Nature Considering the requirements for ideal solution behavior, Section 13.6.1, we examine next the effect of the molecular characteristics of the mix
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453
ture components on its deviation from such behavior. More specifically, we will examine their effect on: a.the type, and b.the magnitude, of this deviation. a. Types of Deviations from Ideal Solution (Raoult's Law) Behavior Consider a binary mixture where the intramolecular (11, and 22) forces are stronger than the intermolecular (12) ones. As a result, the presence of component 2 increases the 'escaping' tendency of component 1, as compared to its value at the same composition, but in an ideal solution where all forces are equal; and the same is the case with component 2. Both activity coefficients are, thus, greater than one. In such systems therefore (assuming ideal vapor behavior):_ *the partial pressure of component_i: P;(real) = X;Y; Pt * is greater than its counterpart: P; (ideal) = X; Pt, had i been in an ideal solution. We refer to them, appropriately, as demonstratingpositive deviations from Raoult's law and an example is the system presented in the upper part of Figure 13.2. These systems represent the typical by far  case. (And, they claim, that segregation is a human invention!). There are a few systems, however, where 'integration' does occur, i.e. the 12 forces are stronger than the 1l and 22 ones. Such systems exhibit strong hydrogen bonding between the 1 and 2 molecules (i.e. association, Section 7. 7), and their activity coefficient values are less than one. They are referred to, consequently, as demonstrating negative deviations from Raoult's law. A typical example is presented in the lower part of Figure 13.2. These positive and negative deviations from Raoult's law are better demonstrated in Figure 13.3, where the total pressure over an equilibrium solution is plotted versus the liquid composition for isothermal conditions. Starting with ideal solution behavior, we note that since the total pressure is the sum of the two partial pressures: P(ideal)
= x1Pt +x2 P{ = P{ +(Pt P{)x1
P(ideal) thus is a linear function of x 1 as shown in the Figure, line 1.
Turning to positive deviations from Raoult' s law, we note that the total pressure:
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(a)
p
x,
p
(b)
x,
1.0
Figure 13.3 Positive (a) and negative (b) deviations from Raoult's law.
is larger than that corresponding to the ideal solution value, as shown at the top of the Figure. Notice that in the case of the upper curve  line 3  the deviation is so large, that P goes through a maximum. At this point we have an azeotrope, where xi = Yi· Negative deviations are demonstrated in the lower part of the Figure. It is easily seen that here P(real) is less than P(ideal ), and there is an azeotrope for curve (3) but not for (2). Similar curves, but with opposite direction obviously, are obtained if we plot Tversus x 1 for isobaric conditions. This is demonstrated with the system propanolwater at 1 atm (Gmehling et al, 1981) presented in Figure 13.4, where the phases present are also shown. Notice the presence of azeotrope at the minimum temperature. From the practical point of view, this azeotrope represents the maximum composition of propanol that can
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Low Pressure VaporLiquid Equilibrium
100.0 ~x, ~y,
g
95.0
(a)
..... 90.0
L 85.0 0.00
0.20
0.40
x,, y,
0.60
0.80
1.00
1.0
0.8
0.6
:>;
(b) 0.4
0.2
x,
Figure 13.4 Txy {a) and yx (b) diagrams for the 1propanol(l) water(2) system at 1 atm. Lines for identification only. L: Liquid; V: Vapor.
be obtained by distillation at this pressure. So far we have considered deviations from ideality resulting from differences among inter(i:i) and intra(ii andjj') molecular forces only. An other factor that can play an important role is the difference in free volume of the mixture components. The simplest way to define this quantity (Elbro et al, 1990) is in terms of what Bondi (1968) calls 'empty volume':
v1 = v v·
where Vis the molar volume and v• is the molar hardcore volume, calculated from the van der Waals volumes given by Bondi. Such differences are more pronounced in systems where there are large differences in molar volumes, such as polymersolvent ones, and lead to negative deviations from Raoult's law as shown with a typical case in FiguFe 13.5 (Iwai and Arai, 1989).
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•
4
0
0.1
0.60
0.70
0.80
0.90
1.00
rather than the mole fraction is used.)
b. Magnitude of Deviations from Raoult's Law The effect of chemical dissimilarity of the mixture components on the size of their activity coefficients is demonstrated with the infinite dilution values presented in Figure 13.6 (Pierotti et al, 1959). Consider, for example, the binary systems of aliphatic ketones with nheptane. The polar character of ketones  originating from the presence of the CO group  decreases with increasing values of Nc, the number of carbon atoms in the ketone: they become progressively more hydrocarbonlike, which explains the observed decrease in the infinite dilution activity coefficient values of ketones in heptane as Nc increases. The same phenomenon, but with the opposite effect, occurs in the case of nalcohols in water, where the main factor is the hydrogen bonding among the OH groups. Again, as the chain length increases, hydrocarbon behavior is approached and the hydrogen bonding effect diminishes. Here however the alcohols become progressively more dissimilar to water, which explains the observed increase of the infinite dilution activity coefficient values of the latter in the former, as Nc increases. Actually, this dissimilarity leads to partial miscibility with higher alcohols.
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• • • • • Water in n Alcohols Primary Alcohols in Water Q..C..Il.IW Aliphatoc Ketones in nHeptane ~
t
= 25
•c
10
Figure 13.6 The effect of chemical dissimilarity between the components of a mixture on the activity coefticients (lines for identification only).
13.7.3 Summary We summarize next our qualitative observations about the activity coefficient up to this point: 1. The activity coefficient is a measure of the deviation from ideal solution behavior, the 'correction' factor to Raoult's law. 2.As the composition of a given component decreases, its activity coefficient's deviation from one increases. 3 .As the dissimilarity among the components of a mixture increases, so does the deviation from ideality (and, from the value of one, for y's). 4.Dissimilarity in the physical and chemical forces has the most pronounced effect, followed by dissimilarity in free volumes. The latter becomes dominant in solventpolymer solutions. 5.Most systems by far demonstrate positive deviations from Raoult's law (except solventpolymer ones). 6.Typical maximum values of gamma for systems with positive deviations from Raoult's law: *completely miscible: 1.1  10 * partially miscible: 20  200 *very immiscible: 200 very high values. (The infinite dilution activi
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ty coefficient, for example, of benzene in water is of the order of 20,000 with a solubility in the parts per million.) 7. For negative deviations from Raoult' s law, typical minimum values for gamma are between 0.9 and 0.5. Still smaller minimum values can be observed for the solvent activity coefficient in solventpolymer systems (Figure 13.5).
13.8 Activity Coefficients: The Effect of Pressure and Temperature 13.8.1 Pressure The effect of pressure on the activity coefficient  at a constant temperature and composition is given by:
[alnr;] ap
T,x
(13.8.1)
where V;E is the partial molar excess volume of component i in the liquid phase. At the low pressures of interest here, the partial molar excess volume is rather small as we saw in Example 11.1 and, consequently, so is the effect of pressure on the activity coefficient. Consider for example the case of water(l) in its mixture with methanol(2) at the conditions of Example 11.1: x 1 = 0. 7; T = 25°C; P = 1 atm, where its partial molar excess volume is 0.3 cm3 /mol. Assuming this value to be independent of pressure for small pressure ranges, then the activity coefficient of water at some pressure P 2 , say 5 atm, relative to its value at 1 atm is given by: y 1[5 atm] = 1.00006 y 1[1 atm] a negligible effect. (It may become important, however, at high temperatures and pressures.)
13.8.2 Temperature The effect of temperature on the activity coefficient  at a constant pressure and composition is given by:
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Low Pressure VaporLiquid Equilibrium
WOW Acetone in nC 7 EtOH in Water a::xxlO nC 7 in MEK ti.i..ti MEK in MeOH 0.0.0.00 Acetone in MeOH ~
7.0 6.0 5.0 4.0 3.0 2.0 1.0
0.0 L..L.Ju....l...W..~.L.J....L..L.JL.J....W..~J....L.L..L.l....1....1....W...J...J....L..L.Ju....l....I....1.~.J...J...L.J....JL..J 380 370 360 350 340 330 320 310 300 290
T (K)
Figure 13.7 Variation of infinite dilution activity coefficients with temperature (Pierotti et al, 1959). Lines for identification only.
[alnrl] ar
= P,x
ii~E RT 2
(13.8.2)
The variation thus of the activity coefficient with temperature depends on the excess enthalpy of mixing, which reflects the difference between inter and intramolecular forces. As a result, the activity coefficient of a mixture component can be a strong function of temperature as shown in Figure 13.7, the effect being determined by the other components present. This explains, for example, the large difference in the variation of y 00 with temperature for acetone in nC 7 , as compared to that in methanol. We conclude, therefore, that  in the typical nonideal solution  the activity coefficient: * is a strong function of composition, * can be a strong function of temperature, and * is a very weak function of pressure. We will discuss later (Sections 13.10 through 13.14) the approach used in developing analytical relationships of the activity coefficient as a function Of composition and temperature. But first, let us examine how the accuracy of experimental activity coefficients is evaluated.
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13.9 Thermodynamic Consistency of Experimental Data 13.9.1 The Interrelationship Among Activity Coefficients We have seen in Chapter 11 that all partial molar properties must satisfy the GibbsDuhem equation, Eq.11.6.3, which when applied to the molar excess Gibbs free energy yields (Problem 13.52): (13.9.1) where yE and sE are the excess molar volume and entropy of the mixture respectively at the specified temperature, pressure and composition. The activity coefficients, thus, of the components of an equilibrium mixture cannot all vary independently of each other: they must satisfy Eq.13.9.1, if they are to be thermodynamically consistent. We examine next how this interrelationship is used to test the consistency of experimental binary data.
13.9.2 Binary Systems Integration of Eq.13.9 .1 for a binary system, from x 1 = 0.0 to x 1 = 1.0, gives (Prausnitz, 1969):
I In Y I In rY 1
o
1
o
1 dx
y2
1 dx
2
I
1
=
=
I RT2 HE dT I RTdxdP dx 1
dxl
o
1
o
dx
yE
I
I
1
(P: constant)
(13.9.2)
(T:constant)
(13.9.3)
Evaluation therefore of the thermodynamic consistency of binary vaporliquid equilibrium data  at constant temperature (isothermal) or constant pressure (isobaric) conditions  requires knowledge of the values of the corresponding excess property as a function of composition. In the typical case such experimental information is not available, and the righthand side of Eqs 13.9.2 and 13.9.3 is set equal to zero. While this is a reasonable assumption for Eq.13.9.3, it may not always be so for Eq.13.9.2, especially for very nonideal mixtures. For isobaric data, therefore, the semiempirical test proposed by Herington (1951)
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Low Pressure VaporLiquid Equilibrium
should be used. For more reliable results, the excess enthalpy can beestimated with a modified UNIFAC model (see Section 13.18.3).
13.9.3 Comments I. Usually the integral in the lefthand side ofEqs 13.9.2 and 13.9.3 is evaluated graphically, as shown in Figure 13.8 for the system acetonecyclohexane. As a result, this method for evaluating the thermodynamic consistency of binary data is referred to as the 'area' test. 2.For isothermal data the 'Consistency Index' (C/), defined by: C/ =
A A
I
p
n
AP +An
I X 100
(13.9.4)
where AP and An are the areas above and below ln(y 1/y2) = 0.0 respectively, becomes the measure of their quality. For isobaric data, the measure of their quality is the quantity (CIJ) with J given by: J = 150 : T;
J1 T = Tmax Tmin
min
2.5
1.5
_......, 0.5
~
?...__.,
s
0.5
1.5
2.5
0~
0.1
0.2
0.3
0.4
Q5
0.6
0.7
OB
0.9
X1
Figure 13.8 The area test for the system acetonecyclohexaneat 25°C.
1.0
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Table 13.2 Classification of Typical Literature Binary VLE Data According to the Area Thermodynamic Consistency Test (Silverman) Type Isothermal Isobaric Number % (CI) (ClJ) Good 147 < 10 60 15 > 20 Unacceptable: area test not possible 16 6 Total 247 100 and Tmin are the temperature extrema in the binary system including the boiling points of the pure components and the azeotrope, if present, while the coefficient 150 is empirical, recommended by Herington on the basis of typical heatofmixing data (Prausnitz). 3.A value of CI less than 5 for isothermal data, and of (CIJ) less than 10 for isobaric ones, suggests good data. 4.The quality of typical literature data is demonstrated in Table 13.2 (Silverman, 1984), which indicates that about 40% of the binary systems tested do not meet the aforementioned test. 5.For more reliable results ln(rJr2) is fitted to a polynomial, which is used to evaluate the C/ value, by integration, and the values of the infinite dilution gammas, by extrapolation to xi = 0.0. 6.It should be kept in mind, also, that the area test is a necessary but not sufficient requirement for good quality data. 7 .For limitations of the area test and other techniques of testing the consistency of VLE data see Prausnitz et al, 1986. In the next Example, we examine the quality of the acetonecyclohexane data of Table 13. E.l. Tmax
13.9.4 Example 13.8 For the system: acetonecyclohexane, Table 13.E.l, determine: a. the consistency index, and b. the infinite dilution activity coefficients. a. The values ofln(y1/y2) are reported in the same Table, and are plotted versus x 1 in Figure 13.8. Drawing the 'best' line through them with a french curve yields the following approximate areas: AP = 47; A11 = 46, both in the same units. Then, C/ = 1.1, which suggests excellent data.
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463
b.From the same Figure: lnr~
==
2.13;
lny;' :: 1.90;
r~
==
8.4
r;' :: 6.7
Comments l.Since the data are isothermal, the C/ index can be used with more confidence in establishing the quality of the data than the (ClJ) quantity, if they were isobaric. 2.1sothermal data are desirable for an other very important reason. Pressure, as we have seen, bas a very small effect on low pressure activity coefficients and, consequently, such data reflect only the nonideality of the solution. Isobaric data, on the other hand, because of the significant temperature effect that may be involved, reflect this effect as well. They are obviously more desirable for distillation calculations. 3.Isothermal data, therefore, are useful in the study of solution nonideality; isobaric, for distillation calculations. 4.Tbe reliability of the area test is facilitated, in this Example, by the relatively large number of available data. In general, it is recommended that  at least  15 data points are taken. 5. There is obviously an uncertainty in the gamma infinity values. It would be probably larger, however, if plots of gamma versus composition were used.
13.10 Activity Coefficients and Composition: Analytical Expressions We have already seen that the activity coefficient is a strong function of composition(Section 13.7.1, Table 13.E.1 and Figure 13.2). We consider therefore in this Section the procedure followed in developing analytical expressions (or models) for the activity coefficient as a function of composition. Several such expressions, essential for modelling of separation processes, are then considered in Sections 13.11 through 13.14.
13.10.1 Procedure The procedure involved consists of two basic steps: 1. Using phenomenological arguments, from purely empirical to those based on statistical mechanics, an expression for the molar excess Gibbs
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free energy GE as a function of composition is developed. 2.The exact relationship:
RTln .
r,
=
[o(NGE)l aN. 1
(11.13.6) T,P,~
is then used to arrive at an expression for the activity coefficient.
13.10.2 Comments l.Development of a model for G E requires an understanding of the different effects involved in the mixing of molecules and, of course, an analytical representation of these effects. As a starting point we note that, since: aE
= HETsE = uE+pvETsE
where HE, sE, uE and yE are respectively the molar excess enthalpy, entropy, internal energy and volume of mixing, the mixing process at constant temperature T and pressure P involves three effects: a.an energetic one, resulting from the difference in intermolecular energies between like and unlike molecules; b.an entropic one, resulting from a lack of complete randomness in the distribution of the molecules in the mixture, caused mainly by differences in the size and shape of the molecules; and c.one caused by an increase or decrease in volume as a result of the mixing process, caused by differences in both, intermolecular energies and molecular sizes and shapes. 2. The first effect tends to be the dominating factor in mixtures of molecules that are not very different in size and the early investigators in the field: van Laar, Hildebrand, Scatchard, Wohl, etc., assumed that: sE = yE = 0. This leads to the concept of 'regular' solution, first introduced by Hildebrand. 3.0n the other hand, for mixtures involving molecules of very different size, such as polymersolvent ones, HE can be much smaller than sE. The assumption of HE = 0, leads to the concept of 'athermal' solution and resulted in the FloryHuggins expression for polymeric solutions (Prausnitz et al), where yE = 0 was also assumed. 4.We will classify the models for GE and, hence, for the activity coefficient, into two types: * the 'Wohltype' that includes the traditional expressions of van Laar,
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Low Pressure VaporLiquid Equilibrium
Margules, and RedlichKister and are considered purely empirical; and * the 'local composition' type, that includes the Wilson, NRTL, and UNIQUAC expressions and are considered semiempirical. The Wohltype models find only limited application today. They are presented, however, because they represent a simple example of how expressions for G E are developed. The obtained expressions for the activity coefficient are also easier to use than those of the local composition models. S.None of these expressions accounts for differences in free volumes (Section 13.7 .2), which can be important for mixtures of molecules that are very different in size. They have been included, thus, for polymersolvent systems (Oishi and Prausnitz, 1978; lwai and Arai; Elbro et al); and for gassolvent ones (Antunes, 1983; Nocon et al, 1983).
13.11 The Wohl Type Expressions 1:3.11.1 The General Expression for a Binary System Wohl (1946) expressed the molar excess Gibbs free energy in a binary solution in terms of a power series of the 'effective' volume fractions of the two components, z 1 and z2 (Prausnitz):
GE
(13.11.1) where:
z1 =
xJqJ ; XJql+Xzqz
Zz
x2q2
= xJqJ+xzqz
q;: measure of the size of molecule i, or of its 'sphere of influence', in the solution. a: interaction parameter, with a physical significance similar to that of the virial coefficients. For example, a 12 reflects the interaction between two molecules, one of component 1 and one of 2; a 122 , among three molecules, one of 1 and two of 2, etc. The following two expressions, that of van Laar and that of Margules, can be obtained as special cases of Eq .13 .11.1, even though they were developed earlier.
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13.11.2 The van Laar Equation For a binary system whose components are not too dissimilar chemically but have different molar volumes, we can assume that interaction coefficients involving more than two molecules can be neglected. Then, the Wohl expression becomes:
GE
2al2xlx2ql q2
RT From this equation, the following expressions are obtained for lny1
A
r 1 and y2 :
= ::
(13.11.3)
B lny2 = ....,
(13.11.4)
(1 +(Ax 1)/(Bx2)] 2 (1 +(Bx2)/(Axi)]2
where: A = 2q 1 a 12 ; B
(13.11.2)
=
2q2 a12
~Actually van Laar  a student and, later, colaborator of van der Waals  developed his expression early in the 20th century starting with the van der Waals equation of state (see Prausnitz).
13.11.3 The Margules Equation If the molecular sizes of the components of a binary system are not too different, we can assume that q 1 = q2 = q and, neglecting interactions involving more than three molecules, obtain the following expressions for the activity coefficients:
A 'xi+ B'xi
(13.11.5)
(A 1 +~B')x~B 1x~
(13.11.6)
lny 1 lny2
=
=
where: A' = q(2a 12 + 6a 112  3a 122); B' = q(6a 122  6a 112) Eqs 13.11.5 and 13.11.6 are referred to as the 3suffix Margules equation (twoparameter expression). For a detailed discussion of the van Laar and Margules expressions, and a comparison of their performance in correlating binary data, see Prausnitz.
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Low Pressure VaporLiquid Equilibrium
13.11.4 Evaluation of the Parameters Evaluation of the parameter values in the van Laar and Margules expressions requires, of course, experimental activity coefficient data for the binary system involved. To this purpose all data are regressed to obtain the parameter values that 'best fit' them. They can, of course, be also obtained from a pair of activity coefficient data by solving a system of two equations with two unknowns. Thus, if for a given x 1 the corresponding activity coefficients are y 1 and y2 , we obtain the following expressions for the van Laar parameters: (13.11. 7)
2
1 lny1] B = [1+ x _ _ lny2 x2 lny2
(13.11.8)
Expressions for the Margules parameters can 'be developed in a similar fashion. The two equations, in spite of the different assumptions involved in their development from Wohl 's expression (Eq.13.11.1), give very similar results, because of their purely empirical character and the fact that they contain the same number of adjustable parameters. To develop some familiarity in working with nonideal systems consider the following Example.
13.11.5 Example 13.9 Using the single experimental datum for the system acetone(l)cyclohexane(2), Table 13.E.l: P = 253.9 mm Hg; x 1 = 0.4235;y 1 = 0.6800and t = 25°C, calculate the equilibrium pressure and vapor composition values at: x 1 = 0.0115; 0.1775; 0.5760; 0.7390; 0.9625, all at 25°C, and compare them to the experimental data. Assume, as in Example 13.1, ideal vapor behavior and use the pure compound vapor pressures given there (instead of the Antoine equation). In this case: (A)
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Since x 1 and the pure compound vapor pressures are given, the values for the activity coefficients at each composition are needed. To this purpose: l.From the given datum point calculate the corresponding y 1 and y2 values using Eq.13.5.1: y 1 = 1. 769; y 2 = 1.446 (see Example 13.1). 2. Use these values to calculate the van Laar parameters from Eqs 13.11. 7 and 13.11.8:A = 2.016; B = 1.683. 3.Calculate y 1 and y2 at each composition, using these parameter values and Eqs 13.11.3 and 13.11.4. Table 13.E.9 presents the obtained y 1 and y 2 values and the corresponding errors. Values of P andy!> calculated from Eqs (A) and (B) respectively, and the corresponding% errors and deviations .1y1 and .1P are also included. Comments l.Eq.(A) determines the pressure at which the liquid at the given composition will start boiling at 25°C, i.e. the bubble point pressure; and Eq.(B) provides the composition of this bubble. 2. The most important quantity for distillation design purposes is the predicted vapor phase composition. 3.Errors in y 1 are usually reported in terms of .1y = Y(calc) Y(exp)• because the percent error depends on whetherit is based on the component with the smallest or largest y. In this case, for example, and at x 1 = 0.0115, while the error in y 1 is 9.6%, it is only 2.1% for Y2 Similarly to avoid masking the error at higher
Table 13.E.9 Calculated Values for the System AcetoneCyclohexane and Comparison with the Experimental Ones at 25°C p Error Error Error Error Xt Yt Y2 Yt % % mmHg % % 2.4 0.1635 9.6. 115.17 0.4 0.0115 7.109 11.8 1.000 3.0 1.9 0.6292 5.0 1.074 2.8 232.12 0.1775 3.571 1.1 0.6928 1.7 3.0 256.53 2.8 1.907 0.5760 1.339 1.4 0.7314 1.0 2.4 2.729 258.41 1.5 0.7390 1.110 1.1 0.9625 1.002 0.5 4.848 15.6 239.92 1.6 0.9262 .1P (mm Hg) .1y1x1000 Xt 18 2.88 0.0115 4.42 0.1775 18 12 2.87 0.5760 8 3.59 0.7390 3.88 0.9625 10 Overall (Abs.) 3.53 13
Low Pressure VaporLiquid Equilibrium
469
values, the difference: liP ~ P(calc)  P(e. (O.OS)x (Yexp) C: Eliminate data points where I Ycal 0.01 Ycalc is calculated from the experimental x values using the method of Fredenslund et al.
can be considered very good, especially if the uncertainty in the multicomponent data is taken into account. It appears that the aforementioned interdependency between the Wilson equation parameters tends to remove some of the uncertainty in the binary data. Even for group D, the overall deviation in y is 0.013 and average deviations of over 0.025 were observed only twice, with the worst case being 0.032. Further support for these findings is provided by Table 13.8. Here, before the binary data were regressed, they were 'screened' according to
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492
Table 13.9 Prediction of Ternary VLE from Binary Data with the Wilson Equation: The Effect of Temperature of the Binary Data System
Parameters evaluated at
Hexanecyclohexanebenzene
70°C
Et acetateethanolwater Methanolcarbon tet. benzene
Avg. Abs. Dev. in Predicted y·1000 at 10°C
15°C
6.6
6.4
70°C
55°C 18.8
70°C 19.0
55°C
34.7°C 2.7
55°C 2.0
25°C 5.0
70°C
2.7
the procedure of Fredenslund et a!, whereby vapor phase compositions are calculated from the experimental x's using a method based on the GibbsDuhem equation. These calculated y's are then used to evaluate the accuracy of the experimental ones. The results suggest that elimination of poor y data in the binary systems does not necessarily improve the obtained predictions.
13.17.3 The Effect of Temperature Since distillation calculations require isobaric data, the temperature range of the multicomponentpredictions may often be different from that ofthe binary systems involved. We have already seen in Section 13.13.2 that, even though the parameters in the local composition models are functions of temperature (Figure 13.10), good correlation of binary data is obtained by considering them to be temperatureindependent (Table 13.4). This assumption can be used, also successfully, in multicomponentpredictions as suggested by the quality of the results presented in Table 13.9. We should keep in mind, however, that the temperature range should be reasonable, typically below 3040°C.
13.17.4 The Effect of System Type Table 13.10 summarizes the prediction results of Silverman in terms of the types of multicomponent systems involved:
Low Pressure VaporLiquid Equilibrium
493
Table 13.10 Multicomponent Prediction Results as a Function of the System Type Number Type .1y·1000 9 Systems with four(7J and five(2) components 7.6 10 Systems including hydrocarbons only 5.4 36 Systems containing alcohols 11.6 13 Systems containing water 14.5 73 All systems 10.6
* number of components, and
* degree of nonideality.
Notice that: 1.In support of the local composition models' contention that only binary parameters are needed for multicomponent predictions, the nine four and fivecomponent systems are predicted as well as  actually better than the rest. 2.As expected, the degree of nonideality effects the obtained accuracy: systems containing hydrocarbons only give the best results, while those containing water, the poorest. This also explains the better prediction results for the four and fivecomponent systems, mostly hydrocarbon ones.
13.17.5 Example 13.12 Evaluate the performance of the UNIQU AC model in the prediction of the vaporliquid equilibrium behavior for the system carbon tetrachloride(1)benzene(2)ipropyl alcohol(3) at 1.013 bar by comparing your results to the experimental data presented in Table 13.E.12 (Hala et a!, 1968, pp.493494). The required UNIQUAC binary parameters are included in the same Table and were obtained by regressing the corresponding VLE data: at 343.15 K for the 12 and 13 binaries (Hirata et a!, 1975); and at 760 mm Hg for the 23 one (see Example 13.10). The results, obtained with the Program VLEGPHI of Appendix E, are presented in the same Table. Notice the good quality of the predicted values with overall average absolute deviations of: P = 1.6%; ..1y 1 = 0.004; ..1y2 = 0.007.
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Table 13.E.12 Information and Results for Example 13.12 L1P% T,K Point L1yl x2 xl Yt,exp 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
346.6 345.1 344.4 347.0 348.6 346.0 346.6 344.5 344.8 347.8 347.0 344.6 344.5 346.3 343.5 346.6 343.8 344.0 346.0 346.4
0.032 0.071 0.301 0.429 0.424 0.092 0.058 0.190 0.151 0.106 0.270 0.256 0.203 0.386 0.302 0.490 0.392 0.594 0.713 0.830
0.362 0.430 0.054 0.528 0.555 0.734 0.224 0.651 0.373 0.143 0.664 0.425 0.172 0.552 0.182 0.448 0.213 0.227 0.236 0.125
4.18 1.43 0.81 2.03 1.21 1.53 0.79 2.26 1.90 4.58 0.64 3.00 0.20 1.86 0.28 0.69 2.29 1.66 1.15 0.10
0.054 0.100 0.471 0.416 0.420 0.096 0.112 0.190 0.194 0.203 0.269 0.276 0.315 0.369 0.391 0.459 0.433 0.542 0.641 0.741
0.004 0.004 0.005 0.001 0.010 0.002 0.004 0.003 0.001 0.002 0.003 0.006 0.002 0.002 0.000 0.001 0.002 0.009 0.011 0.013
Y2,exp
L1y2
0.470 0.472 0.071 0.437 0.487 0.627 0.348 0.545 0.404 0.220 0.550 0.384 0.223 0.456 0.203 0.373 0.209 0.188 0.201 0.106
0.005 0.011 0.003 0.018 0.011 0.013 0.006 0.007 0.001 0.017 0.011 0.001 0.006 0.009 0.002 0.003 0.003 0.002 0.001 0.003
r(1) = 3.39 r(2) = 3.19 r(3) = 2. 78 q(l) = 2.91 q(2) = 2.40 q(3) = 2.51 .:1u12 /R = 69.99 L1u23 IR = 231.52 L1u31 IR = 76.35, all in K .:1u21 1R = 50.65 L1u32 1R = 10.61 L1u 13 /R = 329.55,all inK
13.18 Estimation of VaporLiquid Equilibrium As expected, often not all the necessary for design purposes VLE data are available. Consider, for example, the case where vaporliquid equilibrium information is needed for the design of a fractionation system to recover component A from a mixture containing, in addition, components B and
c.
If experimental VLE data for the three binary systems: AB, BC and CA are available, they can be used  as we have seen  for the prediction of the ternary behavior. Reliable results should be expected, and the availability of compilations of a large number of binary data (Hal a et al;
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495
Hirata et al; and especially the series of volumes by Gmehling and coworkers) makes this approach very attractive. In the typical case, however, some  or even all  of the necessary binary data may not be available. In such cases we resort to estimation techniques.
13.18.1 The GroupContribution Concept Estimation techniques rely on the groupcontribution approach that has found success with pure component properties (critical constants, heat capacities, etc.). The basic assumption behind this approach is that the property of a fluid can be approximated by the sum of the contributions of the functional groups of its molecules. Consider, for example, the system nhexane  acetonitrile. The deviation from Raoult's law of this system  i.e. its activity coeft1cients  is described according to this concept, by: a.the size and shape of the functional groups (CH 3 , CH 2 , and CN) making ~p these molecules, and b. the differences in interaction energies between them: * [(CH 2/CN)  (CH 2/CH2)], and * [(CH2/CN) (CN/CN)]. (The groups: CH 3 and CH 2 are assumed, for simplicity, to have the same interactions with other groups; thus the CH 3 /CN interaction is equal to the CH2/CN one). Notice, now, that if: * a groupcontribution model for the activity coefficient, and * experimental VLE data for this system, are available, they can be used to determine the values of the two bracketed parameters; and, consequently, the VLE behavior of any alkanenitrile mixture, binary or multicomponent, can be then predicted. Several groupcontribution models have been proposed through the years (Pierotti et al, 1959; Derr and Deal, 1969; Fredenslund et al, 1975 etc.). The most recent and promising one, is the UNIF AC model of Fredenslund et al.
13.18.2 The UNIFAC Model This model is based on the UNIQUAC equation and uses the same expression for the activity coefticient, Eq.13.12.6. The combinatorial part of the activity coefficient is also the same, Eq.
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13.12. 7. Only the pure component properties r; and Q; enter into this equation and are calculated as the sum of the group volume and area parameters, Rk and Qk (see Section 13.12.4 for the method of evaluating these properties): r; =
L
v~>Rk and
Q;
= L vZ)Qk
(13.18.1)
where vkfpr + A3pr (14.5.4) 2 3 AW = a. + b. are given by the authors along with values for the acentric factor "';· Through use of experimental data the correlation for the liquid phase fugacity coefficient was extended to supercritical components, and values for their 'liquid' molar volumes were established. In applying hence this method, the parameter values given by the authors should be used. The same applies to the solubility parameter values.
14.6.2 Comments l.The use of the regular solution theory for the activity coefficient restricts the applicability of the method to nonpolar/weaklypolar systems. 2.The use of the corresponding states principle approach for the evaluation of the standard state fugacity provides additional support to this restriction. 3. The evaluation, finally, of the necessary parameters from experimental data imposes restrictions on the type of mixture components. 4.The method is applicable to: a.Hydrocarbon systems, except methanecontaining ones, for: * the reduced temperature range of 0.5 to 1.3, based on the pure component critical temperature, and *pressures up to 2000 psia, but not to exceed 0.8 of the critical pressure of the system (see Figure 14.6). b.Hydrogen and methane containing systems, for: *temperatures from 1 00°F to about 0. 93 of the pseudoreduced critical
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temperature of the equilibrium liquid mixture, but not to exceed 500°F; * pressures up to about 8000 psia; * concentration up to 20% of other dissolved gases in the liquid. 5. Within these restrictions, the average absolute deviation for 1345 data points equals 7.9%, as compared to 7.7% for the BenedictWebbRubin (BWR) equation. The method fails, however, in the convergence region as demonstrated in Figure 14.6. 6.An extension of the method, beyond the aforementioned temperature and pressure ranges, is given by Grayson and Streed (1963).
14.7 The Equation of State Approach The methodology for solving the vaporliquid equilibrium problem using an equation of state has been outlined in Section 13.4 and involves expressing the fugacity coefficients of the mixture components ~; in the liquid (!) and vapor (v) phases through an equation of state (EoS). The equilibrium ratio K; is then calculated from: K. = I
~ X
i
A
=
l
1>; l' 1>; A
(14. 7 .1)
Cubic EoS are the most commonly used equations of state for high pressure VLE calculations. They are relatively simple to use and provide reasonable accuracy. We confine, therefore, our discussion to such EoS and proceed to: !.examine the information needed to carry out VLE calculations using them; 2.comment on their application and the obtained results; 3.comment on their application to polar systems.
14.7.1 Required Information Consider first a twoparameter cubic EoS, say the SoaveRedlichKwong (SRK, Soave, 1972). Eq.11.9.6 gives the expression for the fugacity coefficient of component i in a mixture of given composition at some temperature T and pressure P with this EoS, and requires values for all: * a;;'s and b;'s, obtained through the critical properties and acentric
High Pressure VaporLiquid Equilibrium
523
factor of the mixture components and the system temperature; * ai)' s(i *j), obtained through au, aii, and the interaction coefficients ki). The mformation, thus, needed is: * Tc, Pc• and l.> for each component; and * interaction coefficients for all the pairs of the mixture. The same information is needed for all twoparameter EoS, such as the PR (Peng and Robinson, 1976). Threeparameter ones, such as the vdW711 (Watson et al, 1986), require also critical compressibility factor values for all the components.
14.7 .2 Comments 1.The Soavetypecubic EoS, discussed in Chapter 10, are applicable to nonpolar/weakly polar systems only. 2.Accurate prediction of pure component vapor pressures by the EoS used is essential for good multicomponentmixture results. This requires, as discussed in Section 10.6: a.accurate Tc, Pc, and l.> values, and b. vapo.r pressures (P s ) that are not very low. (While the performance of the various EoS at low P s levels varies, caution should be exercised at very low ones.) 3.Satisfactory predictions of hydrocarbon mixtures can be obtained without the use of interaction coefficients (Graboski and Daubert, 1978; Tsonopoulos and Heidman, 1986). When significant differences in the size and type of the molecules exist, however, typically in methanecontaining systems, interaction coefficients are needed for accurate results (Chao and Lin, 1986; Czerwienski et al, 1988). The last authors, for example, report that best results for the binary systems: methane with propane, cyclohexane and ndecane, are obtained using interaction coefficients of 0.020, 0.022, 0.032 respectively in the SRK EoS. 4.Interaction coefficients are also needed in the presence of nonhydrocarbons, such as H 2 , C02 , H 2S, N2 , etc. Typical results for C02 containing binary mixtures are presented in Table 14.2. 5.Prediction accuracy is sensitive to the value of the interaction coefficients as shown by the typical results of Figure 14.7 for the system C02nbutane. For explanation, see Section 11.9.2. 6.Even though the interaction coefficients are considered temperature and pressure independent to a first approximation, they are actually functions of temperature as Figure 14.8 suggests (see also Kato et al, 1981, and Czerwienski et al).
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Table 14.2 Correlation of C0 2 (1)  Solvent (2) Binaries with the SRK and PR EoS SRK Solvent
Range
k12
PR
AAE% 0 > in p
KI
0.136 0.138
0.4 0.9
1.8 5.2
526 235
0.132 0.129
2.0 3.7
311411
482
0.142
iPentane
277377
189
nHexane
293313
nHeptane
k12
AAE% 0 > in p
KI
0.133 0.135
0.5 0.9
1.9 5.3
3.2 7.9
0.129 0.124
1.8 3.8
3.0 7.9
1.3
3.8
0.136
1.1
4.0
0.131
2.9
5.0
0.122
2.5
3.8
477
0.117
5.5
5.4
0.111
5.3
5.3
310477
2133
0.110
1.9
2.5
0.100
2.3
2.5
nDecane
277510
1188
0.131
5.2
3.7
0.115
5.8
3.7
Benzene
298313
978
O.Q78
2.2
2.6
0.077
2.1
2.5
Cyc1ohexane
473533
13129
0.128
1.9
8.0
0.105
1.9
7.8
Toluene
311477
3152
0.113
5.0
8.7
0.106
4.8
8.2
Decalin
348373
1071
0.190
6.0
0.180
6.1
'nK)
P(bar)
Ethane
250 222289
1321 757
Propane
244266 253273
nButane
Overall Absolute Error % (I)
3.0
4.8
3.0
4.7
Average absolute error %.
7 .For systems containing C02 , H2S, N2 , and CO with hydrocarbons, Graboski and Daubert (1979) give generalized correlations for the estimation of interaction coefficients in the SRK EoS; and Kato et al for C02  nalkane systems with the PR EoS. 8.For H2/hydrocarbon systems, see: Graboski and Daubert; Lin (1980) and, especially, Tsonopoulos and Heidman. 9.For some problems encountered in using the SRK EoS, see Gray (1979). lO.The SRK and PR EoS provide similar correlation results as suggested, for example, by Table 14.2 for the C02 binaries. 11.The vdW711, tPR and tvdW EoS give equally good results with these EoS, as Table 14.3 indicates for the first one.
525
High Pressure VaporLiquid Equilibrium
70
.,. '
• • • • • experimental  _  PR k 12 =0.0   PR k 12 =0.131
I •I
I
I
60 /
I • I I
I
(optimum value for this temperature)
,,
// I
I
I I
/
50
I
/
...0
.a 40
/
a..·
/
,
/
I I I
/
/
I I
I
I
30
I I
20
I
I
/
0~~~LLLLLLL~
0.0
0.1
0.2
0.3
0.4
0.5
x,,y,
0.6
0. 7
0.8
0.9
1.0
Figure 14.7 The phase envelope for the C02  nbutanesystem at 310.9K (exp. data from Knapp et al, 1982, p.607).
0.16 X
0.15 X
0.14 X
'"'0.13
X
~
X
X
0.12
X
X
X
0.11 X
0.10
200
250
300
350
T, K
400
450
500
Figure 14.8 Variation of kij with temperature for the system C02  npentaneand the tPR EoS.
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Applied Chemical Engineering Thermodynamics
Table 14.3 Performance of the vdW711, PR, and SRK EoS in the Correlation of VLE Data and the Prediction of Saturated Liquid Volumes (Czerwienski et al) %Error in P
Lly·lOOO
System
T(K)
C0 2C 3
277/344
9/65
10
11
11
C0 2nC4
311/411
4171
8
9
C0 2nC 5
278/378
2/95
9
C02nC 10
278/378 3/153
C02nC20
310/373
c,c 3
vdW SRK
PR
vdW
1.66
1.44 1.79 17.0
6.7
3.7
12
1.44
1.29 1.86 15.2
6.0
3.3
11
10
3.83
3.73 3.78 12.8
2.3
2.6
2
3
2
4.61
5.25 4.41
14.7
6.7
2.2
5175
0
0
0
2.84
2.72 2.93 53.4
37.1
1.6
278/344
7/99
12
15
13
1.40
1.35 1.46 12.1
4.6
5.3
C 1nC6
294/344
7/99
6
5
6
3.92
4.16 4.10
5.2
1.1
c,nc,o
311/372 1/357
3
2
3
4.12
4.06 4.02 13.0
6.0
5.3
C3nC 10
278/478
2
2
2
2.27
2.21 2.24 17.8
4.9
3.6
6
6
7
2.9
2.9
3.0
8.8
3.2
Overall
P(atm) SRK · PR vdW SRK
%Error in V
2/68
PR
6.5
18.1
Note. vdW stands for the vdW711 EoS; V for saturated liquid volume; Lly represents the average absolute deviation in vapor phase mole fraction; % error is the average absolute one.
12.The same can be expected from other EoS, cubic and noncubic, as suggested by the results of Chao and Lin, summarized in Table 14.4. 13.As expected from our discussion on the prediction of saturated liquid volumes of pure compounds in Section 10.6, the results for mixtures should vary among different EoS. The SRK EoS, thus, gives very poor results while the translated ones, such as the vdW711, provide reasonable accuracy (Table 14.3). 14.The SRK and PR EoS have been also applied to waterhydrocarbon systems by adjusting the value of a;; in the attractive term to reproduce water's vapor pressure. The PR EoS gives better results for the water solubility and K values in the hydrocarbonrich phase, using interaction coefficients for all hydrocarbonwater pairs. Prediction of hydrocarbon solubility in the waterrich phase however, using the same interaction coefficients, gives poor results. For details, and a method for estimating interaction coefficients in such systems, see Tsonopoulos and Heidman.
527
High Pressure VaporLiquid Equilibrium
Table 14.4 Performance of Several EoS in the Correlation of Binary VLE Data Overall Avg. Abs. Error% in K 1 and K2 N(l)
SRK
PR
CCOR(2)
H2/solvent
18
6.9
6.4
5.5
CH4/"
21
4.2
C02/"
17
4.5
System
4.1
BACK(3)
3.8
3.5
4.2
4.1
(1) Number of systems (2) Cubic chain of rotators, Kim et al (1986) (3) Chen and Kreglewski (1977)
Table 14.5 Data Correlation of C02  Solvent Binary VLE Systems Using the tPR EoS with One (k;) and Two (kij, dij) Interaction Coefficients Solvent T, K Avg. Abs. Error % in P kr k;. and dr 311 nC 4 1.5 0.4 nC 5 294 1.7 0.7 nC 7 311 2.2 1.9 2.3 323 0.4 nC2o 1.7 0.5 473 nC24 nC 32 5.6 1.1 398 373 7.5 1.6 nC44 15. To improve the description of binary VLE data  over that obtained with the use of an interaction coefficient kij in the combining rule for aij  the typical approach involves the introduction of a second interaction coefficient by writing Eq .11. 7.20 as: bij
=
0.5(b;+bj)(ldij)
(14.7.2)
This provides increased flexibility in fitting binary data, important when large differences in molecular size are involved, as suggested by the typical results presented in Table 14.5. The obtained kij and dij values are not easily generalized, as indicated
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Applied Chemical Engineering Thermodynamics
by Tsonopoulos and Heidman for H2, COr, and waterhydrocarbon systems. Gasem and Robinson (1985) did obtain generalized kij and dij expressions, however, for C02  nalkanes. 16.From the noncubic EoS, the BWR is still used for systems where sufficient data exist for the evaluation of the required parameters, mostly hydrocarbons. Good results are obtained, but the calculations are more cumbersome and computertime consuming than those with cubic EoS. 17 .Evaluation of kij values from binary VLE data for the equations of state discussed in Chapter ~ 0 can be accomplished using Program EOSKU of Appendix E.
14.7.3 Polar Systems Use of the equation of state approach in the correlation and prediction of VLE behavior of polar systems has the obvious advantage of eliminating the need for estimating the vapor phase imperfection, usually done with the virial equation. This advantage becomes very important at high pressures. First, the virial equation truncated after B may not be applicable at these pressures. Second, even if it is applicable, the uncertainties involved in estimating second virial coefficients, especially cross ones, may lead to significant errors in the obtained vaporphase fugacity coefficients of the mixture components at these pressures (see Problem 14.58). Finally, the two approaches may not converge in the critical region. Application of cubic EoS to polar mixtures requires resolution of two main problems. l.The simple Soavetype expressions for a (Eq.l0.4.2) cannot provide satisfactory prediction of vapor pressures of polar compounds (why?). 2.The 'conventional' mixing rules for parameters a and b: Eqs 11.7.15 through 11.7 .18, 11.7.23 and 14.7 .2, are not typically sufficient for polar mixtures. The first problem is easily dealt through use of compoundspecific parameters for polar compounds, as discussed in Section 10.6. The second problem has been the subject of extensive study in the last few years without a completely satisfactory resolution yet. The following comments represent a very brief account of this effort. 1. The introduction of new mixing rules in polar mixtures has followed two main approaches: a.incorporationoflocal composition GE models into the mixing rule for a in cubic EoS, so as to account for the nonrandomness in such mixtures
High Pressure VaporLiquid Equilibrium
529
on the local level, as done for example by: Huron and Vidal (1979); Gupte et al (1986); Kalospiros et al (1991). This approach leads to the socalled EoS/GE models; b. use of densitydependent(DD) mixing rules (Mollerup, 1981; Whiting and Prausnitz, 1982; Danner and Gupte, 1986). 2.The first approach has the disadvantage that the expression for the second virial coefficient, obtained from the EoS/ GE model as the density approaches zero, does not follow the quadratic mixing rule dictated by statistical mechanics. To overcome this, DD mixing rules have been suggested, that provide an arbitrary interpolation between this theoretically required quadratic mixing rule at zero density and some other type of mixing rule at higher densities. This approach has found limited application however. 3.Kalospiros et al have applied the methodology of Gupte et al, combining the modified vdW711 EoS (Androulakis et al, 1989) with the UNIQUAC GE expression, to polar systems. Excellent results were obtained in both, the correlation of binary systems and the prediction of ternary VLE behavior from binary data. The latter is demonstrated in Figure 14.9. Notice the improvement over the results obtained with the conventional mixing rules. The method cannot be applied in the presence of supercritical components. 4.The procedure of Michelsen (1990), using again an EoS/GE model, appears to resolve the problem of supercritical components, however, and is expected to greatly facilitate the application of the equation of state approach to polar systems.
14.8 Bubble Point, Dew Point, and Flash Calculations High pressure VLE calculations are carried out using K values obtained with the aforementioned methods, in a fashion similar to that used for low pressure ones.
14.8.1 Bubble and Dew Points As we have seen in the previous Chapter in the first case, B.P., a liquid of specified composition: x 1, x2 , ••• is given, and we calculate: *the B.P. temperature at a given pressure, or
530
Applied Chemical Engineering Thermodynamics 1.0
0.8 .>!
[][][][][] y, "'"'"'"'"' y Y2 ooooo 3
~ 0.6
(a)
[]
0.4
0.2
0.0 I L    L  . . . . L .  '     L   '   L    L  . . . . L .  L  _ . 1 0.4 0.8 1.0 0.0 0.2 0.6 Yexp
1.0
[][][][][] y, 0.8
"'"'"' "'"' y Y2 ooooo 3
.>!
~ 0.6
(b)
[]
%'06 B
0.4
"'
0
[]
oKB 0
0.2
0.0 0.0
.00.
""" []Iii 1il
QJ
0.2
0.4
0.6
0.8
1.0
Yexp
Figure 14.9 Comparison of VLE predictions from binary data: ethanol(l) benzene(2) nheptane(3)at 180 mm Hg. (a) vdW711/UNIQUACmodel; (b) vdW7111 conventional mixing rules (Kalospiros et al, 1991).
High Pressure VaporLiquid Equilibrium
531
*the B.P. pressure, at a given temperature. The iteration is carried out until the sum, over all mixture components, of calculated y/s is equal to unity: (14.8.1) In the second case, D.P., a vapor of specified composition: y 1 , y2 , ••• is given, and we calculate: *the D.P. temperature at a given pressure, or *the D.P. pressure at a given temperature. The iteration is carried out until, here, the sum of calculated X;'s is equal to unity: (14.8.2) Bubble and dew point calculations are presented in the next two Examples with the simplest approach, the use of K values from Eqs 14.5.1 through 14.5.5. They can be also carried out with the computer subroutines, using the SRK EoS, given by Daubert (1985), that can be easily adapted to other EoS; and for bubble point pressure calculations, using the EoS of Chapter 10, with the Program VLEEOS presented in Appendix E, which can be easily modified to perform B.P. temperature calculations.
14.8.2 Example 14.1 Calculate the B.P. pressure at 27°C for the following mixture: Propane( I): 0.40 moles ibutane(2): 0.30 moles nbutane(3): 0.30 moles. This is a trialanderror calculation: l.Assume P; 2.Get K's from Eqs 14.5.1 through 14.5.5; 3.Check if: K1x1 + Kzh + K3x3 = 1.0 (within 0.01); 4.If not, assume an other pressure and repeat. The obtained B.P. pressure is 5.9 bar.
14.8.3 Example 14.2 Calculate the D.P. pressure for the mixture of Example 14.1, again at 27°C.
532
Applied Chemical Engineering Thermodynamics
In this case we seek the pressure where:
y 1 + Yz + y3 = 1.0 (within 0.01) Kl Kz K3 Using a similar trialanderror procedure we obtain: P
= 4.5 bar.
14.8.4 Example 14.3 Using the data of Table 14.E:3 for the system carbon dioxide( I) nbutane(2): a.evaluate the optimum k 12 value for the PR EoS; b.using the optimum value at 310.93 K of k12 = 0.131 (Figure 14.7), calculate the B.P. pressures and vapor compositions at the given liquid compositions and temperature and compare them with those obtained in part (a). a. Using the given data and the Program EOSKIJ of Appendix E, the value of k12 in pressure; and of 0.008 in vapor phase mole fraction. Detailed results are presented in Table 14.E.3.
= 0.148 is obtained with an average absolute error of 0.4%
b. Use of k 12 = 0.131 in the Program VLEEOS yields an average absolute error of 1.3% in P and 0.032 in y. The results are shown graphically in Figure 14.E.3 and demonstrate the sensitivity of VLE predictions to the k 12 value (Section 14.7.2).
Table 14.E.3 Experimental and Calculated Results for the System Carbon Dioxide (1)  nButane (2) at 377.59 K (Knapp et a!, p.607)
P (bar)
xi
Y1
17.237 20.684 24.131 27.579 31.026 34.473 41.368 48.263 55.158 62.052 68.947 72.394 75.152
0.004 0.028 0.052 0.076 0.100 0.124 0.174 0.223 0.274 0.328 0.393 0.434 0.498
0.026 0.150 0.244 0.313 0.367 0.411 0.480 0.524 0.552 0.568 0.568 0.552 0.498
peale
(bar) Error %
17.381 20.838 24.286 27.720 31.137 34.531 41.509 48.177 54.858 61.517 68.663 72.394 75.152
0.83 0.76 0.64 0.51 0.36 0.17 0.34 0.18 0.54 0.86 0.41 0.00 0.00
Yi.ca/c
Yl.ca/cYl
0.025 0.148 0.239 0.308 0.361 0.404 0.468 0.510 0.537 0.553 0.556 0.548 0.498
0.001 0.002 0.005 0.005 0.006 0.007 0.012 0.014 0.015 0.015 0.012 0.004 0.000
High Pressure VaporLiquid Equilibrium
533
70
60 50 1...
0
.a 40
cL 30
20 10
• • • • • experimental _ _ PR k 12 =0.148 (optimum value for this temperature} · PR k 12 =0.131 (optimum value for T=310.9 K}
Figure 14.E.3 Bubble point pressures and compositions for the system: C02(1)nC4(2) at 377.59 K.
14.8.5 Flash Calculations Consider a mixture of F moles with mole fractions of its components equal to z 1, z2 , ... available at a temperature 1f and a pressure P1 . This mixture is expanded through a throttling valve flashed as it is referred to  into a vessel where the temperature is T and the pressure P (Figure 14.10). Our objective is to calculate, assuming that equilibrium has been reached: 1. the amount of liquid L and vapor V, and 2.the compositions of the two phases. The information given is:
*F * Zl, Zz, ...
* Tand P. The information sought is: *Land V * the liquid phase compositions: xi> x2 , .. . * the vapor phase compositions: y 1, Yz, .. .
534
Applied Chemical Engineering Thermodynamics
V moles
y1 ' y2 ' ...
T,P F moles
Tt , Pt z1 , z2' ...
L moles Figure 14.10 Schematic diagram for a flash process.
From a material balance on component i: Fz;
=
LX; + Vy;
(14.8.3)
We also have: Y; = K; X;, and L =F V Introduction of the last two relationships into Eq.14.8.3 and solution for X; yields:
(14.8.4) Similarly: FK;Z;
Y;
=
F + V(K; 1)
(14.8.5)
To evaluate V from either of the last two equations we iterate until: Qx or:
=LX; 1 = 0
Qy =
LY; 1
=0
Determination of the X; Is and Y; Is is straightforward.
(14.8.6) (14.8.7)
535
High Pressure VaporLiquid Equilibrium
Comments 1. The procedure for this type of flash calculation is demonstrated in the next Example using Eqs 14.5.1 through 14.5.5. 2.Evaluation of V can be facilitated by using instead of Eq.14.8.6 or 14.8.7, their difference Q (Rachford and Rice, 1952): Q = E0.6
0.4
D D
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
Yexp
Figure 14.11 Prediction of vaporliquid equilibrium with the PR EoS for the system: C2 (1) C3(2) nC4 (3) at 260, 270 and 280 K using kij = 0.0 for all pairs. (Data from Clark and Stead, 1988.) 1800
1700
T=344.3 K
D~
0 DO 0
1600
D 0
D
0
'iii
0
0
a. D
cL 1500
D 0
D 0
0
DDDDD Exp.
1400
_ _ Pred.
D D
1300 0.6
D
0.7
0.8
mole fraction C0 2
0.9
1.0
Figure 14.12 The phase envelope for the system C0 2nC4 nC 10 using kij values for the two C02 alkane binaries, obtained from the corresponding data. Experimental data from Nagarajan et al (1990).
538
Applied Chemical Engineering Thermodynamics
c.Prediction of vapor and liquid volumes from flashes, even with the translated EoS, does not always provide satisfactory results. d.Results become poorer as the critical point is approached, even in the correlation of binary data (Figures 14.7 and 14.E.3). 9.The most extensive compilation of high pressure binary VLE dataand references for multicomponent ones  is presented by Knapp et al. The data are correlated with several EoS. For theoretical background, see the excellent book of Prausnitz et al (1986). 10. Successful application of equations of state to polar systems is under development in the last decade with encouraging results. They have not reached as yet, however, the level of simplicity and performance required for general acceptance.
References Androulakis, J., Kalospiros, N., Tassios, D., 1989. Fluid Phase Equilibria, 45, 135. Chao, K.C., Lin, H.M., 1985./UPAC Conference on Phase Equilibria, Paris. Chao, K.C., Seader, J.D., 1961.AIChEJ., 7, 598. Chen, S.S., Kreglewski, A., 1977. Ber. Bunsenges. Phys. Chem., 81, 1048. Clark, A.Q., Stead, K., 1988. J. Chem. Thermodynamics, 20, 413. Clifford, I.L., Hunter, I.E., 1933. J. Phys. Chem., 37, 101. Czerwienski, G., Tomasula, P., Tassios, D., 1988. Fluid Phase Equilibria, 42, 63. (Reprinted by permission.) Danner, R.P., Gupte, P.A., 1986. Fluid Phase Equilibria, 29, 415. Daubert, T.E., 1985. Chemical Engineering Thermodynamics, McGrawHill, New York. DePriester, C.L., 1953. Chem. Eng. Progr. Symp. Ser., 7, 49. Dowling, G.R., Todd, W.G., March 1973. Chemical Engineering, 115. Gasem, K.A.M., Robinson, R.L., 1985. National AIChE Meeting, Houston, Texas. Graboski, M.S., Daubert, T.E., 1978; 1979. Ind. Eng. Chem. Process Des. Dev., 17, 448; 18, 300. Gray, R., 1979. In Advances in Chemistry Series, K.C. Chao and R.L. Robinson, Editors, No 182, ACS, Washington, D.C. Grayson, H. G., Streed, C.W., June 25,1963. Paper presented at the 6th World Petroleum Congress, Frankfurt/Main, Germany. Gupte, P.A., Rasmussen, P., Fredenslund,Aa., 1986./nd. Eng. Chem. Fundam., 25, 636.
High Pressure VaporLiquid Equilibrium
539
Hadden, S.T., 1953. Chern. Eng. Progr. Symp. Ser., No.7, 53. Hadden, S.T., Grayson, H.G., 1961. Hydrocarbon Processing and Petroleum Refiner, 40, 9, 207. Henley, E.S., Seader, J.D., 1981. EquilibriumStage Separation Operations in Chemical Engineering, Wiley, New York. Hildebrand, J.H., Scott, R.L., 1964. The Solubility of Nonelectrolytes, Dover, New York. Huron, M.J., Vidal, J., 1979. Fluid Phase Equilibria, 3, 255. Kalospiros, N., Misseyannis, G., Androulakis, J., Tassios, D., 1991. Fluid Phase Equilibria, 64, 173. (Reprinted by permission.) Kato, K., Nagahama, K., Hirata, H., 1981. Fluid Phase Equilibria, 1, 219. Kesler, M.G., Lee, B.J., Fish, M.l., Hadden, S.T., May 1971. Hydrocarbon Processing, 257. Kim, H., Lin, H.M., Chao, K.C., 1986. Ind. Eng. Chern. Fundam.,25, 15. King, C.J., 1981. Separation Processes, McGrawHill, New York. Knapp, H., Doring,R., Oellrich, L., Plocker, U., Prausnitz, J.M., 1982. VaporLiquid Equilibria for Mixtures of Low Boiling Substances, DECHEMA Chemistry Series, Vol. VI, Frankfurt/Main, Germany. Lin, H.M., 1980. Ind. Eng. Chern. Process Des. Dev., 19, 501. Maddox, R.N., Erbar, J.H., Feb. 2, 1981. Oil & Gas J., p.74. McCracken, P.G., Smith, J.M., 1956. AIChE J., 2, 498. Michelsen, M.L., 1982. Fluid Phase Equilibria, 9, 1 and 21. Michelsen, M.L., 1990. Fluid Phase Equilibria, 60, 47 and 213. Mollerup, J., 1981. Fluid Phase Equilibria, 1, 121. Nagarajan, N., Gasem, K.A.M., Robinson, R.L., 1990. J.C.E.D., 35, 228. Pak, S.C., Kay, W.B., 1972. Ind. Eng. Chern. Fundam., 11, 225. Peng, D.Y., Robinson, D.R., 1976. Ind. Eng. Chern. Fundam., 15, 59. Prausnitz, J.M., Lichtenthaler, R.N., de Azevedo, E.G., 1986. Molecular Thermodynamics of FluidPhase Equilibria, PrenticeHall, Englewood Cliffs, N.J. Rachford, H. H., Jr., Rice, J.D., 1952(0ctober). J. Petrol. Techno[., 4, 19. Scheibel, E.G., Jenny, E.F., 1945. Ind. Eng. Chern., 37, 80. Soave, G., 1972. Chern. Eng. Sci., 27, 1197. Tsonopoulos, C., Heidman, J.L., 1986. Fluid Phase Equilibria, 30, 391. Van Konynenburg, P.H., Scott, R.L., 1980. Trans. Roy. Soc., London, 298A, 495. Watson, P., Cascella, M., May, D., Salerno, S., Tassios, D., 1986. Fluid Phase Equilibria, 27, 35. Whiting, W.B., Prausnitz, J.M., 1982. Fluid Phase Equilibria, 9, 119.
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Applied Chemical Engineering Thermodynamics
Problems 14.1.a.Is there a difference in the definition of the critical point of a pure compound and of a mixture? b. Discuss the differences in behavior of these two types of fluids in the critical region. 14.2 With reference to mixture behavior at high pressures: a.can isothermal reduction of pressure for a saturated vapor lead to the formation of liquid? b.the same, but for isobaric temperature increase. c. what is retrograde condensation? Is it of any practical importance? Give an example. d. why are K values equal to one at the critical point? 14.3 Can changing the operating pressure of a distillation column effect the purity of the obtained products? Explain. 14.4 With reference to natural gas: a. what is its contribution to the world energy supply? b. when is retrograde condensation important? c. what are 'gas hydrates'? 14.5 Prepare the Pxy diagram for a binary system at three temperatures and show the critical points if present. 14.6 What are the main approaches used in high pressure VLE calculations? 14.7 Consider the mixture C1nC5 at 300 K and 50 atm. What problems do you see in using the gammaphi approach to describe the VLE behavior of this system at these conditions? 14.8.a.A vapor mixture of propane(l), ibutane(2), and nbutane(3), leaving the top of a distillation column with composition: Yt = 0.7; Y2 = 0.2; Y3 = 0.1 is to be condensed in a total condenser. What is the minimum pressure that the condenser should operate at, so that the liquid leaves at 30°C? b.Discuss any assumptions you make and estimate their quantitative effect on the accuracy of your answer. c.Is there a better method? If yes, which one and what accuracy would you expect? 14.9 With reference to the previous problem, assuming a 0.2 bar pressure drop between the condenser and the top of the column and that the vapor and liquid on the top tray of the distillation column are in equilibrium with each other, determine: a. the temperature at that tray; b. the composition of the liquid leaving the tray. 14.10 If a partial condenser were used in Problem 14.8, determine: a. the pressure it should operate at, so that half of the vapor is condensed; b. the composition of the vapor leaving the condenser.
High Pressure VaporLiquid Equilibrium
541
The liquid condensate is at 30°C as in Problem 14.8. 14.11 With reference to Example 14.4, plot: a. the amount of vapor, and b. its propane composition, as a function of the drum operating pressvre and comment on the purity of the obtained vapor. Operating temperature: 27°C. 14.12 At 40 atm and 333 K a liquid phase consisting of propane and a nonvolatile hydrocarbon is in equilibrium with a vapor phase containing 85% (mole) nitrogen and the rest propane. Neglecting the solubility of nitrogen in the liquid phase, find the mole % of propane in the liquid. 14.13 The overhead vapor stream from a fractionating column going to the condenser has the following composition (molar): ethylene = 0.20; propane= 0.30; nbutane = 0.50 It is desired to operate the condenser so that 70% (mole) of the total stream is liquified in the condenser. If the temperature in the condenser is 40°C, what is: a. the required pressure? b. the composition of each phase leaving the condenser? 14.14 A system consisting of methane, propane, and a nonvolatile hydrocarbon is in equilibrium at l00°F and 600 psia. If the vapor phase contains 94.0% (mole) methane and the rest propane, what is the composition of the liquid phase? 14.15 The liquid feed to a fractionating column,has the molar composition: C2 = 0.15; C3 = 0.40; nC4 = 0.45 If the temperature and pressure in the column at the point of introduction of the feed are respectively 50°C and 14 atm, what mole % of the feed is vaporized and what is the composition of the vapor? 14.16 A gas stream at 25°C and 1 atm, consisting of 20% (mole) nitrogen and the rest carbon dioxide, is separated into practically pure C02 with 95% recovery. The product streams leave the separation process at 1 atm and 25°C, the temperature of the surroundings. Find the minimum amount of work required in J/mol of C02 recovered. 14.17 Discuss the advantages and disadvantages in using the: a.KCharts; b.ChaoSeader;and c.Equations of State; approaches in carrying out VLE calculations. 14.18 What is the 'solubility parameter'? How can it be calculated from vapor pressure data? 14.19.a.Could you use the regular solution theory for low pressure VLE calculations? b. What information would you need'? c. What would the restrictions be? 14.20 In using cubic EoS for high pressure VLE calculations: a. what information is needed? b. when are interaction coefficients needed? c.how are they evaluated?
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Applied Chemical Engineering Thermodynamics
14.21 Discuss the advantages and disadvantages of the EoS approach over the y  c/> one in the correlation and prediction of VLE. Consider: a.low pressures; and b. high pressures, including the critical region; for nonpolar and polar mixtures. 14.22 Experimental data for the system methanolhexane are to be correlated with the SRK EoS. a. What changes, if any, are necessary to this purpose in the SRK EoS? b.How would you obtain the best results? 14.23 The feed to a butanepentane splitter has the following composition: iC4 = 0.06; nC4 = 0.17; iC5 = 0.32; nC5 = 0.45 For a 95% recovery of nC4 it is estimated that the distillate will also contain: 0.0595 moles of iC4 , 0.016 moles of iC5 and 0.0059 moles of nC 5 per mole of feed. Using a condensate temperature that you would recommend in this case, determine: a. the operating pressure for a total condenser; b. the composition of the liquid leaving the top tray; c. the bottoms temperature and composition. Make any assumptions you need, but justify them. 14.24 In Example 14.3 it is stated that use of k12 = 0.131 gives poorer prediction of vapor phase compositions than use of k12 = 0.148. Yet it appears from Figure 14.E.3 that the curve (dashed) corresponding to k12 = 0.131 is closer to the experimental dew point curve than that for k12 = 0.148. Determine which k12 gives the best values for K1 and K2 • 14.25 Is the statement made in part (a) of Example 14.4 with reference to the D.P. pressure: "Below it, we will have superheated vapor", correct? Explain qualitatively. 14.26 What are the most important items you have learned in this Chapter? Give a possible industrial application in each case. 14.51 Using the data of Table 14.P.1 calculate: a. the value of the interaction coefficient for the SRK, PR, and vdW 711 EoS at each temperature; b.repeat using all data points together; c. explain any differences in the interaction coefficient values. 14.52 Prepare a flow diagram for a bubble point pressure calculation with the ChaoSeader method. 14.53 Calculate the total pressure and vapor phase composition for the system C1(1) C02(2) at 230 K, x1 = 0.170 and x1 = 0.543, using the ChaoSeader method. Compare your findings with the experimental values given in Table 14.P.l. 14.54 Explain in terms of intermolecular forces the following observation: At 280 K and 70 bar total pressure, the K value for hexane in the methanehexane system is much larger than that in the hydrogen hexane one at the same tempe
High Pressure VaporLiquid Equilibrium
543
Table 14.P.1 VaporLiquid Equilibrium Data for Methane(1) Carbon Dioxide(2). (Knapp et al, p.401). J{K)
P (bar)
xl
230 230 230 230 230 230 230 230 250
8.916 20.265 40.529 55.728 62.821 68.900 69.994 71.403 20.265
0.000 0.050 0.170 0.318 0.394 0.534 0.543 0.584 0.010
Yt 0.000 0.525 0.728 0.764 0.762 0.751 0.730 0.716 0.104
T(K)
P (bar)
xl
250 250 250 250 270 270 270 270 270
30.397 50.662 70.926 79.539 31.947 37.013 42.140 58.575 80.633
0.053 0.166 0.326 0.405 0.000 0.018 0.040 0.113 0.260
Yt 0.361 0.575 0.615 0.564 0.000 0.108 0.190 0.353 0.411
rature and pressure. At 280 K and 1 bar, however, the two K values are about the same. 14.55 Develop the second virial coefficient expression for a binary mixture and the SRK EoS. 14.56 The mixture in Problem 14.13 is to be condensed completely at 30°C. Calculate the minimum pressure required to this purpose, using: a.Eqs 14.5.1 through 14.5.5; and b. an EoS of your choice. Which answer is more accurate? Why? 14.57.a.Establish the reliability of Eqs 14.5.1 through 14.5.5 in predicting the K values of the hydrocarbons involved, using a method of your choice. b.Demonstrate the impact of this uncertainty on a real problem of your choice again. 14.58 Correlate the data for NH 3H 2 0 of Clifford and Hunter (1933) at all temperatures given, using: a. the gammaphi approach; b.the EoS/GE approach. 14.59 Answer Problem 14.25 quantitatively.
15 Chemical Reaction Equilibrium
The petrochemical industry was born in the early 1900's when Thomas Edison's electricity was becoming popular and the use of kerosene was on its way down. With a decreasing kerosene market, chemical engineers in the major oil companies began studying means for usefully converting the excess residue remaining after gasoline was distilled from petroleum. The solution to the problem came as early as 1913, when William Burton of the Standard Oil Company (Indiana) put into operation a series of thermal "cracking • units known as "Burton Stills ". This was the beginning ofa methodfor breaking the larger carbon molecules into smaller ones through a heat process, a process for making chemicals such as ethylene, propylene, butylene, and other olefins from crude oil. Then a· new technique was developed known as "catalytic cracking" and before long there was an abundance of the petrochemicals needed for elastomers, plastics, and syntheticfibers. Ethylene dominates all the other petroleum building blocks and one of its first uses was as ethylene glycol, the antifreeze used in cars. Another automotiveassociated use was tetraethyllead (TEL), a gasoline additive to reduce engine knock and increase fuel efficiency. Soon after commercialization of ethylene glycol and TEL came a whole stream of ethylene derivatives such as styrene and polyethylene. Other major compounds made available from the cracking of liquid petroleum were the elastomers, synthetic fabrics, and the plastics. ("Ten Greatest Achievements of Chemical Engineers", American Institute of Chemical Engineers (AIChE), 1983.)
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Applied Chemical Engineering Thermodynamics
15.1 Introduction Chemical reactors Tepresent the cornerstone of the Chemical and Petroleum Industries. Thus, of the ten greatest achievements of Chemical Engineers  published by the American Institute of Chemical Engineers in celebrating its 75th anniversary at least six involve chemical reactions: * Synthetic rubber *Drugs *Plastics * Synthetics replacing the silkworm * Petrochemicals (see previous page) * Fertilizers Application of a chemical reaction for industrial purposes must consider two key questions: l.How far will the reaction go if sufficient time to reach equilibrium is provided, i.e. what is the equilibrium conversion? and 2.How long will it take for a certain level of conversion  up to the equilibrium value, of course  to be attained? The first represents the subject of Thermodynamics; the second, that of Kinetics, and both must be considered in the design of a reactor. Furthermore the two are often in conflict. For example in the case of the oxidation of so2 to so3 for the production of sulfuric acid, low temperatures favor high conversions but the kinetics is very slow, and the reverse is the case at high temperatures. Thus  as it is often the case in our profession  an optimum must be found, based on economic considerations.
15.2 Objective and Approach The thermodynamic basis for calculating chemical reaction equilibrium was developed in Section 12.5. There we showed that if a system of components A1, A2 , ... , Ak, reacting according to the equation:
v1A 1 + v:zA 2 + ... + v_0k = 0 where v1, v2 , ... , vk, represent the stoichiometric numbers, reaches equilibrium at constant temperature and pressure, then:
547
Chemical Reaction Equilibrium
InK=
.1Go
RT
(12.5.10)
where:
(12.5.11) di is the activity of component i in the mixture, and .:1 G 0 is the Gibbs
free energy of the reaction  the free energy of the products ( G/) minus thatofthe reactants (Gr0 ) both evaluated at the standard state, t.e. at the temperature of the reaction and a pressure of one atmosphere  referred to therefore as standardfree energy:
.1G 0 = Gp o Go =~ (12.5.12) r L v.G.o 1 1 Our objective in this Chapter is to demonstrate how these equations with their rather abstract quantities  can be used in determining the equilibrium conversion of a reacting system, which  as outlined in the Introduction  represents the contribution of thermodynamics to the application of chemical reaction equilibrium in the Chemical and Petroleum Industries. More specifically our objective is to examine how: * temperature, * pressure, and * feed composition effect the equilibrium conversion of the reacting system. We start with the evaluation of the standard free energy of the reaction as a function of the system temperature T  remember it is a function of T only  so that the equilibrium constant K can be evaluated at the temperature of interest. We proceed with the determination of the equilibrium conversion from the known value of the equilibrium constant K, considering first single reactions taking place in a single phase, and examine the factors that effect this conversion. We then move to the more complex cases: reactions taking place in heterogeneous systems and multiple reactions. To this purpose we consider first the Phase Rule and the Duhem theorem  as they apply to reacting systems  and discuss the methodology for identifying the number of independent reactions for the formation of a given equilibrium mixture. Calculation of equilibrium conversions for heterogeneous systems and for multiple reactions is then outlined mostly through some typical examples. We summarize our findings in the Concluding Remarks.
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Applied Chemical Engineering Thermodynamics
15.3 The Dependency of K on Temperature 15.3.1 The Problem According to Eqs 12.5.10 and 12.5.12 evaluation of the equilibrium constant K for a given reaction  at a specified temperature T  requires values for the standard free energies of the components of the reaction, also at T. The latter can be replaced by the standard free energy of formation, ~ G/, of each component from its elements, as we will see in Example 15.1. Such values are available for a large variety of compounds, but at 298.2 K and the ideal gas state at P = 1 atm (or 1 bar). Values for selected compounds are presented in Table 15.1; for other compounds, see the same reference, Reid et a1 (1987), or the references given by Kyle (1984). We proceed, therefore, to describe a method for evaluating K values at temperatures other than 298.2 K from the free energy of formation values of the components of the reaction at 298.2 K. To this purpose, we: !.demonstrate that the variation of K with T depends on the standard enthalpy of the reaction LiH o, which itself is a function of temperature; 2.establish the functional relationship of LiH 0 to T; and 3.develop the expression of K as a function of temperature.
15.3.2 The Variation of K with the Enthalpy of the Reaction Starting with the definition of the Gibbs free energy (Eq.1.6.5): G
=
HTS
(1.6.5)
we will demonstrate that: dlnK(T)
LiH 0 (T)
dT
RTz
(15.3.1)
where the symbolism (T) is introduced to emphasize the temperature dependency of K and LiH o. The standard enthalpy of the reaction, LiH 0 , represents  like the free energy  the difference in enthalpy between products (H/) and reactants (Hr0 ), both evaluated at the standard state. It is calculated  again like the free energy  from the enthalpies of formation of the reaction components at the standard state, LiH/=
549
Chemical Reaction Equilibrium
Table 15.1 Standard Enthalpy and Free Energy of Formation for Selected Compounds(!) at 298.2 K in the Ideal Gas State and, for JiG/, at 1 atm (kJ/mol)
Compound Methane Ethane Ethylene Acetylene Propane Propylene nButane iButane nPentane nHexane nHeptane nOctane Water H2S NH 3 NO N02
74.52 50.519 83.82 31.954 52.510 68.440 228.20 210.68 104.68 24.39 19.71 62.15 125.79 16.70 134.18 20.76 146.71 8.765 166.92 0.08122 187.80 7.977 16.00 208.75 241.814 228.59 20.63 33.44 45.898 16.400 90.250 86.567 33.18 51.328 so2 296.84 300.12 so3 395.72 370.95 co 110.53 137.15 C02 393.51 394.40 Methanol 200.94 162.32 Ethanol 234.95 167.85 (1) Daubert, T.E., Danner, R.P. (Eds.) 1992. Physical and Thermodynamic Properties of Pure Chemicals. Data Compilation, Hemisphere Publishing Corp., New York (extant 1992).
J1H 0
=
Hp 0 H1' 0
= ~V.H. 0 = ~v.JiH
L...J,,
L...J,
f;
0
(15.3.2)
Such values are available for a large variety of compounds, but at 298.2 K and the ideal gas state. Values for selected compounds are presented in Table 15.1. For other compounds, see the same reference, Reid et al, or the references given by Kyle. Differentiation of Eq.1.6.5 yields: dG
=
dHTdSSdT
=
dU+PdV+ VdPTdSSdT
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Applied Chemical Engineering Thermodynamics
From the first law for a reversible change, TdS = dU consequently:
+
PdV, and
dG = VdPSdT
Since we are interested in equilibrium attained at constant pressure: dG dT
= s
and, expressing S through Eq.1.6.5, we obtain: GH T
dG dT
Rearrangement and division by (RT) yields: G 1 dG R T dT  RT2
=
d(GIRT) dT
H RT2
It follows that: d(!!G 0 /RT) _ !!Ho  RT2 dT
(15.3.3)
Introduction of Eq.12.5.10 into Eq.15.3.3leads to the desired expression for the variation of K with !!H 0 , Eq.15.3.1. Eq.15.3.1 represents the starting point for establishing the temperature dependency of the equilibrium constant K. To this purpose we must first establish the temperature dependency of !!H o on T.
15.3.3 The Temperature Dependency of ~H 0 To develop the temperature dependency of !!H 0 on T we take advantage of the fact that enthalpy is a state function and, following the scheme depicted in Figure 15.1, we write Eq.15.3.2 as: !!H 0 (T)
= H/(T) H/(T) = (H/(T) H/(T0 )) +
+(Hp 0 (T0 ) H/ (T0 )] +(H/ (T0 ) Hr o (T)]
(15.3.4)
where T0 = 298.2 K and the subscripts p and r refer to products and reactants respectively. It can be easily shown (Problem 15.5, for example) that the first and third bracketed terms are given by:
551
Chemical Reaction Equilibrium
where a;, b;, C;, and d; are the constants in the temperature dependency expression of the heat capacity of component A; (such as Eq.2.6.7 for an ideal gas). The second bracketed term in Eq.15.3.4 represents, of course, the standard enthalpy of the reaction at T0 = 298.2 K, t1H 0 (T0). After some algebra Eq.15.3.4 becomes: t1H 0 (n
=
s
s
s
2 +_:_T 3 +_!_T 4 t1H0 +Sa T+....!!T 2 3 4
(15 .. 3 5)
where:
Note. For liquids and solids a fifth term in the temperature dependency of the heat capacity is often needed (see, for example, Appendix C), and Eq.15.3.5 and the ensuing Eqs 15.3.6 and 15.3. 7 must be modified accordingly.
Reactants at T
Products at T
Reactants r~ Products at To at To Figure 15.1 Scheme for the evaluation of the standard enthalpy of a reaction at temperature T from its value at T0 •
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Applied Chemical Engineering Thermodynamics
15.3.4 The Expression for K(T) Introduction of Eq.15.3.5 into Eq.15.3.1 yields, after integration between T0 and T, the expression for K(T ): lnK(T)
=
l
LiHo[ 11 +In+ Sa T lnK(T0 )  
R
T T0
R T0
+Sb(TT.)+ Sc(T2r,2)+!!__(T3r,3) 0 0 0 2R
6R
12R
(15.3.7)
where K(T0 ) is the value of K at 298.2 K, given by Eqs 12.5.10 and 12.5.12:
LV; L!Gfi o(To) RT0
(15.3.8)
15.3.5 Comments 1. Evaluation, therefore, of the equilibrium constant for a given reaction as a function of temperature requires: a.Standard free energy and enthalpy offormation values of the reaction components at 298.2 K (fable 15.1). b.Heat capacities as functions of temperature for the reaction components at 1 atm (Appendix C). 2.If heat capacity data are not available, we can assume that LiH o is independent of temperature and integrate Eq .15. 3.1: lnK(T) = lnK(T0 )

t1Ho[_!_ __!_] R
T
T0
(15.3.9)
Eq.15.3.9 gives good results for reasonable distances from T0 , as Example 15.2 will demonstrate. 3.Standard enthalpies of formation are calculated from calorimetric measurements, mostly combustionreactions, and those for the free energy through the relationship: where LiS o is the standard entropy change associated with the reaction.
553
Chemical Reaction Equilibrium
4.The latter may be obtained from the absolute entropies of the pure species involved, calculated from Eq.3.12.3: S(D=S(T=O)+
(Cp) iiH (Cp) iiH (C ) dT + JT• dT+v + J Jo1j __!_:_dT+f T.T Tv 7jT Ij T 1
8 T __
(15.3.10) where: * from the third law, S(T=O) = 0; *the subscripts/and v stand for fusion and vaporization respectively, and s, 1, g for solid, liquid and gas; * no solidphase transitions are involved. Details are given by Lewis et al (1961) and by Denbigh (1981). See also Section 16.9.2 and Problem 15.58. 5.If no data on free energies and enthalpies of formation are available, determination of K at two temperatures is required to establish the temperature dependency forK, of the type given by Eq.15.3.9 (see Problem 15.7). 6. Values of the free energy of formation as a linear function of temper~ture can be estimated by group contribution techniques (Reid et al). The uncertainties, however, in the obtained values  combined with the sensitivity of the equilibrium conversion to the value of the standard free energy of the reaction (see Problem 15.28) make such estimates useful for examining the feasibility of a given reaction only. 7 .Free energy and enthalpy of formation values are typically available at 25°C and in the ideal gas state. Evaluation of their values at other conditions, such as in the liquid phase, is demonstrated in Example 15.3. 8.In Example 15.4 we apply our knowledge of chemical reaction equilibrium to one of the most important reactions, the combustion of fossil fuels, and examine the amount of useful energy that can be produced.
15.3.6 Example 15.1 Consider the reaction: CH4 + 202 .,. C02 + 2H 20 (1)
(2)
(3)
(4)
Show that the standard free energy of this reaction: .::1G 0
= G 0 (3) + 2G 0 (4) G 0 (l) 2G 0 (2)
(A)
can be expressed in terms of the standard free energies of formation of the compounds involved: (B)
Applied Chemical Engineering Thermodynamics
554
For each compound the free energy can be calculated from the free energies of its elements and the corresponding free energy of formation, as follows: G 0 (1) = G 0 (C) + 2G 0 (H 2) + L1G/(1) G 0 (2) = G 0 (02) + 0.0 G 0 (3) = G 0 (Cj + G 0 (02) + L1G/(3) G 0 (4) = G 0 (H2) + 0.5G 0 (02) + L1G/(4) Introductionofthese expressions into Eq.(A) gives Eq.(B). The same procedure can be used for the standard enthalpy of the reaction.
15.3. 7 Example 15.2 For the reaction: H 2 + (1/2) 0 2 .,.. H20, calculate K as a function of temperature from 300 K to 1500 K, using: a . .1H 0 = f(T) b.L1H 0 =constant, equal to its value at T0 = 298.2 K. The following information is obtained from Table 15.1 and from Appendix C. Comp.
L1G/(T0) L1H/(T0)
b·103
c·105
28.36
4.943
0.920
6.142
28.66 32.24
2.380 1.908
2.008 1.057
11.500 3.602
a
d·109
kJ/mol H2 02 H20
0 0 228.59
0 0 241.814
a.From Eqs 15.3.6 and 15.3.8: Sa= 10.450;Sb = 1.845·103 ; Sc = 0.973·105 ; Sd = 3.994·109
L1H0 = 238.69kJ/mol, and lnK(T0) = 228590/(RT0 ) = 92.20 Hence from Eq.15.3. 7: lnK=92.20+28710[_!. _!_]1.257ln[ T]1.110·10 4 (TT0 )+ T
T0
T0
+ 1.951· 10 7 (T 2  T02)  4.003·10 11 (T 3  T03 )
(A)
b. Using .1H 0 = 241.814kJ/mol =constant in Eq.15.3.9: InK= 92.20+29085[.!_ T
_!_] T 0
(B)
The results for the two methods are presented in Table 15.E.2. As expected, the errors are low at small distances from298.2K and become larger with increasing ones.
555
Chemical Reaction Equilibrium
Table 15.E.2 Variation of InK with Temperature, Example 15.2, Using Eqs (A) and (B) T (K)
lnK(A)
lnK(B)
% Error
300 500 750 1000 1250 1500
91.61 52.70 33.07 23.17 17.19 13.19
91.61 52.83 33.44 23.75 17.93 14.05
0.0 0.2 1.1 2.4 4.1 6.6
15.3.8 Example 15.3 The standard free energy and enthalpy of formation of water vapor at 29 8. 2 K are 228.59and 241.814kJ/mol respectively. What are the corresponding values for liquid water at the same conditions? Notice that: where (l) and (v) stand for liquid and vapor respectively. If we now follow a scheme similar to that of Figure 15.1, noting that the superscript (0 ) implies P "' 1 atm, i.e. G 0 (l) "'G(l,l) and G 0 (v) "'G(v,l), we have: G(l,l) G(v,l)"' [ G(l,l) G(l,Ps)) + [ G(l,Ps) G(v,Ps)] + + [ G(v,P')  G(v,l)] where p• "'3.166 kPa, is the vapor pressure of water at 25°C. Since at constant temperature: dG "' VdP, where Vis the molar volume, we obtain for the first and third bracketed terms, respectively: First term, V"' 18 cm3 /mol: G(l,l) G(l,Ps)"' V(l Ps)"' 1.77 J/mol Third term, V "'RTIP, valid in these low pressures: G(v,Ps) G(v,l) "'RTln(Ps!I) "'8591.25 J/mol
The second bracketed term is, of course, equal to zero (why?). Thus: .tiG/(l) "'237.18kJ/mol
Similarly, the calculation of .tiH/(1) leads to three terms: the first and third that represent the effect of pressure  in the range P s to I atm  on liquid and vapor enthalpies, which are negligible in this pressure range; and the second, that represents the enthalpy difference between saturated liquid and vapor at 25°C, i.e. the negative of the enthalpy of vaporization of water at 25°C, equal to 44.0 kJ /mol. Hence: .tiH/(l) "'285.81 kJ/mol
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Applied Chemical Engineering Thermodynamics
15.3.9 Example 15.4 One of the most important chemical reactions involves the combustion of fossil fuels, since over 90% of the energy used in the world comes from them. Assuming that the fuel under consideration is propane, where the combustion reaction is: determine: a. the maximum amount of useful energy, i.e. the exergy, that can be produced in a flow process at 25°C and 1 atm, and b. the amount of exergy obtained as a function of temperature, from 400 K to 1500 K, by using the enthalpy of combustion also in a flow process. a. The maximum obtainable work is given by Eq.5.6.1: (A)
where S and Hare the entropy and enthalpy of the system in its initial state, and the subscript (0 ) refers to the final state where the temperature and pressure are those of the surroundings, taken as 298.2 K and 1 atm. Since the combustion takes place at atmospheric pressure and 298.2K, the initial state of the system here is the reactants at these conditions; and the final, the products also at the same conditions. Eq.(A) becomes, therefore: ~
= [G/G/) = LlG
0
Using the free energy of formation data from Table 15.1, noting that water is in the liquid state (see Example 15.3):
4> = [3(394.40)+ 4(237.18) (24.39)]= 2107.5kJ/mol Note. The exergy obtained from a chemical reaction, is referred to as chemical exergy (Section 5.7.4). b. In this case water is in the vapor phase and, using enthalpies of formation and heat capacity data from Table 15.1 and Appendix C respectively, we obtain the following expression for the heat of the reaction from Eqs 15.3.5 and 15.3.6a: L1H 0 (T)
s
L1H 0 (T0 )+Sa(TT0 )+ ;(T 2 T02 )+
s
s
+ _:( T3 To3)+ .!.(T4 To4) 4 3
where: LlH 0 (T0 ) = 2045.8kJ/mol, and fromEq.15.3.6b:
sa= 43.413; sb
= 4.139·10' 2 ;
sc
= 9.476·10' 5 ;
sd
= 7.159·10" 8 •
A plot of LlH (T) vs. Tis presented in Figure 15.E.4. To calculate next the exergy that can be obtained, we set up a Carnot cycle that absorbs the enthalpy of reaction at T and rejects heat at T0 (Eq.5.6.3): 0
557
Chemical Reaction Equilibrium
~ = dW(T)[l ~] A plot of exergy versus Tis shown also in Figure 15.E.4, along with the chemical exergy at 298.2 K. Comments l.lt is apparent from Figure 15. E. 4 that use of the free energy gives the largest amount of useful energy (chemical exergy). This requires, however, that the reaction is carried out in a fuel cell. While research in this subject has been going on for a long time, the obtained efficiencies are still low, typically about 20 40%. In addition, high initial costs combined with the perceived risk element in this stilldeveloping technology, have led to very limited application of fuel cells (Parkinson, 1986). 2.If we could use all the exergy obtained from the enthalpy of combustion at high temperatures, we could still recover about 80% of the chemical exergy at 298.2 K. 3.In practice however, even with modem power plants, we obtain only about 45% of the maximum amount based on the enthalpy of combustion (see Section 3.15).
2500
2000
t< []
0
[]
[]
[]
[]
[]
[]
[]
[]
[]
[]
0
0
E
:::; 1500 ~
A
A
A
A
A
A
A
A
.g.
'0 0
A A
1000 A
I
oocoo 6H 0 AAAAA !II
;Y;P)v
K = where: V = ""V.
L..J
I
.
'
Ktj}
=
1
= Kq,KyPv
11 A.vl 'PI
(15.4.2)
. Ky = 11Y·vi I
'
Since fugacity coefficients can be determined as functions of composition through an equation of state  such as the virial  the equilibrium constant K has thus been expressed in terms of composition.
15.4.2 Evaluation of the Equilibrium Conversion As we have seen in Section 12.5.3, the number of moles of component i of the reaction mixture at equilibrium is given by:
N;
=
N; o
+ v;e
(15.4.3)
where e is the equilibrium value of the progress of the reaction variable and the superscript (0 ) denotes the value of N; at the beginning of the reaction.
559
Chemical Reaction Equilibrium
Thus, the total number of moles present at equilibrium is: (15.4.4) and the corresponding mole fractions are given by:
IV;
Y; = N
All mole fractions are, therefore, expressed in terms of a single variable,
e, which can be evaluated from the known value of K. The equilibrium conversions are then calculated from Eq.15.4.3.
Comments 1.At low pressures K.p can be set equal to 1, and the fact that most gasphase reactions take place at relatively high temperatures for kinetic purposes, provides additional support for this assumption. 2.At higher pressures K.p can be substantially different from unity even at high temperatures, as we will see with the case of the ammonia synthesis (Example 15.7). In such cases K
B P)b
=
K K y
p 3 3
A plot of lnP3 versus (1/T) is presented in Figure 15.E.12. Decomposition will take place to the left of the line only. Thus at 1000 K: K = 0.052, and the equilibrium partial pressure of C02 over the decomposing CaC03 is 0.052 atm. If P 3 > 0.052 atm, no decomposition will take place.
15.8.3 Example 15.13. Solid + Gas .,.. Solid + Gas A mixture containing20% CO and 80% N2 passes over FeO reducing it to metallic iron: FeO +CO .. Fe+ C02
(1)
(2)
(3)
(4)
Assuming that equilibrium is reached, calculate the amount of iron produced per minute  at 1000 K and a pressure of 1 atm  given a gas mixture flow rate of 100 moles per minute. K = 0.52. Eq.15.8.3 becomes:
K=Ky
Y4 Y2
where: y 2 = (20 e)IN; y4 = e/N; N = (20 e) + e + 80 = 100 Hence: K = _e_, e = 6. 84, and 6. 84 moles of Fe are formed per minute. 20e
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Chemical Reaction Equilibrium
10.0
20.0
cS u
IQ_
c
30.0
40.0
50.0 L._....__....L__.____L___J_L_..J....__!___J_____l_c__J.__....__.....J___.___J_L...J 2.4 2.6 1.2 1.4 1.6 1 .8 2.0 2.2 0.8 1.0
1/T
*
103
Figure 15.E.12 Equilibrium partial pressure of C02 from the decomposition of CaC03 (p, atm; T, K).
15.9 Multiple Reactions 15.9.1 The Problem We consider next systems where more than one reactions take place, which is often the case in gasphase reactions. The thermodynamic analysis of such systems is straightforward for, as we saw in Section 12.6, each reaction is in equilibrium. Thus for a system involving r reactions, there are r progress of the reaction variables with equilibrium values of: e 1, e1 , ... , er, through which all compositions are expressed. Determination thus of the equilibrium conversion requires the calculation of the r values of e, by solving a system of r equations with r unknowns. And this is where the problem lies. The equations are very nonlinear and solution of the resulting system becomes difficult. For r = 2, a trialanderror or a graphical solution can be easily determined. For r = 3, a graphical solution is possible, but somewhat tedious. If repeated
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calculations are required, therefore, this approach becomes impractical. In such cases, or for r > 3, a solution using Lagrange multipliers must be used. We will demonstrate the graphical method for systems involving two and three reactions in the following three Examples. These cover the majority of industrial applications, since systems involving more than three reactions are relatively rare. For the approach that uses Lagrange multipliers, see among others: Smith and Van Ness; Daubert; or Oliver et al (1962).
15.9.2 Example 15.14 One of the most important industrial reactions is the catalytic production of the synthesis gas 1
I :
CH4 + HzO """ CO + 3Hz
An additional reaction takes place to a significant extent: CO + H20 """ C02 + Hz Starting with a ratio of 5 moles of steam to 1 mole of methane, i.e. with an excess of steam for reasons that will be discussed in Example 15.16, calculate the equilibrium composition of the resulting mixture at 600°C and 1 atm. At this temperature, the equilibrium constants for the first and second reaction are respectively (Hougenet al, 1959): K 1 = 0.574; Kz = 2.21. Let component 1, CH4 ; 2, H20; 3, CO; 4, H2 ; and 5, C02 ; and e1, ez the values of the progress of the reaction variables for the first and second reaction respectively at equilibrium. Then: N 1 = 1  e1 Nz = 5 e 1  ez N3 = et ez N4 = 3e1 + e2 Ns = ez N = 6 + 2e1 At P = 1 atm, K = Ky for both reactions. Hence:
K1
Y3 Y/
(e 1 e2 )(3e 1 +e2 ) 3
Y1 Yz
N 2 (1e 1)(5e 1 e2 )
= 0.574 = 2.21
(A)
(B)
We have thus a system of two equations with two unknowns, which we will solve with the graphical method as follows:
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Chemical Reaction Equilibrium
1.Assume a value for e 1 and determine the corresponding value of e2 from Eq.(A) by trialanderror, since it is a fourth degree equation. 2.For the same value of e 1, calculate e2 by solving Eq.(B) which is quadratic. 3.Repeat, using a new value for e 1• 4.Plot the obtained values of e2 vs. e 1 for the two equations. The intersection is the solution, giving: e 1 = 0.912 and e2 = 0.633. Hence: y 1 = 0.0112,y2 = 0.4416,y3 = 0.0357,y4 = 0.4306,y5 = 0.0809.
15.9.3 Example 15.15 With reference to the case presented in Example 15.14, a third possible reaction is: 2CO
.c +
C02
with a value of K3 = 8.14 at 600°C. This reaction is undesirable because the carbon deposits on the catalyst reducing its activity. Considering the conditions of Example 15.14, will there be any carbon deposition? The value of K3 for this reaction, calculated from the results of Example 15.14 using Eq.15.8.3, is:
K3 = y51y32 = 63.5 which is much larger than the equilibrium value of 8.14. No carbon, therefore, will be formed(why?).
15.9.4 Example 15.16 An examination of the second reaction of Example 15.14: CO + H20 ""' C02 + H 2 indicates that an excess of water shifts the reaction to the right, increasing the amount of C02 • This increase, in tum, shifts the third reaction: 2CO.,. C
+ C02
to the left, prohibiting the undesirable formation of carbon. At what steam to methane ratio will carbon be formed? Following Example 15.14, we replace the 5 moles of H20 by z, the unknown number of moles of steam per mole of methane, and introduce an additional equation resulting from the formation of carbon. Thus we have the following system of three equations with three unknowns:
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(A)
(B)
= 8.14
(C)
where: N = z + 1 + 2e 1• This system can be solved graphically for the three unknowns: e 1 , e2 , and z, by finding the point of intersection of three lines in space. To this purpose: l.Assume a value of z; 2.Plot e2 versus e 1 for the three Eqs (A), (B), and (C); 3.Repeat until the obtained from the three lines triangle becomes a point. This is demonstrated in Figure 15.E.16 and gives: z = 1.38; e 1 = 0.539; and e2 = 0.227.
15.10 Concluding Remarks 1. The equilibrium conversion for a given reaction is determined from knowledge of its equilibrium constant K, which in turn is evaluated in the typical case from data on the: a. standard free energy and enthalpy of formation at 298.2 K; and b.heat capacity as a function of temperature at P = 1 atm; of the reactants and products. 2.In addition, for gasphase reactions an equation of state is needed to estimate the fugacity coefficients of the mixture components; and for liquidphase ones, parameter values of some model for the activity coefficient. 3 .Because of the logarithmic relationship involved the value of K, and therefore that of the equilibrium conversion, for a reaction is very sensitive to the values used for the free energy and enthalpy of formation (Problem 15.28). 4.Since lnK = I1G 0 /RT, the feasibility of a reaction is determined by the value of 11 G 0 • The following approximate criteria can be used: I1G 0 < 0: Very promising, high conversion; 0 Promising, but low conversion; 0 < 11 G < 4:
577
Chemical Reaction Equilibrium 0.80
(A) 0.50
z
2.00
e2
(C)
0.40
(B) 0.20
0.00 0.00
0.20
0.40
0.50
0.80
1.00
e1
0.80
(A)
=
z
1.00
0.50
(C)
e2 0.40
0.20
(B) 0.00 0.00
0.20
0.40
0.60
0.80
1.00
e1
0.80
z
=
(A)
1.38
0.50
(C)
e2 0.40
0.20
(B)
0.00 0.00
0.20
0.40
0.60
0.80
1.00
e1
Figure 15.E.16 Graphical solution of Example 15.16.
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Ll G 0 < 40: Doubtful, but it warrants further examination; LlG 0 > 40: Very unfavorable; where Ll G 0 is in kJ /mol. The term 'feasible' is used here in the practical sense for, of course, every reaction is feasible even·ifthe equilibrium conversion is of the order of 1020 . 5.The fact that a reaction is thermodynamically feasible does not necessarily mean that it will indeed take place. A catalyst is often needed for the reaction to attain significant conversions. 6.Keep in mind, however, that while equilibrium conversions can be determined  or at least estimated  in the majority of cases without resort to experimental work, this is not the case with finding the appropriate catalyst. Even though great progress has been made in identifying promising catalysts  since the days when 1000 catalysts were tried for the ammonia synthesis  extensive experimental work is still required. 7 .Catalysts can be also used to suppress undesirable, even highly favored thermodynamically, sidereactions. Thus in the methanol synthesis, as an example:
4